3.13 Electronics (A-level only) — revision question pack

18 specification points · notes, questions, answers and worked methods

Checked against AQA 7408 section 3.13. Review basis: the qualification registry sourced from the AQA A-level Physics (7408) specification; registry verification recorded 11 July 2026.

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Answer all questions in the spaces provided.

3.13.1.1 · MOSFET (metal-oxide semiconducting field-effect transistor)

Explanation

The required MOSFET is an n-channel enhancement-mode device with gate, drain and source terminals. An insulating oxide separates the gate from the conducting region, giving a very high input resistance and negligible steady gate current.
The gate–source potential difference $V_{\mathrm{GS}}$ controls drain current.
Below threshold $V_{\mathrm{th}}$ , the channel is absent and the device is off; above threshold, a channel forms and drain current can flow.
Output characteristics show $I_{\mathrm D}$ against $V_{\mathrm{DS}}$ for different gate voltages, including $I_{\mathrm{DSS}}$ where specified.
As a switch, a voltage applied at the high-resistance gate controls a much larger load current.

Simplified n-channel enhancement MOSFET structure showing the insulated gate, drain and source.

Worked example

A MOSFET has threshold voltage $2.5\,\text{V}$ . State its switching state for $V_{\mathrm{GS}}=1.8\,\text{V}$ and for $V_{\mathrm{GS}}=4.0\,\text{V}$ .

1.Compare each gate–source potential difference with $V_{\mathrm{th}}$ .
2. $1.8\,\text{V}<2.5\,\text{V}$ , so no conducting channel forms.
3. $4.0\,\text{V}>2.5\,\text{V}$ , so an n-channel forms and the device can conduct.

Answer: The device is off at $1.8\,\text{V}$ and on at $4.0\,\text{V}$ .

Common mistakes

Don't treat the gate as a current-operated input and expect a large steady gate current.
Don't interchange the drain, source and gate labels on the device or characteristic.
Don't claim $V_{\mathrm{GS}}=V_{\mathrm{th}}$ guarantees a fully conducting switch rather than the onset of channel formation.

Exam tip

When reading characteristics, identify the selected $V_{\mathrm{GS}}$ curve before taking $I_{\mathrm D}$ or $V_{\mathrm{DS}}$ .

Tier 1 · Easy

State the condition involving $V_{\mathrm{GS}}$ and $V_{\mathrm{th}}$ for an enhancement-mode n-channel MOSFET to begin conducting.
[1 mark]
Total for this question: 1
State what is meant by $I_{\mathrm{DSS}}$ for a MOSFET.
[1 mark]
Total for this question: 1

Tier 2 · Standard

A $9.0\,\text{V}$ moisture alarm has a sensor of resistance $R$ from the positive supply to a MOSFET gate and a $1.5\,\text{M}\Omega$ resistor from the gate to $0\,\text{V}$ . The MOSFET threshold voltage is $3.0\,\text{V}$ . Determine $R$ when the MOSFET first switches on.
[3 marks]
Total for this question: 3
The gate of an n-channel MOSFET is connected to a $6.00\,\text{V}$ supply through a $22.0\,\text{M}\Omega$ resistor. Although the normal steady gate current is negligible, a fault current of $30.0\,\text{nA}$ in the gate circuit flows through this resistor. The source is at $0\,\text{V}$ . Calculate $V_{\mathrm{GS}}$ .
[2 marks]
Total for this question: 2
A logic output connected to the gate of an n-channel enhancement-mode MOSFET alternates between $0.30\,\text{V}$ and $4.80\,\text{V}$ . Its source terminal is held at $0\,\text{V}$ ; replacement MOSFETs have threshold voltages between $1.7\,\text{V}$ and $2.4\,\text{V}$ . Determine the guaranteed switching state at each logic level and explain why the logic output supplies negligible steady power to the gate.
[3 marks]
Total for this question: 3

Tier 3 · Hard

A MOSFET controls a $120\,\Omega$ load from a $9.0\,\text{V}$ supply. At the applied gate voltage its output characteristic is approximated by $I_{\mathrm{D}}=48\,\text{mA}$ for $V_{\mathrm{DS}}\geq2.0\,\text{V}$ . Use load-line reasoning to determine the operating values of $V_{\mathrm{DS}}$ , load power and MOSFET power. Deduce whether the MOSFET is acting as an efficient fully-on switch.
[4 marks]
Total for this question: 4
Explain how the simplified structure of an n-channel enhancement-mode MOSFET gives both a very high input resistance and voltage-controlled switching.
[4 marks]
Total for this question: 4
When fully on, a MOSFET switch behaves as a fixed resistance of $6.0\,\Omega$ between drain and source. It connects a load of resistance $42\,\Omega$ to a $24\,\text{V}$ supply. Calculate the current, the power dissipated in the MOSFET, and the percentage of the supply power that is wasted in the MOSFET.
[3 marks]
Total for this question: 3
For one fixed value of $V_{\mathrm{GS}}$ , read-offs from a MOSFET output characteristic give $I_{\mathrm D}=0.16\,\text{A}$ at $V_{\mathrm{DS}}=0.50\,\text{V}$ , $0.29\,\text{A}$ at $1.0\,\text{V}$ , and $0.34\,\text{A}$ at both $2.0\,\text{V}$ and $4.0\,\text{V}$ . A student says that the drain current is independent of $V_{\mathrm{DS}}$ whenever the MOSFET is on. Calculate the percentage increase in drain current from $0.50\,\text{V}$ to $2.0\,\text{V}$ . Discuss whether the student is correct.
[4 marks]
Total for this question: 4
A MOSFET gate is connected to a control output through a $4.7\,\text{M}\Omega$ resistor. An ideal voltmeter, which draws no current, is used for both measurements. The potential difference across the resistor is $4.7\,\mu\text{V}$ and the gate-source potential difference is $4.0\,\text{V}$ . At this setting the drain current is $0.24\,\text{A}$ . Treat the resistor current as steady gate leakage. Determine the gate current, the MOSFET input resistance and the ratio $I_{\mathrm D}/I_{\mathrm G}$ .
[3 marks]
Total for this question: 3

3.13.1.2 · Zener diode

Explanation

A Zener diode has an anode and cathode and is used in reverse bias. Its characteristic has a sharp breakdown at the Zener voltage $V_{\mathrm Z}$ .
Beyond this point, a large change in reverse current causes only a small change in voltage, provided current remains above the typical minimum operating value.
A series resistor is essential because it limits current and drops the remaining supply voltage.
The nearly constant diode voltage can serve as a constant-voltage source or reference voltage.
The specification does not require a full stabiliser treatment, but characteristic questions may require identification of forward conduction, reverse leakage, breakdown voltage and minimum operating current.

The Zener characteristic has a sharp reverse-breakdown region at

V_{\mathrm Z}

Worked example

A $6.8\,\text{V}$ Zener reference is connected to a $12.0\,\text{V}$ supply through $330\,\Omega$ . Calculate the Zener current when no load is connected.

1.The resistor potential difference is $12.0-6.8=5.2\,\text{V}$ .
2.Use $I=V/R$ for the series resistor.
3. $I=5.2/330=1.58\times10^{-2}\,\text{A}$ .

Answer: The Zener current is $15.8\,\text{mA}$ .

Common mistakes

Don't place the Zener in forward bias when using the reverse-breakdown voltage reference.
Don't omit the series resistor, because this provides no limit on breakdown current.
Don't read the reverse-breakdown voltage from the forward-conduction branch of the characteristic.

Exam tip

On a characteristic, label anode, cathode and reverse-bias polarity before identifying $V_{\mathrm Z}$ .

Tier 1 · Easy

State the bias direction and operating region used when a Zener diode provides a constant-voltage reference.
[1 mark]
Total for this question: 1
State the purpose of the resistor connected in series with a reverse-biased Zener diode.
[1 mark]
Total for this question: 1

Tier 2 · Standard

A no-load voltage reference uses a $5.6\,\text{V}$ Zener diode on a $12.0\,\text{V}$ supply. Calculate the series resistance required for a Zener current of $16\,\text{mA}$ .
[3 marks]
Total for this question: 3
Explain how the reverse-bias characteristic of a Zener diode shows that it can provide a reference voltage.
[3 marks]
Total for this question: 3
A loaded reference uses a $7.5\,\text{V}$ Zener diode connected to a $14.0\,\text{V}$ supply through a $240\,\Omega$ series resistor, with a load in parallel with the diode. While the diode remains in breakdown, determine the power dissipated by the series resistor and deduce the smallest of the $0.125\,\text{W}$ , $0.25\,\text{W}$ and $0.50\,\text{W}$ ratings that is suitable.
[2 marks]
Total for this question: 2

Tier 3 · Hard

A $6.2\,\text{V}$ Zener circuit is supplied from a fixed $14.0\,\text{V}$ source. The parallel load can draw up to $18\,\text{mA}$ and the Zener needs at least $5.0\,\text{mA}$ . Determine the largest suitable series resistor from $330\,\Omega$ , $360\,\Omega$ and $390\,\Omega$ , and then determine minimum safe power ratings for the resistor and Zener when the load is disconnected.
[5 marks]
Total for this question: 5
A $6.8\,\text{V}$ Zener reference is supplied from a $14.0\,\text{V}$ source through a fixed $330\,\Omega$ resistor. The Zener diode needs at least $5.0\,\text{mA}$ to remain in breakdown. Determine the maximum load current that the reference can supply. Explain what happens to the output potential difference if the load current is increased beyond this value.
[4 marks]
Total for this question: 4
The graph of a Zener diode's reverse characteristic is a straight line through the read-off points $(8.0\,\text{mA},6.05\,\text{V})$ and $(28.0\,\text{mA},6.25\,\text{V})$ . The reverse-biased diode is connected to an $11.0\,\text{V}$ supply through $180\,\Omega$ , with no load. Use the graph readings and the resistor relation to determine the operating current and Zener potential difference.
[4 marks]
Total for this question: 4
A $6.2\,\text{V}$ Zener diode is modelled as a constant-voltage source while it is in breakdown. It is connected to a variable supply through a $270\,\Omega$ series resistor. The Zener current is increased from $8.0\,\text{mA}$ to $26\,\text{mA}$ . State the change in Zener potential difference. Determine the supply potential at the second current and the increase in power dissipated by the Zener diode.
[3 marks]
Total for this question: 3
A $4.7\,\text{V}$ Zener reference has a fixed $270\,\Omega$ series resistor and a $1.20\,\text{k}\Omega$ load connected across the diode. The Zener current must remain between $3.0\,\text{mA}$ and $24\,\text{mA}$ . Determine the range of supply potentials for which the reference operates within these limits, and calculate the greatest Zener power in that range.
[4 marks]
Total for this question: 4

3.13.1.3 · Photodiode

Explanation

A photodiode produces a current when radiation creates charge carriers. Its current–voltage characteristic shifts with illumination, and a spectral-response curve shows that sensitivity depends on wavelength.
In the required photoconductive mode, the device is reverse biased and used as a detector in an optical system; greater incident intensity produces a larger reverse photocurrent over its useful range.
A photodiode can also be coupled to a scintillator.
An atomic particle deposits energy in the scintillator, which emits a flash; the photodiode converts that flash into an electrical pulse.
Examiners may ask for interpretation of characteristic or spectral-response graphs and the complete particle-to-light-to-electrical detection chain.

A photodiode's spectral response varies with wavelength and peaks over a limited range.

Worked example

Explain how a photodiode and scintillator detect an atomic particle.

1.The particle deposits energy in the scintillator.
2.The scintillator converts some deposited energy into a flash of light.
3.The reverse-biased photodiode detects the flash as a current pulse.

Answer: The detection chain is particle energy, scintillator light pulse, then photodiode electrical pulse.

Common mistakes

Don't describe the photodiode as changing resistance like an LDR rather than producing a photocurrent.
Don't use forward bias when describing the required photoconductive detector mode.
Don't say the atomic particle directly creates the photodiode pulse and omit the scintillator flash.

Exam tip

For graph interpretation, quote both the wavelength region and the relative response rather than merely naming the peak.

Tier 1 · Easy

Name the operating mode of a photodiode that has an external reverse-bias voltage.
[1 mark]
Total for this question: 1
State the type of graph used to show how a photodiode's sensitivity varies with wavelength.
[1 mark]
Total for this question: 1

Tier 2 · Standard

A photodiode has responsivity $0.42\,\text{A W}^{-1}$ at the wavelength used. Radiation of power $8.0\,\mu\text{W}$ reaches it and its photocurrent passes through a $180\,\text{k}\Omega$ resistor. Calculate the magnitude of the resistor voltage.
[3 marks]
Total for this question: 3
At its most sensitive wavelength a photodiode has responsivity $0.50\,\text{A W}^{-1}$ . Its spectral-response curve has relative response $0.60$ at $900\,\text{nm}$ . Calculate the photocurrent produced at $900\,\text{nm}$ by $12\,\mu\text{W}$ of radiation.
[3 marks]
Total for this question: 3
A photodiode has maximum responsivity $0.45\,\text{A W}^{-1}$ . At $650\,\text{nm}$ its relative spectral response is $0.80$ , and a monochromatic source produces $4.32\,\mu\text{A}$ . The same source power at an unknown wavelength produces $2.70\,\mu\text{A}$ . The unknown wavelength is one of $450\,\text{nm}$ , $750\,\text{nm}$ and $900\,\text{nm}$ . Their relative responses are $0.20$ , $0.70$ and $0.50$ respectively. Determine the unknown wavelength.
[3 marks]
Total for this question: 3

Tier 3 · Hard

A particle deposits energy in a scintillator, producing a $25\,\text{ns}$ optical pulse of constant power $4.0\,\mu\text{W}$ . A reverse-biased photodiode has responsivity $0.48\,\text{A W}^{-1}$ and its readout converts the photocurrent using $270\,\text{k}\Omega$ . Explain how the particle is detected, then calculate the voltage-pulse magnitude and the number of charge carriers in the photocurrent. Use $e=1.60\times10^{-19}\,\text{C}$ .
[5 marks]
Total for this question: 5
A spectrometer sends separate narrow pulses at $520\,\text{nm}$ and $850\,\text{nm}$ through an optical filter to the same reverse-biased photodiode. The responsivities are $0.28\,\text{A W}^{-1}$ and $0.56\,\text{A W}^{-1}$ , and the measured photocurrents are $3.36\,\mu\text{A}$ and $8.40\,\mu\text{A}$ respectively. The filter transmits $75\%$ at $520\,\text{nm}$ and $60\%$ at $850\,\text{nm}$ . Determine the ratio of the source powers $P_{850}/P_{520}$ before the filter.
[5 marks]
Total for this question: 5
At a reverse bias of $6.0\,\text{V}$ , a photodiode characteristic gives a dark reverse current of $0.20\,\mu\text{A}$ and a total reverse current of $5.60\,\mu\text{A}$ during a $12.0\,\mu\text{s}$ scintillator flash. Determine the charge carried by the flash-induced photocurrent and the number of charge carriers. Use $e=1.60\times10^{-19}\,\text{C}$ .
[3 marks]
Total for this question: 3
A reverse-biased photodiode has maximum responsivity $0.62\,\text{A W}^{-1}$ . Its relative spectral responses are $0.35$ at $480\,\text{nm}$ and $0.80$ at $720\,\text{nm}$ . Radiation of powers $18\,\mu\text{W}$ and $7.0\,\mu\text{W}$ at these respective wavelengths reaches the detector simultaneously. Determine the total photocurrent and the magnitude of the potential difference it produces across an $120\,\text{k}\Omega$ resistor.
[4 marks]
Total for this question: 4
Each atomic particle entering a scintillator produces a light pulse of energy $3.6\times10^{-13}\,\text{J}$ at a wavelength where a reverse-biased photodiode has responsivity $0.44\,\text{A W}^{-1}$ . During one counting interval the integrated photodiode current corresponds to charge $4.752\times10^{-10}\,\text{C}$ . Explain the detection chain and determine the number of particles detected. Assume every scintillator pulse reaches the photodiode.
[4 marks]
Total for this question: 4

3.13.1.4 · Hall effect sensor

Explanation

A Hall effect sensor responds to magnetic field and supplies an electrical output suitable for control or measurement. The internal operating principle is not required.
In an attitude monitor, fixed Hall sensors detect the position of magnets attached to a moving or rotating system, allowing orientation to be inferred.
In a tachometer, a magnet or magnetic sector passes the sensor once or several times per revolution, producing output pulses.
The pulse frequency is therefore the rotation frequency multiplied by the number of pulses per revolution.
Examiners expect the application chain—magnetic target, changing sensor output, pulse counting or calibration, then attitude or speed—not a derivation of Hall voltage.

A wheel-mounted magnet passing a Hall sensor produces one pulse per passage.

Worked example

A wheel carries four equally spaced magnets. A Hall sensor records $120$ pulses each second. Calculate the wheel speed.

1.Four magnets produce four pulses per revolution.
2.Rotation frequency $=120/4=30\,\text{rev s}^{-1}$ .
3.Convert to revolutions per minute: $30\times60=1800\,\text{rev min}^{-1}$ .

Answer: The wheel speed is $1.8\times10^3\,\text{rev min}^{-1}$ .

Common mistakes

Don't equate pulse frequency with rotation frequency when several magnets produce pulses each revolution.
Don't explain attitude monitoring without linking magnet position to the Hall-sensor output.
Don't give a detailed charge-carrier derivation even though the sensor's operating principle is not required.

Exam tip

In a tachometer calculation, state the number of pulses per revolution before converting pulse rate to rotation rate.

Tier 1 · Easy

The magnetic flux density through a Hall element doubles without changing its current or orientation. State the effect on its Hall voltage.
[1 mark]
Total for this question: 1
Explain how a Hall effect sensor and a magnet fixed to a moving control surface can be used to monitor the surface's attitude.
[2 marks]
Total for this question: 2

Tier 2 · Standard

A Hall sensor has a zero-field output of $2.500\,\text{V}$ and sensitivity $45\,\text{mV T}^{-1}$ . Calculate its output for a flux density of $-0.18\,\text{T}$ .
[3 marks]
Total for this question: 3
A Hall attitude monitor gives $1.800\,\text{V}$ at $-30.0^\circ$ and $3.000\,\text{V}$ at $+30.0^\circ$ , with a linear response between these angles. Calculate the attitude represented by an output of $2.460\,\text{V}$ .
[3 marks]
Total for this question: 3
A Hall tachometer has seven equally spaced magnets on its rotor. The sensor records $280$ pulses during a $2.50\,\text{s}$ interval. Determine the mean rotational speed in $\text{rev min}^{-1}$ and the mean time interval between successive pulses.
[3 marks]
Total for this question: 3

Tier 3 · Hard

A Hall sensor's output changes by $0.180\,\text{V}$ in a calibration field of $60\,\text{mT}$ . In a speed monitor its peak output change is $0.120\,\text{V}$ . Six identical magnets on a wheel each produce one pulse, and the pulse frequency is $180\,\text{Hz}$ . Determine the peak magnetic flux density at the sensor and the wheel speed in revolutions per minute. Explain how the field calculation uses the Hall effect.
[5 marks]
Total for this question: 5
A turbine shaft rotates at $840\,\text{rev min}^{-1}$ . A Hall tachometer produces one pulse whenever a magnet on the shaft passes the sensor, and its pulse frequency is $210\,\text{Hz}$ . Determine the number of equally spaced magnets on the shaft. State one advantage of using more magnets per revolution.
[3 marks]
Total for this question: 3
A Hall tachometer rotor has $12$ equally spaced magnet positions, but one magnet has detached. In each repeating pulse pattern there are ten intervals of $2.00\,\text{ms}$ and one interval of $4.00\,\text{ms}$ . A controller measures the mean pulse frequency over many revolutions and divides it by $12$ to obtain the rotation frequency. Determine the true rotor speed and the speed displayed by the controller, both in $\text{rev min}^{-1}$ . Calculate the percentage error in the displayed speed.
[4 marks]
Total for this question: 4
Two Hall sensors monitor the attitude $\theta$ of a control surface. Their calibrated outputs are $V_1=2.10+0.0060\theta$ and $V_2=2.26-0.0060\theta$ , with potentials in volts and $\theta$ in degrees. In use, the outputs are $2.244\,\text{V}$ and $2.116\,\text{V}$ . Determine the attitude. Explain why using the difference $V_1-V_2$ rejects a later equal drift added to both sensor outputs.
[3 marks]
Total for this question: 3
A Hall tachometer uses nine equally spaced magnets. A rotor is intended to turn steadily at $1200\,\text{rev min}^{-1}$ . Calculate the pulse count expected during a $0.400\,\text{s}$ gate. The recorded count is $76$ pulses with an uncertainty of one pulse. Deduce whether the count is consistent with the intended speed. Determine the minimum gate time needed for a one-pulse uncertainty to correspond to no more than $10\,\text{rev min}^{-1}$ .
[4 marks]
Total for this question: 4

3.13.2.1 · Difference between analogue and digital signals

Explanation

An analogue signal varies continuously, whereas digital data uses discrete levels; binary uses two voltage levels representing 0 and 1. Sensors commonly produce analogue data.
Analogue-to-digital conversion samples the signal at regular times and quantises each sample to one of $2^n$ levels for an $n$ -bit code. Higher sampling rate preserves faster time variation; more bits reduce quantisation error and improve amplitude resolution, but both increase data rate.
Pulse code modulation transmits the resulting codes as pulses. Digital signals can be regenerated from noisy inputs if levels remain distinguishable, whereas analogue noise accumulates.
Digital sampling nevertheless loses information through sampling and quantisation.
Only recognition of binary numbers 1–10 is required, not binary arithmetic.

Regular samples of a continuous analogue signal are assigned to discrete quantised levels.

Worked example

Audio is sampled at $12\,\text{kHz}$ using $10$ bits per sample. Calculate the bit rate and the number of quantisation levels.

1.Bit rate $=(12\times10^3)(10)=1.20\times10^5\,\text{bit s}^{-1}$ .
2.An $n$ -bit sample has $2^n$ possible levels.
3.Number of levels $=2^{10}=1024$ .

Answer: The bit rate is $120\,\text{kbit s}^{-1}$ and there are $1024$ levels.

Common mistakes

Don't claim a digital recording exactly reproduces the analogue input and ignore quantisation.
Don't say extra bits per sample improve time resolution rather than amplitude resolution.
Don't claim noise has no effect on digital transmission even when it moves a pulse across the decision threshold.

Exam tip

A compare question should balance regeneration and noise resistance against sampling loss, quantisation error and increased data rate.

Tier 1 · Easy

State one difference between an analogue signal and a binary digital signal.
[1 mark]
Total for this question: 1
Calculate the number of bits contained in three bytes.
[1 mark]
Total for this question: 1

Tier 2 · Standard

An $8$ -bit converter divides an input range from $0$ to $4.8\,\text{V}$ into equal quantisation intervals. Calculate the interval width and the maximum quantisation error.
[3 marks]
Total for this question: 3
A regenerator treats received levels above $2.5\,\text{V}$ as binary $1$ and levels below $2.5\,\text{V}$ as binary $0$ . Four successive received levels are $4.4\,\text{V}$ , $0.8\,\text{V}$ , $3.2\,\text{V}$ and $1.7\,\text{V}$ . The first level received is the most significant bit. Determine the recovered decimal number.
[2 marks]
Total for this question: 2
A telemetry link can carry $36\,\text{kbit s}^{-1}$ and sends $6.0\times10^3$ samples each second. Each sample uses the greatest whole number of bits that the link can carry. For a converter range of $0$ to $3.2\,\text{V}$ , determine the bits per sample, the number of quantisation levels and the quantisation interval.
[3 marks]
Total for this question: 3

Tier 3 · Hard

A sensor signal contains frequencies up to $4.0\,\text{kHz}$ . It is sampled at $10\,\text{kHz}$ with $14$ bits per sample for $45\,\text{s}$ . Explain whether the sampling rate is sufficient, calculate the uncompressed data size in bits and bytes, and discuss one consequence of reducing the resolution to $8$ bits.
[5 marks]
Total for this question: 5
Explain why regenerators can prevent noise accumulating along a digital temperature-data link but cannot recover information lost when the original analogue sensor signal was sampled and quantised.
[5 marks]
Total for this question: 5
An analogue input is either a $1.0\,\text{kHz}$ cosine signal or a $7.0\,\text{kHz}$ cosine signal of the same amplitude. Both are offered to a converter that samples at $8.0\,\text{kHz}$ . State the minimum sampling rate the Nyquist condition requires for each signal, and explain why the $7.0\,\text{kHz}$ signal sampled at $8.0\,\text{kHz}$ cannot be distinguished from the $1.0\,\text{kHz}$ signal sampled at the same rate.
[3 marks]
Total for this question: 3
A digital link represents binary $0$ by $0.40\,\text{V}$ and binary $1$ by $4.60\,\text{V}$ ; its decision level is $2.50\,\text{V}$ . In a worst-case model, each cable section moves either received level by $0.15\,\text{V}$ towards the decision level. Determine the greatest number of consecutive sections possible without regeneration before either bit can be misidentified. Explain why placing a regenerator after every fourth section prevents this degradation from accumulating through a long link. A level that arrives exactly at the decision level cannot be identified reliably.
[4 marks]
Total for this question: 4
A converter must encode a sensor signal spanning $0$ to $3.3\,\text{V}$ and containing frequencies up to $5.0\,\text{kHz}$ . The uncompressed link can carry at most $96\,\text{kbit s}^{-1}$ . For this question, an $n$ -bit converter divides the range into $2^n$ equal quantisation intervals, and the maximum quantisation error is half an interval. The design requires a maximum quantisation error no greater than $4.0\,\text{mV}$ and a sampling rate that meets the Nyquist condition. Determine the minimum whole number of bits per sample and the minimum bit rate that satisfies both requirements. Deduce whether the link can carry the data.
[4 marks]
Total for this question: 4

3.13.3.1 · LC resonance filters

Explanation

Only parallel LC resonance is required. Energy alternates between the capacitor's electric field and the inductor's magnetic field at resonant frequency $f_0=1/(2\pi\sqrt{LC})$ ; derivation is not required.
The mechanical analogy is a mass–spring oscillator: inductance corresponds to mass and capacitance to the spring property. The required energy or voltage response has a peak at $f_0$ ; a current-response curve is not required.
Bandwidth $f_B$ is measured between the two 50% energy points, and quality factor is $Q=f_0/f_B$ .
A high Q gives a narrow, selective response.
Inductance must be in henries and capacitance in farads for frequency in hertz.

The parallel LC energy response peaks at

f_0

and has bandwidth

f_B

at the 50% energy points.

Worked example

A parallel LC filter has $L=1.5\,\text{mH}$ and $C=4.7\,\text{nF}$ . Calculate its resonant frequency.

1.Convert: $L=1.5\times10^{-3}\,\text{H}$ and $C=4.7\times10^{-9}\,\text{F}$ .
2.Use $f_0=1/(2\pi\sqrt{LC})$ .
3. $f_0=1/[2\pi\sqrt{(1.5\times10^{-3})(4.7\times10^{-9})}]$ .

Answer: $f_0=6.0\times10^4\,\text{Hz}$ to two significant figures.

Common mistakes

Don't use a series-resonance arrangement even though only parallel resonance is required.
Don't read $f_B$ as one side of the peak instead of the full separation of the 50% energy points.
Don't label inductance as the spring analogy and capacitance as the mass analogy.

Exam tip

On a response graph, mark both 50% energy intersections before calculating $f_B$ and then $Q$ .

Tier 1 · Easy

Calculate the resonant frequency of an LC filter containing $L=2.0\,\text{mH}$ and $C=8.0\,\text{nF}$ .
[2 marks]
Total for this question: 2
State the electrical quantity corresponding to mass in the mass-spring analogy for an LC circuit.
[1 mark]
Total for this question: 1

Tier 2 · Standard

A tuned circuit has a response peak at $150\,\text{kHz}$ . Its lower and upper half-power frequencies are $143\,\text{kHz}$ and $157\,\text{kHz}$ . Determine its bandwidth and Q-factor.
[3 marks]
Total for this question: 3
A spectrum analyser retunes a parallel LC input filter from $180\,\text{kHz}$ to $225\,\text{kHz}$ while keeping its inductance constant. Determine the percentage change in capacitance.
[3 marks]
Total for this question: 3
A parallel LC filter contains an inductor labelled $1.00\,\text{mH}\pm2\%$ and a capacitor labelled $100\,\text{pF}\pm5\%$ . Calculate the highest possible resonant frequency allowed by these tolerances.
[3 marks]
Total for this question: 3

Tier 3 · Hard

A receiver uses a $0.75\,\text{mH}$ inductor in an LC filter tuned to $620\,\text{kHz}$ . Its measured half-power frequencies are $609\,\text{kHz}$ and $631\,\text{kHz}$ . Calculate the required capacitance and Q-factor. The wanted transmission occupies $18\,\text{kHz}$ centred on resonance; discuss whether the measured bandwidth is suitable.
[5 marks]
Total for this question: 5
Explain the energy transfers during free oscillation of an ideal parallel LC circuit whose capacitor is initially charged.
[4 marks]
Total for this question: 4
A receiver must tune continuously from $0.55\,\text{MHz}$ to $1.60\,\text{MHz}$ using one fixed inductor and a parallel variable capacitor whose range is $20\,\text{pF}$ to $365\,\text{pF}$ . Available inductors are $100\,\mu\text{H}$ , $220\,\mu\text{H}$ and $270\,\mu\text{H}$ . Determine the only suitable inductor and the capacitance needed at each end of the tuning range.
[5 marks]
Total for this question: 5
An ideal parallel LC circuit contains a $68\,\text{nF}$ capacitor initially charged to $12\,\text{V}$ and a $4.7\,\text{mH}$ inductor carrying no current. Determine the initial stored energy and the resonant frequency. Calculate the capacitor potential difference at the instant its stored electric-field energy has fallen to one quarter of its initial value.
[3 marks]
Total for this question: 3
A parallel tuned circuit contains a fixed $3.3\,\text{mH}$ inductor and is first connected to a $220\,\text{pF}$ capacitor. Its Q-factor is $85$ . The capacitor is then replaced by an $82\,\text{pF}$ component. Assume the Q factor is unchanged by the retuning. Calculate both resonant frequencies and both bandwidths. Explain the effect of the replacement on tuning and selectivity in absolute frequency terms.
[4 marks]
Total for this question: 4

3.13.3.2 · The ideal operational amplifier

Explanation

An operational amplifier is a system building block with inverting and non-inverting signal inputs, an output, and positive and negative power-supply connections. An ideal op amp has infinite open-loop gain and infinite input resistance, so no current enters either signal input.
For a real open-loop device, $V_{\mathrm{out}}=A_{\mathrm{OL}}(V_+-V_-)$ until the output reaches a supply limit.
Used as a comparator, the sign of $V_+-V_-$ drives the output towards the positive or negative saturation level.
A tiny differential input can therefore switch a large output.
Feedback amplifier rules and virtual earth belong to the following configurations; they should not be assumed for an open-loop comparator.

An operational amplifier has two signal inputs, an output and two power-supply connections.

Worked example

An open-loop op amp has $A_{\mathrm{OL}}=2.0\times10^5$ , $V_+=1.001\,\text{V}$ and $V_-=1.000\,\text{V}$ . Calculate the unsaturated output.

1.Find the differential input: $V_+-V_-=0.001\,\text{V}$ .
2.Use $V_{\mathrm{out}}=A_{\mathrm{OL}}(V_+-V_-)$ .
3. $V_{\mathrm{out}}=(2.0\times10^5)(0.001)=200\,\text{V}$ before applying supply limits.

Answer: The calculated open-loop value is $+200\,\text{V}$ , so a practical device on lower supply rails saturates positively.

Common mistakes

Don't reverse the differential input and predict negative output when $V_+>V_-$ .
Don't assume a real op amp can produce the calculated open-loop output beyond its supply rails.
Don't apply the virtual-earth rule to an open-loop comparator with no negative feedback.

Exam tip

For a comparator, determine the sign of $V_+-V_-$ first, then state the corresponding saturation direction.

Tier 1 · Easy

State the current entering either input terminal of an ideal operational amplifier.
[1 mark]
Total for this question: 1
State the open-loop voltage gain of an ideal operational amplifier.
[1 mark]
Total for this question: 1

Tier 2 · Standard

An ideal operational amplifier has negative feedback, its non-inverting input is connected to $0\,\text{V}$ , and its output is not saturated. State the potential of the inverting input and explain why that point is called a virtual earth.
[3 marks]
Total for this question: 3
An ideal operational amplifier is connected as a comparator with supply potentials of $0\,\text{V}$ and $+12\,\text{V}$ . Its input potentials are $V_+=2.10\,\text{V}$ and $V_-=2.40\,\text{V}$ . Determine its output potential.
[2 marks]
Total for this question: 2
A temperature sensor produces $20.0\,\text{mV}$ per $^\circ\text{C}$ , and its output is $0\,\text{V}$ at $0^\circ\text{C}$ . It is to operate an ideal op-amp comparator from $\pm9.0\,\text{V}$ supplies. A $1.20\,\text{V}$ reference is available. Determine the switching temperature and state the input connections needed for the output to switch from negative to positive saturation as temperature rises.
[2 marks]
Total for this question: 2

Tier 3 · Hard

The non-inverting terminal of an ideal op amp is earthed. A $+0.80\,\text{V}$ source is connected to the inverting node through $40\,\text{k}\Omega$ , and negative feedback connects the output to that node through $200\,\text{k}\Omega$ . The supply rails are $\pm6.0\,\text{V}$ . Use the ideal-op-amp rules to determine the output voltage and verify that the assumed linear operation is possible.
[5 marks]
Total for this question: 5
A real open-loop operational amplifier has gain $2.5\times10^5$ and supply rails at $\pm8.0\,\text{V}$ . Its inverting input is fixed at $1.800000\,\text{V}$ . Determine the range of non-inverting input potentials for which the calculated output is unsaturated, and the output when $V_+=1.800020\,\text{V}$ .
[5 marks]
Total for this question: 5
Two ideal op-amp comparators form a temperature-window detector and use $\pm6.0\,\text{V}$ supplies. A sensor produces $V_{\mathrm S}=0.030T+0.20$ , where $V_{\mathrm S}$ is in volts and $T$ is in $^\circ\text{C}$ . Comparator A has $V_+=V_{\mathrm S}$ and $V_-=1.40\,\text{V}$ . Comparator B has $V_+=2.60\,\text{V}$ and $V_-=V_{\mathrm S}$ . Determine the temperature window in which both outputs are positive. State each comparator output below, inside and above this window. Ignore the outputs exactly at the switching temperatures.
[4 marks]
Total for this question: 4
An ideal op amp is used as a comparator on $\pm7.5\,\text{V}$ supplies. During a $12.0\,\text{s}$ test, $V_-=1.55\,\text{V}$ . For $0\leq t\leq6.0\,\text{s}$ , $V_+=0.50+0.30t$ ; for $6.0\leq t\leq12.0\,\text{s}$ , $V_+=4.10-0.30t$ , where potentials are in volts and $t$ is in seconds. Determine both switching times, describe the output over the test and calculate the fraction of the test spent at positive saturation.
[4 marks]
Total for this question: 4
A sensor connected to the non-inverting input of an ideal comparator has nominal output $V_{\mathrm S}=0.90+0.055T$ , with $T$ measured in $^\circ\text{C}$ and $V_{\mathrm S}$ measured in volts. At any temperature its actual output may differ from the nominal value by $\pm0.080\,\text{V}$ . The inverting input is held at $2.55\,\text{V}$ and the supply rails are $\pm10.0\,\text{V}$ . Determine the temperature range that guarantees negative saturation, the range that guarantees positive saturation and the interval in which either output is possible. State why connecting the sensor to an ideal input does not alter its calibration.
[5 marks]
Total for this question: 5

3.13.4.1 · Operational amplifier: inverting amplifier configuration

Explanation

In an inverting amplifier, the non-inverting input is earthed and negative feedback holds the inverting input at approximately $0\,\text{V}$ . This is a virtual earth: it has earth potential but is not directly connected to earth.
Infinite input resistance means no current enters the op amp, so the current through $R_{\mathrm{in}}$ must flow through feedback resistor $R_f$ .
Kirchhoff analysis gives $V_{\mathrm{out}}/V_{\mathrm{in}}=-R_f/R_{\mathrm{in}}$ .
The minus sign means the output is inverted, or $180^\circ$ out of phase.
The derivation, virtual-earth reasoning and gain calculations are all required; real outputs may clip at supply limits.

Inverting amplifier with an earthed non-inverting input and feedback to the virtual-earth node.

Worked example

An inverting amplifier has $R_{\mathrm{in}}=12\,\text{k}\Omega$ , $R_f=72\,\text{k}\Omega$ and $V_{\mathrm{in}}=0.35\,\text{V}$ . Calculate the output.

1.Use $V_{\mathrm{out}}/V_{\mathrm{in}}=-R_f/R_{\mathrm{in}}$ .
2.Gain $=-72/12=-6.0$ .
3. $V_{\mathrm{out}}=(-6.0)(0.35)=-2.10\,\text{V}$ .

Answer: $V_{\mathrm{out}}=-2.1\,\text{V}$ .

Common mistakes

Don't call the virtual-earth node a direct physical connection to earth.
Don't allow current to enter the ideal inverting input instead of directing it through $R_f$ .
Don't drop the minus sign and miss the phase reversal.

Exam tip

For a derivation, state virtual earth and zero input current before applying Kirchhoff's current law.

Tier 1 · Easy

An inverting amplifier has $R_{\text{in}}=10\,\text{k}\Omega$ , $R_f=47\,\text{k}\Omega$ and $V_{\text{in}}=+0.60\,\text{V}$ . Calculate $V_{\text{out}}$ .
[2 marks]
Total for this question: 2
An inverting amplifier uses a $24\,\text{k}\Omega$ input resistor and a $180\,\text{k}\Omega$ feedback resistor. Calculate its closed-loop voltage gain.
[1 mark]
Total for this question: 1

Tier 2 · Standard

The non-inverting input of an ideal operational amplifier is earthed. Its input and feedback resistors are $12\,\text{k}\Omega$ and $68\,\text{k}\Omega$ . Derive the closed-loop gain from currents at the inverting input, then determine the output for $V_{\text{in}}=-0.45\,\text{V}$ .
[3 marks]
Total for this question: 3
An inverting amplifier must convert an input of $-0.28\,\text{V}$ into an output of $+2.52\,\text{V}$ . Its input resistor is $18\,\text{k}\Omega$ . Determine the required feedback resistance.
[2 marks]
Total for this question: 2
An inverting amplifier has voltage gain $-10.0$ and must produce an output peak of $4.00\,\text{V}$ . The output drives no external load, so its current is only the current in the feedback resistor. The maximum permitted output current is $0.500\,\text{mA}$ . Determine the minimum input resistance.
[2 marks]
Total for this question: 2

Tier 3 · Hard

An inverting amplifier has $R_{\text{in}}=15\,\text{k}\Omega$ and $R_f=120\,\text{k}\Omega$ . Its output saturates at $\pm8.0\,\text{V}$ . A sinusoidal input has peak voltage $1.4\,\text{V}$ . Determine the ideal output peak, describe the actual output, and calculate the largest input peak that would avoid clipping.
[5 marks]
Total for this question: 5
A sensor of emf $0.600\,\text{V}$ and internal resistance $4.00\,\text{k}\Omega$ drives an inverting amplifier through a $20.0\,\text{k}\Omega$ input resistor. The feedback resistance is $120\,\text{k}\Omega$ . Determine the output potential, accounting for source loading.
[3 marks]
Total for this question: 3
An inverting amplifier must accept an input peak of $0.420\,\text{V}$ without its output exceeding $4.50\,\text{V}$ . The nominal input resistor is $20.0\,\text{k}\Omega$ and available feedback resistors are $180\,\text{k}\Omega$ and $220\,\text{k}\Omega$ . Every resistor may differ from its nominal value by $\pm5\%$ . Determine the feedback resistor that gives the greatest gain while guaranteeing no clipping.
[4 marks]
Total for this question: 4
An inverting amplifier uses $R_{\text{in}}=32.0\,\text{k}\Omega$ . Its feedback path contains resistor $R_A$ and a switch that can add resistor $R_B$ in series with $R_A$ . The required closed-loop gains are $-3.25$ with the switch open and $-7.75$ with it closed. Determine $R_A$ and $R_B$ . For an input peak of $0.360\,\text{V}$ , calculate both output peaks and the peak current in the feedback path for each setting.
[5 marks]
Total for this question: 5
An inverting amplifier has $R_{\text{in}}=32.0\,\text{k}\Omega$ , $R_f=192\,\text{k}\Omega$ and symmetrical output limits of $\pm5.70\,\text{V}$ . For a constant input potential, the feedback resistor must dissipate no more than $0.180\,\text{mW}$ . Determine the greatest permitted magnitude of input potential and state which restriction sets it.
[4 marks]
Total for this question: 4

3.13.4.2 · Operational amplifier: non-inverting amplifier configuration

Explanation

In a non-inverting amplifier, the signal is applied to the non-inverting input, so output and input have the same polarity. Negative feedback makes the two input potentials approximately equal while the output remains unsaturated.
The resistor divider between output and earth sets the inverting-input potential, leading to $V_{\mathrm{out}}/V_{\mathrm{in}}=1+R_f/R_1$ . Derivation is not required by the specification, but calculations using the formula are.
The gain is positive and cannot be less than one.
Infinite ideal input resistance means the signal source supplies negligible current.
A calculated output must still be checked against the real device's supply-limited output range.

Non-inverting amplifier with a feedback divider from output to the inverting input.

Worked example

A non-inverting amplifier has $R_f=36\,\text{k}\Omega$ , $R_1=9.0\,\text{k}\Omega$ and input $0.40\,\text{V}$ . Calculate the output.

1.Use gain $=1+R_f/R_1$ .
2.Gain $=1+36/9.0=5.0$ .
3. $V_{\mathrm{out}}=(5.0)(0.40)=2.0\,\text{V}$ .

Answer: $V_{\mathrm{out}}=+2.0\,\text{V}$ .

Common mistakes

Don't omit the 1 and use gain $R_f/R_1$ .
Don't add a minus sign even though the signal enters the non-inverting input.
Don't report an output beyond the supply rails without identifying saturation.

Exam tip

The configuration name predicts the sign: a non-inverting amplifier has positive closed-loop gain.

Tier 1 · Easy

A non-inverting amplifier has $R_1=10\,\text{k}\Omega$ , $R_f=39\,\text{k}\Omega$ and input voltage $0.80\,\text{V}$ . Calculate its output voltage.
[2 marks]
Total for this question: 2
A non-inverting amplifier has $R_1=15\,\text{k}\Omega$ and $R_f=105\,\text{k}\Omega$ . Calculate its voltage gain.
[1 mark]
Total for this question: 1

Tier 2 · Standard

In a non-inverting amplifier, $R_1=8.2\,\text{k}\Omega$ joins the inverting input to earth and $R_f=33\,\text{k}\Omega$ joins the output to that input. Use the feedback-divider potential to derive the gain and calculate the output when $V_{\text{in}}=0.50\,\text{V}$ .
[4 marks]
Total for this question: 4
A pressure sensor supplies $0.25\,\text{V}$ to a non-inverting amplifier and requires an output of $2.00\,\text{V}$ . The resistor from the inverting input to earth is $12\,\text{k}\Omega$ . Determine the feedback resistance.
[2 marks]
Total for this question: 2
A non-inverting amplifier has $R_1=18.0\,\text{k}\Omega$ between its inverting input and earth and $R_f=72.0\,\text{k}\Omega$ in the feedback path. Its output is $+7.50\,\text{V}$ . Determine the input potential and the current in the feedback divider.
[3 marks]
Total for this question: 3

Tier 3 · Hard

A non-inverting amplifier uses $R_1=18\,\text{k}\Omega$ and $R_f=82\,\text{k}\Omega$ . Its output saturates at $\pm11.5\,\text{V}$ . Calculate the maximum peak-to-peak sinusoidal input that can be amplified without clipping, and state the corresponding output peak-to-peak voltage.
[5 marks]
Total for this question: 5
A non-inverting amplifier saturates at $\pm13.0\,\text{V}$ . Its input can have a peak potential of $0.900\,\text{V}$ and the resistor from the inverting input to earth is $10.0\,\text{k}\Omega$ . Available feedback resistors are $120\,\text{k}\Omega$ , $140\,\text{k}\Omega$ and $150\,\text{k}\Omega$ . Determine the greatest available resistance that avoids saturation.
[3 marks]
Total for this question: 3
A non-inverting amplifier is designed for voltage gain $7.50$ . When its output is $9.00\,\text{V}$ , the intended current through the feedback divider is $50.0\,\mu\text{A}$ . Determine the ideal values of the resistance $R_1$ from the inverting input to earth and the feedback resistance $R_f$ . Only $150\,\text{k}\Omega$ and $160\,\text{k}\Omega$ are available for $R_f$ , while the ideal $R_1$ is available. Select the closer feedback resistance and calculate the percentage error in gain.
[4 marks]
Total for this question: 4
Two identical non-inverting amplifiers are connected in cascade. Their combined voltage gain must be $64.0$ , and each has $R_1=16.0\,\text{k}\Omega$ from its inverting input to earth. Determine the feedback resistance for each stage. For an input peak of $75.0\,\text{mV}$ , calculate the intermediate and ideal final output peaks, then state the actual final peak if both amplifiers are limited to $\pm4.50\,\text{V}$ .
[5 marks]
Total for this question: 5
A non-inverting amplifier has $R_1=14.0\,\text{k}\Omega$ and $R_f=91.0\,\text{k}\Omega$ . A sinusoidal input has peak potential $0.720\,\text{V}$ . The output can approach no closer than $0.850\,\text{V}$ to either supply rail. Determine the smallest magnitude of the symmetrical supply rails that permits an undistorted output.
[3 marks]
Total for this question: 3

3.13.4.3 · Operational amplifier: summing amplifier configuration

Explanation

A summing amplifier uses several input resistors feeding one virtual-earth inverting node. Because no current enters the ideal input, the algebraic sum of input currents flows through the feedback resistor.
The required result is $V_{\mathrm{out}}=-R_f(V_1/R_1+V_2/R_2+V_3/R_3+\ldots)$ ; derivation is not required.
Each resistor ratio sets that input's weighting, and positive and negative voltages must retain their signs.
The related difference-amplifier configuration produces $V_{\mathrm{out}}=(V_+-V_-)R_f/R_1$ for the specified matched ratios, also without derivation.
Both configurations remain limited by output saturation, so the calculated ideal value may be clipped.

Three weighted inputs feed the virtual-earth summing node of an inverting summing amplifier.

Worked example

A summing amplifier has $R_f=40\,\text{k}\Omega$ . Inputs $0.60\,\text{V}$ and $-0.20\,\text{V}$ use $20\,\text{k}\Omega$ and $10\,\text{k}\Omega$ . Calculate the output.

1.Use $V_{\mathrm{out}}=-R_f(V_1/R_1+V_2/R_2)$ .
2.The signed current terms are $0.60/20=0.030$ and $-0.20/10=-0.020$ .
3. $V_{\mathrm{out}}=-40(0.010)=-0.40\,\text{V}$ .

Answer: $V_{\mathrm{out}}=-0.40\,\text{V}$ .

Common mistakes

Don't add input voltage magnitudes and lose cancellation from a negative input.
Don't use one common input resistance in the calculation when the inputs have different $R_i$ values.
Don't reverse the difference-amplifier subtraction and change the output sign.

Exam tip

For a weighted sum, calculate each signed $V_i/R_i$ term separately before multiplying by $-R_f$ .

Tier 1 · Easy

A summing amplifier has $R_f=20\,\text{k}\Omega$ . Two inputs of $0.30\,\text{V}$ and $0.50\,\text{V}$ are each connected through $10\,\text{k}\Omega$ . Calculate the output voltage.
[2 marks]
Total for this question: 2
A difference amplifier has matched resistor ratios $R_f/R_1=6.0$ . The source signal $V_1=0.25\,\text{V}$ feeds the inverting input and $V_2=0.70\,\text{V}$ feeds the non-inverting input. Use $V_{\text{out}}=(V_2-V_1)R_f/R_1$ to calculate the output potential.
[1 mark]
Total for this question: 1

Tier 2 · Standard

A summing amplifier has $R_f=60\,\text{k}\Omega$ . Inputs $+0.90\,\text{V}$ , $-1.20\,\text{V}$ and $+0.50\,\text{V}$ are connected through $15\,\text{k}\Omega$ , $30\,\text{k}\Omega$ and $20\,\text{k}\Omega$ , respectively. Determine $V_{\text{out}}$ .
[3 marks]
Total for this question: 3
A summing amplifier has $R_f=90\,\text{k}\Omega$ . One input is $+0.40\,\text{V}$ through $30\,\text{k}\Omega$ and a second input $V_2$ is connected through $45\,\text{k}\Omega$ . The output is $-0.60\,\text{V}$ . Determine $V_2$ .
[2 marks]
Total for this question: 2
A three-input summing amplifier has $R_f=40\,\text{k}\Omega$ , with $V_1$ connected through $10\,\text{k}\Omega$ , $V_2$ through $20\,\text{k}\Omega$ and $V_3$ through $40\,\text{k}\Omega$ . Each input is switched independently between $0\,\text{V}$ and $-1.0\,\text{V}$ . Determine the three input settings that produce $V_{\text{out}}=+5.0\,\text{V}$ .
[3 marks]
Total for this question: 3

Tier 3 · Hard

A summing amplifier has $R_f=100\,\text{k}\Omega$ and saturates at $-10.5\,\text{V}$ on negative output. Inputs $+1.20\,\text{V}$ , $-0.60\,\text{V}$ and $+1.50\,\text{V}$ pass through $12\,\text{k}\Omega$ , $30\,\text{k}\Omega$ and $20\,\text{k}\Omega$ . Calculate the ideal output, state the actual output, and find the value to which the third input must be reduced to put the amplifier just at the saturation boundary.
[6 marks]
Total for this question: 6
A two-channel audio summing amplifier has $R_f=120\,\text{k}\Omega$ . With only $V_1=+0.50\,\text{V}$ applied, the output is $-2.00\,\text{V}$ . When $V_2=-0.20\,\text{V}$ is also applied, the output becomes $-0.80\,\text{V}$ . Determine the two input resistances.
[3 marks]
Total for this question: 3
A two-sensor summing amplifier uses a feedback resistance of $180\,\text{k}\Omega$ . Sensor $A$ ranges from $0$ to $+0.600\,\text{V}$ and sensor $B$ ranges from $0$ to $+1.20\,\text{V}$ . At their maximum inputs, the contribution from sensor $A$ to the magnitude of the output must be twice the contribution from sensor $B$ . The combined output must then be just at the negative saturation boundary of $-9.00\,\text{V}$ . Determine the two input resistances $R_A$ and $R_B$ .
[4 marks]
Total for this question: 4
A summing amplifier is to be used as a four-bit digital-to-analogue converter. A logic $1$ supplies $+0.800\,\text{V}$ and logic $0$ supplies $0\,\text{V}$ . The feedback resistance is $160\,\text{k}\Omega$ , and the least significant bit must change the ideal output by $0.400\,\text{V}$ . Design the four input resistances to give binary weighting, stating them from most significant bit to least significant bit. The negative output limit is $-4.50\,\text{V}$ ; identify the first input code, in ascending binary order, that saturates the amplifier.
[5 marks]
Total for this question: 5
A summing amplifier must convert a sensor potential that ranges from $+0.200\,\text{V}$ to $+1.00\,\text{V}$ into an output that changes linearly from $+6.00\,\text{V}$ to $-6.00\,\text{V}$ . A constant reference input of $-0.750\,\text{V}$ is available and $R_f=180\,\text{k}\Omega$ . Determine the input resistance for the sensor and the input resistance for the reference.
[5 marks]
Total for this question: 5

3.13.4.4 · Real operational amplifiers

Explanation

Real operational amplifiers depart from the ideal model: open-loop gain and input resistance are finite, output voltage and current are limited, and response depends on frequency. A frequency-response curve shows gain approximately constant at low frequency, then falling above a corner frequency.
For a given device, gain multiplied by bandwidth is approximately constant, so increasing closed-loop gain reduces usable bandwidth.
Signals above that bandwidth are attenuated and phase shifted.
The specification requires limitations, the response curve and the gain–bandwidth relationship; a detailed slew-rate treatment is not required.
Examiners may ask for a bandwidth calculation or for an explanation of why a high-gain circuit cannot amplify high-frequency signals faithfully.

A real op amp's gain falls above its bandwidth; gain multiplied by bandwidth is approximately constant.

Worked example

An op amp has gain–bandwidth product $2.4\,\text{MHz}$ . Calculate the bandwidth at closed-loop gain $80$ .

1.Use gain multiplied by bandwidth equals the device constant.
2. $f_B=(2.4\times10^6)/80$ .
3. $f_B=3.0\times10^4\,\text{Hz}$ .

Answer: The bandwidth is $30\,\text{kHz}$ .

Common mistakes

Don't assume a real op amp retains its low-frequency gain at every frequency.
Don't multiply the gain–bandwidth product by gain instead of dividing by gain.
Don't attribute all real-device error to finite input resistance and ignore output limits and frequency response.

Exam tip

On a response graph, read bandwidth at the end of the flat-gain region and link greater gain to smaller bandwidth.

Tier 1 · Easy

An operational amplifier has a gain--bandwidth product of $3.0\,\text{MHz}$ . Calculate its bandwidth when its closed-loop voltage gain is $30$ .
[2 marks]
Total for this question: 2
State two limitations of a real operational amplifier compared with an ideal operational amplifier.
[2 marks]
Total for this question: 2

Tier 2 · Standard

A real operational amplifier has slew rate $1.5\,\text{V}\,\mu\text{s}^{-1}$ . Determine whether it can produce an undistorted sine wave of frequency $80\,\text{kHz}$ and peak output $4.0\,\text{V}$ . Also calculate the greatest frequency allowed by the slew rate at this peak voltage.
[4 marks]
Total for this question: 4
A real operational amplifier has a gain--bandwidth product of $1.8\,\text{MHz}$ . Calculate the greatest closed-loop gain usable at $45\,\text{kHz}$ and discuss whether a closed-loop gain of $60$ is suitable at this frequency.
[2 marks]
Total for this question: 2
A source of emf $0.800\,\text{V}$ and internal resistance $200\,\text{k}\Omega$ drives a real operational amplifier whose input resistance at its input terminals is $800\,\text{k}\Omega$ . Treat these resistances as a potential divider. The closed-loop gain is $10.0$ and the output is within its limits. Determine the actual output and the percentage by which it is below the ideal output for infinite input resistance.
[3 marks]
Total for this question: 3

Tier 3 · Hard

An operational amplifier has gain--bandwidth product $4.0\,\text{MHz}$ and slew rate $0.80\,\text{V}\,\mu\text{s}^{-1}$ . It is used at closed-loop gain $25$ to produce a $60\,\text{kHz}$ sine wave whose ideal output peak is $3.0\,\text{V}$ . Test both frequency limitations and calculate the largest undistorted output peak at this frequency.
[6 marks]
Total for this question: 6
A sensor signal has peak potential $0.180\,\text{V}$ and frequency $100\,\text{kHz}$ . It requires closed-loop gain $40$ . Op amp A has gain--bandwidth product $2.4\,\text{MHz}$ and op amp B has gain--bandwidth product $6.0\,\text{MHz}$ ; both have maximum output peak $8.0\,\text{V}$ . Discuss which op amp is suitable.
[3 marks]
Total for this question: 3
A real operational amplifier has gain--bandwidth product $3.60\,\text{MHz}$ , maximum output peak $9.00\,\text{V}$ and maximum output current $8.00\,\text{mA}$ . It drives a $470\,\Omega$ load with a $0.150\,\text{V}$ peak input at $80.0\,\text{kHz}$ . Ignore the current in the feedback network. Its closed-loop gain can be set to any positive whole number. Determine the greatest gain for which none of the three limits is exceeded.
[4 marks]
Total for this question: 4
A technician measures the frequency response of a real operational amplifier at three closed-loop gains. The measured gain and bandwidth pairs are $25.0$ with $160\,\text{kHz}$ , $80.0$ with $50.0\,\text{kHz}$ , and $125$ with $28.0\,\text{kHz}$ . Assuming a constant gain--bandwidth product, evaluate the measurements, identify the anomalous pair and predict the bandwidth that should have been recorded for that gain.
[4 marks]
Total for this question: 4
Identical op amps each have gain--bandwidth product $2.10\,\text{MHz}$ . They are to be cascaded to provide total voltage gain $1000$ for a signal containing frequencies up to $75.0\,\text{kHz}$ . Determine the minimum number of identical stages required. Calculate the gain and bandwidth of each stage in that arrangement and, if each stage is non-inverting with $R_1=15.0\,\text{k}\Omega$ , determine its feedback resistance.
[5 marks]
Total for this question: 5

3.13.5.1 · Combinational logic

Explanation

A combinational circuit's output depends only on its present inputs. Boolean notation links expressions, truth tables and gate networks: $\overline A$ means NOT A, $A\cdot B$ means A AND B, and $A+B$ means A OR B.
Students must identify and use AND, NAND, OR, NOR, NOT and EOR gates in combinations. EOR outputs 1 only when its two inputs differ.
To construct a circuit from a truth table, form Boolean terms for rows with output 1 and combine them, or simplify an equivalent expression.
To deduce a truth table, evaluate intermediate gate outputs systematically.
Internal gate circuitry is not required.

Worked example

Write a Boolean expression for an output that is 1 only when $A=1$ and $B=0$ , then name the required gates.

1.Invert B to obtain $\overline B$ .
2.AND A with the inverted input.
3.The output is $Y=A\cdot\overline B$ .

Answer: Use a NOT gate on B followed by an AND gate: $Y=A\cdot\overline B$ .

Common mistakes

Don't treat Boolean plus as ordinary arithmetic addition instead of OR.
Don't state that EOR outputs 1 when both inputs are 1.
Don't build a circuit for rows where the truth-table output is 0 rather than the required output-1 rows.

Exam tip

When deducing a multi-gate truth table, add one intermediate-output column for each gate.

Tier 1 · Easy

For the Boolean function $Y=A\cdot\overline{B}$ , state the four values of $Y$ for inputs $AB=00,01,10,11$ in that order.
[2 marks]
Total for this question: 2
Identify the single gate equivalent to the Boolean expression $Y=\overline{A+B}$ .
[1 mark]
Total for this question: 1

Tier 2 · Standard

A two-input circuit must output $1$ for $AB=01$ and $AB=10$ , but $0$ for $AB=00$ and $AB=11$ . Identify the single gate that performs this function and write an equivalent Boolean expression using only AND, OR and NOT.
[3 marks]
Total for this question: 3
A three-input logic system has outputs $0,0,0,1,0,1,0,1$ when $ABC$ runs in binary order from $000$ to $111$ . Deduce a minimal Boolean expression and state the two-gate implementation.
[2 marks]
Total for this question: 2
A fault indicator has inputs $A$ , $B$ and $C$ . Its output is $1$ only when exactly one input is $1$ . Deduce a Boolean expression using only AND, OR and NOT, and state a gate sequence that implements it.
[2 marks]
Total for this question: 2

Tier 3 · Hard

A control output is $Y=(A+B)\cdot\overline{C}$ . Construct its eight-row truth-table output with $ABC$ in ascending binary order from $000$ to $111$ , then describe a NAND-only implementation.
[6 marks]
Total for this question: 6
A circuit must implement $Q=AB+AC$ using NAND gates only. Write an equivalent NAND-only expression and state the minimum number of NAND gates required.
[2 marks]
Total for this question: 2
A logic chain has $X=A\operatorname{\ EOR}B$ and $Y=X\operatorname{\ EOR}C$ . State the eight output values, in order from input 000 to 111, of $Y$ for the input rows from $ABC=000$ to $ABC=111$ in standard binary order. Deduce a single expression for $Y$ in terms of $A$ , $B$ and $C$ .
[3 marks]
Total for this question: 3
Two unsigned two-bit numbers are $X=2A+B$ and $Z=2C+D$ , where $A$ and $C$ are the most significant bits. An output $Y$ must be $1$ only when $X>Z$ . Deduce a Boolean expression for $Y$ using AND, OR, NOT and EOR gates.
[3 marks]
Total for this question: 3
A four-way data selector has data inputs $D_0,D_1,D_2,D_3$ and selector inputs $S_1,S_0$ . Its output equals $D_0,D_1,D_2$ or $D_3$ when $S_1S_0$ is $00,01,10$ or $11$ , respectively. Write a Boolean expression using only AND, OR and NOT that implements the selector. The data inputs are then fixed at $D_0=0$ , $D_1=1$ , $D_2=1$ and $D_3=0$ . Deduce the single gate that can replace the complete selector for this setting.
[4 marks]
Total for this question: 4

3.13.5.2 · Sequential logic

Explanation

Sequential logic has memory, so outputs depend on previous state as well as present inputs. Required counting circuits are binary, BCD and Johnson counters.
Clock pulses trigger state changes; reset establishes a known starting state, and an up/down input selects counting direction where provided. A binary counter progresses through binary outputs, while BCD uses the ten states for decimal 0–9.
A Johnson counter circulates a pattern through its outputs.
A modulo- $n$ counter has $n$ stable states: logic detects the next unwanted state and drives reset, returning the count to zero.
Gates are building blocks; their internal circuitry is not required.

A modulo counter uses output-decoding logic to drive the reset input at the chosen count.

Worked example

A three-bit up-counter is made modulo 6. State the stable sequence and the state decoded to reset it.

1.Modulo 6 needs six stable states representing decimal 0 to 5.
2.The sequence is $000,001,010,011,100,101$ .
3.The next state is decimal 6, binary $110$ , so logic decodes $110$ to reset.

Answer: The counter cycles from $000$ to $101$ and resets when $110$ is detected.

Common mistakes

Don't call a three-bit binary counter modulo 3 rather than recognising its eight possible states.
Don't include the decoded reset state as a stable output of the modulo counter.
Don't confuse BCD's ten decimal-digit states with a full four-bit binary count of sixteen states.

Exam tip

For modulo- $n$ , list states 0 to $n-1$ , then decode state $n$ to operate reset.

Tier 1 · Easy

A three-bit up-counter is reset to $000$ . State its output after five clock pulses, writing the most significant bit first.
[2 marks]
Total for this question: 2
A BCD up-counter initially displays binary $0111$ . State its output after six clock pulses.
[2 marks]
Total for this question: 2

Tier 2 · Standard

State the number of distinct states of a four-stage Johnson counter and of a four-bit BCD counter. A BCD counter receives a $1.20\,\text{kHz}$ clock; determine the rate at which it completes full count cycles.
[4 marks]
Total for this question: 4
Three successive outputs of a four-bit binary counter are $0101$ , $0100$ and $0011$ . State the setting of its up/down control input and predict the next output.
[2 marks]
Total for this question: 2
A four-bit binary up/down counter starts at $1101$ . It counts up for seven clock pulses and then counts down for ten pulses, wrapping between $1111$ and $0000$ when necessary. Determine its final output.
[2 marks]
Total for this question: 2

Tier 3 · Hard

A three-bit up-counter is converted to a modulo- $6$ counter by decoding one state to reset it. Identify the decoded reset state, list the stable count sequence, determine the output after $53$ pulses from reset, and calculate the cycle rate for a $24\,\text{kHz}$ clock.
[6 marks]
Total for this question: 6
A four-stage Johnson counter starts at $Q_3Q_2Q_1Q_0=0000$ . On each pulse, $\overline{Q_0}$ enters $Q_3$ and every other bit shifts one place towards $Q_0$ . Determine the state after $37$ pulses.
[4 marks]
Total for this question: 4
A five-stage Johnson counter starts at $Q_4Q_3Q_2Q_1Q_0=00000$ . At every clock edge the value NOT $Q_0$ is loaded into $Q_4$ , while the stored values advance towards $Q_0$ . An output is formed by $Y=Q_4\operatorname{\ EOR}Q_2$ . For a $30.0\,\text{kHz}$ clock, determine the repeating state sequence, the corresponding repeating pattern of $Y$ and the frequency of $Y$ .
[3 marks]
Total for this question: 3
A four-bit binary up-counter starts at $0000$ and is driven by a $1.60\,\text{MHz}$ clock. The outputs $Q_0$ to $Q_3$ have $Q_0$ as the least significant bit. Determine the frequencies at $Q_0$ and $Q_3$ , and state the complete counter output after $2027$ clock pulses.
[4 marks]
Total for this question: 4
Two cascaded BCD down-counters display a number from $00$ to $99$ . The units counter clocks the tens counter once whenever it changes from $0$ to $9$ . Starting from $42$ , the system receives $137$ clock pulses. Determine the number of clock pulses received by the tens counter, the final two-digit display and the four-bit BCD output of each counter. The input clock frequency is $2.50\,\text{kHz}$ ; also calculate the rate at which the complete two-digit count pattern repeats.
[5 marks]
Total for this question: 5

3.13.5.3 · Astables

Explanation

An astable has no stable state and oscillates continuously, providing a clock pulse train. Pulse rate is frequency, with $f=1/T$ .
Pulse width is the duration of one high or low level as stated. Duty cycle is the high-time fraction of a period, while mark-to-space ratio compares high time with low time.
An external RC network controls the charging and discharging times, so changing resistance or capacitance changes running frequency.
A larger RC time constant usually gives a longer period and lower frequency.
No particular circuit or device, such as a 555 timer, is required; questions must supply any device-specific timing relationship.

An astable clock waveform showing pulse width and period.

Worked example

An astable output is high for $0.80\,\text{ms}$ and low for $1.20\,\text{ms}$ . Calculate frequency, duty cycle and mark-to-space ratio.

1. $T=0.80+1.20=2.00\,\text{ms}$ , so $f=1/T=500\,\text{Hz}$ .
2.Duty cycle $=(0.80/2.00)\times100=40\%$ .
3.Mark-to-space ratio $=0.80:1.20=2:3$ .

Answer: $f=500\,\text{Hz}$ , duty cycle $40\%$ , mark-to-space ratio $2:3$ .

Common mistakes

Don't use only the high pulse width as the complete period.
Don't calculate duty cycle using low time divided by period.
Don't state that increasing the external RC time constant increases running frequency.

Exam tip

Mark the high time and low time on the waveform before finding period, duty cycle or mark-to-space ratio.

Tier 1 · Easy

An astable produces a pulse train of period $2.5\,\text{ms}$ with a high pulse lasting $1.0\,\text{ms}$ . Calculate the frequency and duty cycle.
[2 marks]
Total for this question: 2
An astable runs at $2.0\,\text{kHz}$ and its output remains high for $0.18\,\text{ms}$ in each cycle. Calculate the low time.
[2 marks]
Total for this question: 2

Tier 2 · Standard

In an astable, a capacitor charges through $47\,\text{k}\Omega$ from $V_S/3$ to $2V_S/3$ and then discharges through the same resistance over the reverse threshold interval. The capacitance is $22\,\text{nF}$ . Using exponential charging, determine the period and frequency.
[4 marks]
Total for this question: 4
For one astable circuit the supplied timing relation is $T=1.4RC$ . Determine the capacitance needed for a frequency of $250\,\text{Hz}$ when $R=68\,\text{k}\Omega$ .
[2 marks]
Total for this question: 2
An astable obeys the supplied relation $T=1.4RC$ . At $20\,{}^\circ\text{C}$ , $R=56.0\,\text{k}\Omega$ and $C=33.0\,\text{nF}$ . At a higher temperature its frequency is $8.00\%$ lower and the capacitance is unchanged. Determine the new effective resistance and its percentage increase.
[2 marks]
Total for this question: 2

Tier 3 · Hard

An astable capacitor switches between $V_S/3$ and $2V_S/3$ , with the output high while the capacitor charges. It charges through $82\,\text{k}\Omega$ , discharges through $33\,\text{k}\Omega$ , and has capacitance $15\,\text{nF}$ . Calculate the high time, low time, frequency, duty cycle and mark-to-space ratio, taking each threshold time as $RC\ln2$ .
[6 marks]
Total for this question: 6
An astable is required to run at $3.50\,\text{kHz}$ with duty cycle $60.0\%$ . Its capacitor is $12.0\,\text{nF}$ and its timing equations are $t_{\text{high}}=0.7(R_1+R_2)C$ and $t_{\text{low}}=0.7R_2C$ . Determine $R_1$ and $R_2$ .
[2 marks]
Total for this question: 2
An astable obeys $T=0.7(R_1+2R_2)C$ and $t_{\text{high}}=0.7(R_1+R_2)C$ . The nominal values are $R_1=46.52\,\text{k}\Omega$ , $R_2=9.90\,\text{k}\Omega$ and $C=22.0\,\text{nF}$ . Each resistor has independent tolerance $\pm5\%$ and the capacitor has tolerance $\pm10\%$ . Determine the full possible ranges of frequency and duty cycle.
[4 marks]
Total for this question: 4
An astable has $t_{\text{high}}=0.7(R+R_T)C$ and $t_{\text{low}}=0.7RC$ , where $R=20.0\,\text{k}\Omega$ and $C=10.0\,\text{nF}$ . A thermistor has resistance $30.0\,\text{k}\Omega$ at the lower temperature and $10.0\,\text{k}\Omega$ at the higher temperature. Determine the frequency and duty cycle at each temperature, and explain both changes.
[5 marks]
Total for this question: 5
For an astable, the low interval is $0.7R_2C$ , whereas the high interval is $0.7(R_1+R_2)C$ . Initially $R_1=15.0\,\text{k}\Omega$ , $R_2=10.0\,\text{k}\Omega$ and $C=20.0\,\text{nF}$ . A load dissipates $0.360\,\text{W}$ while the output is high and negligible power while it is low; its maximum permitted average power is $0.240\,\text{W}$ . Determine the initial frequency and average power. Keeping $R_2$ and $C$ unchanged, calculate the greatest allowed $R_1$ and the resulting frequency.
[5 marks]
Total for this question: 5

3.13.6.1 · Principles of communication systems

Explanation

A real-time communication system is represented by a sequence of functional blocks. The input transducer converts information such as sound into an electrical signal.
The transmitter processes that signal into a form suitable for the chosen channel, may modulate a carrier, and supplies enough power. The transmission channel carries the signal and may add attenuation, noise or interference.
The receiver selects the wanted signal, amplifies it and recovers the information. Finally, an output transducer converts the recovered electrical signal into the required form, such as sound.
Only the purpose of each stage is required, not its internal circuit.
The block order and energy or information transformations must remain clear.

Functional blocks of a real-time communication system in signal-flow order.

Worked example

For a live radio broadcast, state the purpose of the transmitter, channel and receiver.

1.The transmitter prepares the information signal, modulates a carrier and launches the signal.
2.The radio channel carries the electromagnetic wave and may add noise or attenuation.
3.The receiver selects the wanted signal and demodulates it to recover the information.

Answer: The three stages prepare and launch, carry, then select and recover the information signal.

Common mistakes

Don't place the receiver before the transmission channel.
Don't call the carrier the original information rather than the wave modified to carry it.
Don't describe detailed circuitry instead of the purpose of each functional block.

Exam tip

For a block-diagram question, use one precise conversion or purpose statement per stage.

Tier 1 · Easy

Place these stages of a real-time communication system in order: receiver, output transducer, transmission channel, input transducer, transmitter.
[2 marks]
Total for this question: 2
State the energy conversion carried out by a microphone at the input and by a loudspeaker at the output of a live audio link.
[2 marks]
Total for this question: 2

Tier 2 · Standard

In a live radio link, explain the different purposes of the transmitter, transmission channel and receiver.
[3 marks]
Total for this question: 3
In a fibre-optic intercom, a test instrument shows the correct electrical audio signal at the receiver output, but the user hears no sound. Identify the faulty functional block and justify why the earlier blocks are not responsible.
[3 marks]
Total for this question: 3
A remote river monitor uses water level as its information source and shows the recovered information as a visible reading at a control station. State the purposes of the input transducer and the output transducer, including the energy or signal conversion performed by each.
[3 marks]
Total for this question: 3

Tier 3 · Hard

A remote weather station sends a continuously updated temperature reading to a control room. Describe a complete real-time communication system for this task, giving the purpose of each block and explaining where unwanted noise can affect the recovered reading.
[5 marks]
Total for this question: 5
During a communication-system test, the signal at the transmitter output is correct but no signal reaches the receiver input. After that fault is repaired, the signal at the receiver input is correct but the receiver output is distorted. Identify the two faulty functional blocks and justify each choice.
[3 marks]
Total for this question: 3
A radio channel attenuates the transmitted signal and adds noise before the signal reaches the receiver. Explain why increasing transmitter power can improve the quality of the recovered information, whereas increasing receiver gain by the same factor cannot restore information already obscured by the channel noise.
[4 marks]
Total for this question: 4
A radio link between two valleys uses an unattended relay on a ridge. Starting with a microphone and ending with a loudspeaker, state the complete functional-block order and explain the separate purposes of the relay receiver and relay transmitter.
[3 marks]
Total for this question: 3
A remotely operated machine needs a control signal from an operator and a separate position signal returned to the operator. Outline the functional-block order of each communication chain and explain why two chains are required.
[3 marks]
Total for this question: 3

3.13.6.2 · Transmission media

Explanation

Transmission media include metal wire, optical fibre and electromagnetic paths using radio or microwave.
Long-wavelength ground waves diffract around Earth's surface; sky waves can be refracted or reflected by atmospheric layers and return beyond the horizon.
Microwaves support line-of-sight terrestrial and satellite links at typical supplied frequencies.
Satellite uplink and downlink frequencies differ so a strong transmitted signal does not de-sense the receiver.
Media must be compared through data rate, cost and security: fibre offers high capacity and confines the signal but is costly to install; metal wire is established but has limited bandwidth and can radiate; radio covers wide areas without cable but is easier to intercept and shares spectrum.

Ground-wave diffraction and sky-wave refraction or reflection can carry radio beyond the horizon.

Worked example

Compare optical fibre with a radio link for data rate and security.

1.Optical fibre has a large available bandwidth and therefore supports a high data rate.
2.Its guided signal remains inside a physical cable, making interception more difficult.
3.Radio avoids cable installation and covers a wide area, but shared spectrum limits capacity and broadcast signals are easier to intercept.

Answer: Fibre normally offers higher data capacity and physical security; radio offers flexible wide-area coverage at lower installation cost.

Common mistakes

Don't claim all radio propagation is line of sight and ignore ground-wave diffraction and sky waves.
Don't use the same satellite frequency for uplink and downlink and omit receiver de-sensing.
Don't call fibre automatically cheapest without considering installation and interface cost.

Exam tip

A compare question should apply the named criteria—data rate, cost and security—to both media.

Tier 1 · Easy

Identify a suitable transmission medium for a high-data-rate link that should be difficult to intercept, and give one reason for your choice.
[2 marks]
Total for this question: 2
Explain why terrestrial microwave relay stations normally require an unobstructed line of sight.
[2 marks]
Total for this question: 2

Tier 2 · Standard

Explain two ways in which a radio transmission can reach a receiver beyond the horizon, and relate each way to the wave behaviour involved.
[4 marks]
Total for this question: 4
A temporary outdoor event needs a communication link installed quickly between mobile staff. Discuss whether a radio link or newly buried copper cable is more suitable, referring to cost and security.
[3 marks]
Total for this question: 3
An emergency beacon must reach receivers beyond the horizon and behind hills without relay stations. It can use either a $200\,\text{kHz}$ carrier or a $5.00\,\text{GHz}$ carrier. Using $c=3.00\times10^8\,\text{m s}^{-1}$ , determine the more suitable carrier and explain the propagation difference.
[3 marks]
Total for this question: 3

Tier 3 · Hard

A signal travels from one ground station to another through a geostationary satellite $3.60\times10^7\,\text{m}$ above Earth. Treat both path sections as vertical and use $c=3.00\times10^8\,\text{m s}^{-1}$ . Calculate the minimum one-way delay, explain why uplink and downlink frequencies differ, and compare this link with optical fibre for data rate and security.
[6 marks]
Total for this question: 6
A hospital must transfer a $750\,\text{MB}$ scan in at most $5.0\,\text{s}$ . Use $1\,\text{byte}=8\,\text{bits}$ . Installed links offer optical fibre at $2.0\,\text{Gbit s}^{-1}$ , microwave radio at $400\,\text{Mbit s}^{-1}$ or copper wire at $200\,\text{Mbit s}^{-1}$ . Determine the suitable medium, including a security reason.
[4 marks]
Total for this question: 4
At a satellite receiver, the wanted uplink signal has power $8.0\times10^{-12}\,\text{W}$ . Leakage from the satellite's downlink transmitter before filtering would be $2.0\times10^{-6}\,\text{W}$ . The residual leakage must be no more than $10\%$ of the uplink power. Two frequency-selective filters reduce the leakage by factors of $1.0\times10^6$ and $1.0\times10^7$ , respectively. Determine the minimum required reduction factor and select the suitable filter.
[4 marks]
Total for this question: 4
Two microwave stations are $70.0\,\text{km}$ apart. Their aerials are $45.0\,\text{m}$ and $80.0\,\text{m}$ above level ground. Use $d=\sqrt{2Rh}$ for the distance to the horizon from height $h$ , with Earth radius $R=6.37\times10^6\,\text{m}$ . Determine whether the direct path is line of sight. A $25.0\,\text{m}$ relay can be placed midway between the stations; determine whether one such relay provides line of sight for both sections.
[5 marks]
Total for this question: 5
Model an ionised layer as horizontal and $225\,\text{km}$ above Earth. The aerial cannot radiate rays at more than $62.0^\circ$ to the vertical. Treat each returned hop as symmetrical about its highest point. Determine the greatest ground distance covered by one hop. A link must span $2.75\times10^3\,\text{km}$ ; determine the minimum number of hops and the angle to the vertical when equal hops cover this distance.
[5 marks]
Total for this question: 5

3.13.6.3 · Time-division multiplexing

Explanation

Time-division multiplexing allows several channels to share one transmission path by assigning each a recurring time slot. A frame contains one slot from each channel in a fixed order, then the pattern repeats.
If each frame carries one new sample per channel, frame rate equals each channel's sampling rate; slot rate is frame rate multiplied by channel count.
Total bit rate is frame rate multiplied by all bits per frame, including synchronisation or control bits.
The receiver uses timing information to separate, or demultiplex, the slots and return samples to their correct channels.
Synchronisation is essential because a one-slot timing error would exchange channel data.

A TDM frame assigns one recurring slot to each channel plus synchronisation information.

Worked example

Eight channels are sampled at $6.0\,\text{kHz}$ using $12$ bits per sample, with $4$ sync bits per frame. Calculate the link bit rate.

1.One frame has $8(12)+4=100$ bits.
2.One sample per channel per frame gives frame rate $6.0\times10^3\,\text{s}^{-1}$ .
3.Bit rate $=(100)(6.0\times10^3)=6.0\times10^5\,\text{bit s}^{-1}$ .

Answer: The required bit rate is $600\,\text{kbit s}^{-1}$ .

Common mistakes

Don't omit synchronisation bits when calculating total bits per frame.
Don't multiply by channel count twice after already counting every channel's bits in a frame.
Don't state that TDM sends every channel continuously rather than allocating recurring time slots.

Exam tip

Write bits per frame first, including overhead, then multiply once by frame rate.

Tier 1 · Easy

Four channels share a time-division multiplexed link. Each frame contains one $8$ -bit slot from every channel. State the number of slots and the number of data bits in one frame.
[2 marks]
Total for this question: 2
A time-division multiplexed system sends $20\,000$ frames each second. Calculate the duration of one frame.
[1 mark]
Total for this question: 1

Tier 2 · Standard

Six signals are each sampled at $8.0\,\text{kHz}$ and represented by $10$ bits per sample. One sample from each signal forms a TDM frame with no overhead. Calculate the frame rate, slot rate and transmitted bit rate.
[4 marks]
Total for this question: 4
Five channels share a $780\,\text{kbit s}^{-1}$ TDM link. Each frame contains one $12$ -bit sample from each channel and $5$ synchronisation bits. Determine the greatest sampling rate per channel.
[2 marks]
Total for this question: 2
A TDM frame consists of one synchronisation bit followed, in order, by equal-length slots for channels $A$ , $B$ and $C$ . One synchronisation bit is lost in transmission, so the receiver no longer identifies the correct start of that frame. Explain the consequence for demultiplexing and the recovered channel data until synchronisation is regained.
[3 marks]
Total for this question: 3

Tier 3 · Hard

A TDM system carries $24$ channels, each sampled at $12.0\,\text{kHz}$ with $12$ bits per sample. Every frame also contains $8$ synchronisation bits. Calculate the frame rate, data-slot rate, total bits per frame and transmitted bit rate. Decide whether a $4.00\,\text{Mbit s}^{-1}$ link has sufficient capacity.
[6 marks]
Total for this question: 6
A TDM link combines four digital channels. Each frame contains one $10$ -bit sample from each channel and $4$ synchronisation bits. The frame rate is $8.0\,\text{kHz}$ . Determine the transmitted bit rate and explain how the receiver separates the channels.
[3 marks]
Total for this question: 3
A TDM link transmits at $512\,\text{kbit s}^{-1}$ . All channels are sampled at $4.00\,\text{kHz}$ , with one $12$ -bit sample from each channel in every frame. Each frame also contains $8$ synchronisation bits. Each analogue channel contains frequencies up to $1.50\,\text{kHz}$ . Determine the number of channels carried and decide whether the sampling rate satisfies the Nyquist condition.
[3 marks]
Total for this question: 3
A TDM system combines three channels with different sampling requirements. Channel $A$ supplies $10$ -bit samples at $12.0\,\text{kHz}$ , channel $B$ supplies $12$ -bit samples at $6.00\,\text{kHz}$ and channel $C$ supplies $16$ -bit samples at $3.00\,\text{kHz}$ . A master frame repeats at $3.00\,\text{kHz}$ and contains $6$ synchronisation bits. Determine the number of sample slots allocated to each channel in a master frame, the total bits per master frame and the transmitted bit rate.
[5 marks]
Total for this question: 5
A $1.12\,\text{Mbit s}^{-1}$ TDM link sends $8000$ frames each second. Every frame carries one $12$ -bit sample from each of six audio channels and $8$ synchronisation bits. Each extra sensor needs one $10$ -bit sample in every frame. Determine the number of spare bits in one frame before sensor data are added and the greatest number of sensors that can be carried at one sample per frame.
[2 marks]
Total for this question: 2

3.13.6.4 · Amplitude (AM) and frequency modulation (FM) techniques

Explanation

Modulation changes a high-frequency carrier according to a lower-frequency information signal. In amplitude modulation, carrier amplitude follows the information while carrier frequency stays fixed.
In frequency modulation, instantaneous carrier frequency varies while amplitude stays approximately constant. From a time graph, rapid oscillations give carrier frequency; the slower envelope repetition in AM, or compression–spreading pattern in FM, gives information frequency.
For maximum information frequency $f_M$ , simple AM bandwidth is $2f_M$ . For FM with maximum deviation $\Delta f$ , bandwidth is $2(\Delta f+f_M)$ .
Detailed modulation circuits and advanced mathematical treatment are not required.
Wider available channel bandwidth permits greater data capacity but fewer non-overlapping channels.

AM varies carrier amplitude, while FM varies carrier spacing according to the information signal.

Worked example

A signal has $f_M=7.0\,\text{kHz}$ . For FM, $\Delta f=22\,\text{kHz}$ . Calculate AM and FM bandwidths.

1.AM bandwidth $=2f_M=2(7.0)=14\,\text{kHz}$ .
2.FM bandwidth $=2(\Delta f+f_M)$ .
3.FM bandwidth $=2(22+7.0)=58\,\text{kHz}$ .

Answer: AM requires $14\,\text{kHz}$ and FM requires $58\,\text{kHz}$ .

Common mistakes

Don't change carrier frequency in an AM sketch instead of changing its amplitude.
Don't use the rapid carrier oscillation count as the information frequency.
Don't quote AM bandwidth as $f_M$ and omit the second sideband.

Exam tip

On a modulation graph, identify carrier frequency from the fast cycles and information frequency from the slow pattern.

Tier 1 · Easy

An AM transmission carries a highest information frequency of $6.0\,\text{kHz}$ . Calculate its minimum bandwidth.
[2 marks]
Total for this question: 2
Identify the modulation technique in which a carrier keeps constant amplitude while its cycle spacing changes with an information signal.
[1 mark]
Total for this question: 1

Tier 2 · Standard

A signal with maximum information frequency $8.0\,\text{kHz}$ frequency-modulates a carrier with maximum deviation $25\,\text{kHz}$ . Calculate the FM bandwidth and the bandwidth if the same information used AM.
[4 marks]
Total for this question: 4
The spectrum of an AM transmission contains lines at $1.431\,\text{MHz}$ , $1.438\,\text{MHz}$ and $1.445\,\text{MHz}$ . Identify the carrier and the two sidebands, infer the highest information frequency, and state the occupied bandwidth.
[3 marks]
Total for this question: 3
The instantaneous frequency of an FM carrier ranges symmetrically from $95.925\,\text{MHz}$ to $96.075\,\text{MHz}$ . The highest information frequency is $12.0\,\text{kHz}$ . Determine the unmodulated carrier frequency, maximum frequency deviation and FM bandwidth.
[3 marks]
Total for this question: 3

Tier 3 · Hard

A broadcast signal has maximum information frequency $15\,\text{kHz}$ . In FM its maximum frequency deviation is $75\,\text{kHz}$ . Calculate the FM and AM bandwidths and hence the greatest ideal number of non-overlapping channels of each type in a $900\,\text{kHz}$ allocation. Explain one signal-to-noise advantage and one channel-usage disadvantage of FM.
[6 marks]
Total for this question: 6
A voltage--time graph of an FM signal shows that the cycle-spacing pattern repeats every $0.50\,\text{ms}$ . The trace contains $125$ complete carrier cycles during one repetition of that pattern. Determine the carrier frequency and the information frequency.
[2 marks]
Total for this question: 2
A telemetry channel has usable bandwidth $136\,\text{kHz}$ . To meet its noise-performance requirement, an FM transmission must have maximum frequency deviation at least three times its highest information frequency. Determine the greatest possible information frequency and the corresponding minimum frequency deviation.
[2 marks]
Total for this question: 2
Two AM transmitters have carrier frequencies $720\,\text{kHz}$ and $746\,\text{kHz}$ . Their highest information frequencies are $9.00\,\text{kHz}$ and $13.0\,\text{kHz}$ , respectively. Determine the occupied frequency range of each transmission and the unused frequency interval between them.
[3 marks]
Total for this question: 3
An FM transmitter is assigned a channel extending $95.0\,\text{kHz}$ either side of its nominal carrier frequency. Its actual unmodulated carrier may drift by up to $7.00\,\text{kHz}$ towards either edge, and its maximum frequency deviation is fixed at $52.0\,\text{kHz}$ . Using $B=2(\Delta f+f_M)$ , determine the greatest permitted information frequency in the worst case and the corresponding transmitted bandwidth.
[2 marks]
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3.13.1.1 · MOSFET (metal-oxide semiconducting field-effect transistor)

Tier 1 · Easy

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01.1	$V_{\mathrm{GS}}>V_{\mathrm{th}}$ .	A conducting channel begins to form when the gate-source potential difference exceeds the threshold value, so the condition is $V_{\mathrm{GS}}>V_{\mathrm{th}}$ .	1
02.1	The drain-source current when $V_{\mathrm{GS}}=0$ (only a very small leakage current in an enhancement-mode device).	$I_{\mathrm{DSS}}$ is the small drain-source current that flows at $V_{\mathrm{GS}}=0$ , when the enhancement-mode device should be off.	1

Tier 2 · Standard

Mark scheme for 3.13.1.1 Tier 2 · Standard
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01.1	$3.0\,\text{M}\Omega$	The insulated gate takes negligible current, so the two resistors form an unloaded potential divider. At threshold, $3.0=9.0\times\frac{1.5}{R+1.5}$ , with resistances in $\text{M}\Omega$ . Hence $R+1.5=4.5$ and $R=3.0\,\text{M}\Omega$ .	3
02.1	$V_{\mathrm{GS}}=5.34\,\text{V}$	The resistor potential difference is $V_R=IR=(30.0\times10^{-9})(22.0\times10^6)=0.660\,\text{V}$ . The gate potential is therefore $6.00-0.660=5.34\,\text{V}$ to three significant figures. Since the source is at $0\,\text{V}$ , $V_{\mathrm{GS}}=5.34\,\text{V}$ .	2
03.1	The MOSFET is off at $0.30\,\text{V}$ and on at $4.80\,\text{V}$ for every stated threshold; the insulated gate gives a very high input resistance, so its steady current and input power are negligible.	Because the source is at $0\,\text{V}$ , the two gate levels are also the two values of $V_{\mathrm{GS}}$ . The low level, $0.30\,\text{V}$ , is below even the smallest possible $V_{\mathrm{th}}$ , so no channel forms. The high level, $4.80\,\text{V}$ , is above even the largest possible $V_{\mathrm{th}}$ , so a channel forms. The oxide layer insulates the gate, giving a very high input resistance and hence negligible steady gate current and input power.	3

Tier 3 · Hard

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01.1	$V_{\mathrm{DS}}=3.24\,\text{V}$ , $P_{\mathrm{load}}=0.276\,\text{W}$ and $P_{\mathrm{MOSFET}}=0.156\,\text{W}$ ; it is not acting as an efficient fully-on switch.	The load line is $V_{\mathrm{DS}}=9.0-I_{\mathrm{D}}(120)$ . Using $I_{\mathrm{D}}=0.048\,\text{A}$ gives $V_{\mathrm{DS}}=9.0-(0.048)(120)=3.24\,\text{V}$ . This is at least $2.0\,\text{V}$ , so the assumed part of the output characteristic is self-consistent. The load power is $P_{\mathrm{load}}=I_{\mathrm{D}}^2R=(0.048)^2(120)=0.27648\,\text{W}=0.276\,\text{W}$ . The MOSFET dissipates $P_{\mathrm{MOSFET}}=I_{\mathrm{D}}V_{\mathrm{DS}}=(0.048)(3.24)=0.156\,\text{W}$ . A fully-on switch should have a very small $V_{\mathrm{DS}}$ and small dissipation, so this operating point is not an efficient fully-on state.	4
02.1	An insulating oxide (silicon dioxide) layer separates the gate from the semiconductor, so negligible steady gate current flows. A positive gate-source potential creates an electric field; above the threshold potential difference this field forms an n-channel between source and drain, allowing drain current. Below threshold the channel is absent, so the device is off.	Link each behaviour to the structure. The silicon dioxide layer is insulating, which accounts for the very high gate input resistance. The gate-source potential difference acts through the oxide by an electric field rather than by a gate current. When $V_{\mathrm{GS}}>V_{\mathrm{th}}$ , the field creates a conducting n-channel joining source and drain, so $I_{\mathrm D}$ can flow. When $V_{\mathrm{GS}}<V_{\mathrm{th}}$ , that channel is absent and the drain-source path is switched off.	4
03.1	$I=0.50\,\text{A}$ ; $P_{\mathrm{MOSFET}}=1.5\,\text{W}$ ; $12.5\%$ of the supply power is wasted in the MOSFET.	The MOSFET and load are in series, so the circuit resistance is $6.0+42=48\,\Omega$ and $I=24/48=0.50\,\text{A}$ . The MOSFET dissipates $P=I^2R_{\mathrm{on}}=(0.50)^2(6.0)=1.5\,\text{W}$ . The supply delivers $P=VI=(24)(0.50)=12\,\text{W}$ , so the wasted percentage is $1.5/12=12.5\%$ . This is already a substantial loss, which is why practical switching MOSFETs have on-resistances far below the load resistance.	3
04.1	The drain current increases by $110\%$ (about $1.1\times10^2\%$ ). The student is incorrect for the whole on-state: $I_{\mathrm D}$ changes at low $V_{\mathrm{DS}}$ , although this characteristic is approximately flat from $2.0\,\text{V}$ to $4.0\,\text{V}$ . accept $112.5\%$ or $113\%$ before rounding to two significant figures	Use the two stated read-offs: the increase is $0.34-0.16=0.18\,\text{A}$ , so the percentage increase is $(0.18/0.16)\times100=112.5\%$ , or $1.1\times10^2\%$ to two significant figures. This substantial rise shows that the claim is not valid across the complete on-state. The equal $0.34\,\text{A}$ read-offs at $2.0\,\text{V}$ and $4.0\,\text{V}$ show why the claim is a useful approximation only on the nearly horizontal part of this particular output characteristic.	4
05.1	$I_{\mathrm G}=1.0\,\text{pA}$ , $R_{\mathrm{in}}=4.0\times10^{12}\,\Omega$ and $I_{\mathrm D}/I_{\mathrm G}=2.4\times10^{11}$ .	The resistor potential difference is given directly, so $I_{\mathrm G}=(4.7\times10^{-6})/(4.7\times10^6)=1.0\times10^{-12}\,\text{A}=1.0\,\text{pA}$ . The ideal voltmeter adds no current. Hence $R_{\mathrm{in}}=V_{\mathrm{GS}}/I_{\mathrm G}=4.0/(1.0\times10^{-12})=4.0\times10^{12}\,\Omega$ . Finally, $I_{\mathrm D}/I_{\mathrm G}=0.24/(1.0\times10^{-12})=2.4\times10^{11}$ .	3

3.13.1.2 · Zener diode

Tier 1 · Easy

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01.1	Reverse bias in the Zener breakdown region.	The nearly constant-voltage part of a Zener characteristic is its reverse-breakdown region, so the diode must be reverse biased beyond the Zener voltage.	1
02.1	It limits the current through the Zener diode.	Once breakdown begins, the Zener voltage changes very little while its current can rise rapidly. The series resistor drops the remaining supply voltage and limits that current.	1

Tier 2 · Standard

Mark scheme for 3.13.1.2 Tier 2 · Standard
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01.1	$400\,\Omega$	The resistor potential difference is $V_R=12.0-5.6=6.4\,\text{V}$ . The no-load resistor current equals the Zener current, $I_R=16\,\text{mA}=0.016\,\text{A}$ . Therefore $R=V_R/I_R=6.4/0.016=400\,\Omega$ .	3
02.1	Beyond the Zener breakdown voltage, a large change in reverse current produces only a small change in voltage. The current must remain above the typical minimum operating value, so the voltage is then nearly constant and can act as a reference.	Identify the reverse-breakdown section of the characteristic. Its steep current variation for little voltage variation means the diode potential difference stays close to $V_{\mathrm Z}$ . This behaviour applies once the reverse current exceeds the typical minimum operating current, allowing $V_{\mathrm Z}$ to serve as a reference.	3
03.1	$P_R=0.176\,\text{W}$ , so use a $0.25\,\text{W}$ resistor.	The resistor potential difference is $14.0-7.5=6.5\,\text{V}$ , so $P_R=V_R^2/R=6.5^2/240=0.176\,\text{W}$ . The $0.125\,\text{W}$ rating is smaller than this dissipation, so the next available rating, $0.25\,\text{W}$ , is required.	2

Tier 3 · Hard

Mark scheme for 3.13.1.2 Tier 3 · Hard
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01.1	$330\,\Omega$ ; use at least a $0.25\,\text{W}$ resistor and a $0.25\,\text{W}$ Zener diode.	At maximum load the resistor must supply $I_{\mathrm{S}}=I_{\mathrm{L}}+I_{\mathrm{Z,min}}=18+5=23\,\text{mA}$ . Its maximum allowed resistance is $R_{\max}=(14.0-6.2)/(0.023)=339\,\Omega$ , so only $330\,\Omega$ is suitable. With the load disconnected, all resistor current enters the Zener: $I=(14.0-6.2)/330=0.0236\,\text{A}$ . The resistor dissipates $P_R=I^2R=(0.0236)^2(330)=0.184\,\text{W}$ , and the Zener dissipates $P_{\mathrm{Z}}=(6.2)(0.0236)=0.147\,\text{W}$ . The next stated rating above each value is $0.25\,\text{W}$ .	5
02.1	The maximum load current is $16.8\,\text{mA}$ . Above this current the Zener current falls below its minimum operating value, so the diode leaves breakdown and the output is no longer maintained at $6.8\,\text{V}$ .	The series-resistor current is $I_R=(14.0-6.8)/330=21.818\ldots\,\text{mA}$ . At the maximum load, the Zener must still receive $5.0\,\text{mA}$ , so $I_{\mathrm{L,max}}=21.818\ldots-5.0=16.818\ldots\,\text{mA}=16.8\,\text{mA}$ to three significant figures. This rounded value is just below the exact limit, so it still leaves at least $5.0\,\text{mA}$ through the Zener. A larger load current leaves less than the minimum Zener current, so the diode is no longer in its breakdown region and cannot hold the output at $6.8\,\text{V}$ .	4
03.1	$I_{\mathrm Z}=26.5\,\text{mA}$ and $V_{\mathrm Z}=6.23\,\text{V}$ accept $I_{\mathrm Z}$ in the range $26.4$ – $26.6\,\text{mA}$ and $V_{\mathrm Z}$ in the range $6.2$ – $6.25\,\text{V}$	Between the two graph points, a $20.0\,\text{mA}$ current change corresponds to a $0.20\,\text{V}$ voltage change, so the straight graph section is $V_{\mathrm Z}=5.97+10.0I_{\mathrm Z}$ when current is in amperes. The resistor gives $11.0=180I_{\mathrm Z}+V_{\mathrm Z}$ . Combining these relations gives $11.0=190I_{\mathrm Z}+5.97$ , so $I_{\mathrm Z}=0.02647\,\text{A}=26.5\,\text{mA}$ and $V_{\mathrm Z}=11.0-(0.02647)(180)=6.23\,\text{V}$ .	4
04.1	$\Delta V_{\mathrm Z}=0\,\text{V}$ , the second supply potential is $13.2\,\text{V}$ and the Zener power increases by $0.11\,\text{W}$ .	In the specified constant-voltage model, $V_{\mathrm Z}$ remains $6.2\,\text{V}$ , so its change is zero. At $26\,\text{mA}$ the resistor potential difference is $IR=(0.026)(270)=7.02\,\text{V}$ and the supply is $6.2+7.02=13.22\,\text{V}=13.2\,\text{V}$ . The Zener power increase is $\Delta P=V_{\mathrm Z}(I_2-I_1)=6.2(0.026-0.0080)=0.1116\,\text{W}=0.11\,\text{W}$ .	3
05.1	$6.57\,\text{V}\leq V_{\mathrm S}\leq12.2\,\text{V}$ and the greatest Zener power is $0.113\,\text{W}$ . accept an upper limit of $12.23$ – $12.24\,\text{V}$ before rounding	While the diode is in breakdown, the load current is $I_{\mathrm L}=4.7/1200=3.9167\,\text{mA}$ . At the lower supply limit the resistor current is $3.9167+3.0=6.9167\,\text{mA}$ , so $V_{\mathrm S,min}=4.7+(0.0069167)(270)=6.5675\,\text{V}=6.57\,\text{V}$ . At the upper limit it is $3.9167+24=27.9167\,\text{mA}$ , giving $V_{\mathrm S,max}=4.7+(0.0279167)(270)=12.2375\,\text{V}=12.2\,\text{V}$ to three significant figures. Maximum diode power occurs at $24\,\text{mA}$ : $P_{\mathrm Z}=(4.7)(0.024)=0.1128\,\text{W}=0.113\,\text{W}$ .	4

3.13.1.3 · Photodiode

Tier 1 · Easy

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01.1	Photoconductive mode.	A reverse-biased photodiode is operating in photoconductive mode; photovoltaic mode uses no external bias.	1
02.1	A spectral-response curve.	A spectral-response curve plots the response or sensitivity of the photodiode against wavelength.	1

Tier 2 · Standard

Mark scheme for 3.13.1.3 Tier 2 · Standard
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01.1	$0.60\,\text{V}$	The photocurrent is $I_{\mathrm{p}}=SP=(0.42)(8.0\times10^{-6})=3.36\times10^{-6}\,\text{A}$ . The resistor voltage is $V=IR=(3.36\times10^{-6})(180\times10^3)=0.6048\,\text{V}$ , which is $0.60\,\text{V}$ to two significant figures.	3
02.1	$3.6\,\mu\text{A}$	The responsivity at $900\,\text{nm}$ is $0.60(0.50)=0.30\,\text{A W}^{-1}$ . Hence $I=SP=(0.30)(12\times10^{-6})=3.6\times10^{-6}\,\text{A}=3.6\,\mu\text{A}$ .	3
03.1	$900\,\text{nm}$	At $650\,\text{nm}$ the responsivity is $0.80(0.45)=0.36\,\text{A W}^{-1}$ , so the source power is $P=I/S=(4.32\times10^{-6})/0.36=12.0\,\mu\text{W}$ . The unknown-wavelength responsivity is $S=(2.70\times10^{-6})/(12.0\times10^{-6})=0.225\,\text{A W}^{-1}$ , giving relative response $0.225/0.45=0.50$ . Of the three stated wavelengths, this is $900\,\text{nm}$ .	3

Tier 3 · Hard

Mark scheme for 3.13.1.3 Tier 3 · Hard
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01.1	The scintillator flash produces a photodiode current pulse; the voltage magnitude is $0.52\,\text{V}$ and the pulse contains $3.0\times10^5$ charge carriers.	The particle excites the scintillator, which emits a flash; the reverse-biased photodiode converts that light pulse into an electrical current pulse. The current is $I=SP=(0.48)(4.0\times10^{-6})=1.92\times10^{-6}\,\text{A}$ . The readout voltage magnitude is $V=IR=(1.92\times10^{-6})(270\times10^3)=0.5184\,\text{V}=0.52\,\text{V}$ . The charge in $25\,\text{ns}$ is $Q=It=(1.92\times10^{-6})(25\times10^{-9})=4.80\times10^{-14}\,\text{C}$ . Hence the number of carriers is $N=Q/e=(4.80\times10^{-14})/(1.60\times10^{-19})=3.0\times10^5$ .	5
02.1	$P_{850}/P_{520}=1.6$	After the filter, $P_{520}=I/S=(3.36\times10^{-6})/0.28=12.0\,\mu\text{W}$ and $P_{850}=(8.40\times10^{-6})/0.56=15.0\,\mu\text{W}$ . Before the filter, the powers were $12.0/0.75=16.0\,\mu\text{W}$ and $15.0/0.60=25.0\,\mu\text{W}$ . Therefore $P_{850}/P_{520}=25.0/16.0=1.5625=1.6$ to two significant figures.	5
03.1	$Q=6.48\times10^{-11}\,\text{C}$ and $N=4.05\times10^8$ carriers	The flash-induced photocurrent is the difference between the illuminated and dark graph readings: $I_{\mathrm p}=5.60-0.20=5.40\,\mu\text{A}$ . The pulse charge is $Q=I_{\mathrm p}t=(5.40\times10^{-6})(12.0\times10^{-6})=6.48\times10^{-11}\,\text{C}$ . Hence $N=Q/e=(6.48\times10^{-11})/(1.60\times10^{-19})=4.05\times10^8$ charge carriers.	3
04.1	$I_{\mathrm p}=7.4\,\mu\text{A}$ and $V=0.89\,\text{V}$ .	The responsivities are $S_{480}=0.35(0.62)=0.217\,\text{A W}^{-1}$ and $S_{720}=0.80(0.62)=0.496\,\text{A W}^{-1}$ . The two photocurrents add, so $I_{\mathrm p}=(0.217)(18\times10^{-6})+(0.496)(7.0\times10^{-6})=3.906\times10^{-6}+3.472\times10^{-6}=7.378\times10^{-6}\,\text{A}=7.4\,\mu\text{A}$ to two significant figures. Hence $V=IR=(7.378\times10^{-6})(120\times10^3)=0.88536\,\text{V}=0.89\,\text{V}$ to two significant figures.	4
05.1	$3.00\times10^3$ particles; each particle produces scintillator light that the photodiode converts into a current pulse.	An incident particle transfers energy to the scintillator, which emits a light pulse; the reverse-biased photodiode converts the light into an electrical current pulse. Responsivity gives charge per optical energy because $S=I/P=Q/E$ . Thus one pulse produces $Q_1=SE=(0.44)(3.6\times10^{-13})=1.584\times10^{-13}\,\text{C}$ . The count is $N=Q_{\mathrm{total}}/Q_1=(4.752\times10^{-10})/(1.584\times10^{-13})=3.00\times10^3$ .	4

3.13.1.4 · Hall effect sensor

Tier 1 · Easy

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01.1	The Hall voltage doubles.	With current and geometry fixed, $V_{\mathrm{H}}\propto B$ . Doubling $B$ therefore doubles $V_{\mathrm{H}}$ .	1
02.1	Moving the surface changes the flux density at the sensor, so the Hall output changes. Calibrating output against angle gives the attitude.	Moving the surface changes the flux density at the fixed Hall sensor, producing a corresponding change in output. A calibration of output against angle then gives the surface's attitude.	2

Tier 2 · Standard

Mark scheme for 3.13.1.4 Tier 2 · Standard
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01.1	$2.492\,\text{V}$	The Hall-voltage change is $\Delta V=(45\times10^{-3})(-0.18)=-8.1\times10^{-3}\,\text{V}$ . Add this signed change to the offset: $V_{\mathrm{out}}=2.500-0.0081=2.4919\,\text{V}$ , giving $2.492\,\text{V}$ .	3
02.1	$+3.0^\circ$	The output changes by $1.200\,\text{V}$ across $60.0^\circ$ , so the sensitivity is $1.200/60.0=0.0200\,\text{V degree}^{-1}$ . Zero attitude corresponds to the midpoint output, $2.400\,\text{V}$ . The measured output is $0.060\,\text{V}$ above this, so the angle is $0.060/0.0200=+3.0^\circ$ .	3
03.1	$960\,\text{rev min}^{-1}$ and $8.93\,\text{ms}$ between successive pulses.	The pulse frequency is $280/2.50=112\,\text{Hz}$ . Seven pulses occur per revolution, so the rotation frequency is $112/7=16.0\,\text{rev s}^{-1}$ , giving $16.0\times60=960\,\text{rev min}^{-1}$ . The mean interval between successive pulses is $1/112=8.93\times10^{-3}\,\text{s}=8.93\,\text{ms}$ .	3

Tier 3 · Hard

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01.1	$40\,\text{mT}$ and $1.8\times10^3\,\text{rev min}^{-1}$	For fixed sensor current and geometry, $V_{\mathrm{H}}\propto B$ . Therefore $B=(0.120/0.180)(60\,\text{mT})=40\,\text{mT}$ . Six pulses are produced per revolution, so the rotation frequency is $180/6=30\,\text{rev s}^{-1}$ . Multiplying by $60\,\text{s min}^{-1}$ gives $30\times60=1.8\times10^3\,\text{rev min}^{-1}$ .	5
02.1	$15$ magnets; more magnets give more pulses per revolution, so the rotation can be measured over a shorter time interval or with finer angular resolution.	The rotation frequency is $840/60=14\,\text{rev s}^{-1}$ . Since pulse frequency equals rotation frequency multiplied by the number of magnets, the number is $210/14=15$ . Adding magnets increases the number of pulses in each revolution, reducing the time or angle between readings.	3
03.1	The true speed is $2.50\times10^3\,\text{rev min}^{-1}$ ; the controller displays $2.29\times10^3\,\text{rev min}^{-1}$ , an $8.33\%$ underestimate.	One complete pulse pattern lasts $10(2.00)+4.00=24.0\,\text{ms}$ and corresponds to one rotor revolution. The true rotation frequency is therefore $1/(24.0\times10^{-3})=41.67\,\text{rev s}^{-1}$ , giving $41.67\times60=2.50\times10^3\,\text{rev min}^{-1}$ . Only $11$ pulses occur per revolution, so the mean pulse frequency is $11/(24.0\times10^{-3})=458.3\,\text{Hz}$ . The controller displays $(458.3/12)(60)=2.29\times10^3\,\text{rev min}^{-1}$ . Its percentage error is $[(2291.7-2500)/2500]\times100=-8.33\%$ , so it underestimates the speed by $8.33\%$ because it assumes the missing twelfth pulse is present.	4
04.1	$\theta=+24^\circ$ ; an equal common drift cancels when one output is subtracted from the other. accept finding $\theta=(2.244-2.10)/0.0060=+24^\circ$ from $V_1$ alone	Subtracting the calibration equations gives $V_1-V_2=-0.16+0.0120\theta$ . The measured difference is $2.244-2.116=0.128\,\text{V}$ , so $0.128=-0.16+0.0120\theta$ and $\theta=0.288/0.0120=+24^\circ$ . If a drift $d$ is added to both outputs, $(V_1+d)-(V_2+d)=V_1-V_2$ , so the inferred attitude is unchanged.	3
05.1	$72$ pulses are expected; the measured count is not consistent with the intended speed; the minimum gate time is $0.67\,\text{s}$ .	The intended rotation frequency is $1200/60=20\,\text{rev s}^{-1}$ , so the expected count is $(20)(9)(0.400)=72$ pulses. The measured interval is $75$ to $77$ pulses, which does not contain $72$ , so the result is not consistent with the intended speed. For gate time $t$ , one pulse corresponds to $60/(9t)\,\text{rev min}^{-1}$ . Requiring $60/(9t)\leq10$ gives $t\geq0.666\ldots\,\text{s}$ , so the minimum time is $0.67\,\text{s}$ .	4

3.13.2.1 · Difference between analogue and digital signals

Tier 1 · Easy

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01.1	An analogue signal varies continuously, whereas a binary digital signal has two discrete levels.	The defining contrast is continuous variation for analogue information and discrete allowed levels for digital information; binary uses two such levels.	1
02.1	$24$ bits	One byte contains eight bits, so three bytes contain $3\times8=24$ bits.	1

Tier 2 · Standard

Mark scheme for 3.13.2.1 Tier 2 · Standard
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01.1	$18.75\,\text{mV}$ and $\pm9.38\,\text{mV}$	An $8$ -bit code gives $2^8=256$ intervals. Their width is $\Delta V=4.8/256=0.01875\,\text{V}=18.75\,\text{mV}$ . Rounding to the nearest level gives a maximum error of half an interval, $\pm\Delta V/2=\pm9.375\,\text{mV}$ , or $\pm9.38\,\text{mV}$ .	3
02.1	Binary $1010$ , which is decimal $10$ .	Applying the decision level gives $1,0,1,0$ , so the recovered code is $1010$ . Its place values are $8,4,2,1$ , hence $1010=8+2=10$ in decimal.	2
03.1	$6$ bits per sample, $64$ quantisation levels and a $50\,\text{mV}$ interval.	The available bits per sample are $(36\times10^3)/(6.0\times10^3)=6$ . This gives $2^6=64$ quantisation levels. The interval is $3.2/64=0.050\,\text{V}=50\,\text{mV}$ .	3

Tier 3 · Hard

Mark scheme for 3.13.2.1 Tier 3 · Hard
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01.1	The rate is sufficient; the recording contains $6.3\times10^6$ bits or $7.9\times10^5$ bytes, and $8$ -bit conversion reduces data rate but increases quantisation error.	The minimum sampling rate is $2f_{\max}=2(4.0)=8.0\,\text{kHz}$ , so $10\,\text{kHz}$ is sufficient. The bit rate is $(10\times10^3)(14)=1.40\times10^5\,\text{bit s}^{-1}$ . In $45\,\text{s}$ the data size is $(1.40\times10^5)(45)=6.30\times10^6$ bits. Dividing by $8$ gives $7.875\times10^5$ bytes, or $7.9\times10^5$ bytes to two significant figures. Reducing to $8$ bits gives only $2^8=256$ levels instead of $2^{14}=16384$ , so it lowers bit rate and storage but increases quantisation steps, quantisation noise and loss of amplitude detail.	5
02.1	A regenerator compares each noisy pulse with decision levels, identifies a 0 or 1, and produces a new pulse at the standard level and shape, so small added noise is removed at each stage. If noise crosses a decision level, a bit error can still occur. Sampling omits changes between sample times and quantisation replaces a range of analogue values by one level; that discarded information is absent from the bit stream, so later regeneration cannot recreate it.	Separate transmission noise from conversion loss. During transmission, two distinct digital levels allow a regenerator to decide which bit was intended and reshape the pulse, provided the noise has not moved it across the decision threshold. Repeating this process prevents small disturbances from building up. During analogue-to-digital conversion, finite sampling leaves the signal between samples unrecorded and finite resolution rounds each sample to a quantised level. Several original values therefore produce the same code. A regenerator receives only that code, so it has no evidence from which to restore the discarded time or amplitude detail.	5
03.1	Nyquist minimum rates: $2.0\,\text{kHz}$ for the $1.0\,\text{kHz}$ signal and $14.0\,\text{kHz}$ for the $7.0\,\text{kHz}$ signal. At $8.0\,\text{kHz}$ the $7.0\,\text{kHz}$ signal is undersampled and every one of its samples coincides with the corresponding sample of the $1.0\,\text{kHz}$ signal, so the two inputs are indistinguishable.	The Nyquist condition requires a sampling rate of at least twice the signal frequency, so the minimum rates are $2(1.0)=2.0\,\text{kHz}$ and $2(7.0)=14.0\,\text{kHz}$ . The $8.0\,\text{kHz}$ rate exceeds the first but not the second. At the sample instants $t=n/8000\,\text{s}$ , $\cos(2\pi(7000)n/8000)=\cos(2\pi n-2\pi(1000)n/8000)=\cos(2\pi(1000)n/8000)$ , so each sample of the $7.0\,\text{kHz}$ signal equals the corresponding sample of the $1.0\,\text{kHz}$ signal. The undersampled $7.0\,\text{kHz}$ input is therefore reconstructed as its $1.0\,\text{kHz}$ alias.	3
04.1	reject $14$ : after $14$ sections both worst-case levels sit exactly at $2.50\,\text{V}$ $13$ consecutive sections; after four sections both levels remain on the correct side of $2.50\,\text{V}$ , and regeneration restores their standard values before the next group.	After $N$ sections the worst-case levels are $0.40+0.15N$ and $4.60-0.15N$ . Correct decisions require $0.40+0.15N<2.50$ and $4.60-0.15N>2.50$ . Both give $N<14$ , so the greatest whole number is $13$ ; the levels are then $2.35\,\text{V}$ and $2.65\,\text{V}$ , each one full section-step from the decision level. After four sections they are $1.00\,\text{V}$ and $4.00\,\text{V}$ . A regenerator identifies the bits and recreates $0.40\,\text{V}$ and $4.60\,\text{V}$ , so the displacement does not carry into the next four-section group.	4
05.1	The converter needs at least $9$ bits per sample and a minimum bit rate of $90\,\text{kbit s}^{-1}$ , so the $96\,\text{kbit s}^{-1}$ link can carry the data.	An error limit of $4.0\,\text{mV}$ requires an interval no greater than $8.0\,\text{mV}$ . Therefore $3.3/2^n\leq0.0080$ , so $2^n\geq412.5$ . Eight bits give only $256$ intervals, whereas nine give $512$ , so the minimum is $n=9$ . Nyquist sampling requires at least $2(5.0)=10\,\text{kHz}$ . The minimum bit rate is therefore $(9)(10\times10^3)=90\times10^3\,\text{bit s}^{-1}=90\,\text{kbit s}^{-1}$ . This is below the $96\,\text{kbit s}^{-1}$ limit, so both requirements can be met.	4

3.13.3.1 · LC resonance filters

Tier 1 · Easy

Mark scheme for 3.13.3.1 Tier 1 · Easy
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01.1	$4.0\times10^4\,\text{Hz}$	Convert to SI units: $L=2.0\times10^{-3}\,\text{H}$ and $C=8.0\times10^{-9}\,\text{F}$ . Then $f_0=1/[2\pi\sqrt{(2.0\times10^{-3})(8.0\times10^{-9})}]=3.98\times10^4\,\text{Hz}$ , giving $4.0\times10^4\,\text{Hz}$ to two significant figures.	2
02.1	Inductance, $L$ .	In the specified analogy, inductance has the inertia-like role and therefore corresponds to mass.	1

Tier 2 · Standard

Mark scheme for 3.13.3.1 Tier 2 · Standard
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01.1	$14\,\text{kHz}$ and $Q=11$	The bandwidth is the full separation of the half-power frequencies: $\Delta f=157-143=14\,\text{kHz}$ . Hence $Q=f_0/\Delta f=150/14=10.7$ , which is $11$ to two significant figures.	3
02.1	The capacitance decreases by $36\%$ .	With $L$ unchanged, $f_0\propto1/\sqrt{C}$ , so $C_2/C_1=(f_1/f_2)^2=(180/225)^2=0.64$ . The new capacitance is $64\%$ of the original value, so it has decreased by $100\%-64\%=36\%$ .	3
03.1	$522\,\text{kHz}$	Since $f_0=1/(2\pi\sqrt{LC})$ , the highest frequency occurs when both component values are smallest. Thus $L_{\min}=0.980\,\text{mH}$ and $C_{\min}=95.0\,\text{pF}$ . Substitution gives $f_{\max}=1/[2\pi\sqrt{(0.980\times10^{-3})(95.0\times10^{-12})}]=5.216\times10^5\,\text{Hz}=522\,\text{kHz}$ to three significant figures.	3

Tier 3 · Hard

Mark scheme for 3.13.3.1 Tier 3 · Hard
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01.1	$C=88\,\text{pF}$ and $Q=28$ ; the $22\,\text{kHz}$ bandwidth passes the full $18\,\text{kHz}$ transmission while retaining some selectivity.	Rearrange the resonance equation: $C=1/(4\pi^2f_0^2L)$ . Substitution gives $C=1/[4\pi^2(620\times10^3)^2(0.75\times10^{-3})]=8.79\times10^{-11}\,\text{F}=87.9\,\text{pF}$ , or $88\,\text{pF}$ . The measured bandwidth is $\Delta f=631-609=22\,\text{kHz}$ , so $Q=620/22=28.2\approx28$ . Because $22\,\text{kHz}>18\,\text{kHz}$ , the full wanted transmission lies within the half-power bandwidth; the response is only $4\,\text{kHz}$ wider, so it retains more selectivity than a much broader filter.	5
02.1	The charged capacitor initially stores electric-field energy. It discharges through the inductor, so current and the inductor's magnetic-field energy increase while the capacitor's energy falls. When the capacitor is momentarily uncharged, current and magnetic energy are maximum. The collapsing magnetic field keeps current flowing and charges the capacitor with reversed polarity, so energy continues to alternate.	Begin with electric-field energy in the charged capacitor and no current. As charge moves, capacitor energy is transferred to the magnetic field of the inductor. At zero capacitor charge the current, and therefore magnetic energy, is greatest. Inductance opposes the change of current, so current persists and recharges the capacitor with the opposite polarity. The sequence then reverses and repeats.	4
03.1	$L=270\,\mu\text{H}$ ; use $C=310\,\text{pF}$ at $0.55\,\text{MHz}$ and $C=36.6\,\text{pF}$ at $1.60\,\text{MHz}$ .	Rearranging the resonance equation gives $L=1/(4\pi^2f_0^2C)$ . To reach $0.55\,\text{MHz}$ without exceeding $365\,\text{pF}$ requires $L\geq1/[4\pi^2(0.55\times10^6)^2(365\times10^{-12})]=229\,\mu\text{H}$ . To reach $1.60\,\text{MHz}$ without going below $20\,\text{pF}$ requires $L\leq1/[4\pi^2(1.60\times10^6)^2(20\times10^{-12})]=495\,\mu\text{H}$ . Only $270\,\mu\text{H}$ lies in this interval. For this inductor, $C=1/[4\pi^2(0.55\times10^6)^2(270\times10^{-6})]=310\,\text{pF}$ and $C=1/[4\pi^2(1.60\times10^6)^2(270\times10^{-6})]=36.6\,\text{pF}$ , both within the available range.	5
04.1	$E=4.90\,\mu\text{J}$ ; $f_0=8.90\,\text{kHz}$ ; $6.0\,\text{V}$ when the stored energy has fallen to one quarter.	Initially the capacitor stores $E=\tfrac12CV^2=\tfrac12(68\times10^{-9})(12)^2=4.896\times10^{-6}\,\text{J}=4.90\,\mu\text{J}$ . The resonant frequency is $f_0=1/[2\pi\sqrt{(4.7\times10^{-3})(68\times10^{-9})}]=8902.6\,\text{Hz}=8.90\,\text{kHz}$ . Since $E=\tfrac12CV^2$ , the energy falls to one quarter when the potential difference halves: $V=12/2=6.0\,\text{V}$ .	3
05.1	$f_1=187\,\text{kHz}$ with $f_{B1}=2.20\,\text{kHz}$ ; $f_2=306\,\text{kHz}$ with $f_{B2}=3.60\,\text{kHz}$ . The circuit tunes to a higher frequency and has a wider bandwidth in hertz, although its Q-factor is unchanged.	Using $f_0=1/(2\pi\sqrt{LC})$ , $f_1=1/[2\pi\sqrt{(3.3\times10^{-3})(220\times10^{-12})}]=1.86789\times10^5\,\text{Hz}=187\,\text{kHz}$ . Similarly, $f_2=1/[2\pi\sqrt{(3.3\times10^{-3})(82\times10^{-12})}]=3.05954\times10^5\,\text{Hz}=306\,\text{kHz}$ . Since $Q=f_0/f_B$ , $f_{B1}=186789/85=2197.5\,\text{Hz}=2.20\,\text{kHz}$ and $f_{B2}=305954/85=3599.5\,\text{Hz}=3.60\,\text{kHz}$ . Reducing capacitance raises resonance; unchanged Q means unchanged fractional bandwidth, so the higher-frequency response spans more hertz.	4

3.13.3.2 · The ideal operational amplifier

Tier 1 · Easy

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01.1	$0\,\text{A}$	An ideal operational amplifier has infinite input impedance, so no current enters either input: $I_+=I_-=0$ .	1
02.1	Infinite.	Infinite open-loop gain is one of the defining characteristics of the ideal operational amplifier.	1

Tier 2 · Standard

Mark scheme for 3.13.3.2 Tier 2 · Standard
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01.1	The inverting input is at approximately $0\,\text{V}$ ; it is held at earth potential by negative feedback and very large gain but is not physically connected to earth.	For an ideal op amp operating with negative feedback, the very large open-loop gain makes $V_+-V_-$ negligible. Thus $V_-=V_+=0\,\text{V}$ . The inverting node is therefore at earth potential, but because it has no direct conducting connection to earth it is described as a virtual earth.	3
02.1	Approximately $0\,\text{V}$ .	The differential input is $V_+-V_-=2.10-2.40=-0.30\,\text{V}$ . The ideal open-loop gain drives the output towards the negative supply for a negative differential input. Here the negative supply rail is $0\,\text{V}$ , so the comparator output is approximately $0\,\text{V}$ .	2
03.1	$60.0^\circ\text{C}$ ; connect the sensor to the non-inverting input and the $1.20\,\text{V}$ reference to the inverting input.	At the switching point the two input potentials are equal, so $T=1.20/(20.0\times10^{-3})=60.0^\circ\text{C}$ . Connect the rising sensor voltage to $V_+$ and the fixed reference to $V_-$ . Below $60.0^\circ\text{C}$ , $V_+<V_-$ and the very large open-loop gain drives the output towards $-9.0\,\text{V}$ . Above $60.0^\circ\text{C}$ , $V_+>V_-$ and the output is driven towards $+9.0\,\text{V}$ .	2

Tier 3 · Hard

Mark scheme for 3.13.3.2 Tier 3 · Hard
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01.1	$-4.0\,\text{V}$ ; this lies within the $\pm6.0\,\text{V}$ rails.	Negative feedback and infinite gain give $V_-=V_+=0\,\text{V}$ , so the inverting node is a virtual earth. The current from the source to the node is $I=(0.80-0)/(40\times10^3)=20\,\mu\text{A}$ . Infinite input impedance means none enters the op amp, so the same current must leave the node through the $200\,\text{k}\Omega$ feedback resistor. Therefore $0-V_{\mathrm{out}}=I(200\times10^3)=(20\times10^{-6})(200\times10^3)=4.0\,\text{V}$ , giving $V_{\mathrm{out}}=-4.0\,\text{V}$ . Since $-4.0\,\text{V}$ lies between the $-6.0\,\text{V}$ and $+6.0\,\text{V}$ rails, the output is not saturated and the linear-feedback assumption is self-consistent.	5
02.1	$1.799968\,\text{V}<V_+<1.800032\,\text{V}$ ; at $1.800020\,\text{V}$ , $V_{\mathrm{out}}=+5.0\,\text{V}$ .	Unsaturated operation requires $\|V_{\mathrm{out}}\|<8.0\,\text{V}$ . From $V_{\mathrm{out}}=A_{\mathrm{OL}}(V_+-V_-)$ , $\|V_+-V_-\|<8.0/(2.5\times10^5)=3.2\times10^{-5}\,\text{V}=32\,\mu\text{V}$ . With $V_-=1.800000\,\text{V}$ , the range is therefore $1.799968\,\text{V}<V_+<1.800032\,\text{V}$ ; at either endpoint the calculated output reaches a supply rail and saturation begins. For $V_+=1.800020\,\text{V}$ , the differential input is $20\,\mu\text{V}$ and $V_{\mathrm{out}}=(2.5\times10^5)(20\times10^{-6})=+5.0\,\text{V}$ , which is inside the supply limits.	5
03.1	For $T<40^\circ\text{C}$ : A is $-6.0\,\text{V}$ and B is $+6.0\,\text{V}$ . For $40^\circ\text{C}<T<80^\circ\text{C}$ : both are $+6.0\,\text{V}$ . For $T>80^\circ\text{C}$ : A is $+6.0\,\text{V}$ and B is $-6.0\,\text{V}$ .	Comparator A changes sign when $V_{\mathrm S}=1.40\,\text{V}$ , giving $T=(1.40-0.20)/0.030=40^\circ\text{C}$ . It is positive above this temperature because then $V_+>V_-$ . Comparator B changes sign when $V_{\mathrm S}=2.60\,\text{V}$ , giving $T=(2.60-0.20)/0.030=80^\circ\text{C}$ . It is positive below this temperature because its fixed $V_+$ then exceeds $V_{\mathrm S}$ at $V_-$ . Therefore the output pair $(A,B)$ is $(-6.0,+6.0)\,\text{V}$ below $40^\circ\text{C}$ , $(+6.0,+6.0)\,\text{V}$ between $40^\circ\text{C}$ and $80^\circ\text{C}$ , and $(+6.0,-6.0)\,\text{V}$ above $80^\circ\text{C}$ .	4
04.1	The output switches positive at $3.50\,\text{s}$ and negative at $8.50\,\text{s}$ ; it is at $-7.5\,\text{V}$ before and after these times and at $+7.5\,\text{V}$ between them. The positive fraction is $0.417$ or $41.7\%$ .	On the rising section, equality of the inputs gives $0.50+0.30t=1.55$ , so $t=3.50\,\text{s}$ . On the falling section, $4.10-0.30t=1.55$ , giving $t=8.50\,\text{s}$ . The ideal open-loop gain drives the output to positive saturation when $V_+>V_-$ and to negative saturation when $V_+<V_-$ . It is therefore positive for $8.50-3.50=5.00\,\text{s}$ , a fraction $5.00/12.0=0.4167=41.7\%$ .	4
05.1	Negative saturation is guaranteed for $T<28.5^\circ\text{C}$ and positive saturation for $T>31.5^\circ\text{C}$ ; either output is possible from $28.5^\circ\text{C}$ to $31.5^\circ\text{C}$ . The ideal input draws no current and therefore does not load the sensor. (accept $28.5$ to $28.55^\circ\text{C}$ and $31.45$ to $31.5^\circ\text{C}$ )	Negative saturation is guaranteed only when even the largest sensor output is below the reference: $0.90+0.055T+0.080<2.55$ . This gives $T<1.57/0.055=28.545\ldots^\circ\text{C}$ , or $T<28.5^\circ\text{C}$ . Positive saturation is guaranteed only when even the smallest output exceeds the reference: $0.90+0.055T-0.080>2.55$ , giving $T>1.73/0.055=31.4545\ldots^\circ\text{C}$ , or $T>31.5^\circ\text{C}$ . Between these limits the uncertainty band crosses $2.55\,\text{V}$ , so either rail output is possible. Infinite ideal input resistance means zero input current, so the comparator does not change the sensor output relation.	5

3.13.4.1 · Operational amplifier: inverting amplifier configuration

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01.1	$-2.8\,\text{V}$	The gain is $-R_f/R_{\text{in}}=-47/10=-4.7$ . Hence $V_{\text{out}}=(-4.7)(0.60)=-2.82\,\text{V}$ , which is $-2.8\,\text{V}$ to two significant figures.	2
02.1	$-7.5$	For an inverting amplifier, $G=-R_f/R_{\text{in}}$ . Therefore $G=-180/24=-7.5$ . The negative sign shows that the output is inverted.	1

Tier 2 · Standard

Mark scheme for 3.13.4.1 Tier 2 · Standard
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01.1	$+2.6\,\text{V}$	Negative feedback makes the inverting input a virtual earth, and no current enters the operational amplifier. Therefore $V_{\text{in}}/R_{\text{in}}=(0-V_{\text{out}})/R_f$ . Rearranging gives $V_{\text{out}}/V_{\text{in}}=-R_f/R_{\text{in}}=-68/12=-5.67$ . Thus $V_{\text{out}}=(-5.67)(-0.45)=+2.55\,\text{V}$ , or $+2.6\,\text{V}$ to two significant figures.	3
02.1	$162\,\text{k}\Omega$	The required gain is $G=V_{\text{out}}/V_{\text{in}}=2.52/(-0.28)=-9.0$ . Since $G=-R_f/R_{\text{in}}$ , $R_f=9.0(18\,\text{k}\Omega)=162\,\text{k}\Omega$ .	2
03.1	$800\,\Omega$	At the current limit, $R_f=V_{\text{out}}/I_{\text{out}}=4.00/(0.500\times10^{-3})=8.00\,\text{k}\Omega$ . Since $\|G\|=R_f/R_{\text{in}}=10.0$ , the minimum input resistance is $R_{\text{in}}=8.00\,\text{k}\Omega/10.0=0.800\,\text{k}\Omega=800\,\Omega$ . A larger resistance pair with the same ratio would draw less output current.	2

Tier 3 · Hard

Mark scheme for 3.13.4.1 Tier 3 · Hard
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01.1	Ideal peak $11.2\,\text{V}$ ; the inverted output is clipped at $\pm8.0\,\text{V}$ ; maximum unclipped input peak $1.0\,\text{V}$ .	The closed-loop gain is $-R_f/R_{\text{in}}=-120/15=-8.0$ . The ideal output peak is therefore $8.0\times1.4=11.2\,\text{V}$ and it is inverted relative to the input. Because $11.2\,\text{V}$ exceeds both saturation magnitudes, the positive and negative peaks flatten at $+8.0\,\text{V}$ and $-8.0\,\text{V}$ . At the clipping boundary, $8.0\|V_{\text{in,peak}}\|=8.0\,\text{V}$ , so the largest unclipped input peak is $1.0\,\text{V}$ .	5
02.1	$-3.00\,\text{V}$	Negative feedback holds the inverting input at virtual-earth potential. The sensor's internal resistance is therefore in series with the amplifier input resistance, so $I=0.600/[(4.00+20.0)\times10^3]=25.0\,\mu\text{A}$ . No current enters the ideal op amp, hence the same current flows through the feedback resistor. Therefore $V_{\text{out}}=-I R_f=-(25.0\times10^{-6})(120\times10^3)=-3.00\,\text{V}$ .	3
03.1	$180\,\text{k}\Omega$ ; its worst-case output peak is $4.18\,\text{V}$ , whereas the $220\,\text{k}\Omega$ option could produce $5.11\,\text{V}$ .	The greatest possible gain magnitude occurs with the feedback resistor $5\%$ high and the input resistor $5\%$ low. For the $180\,\text{k}\Omega$ option, $\|G\|=(180\times1.05)/(20.0\times0.95)=9.947\ldots$ , so the largest output peak is $0.420(9.947\ldots)=4.18\,\text{V}$ , below the limit. For $220\,\text{k}\Omega$ , $\|G\|=(220\times1.05)/(20.0\times0.95)=12.158\ldots$ and the possible output peak is $0.420(12.158\ldots)=5.11\,\text{V}$ . Therefore only the $180\,\text{k}\Omega$ resistor guarantees an unclipped output.	4
04.1	$R_A=104\,\text{k}\Omega$ and $R_B=144\,\text{k}\Omega$ ; output peaks $1.17\,\text{V}$ and $2.79\,\text{V}$ , both inverted; feedback-current peak $11.25\,\mu\text{A}$ in either setting.	With the switch open, $-R_A/R_{\text{in}}=-3.25$ , so $R_A=3.25(32.0\,\text{k}\Omega)=104\,\text{k}\Omega$ . With the switch closed, $-(R_A+R_B)/R_{\text{in}}=-7.75$ , giving $R_A+R_B=248\,\text{k}\Omega$ and $R_B=144\,\text{k}\Omega$ . The output-peak magnitudes are $3.25(0.360)=1.17\,\text{V}$ and $7.75(0.360)=2.79\,\text{V}$ ; both outputs are inverted. The virtual-earth input current is $0.360/(32.0\times10^3)=11.25\,\mu\text{A}$ . No current enters the ideal op amp, so this same peak current flows through whichever feedback resistance is selected.	5
05.1	Greatest input magnitude $0.950\,\text{V}$ ; the output-voltage limit sets it, with feedback-resistor power $0.169\,\text{mW}$ .	The gain magnitude is $R_f/R_{\text{in}}=192/32.0=6.00$ , so the output limit requires $\|V_{\text{in}}\|\leq5.70/6.00=0.950\,\text{V}$ . The power limit gives a maximum feedback current $I=\sqrt{P/R_f}=\sqrt{(0.180\times10^{-3})/(192\times10^3)}=30.62\ldots\,\mu\text{A}$ . The same current flows through the input resistor, so this limit permits $\|V_{\text{in}}\|=I R_{\text{in}}=(30.62\ldots\times10^{-6})(32.0\times10^3)=0.9798\ldots\,\text{V}$ . The voltage limit is therefore more restrictive. At $0.950\,\text{V}$ , $I=0.950/(32.0\times10^3)=29.69\,\mu\text{A}$ and $I^2R_f=0.169\,\text{mW}$ , below the power limit.	4

3.13.4.2 · Operational amplifier: non-inverting amplifier configuration

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01.1	$3.9\,\text{V}$	The gain is $1+R_f/R_1=1+39/10=4.9$ . Therefore $V_{\text{out}}=4.9\times0.80=3.92\,\text{V}$ , giving $3.9\,\text{V}$ to two significant figures.	2
02.1	$8.0$	The non-inverting gain is $G=1+R_f/R_1=1+105/15=8.0$ .	1

Tier 2 · Standard

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01.1	$2.5\,\text{V}$	No current enters the ideal input, so $R_f$ and $R_1$ form an unloaded divider: $V_-=V_{\text{out}}R_1/(R_1+R_f)$ . Negative feedback gives $V_-=V_+=V_{\text{in}}$ . Hence $V_{\text{out}}/V_{\text{in}}=(R_1+R_f)/R_1=1+R_f/R_1$ . The gain is $1+33/8.2=5.02$ , so $V_{\text{out}}=5.02\times0.50=2.51\,\text{V}$ , or $2.5\,\text{V}$ to two significant figures.	4
02.1	$84\,\text{k}\Omega$	The required gain is $G=2.00/0.25=8.0$ . Using $G=1+R_f/R_1$ gives $R_f=(G-1)R_1=(8.0-1)(12\,\text{k}\Omega)=84\,\text{k}\Omega$ .	2
03.1	Input potential $1.50\,\text{V}$ ; divider current $83.3\,\mu\text{A}$ .	The closed-loop gain is $G=1+R_f/R_1=1+72.0/18.0=5.00$ . Hence $V_{\text{in}}=7.50/5.00=1.50\,\text{V}$ . No current enters the ideal inverting input, so $R_f$ and $R_1$ carry the same current. Their series resistance is $90.0\,\text{k}\Omega$ , giving $I=7.50/(90.0\times10^3)=83.3\,\mu\text{A}$ .	3

Tier 3 · Hard

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01.1	Maximum input $4.1\,\text{V peak-to-peak}$ ; output $23.0\,\text{V peak-to-peak}$ .	The gain is $1+82/18=5.56$ . The largest output peak before saturation is $11.5\,\text{V}$ , so the largest input peak is $11.5/5.56=2.07\,\text{V}$ . A peak-to-peak value is twice a peak value, giving an input of $2(2.07)=4.14\,\text{V peak-to-peak}$ , or $4.1\,\text{V peak-to-peak}$ . The corresponding output spans from $-11.5\,\text{V}$ to $+11.5\,\text{V}$ , so it is $23.0\,\text{V peak-to-peak}$ .	5
02.1	$120\,\text{k}\Omega$	For $R_f=120\,\text{k}\Omega$ , $G=1+120/10.0=13.0$ , so the output peak is $13.0(0.900)=11.7\,\text{V}$ , below saturation. The next value, $140\,\text{k}\Omega$ , gives $G=1+140/10.0=15.0$ and an output peak of $15.0(0.900)=13.5\,\text{V}$ , which exceeds $13.0\,\text{V}$ . Therefore $120\,\text{k}\Omega$ is the greatest available resistance that avoids saturation.	3
03.1	Ideal values $R_1=24.0\,\text{k}\Omega$ and $R_f=156\,\text{k}\Omega$ ; select $160\,\text{k}\Omega$ , giving gain $7.67$ and a $2.22\%$ error.	No current enters the ideal op amp, so the divider current flows through both resistors. Their ideal total resistance is $R_1+R_f=9.00/(50.0\times10^{-6})=180\,\text{k}\Omega$ . Since $G=(R_1+R_f)/R_1=7.50$ , $R_1=180/7.50=24.0\,\text{k}\Omega$ and $R_f=180-24.0=156\,\text{k}\Omega$ . With $R_f=150\,\text{k}\Omega$ , $G=1+150/24.0=7.25$ ; with $160\,\text{k}\Omega$ , $G=1+160/24.0=7.666\ldots$ , which is closer to $7.50$ . Its percentage gain error is $(7.666\ldots-7.50)/7.50\times100\%=2.22\%$ .	4
04.1	Each $R_f=112\,\text{k}\Omega$ ; intermediate peak $0.600\,\text{V}$ ; ideal final peak $4.80\,\text{V}$ ; actual final peak $4.50\,\text{V}$ .	Identical stages have equal gains, so each gain is $\sqrt{64.0}=8.00$ . From $G=1+R_f/R_1$ , $R_f=(8.00-1)(16.0\,\text{k}\Omega)=112\,\text{k}\Omega$ . The first-stage peak is $8.00(75.0\,\text{mV})=0.600\,\text{V}$ , within its output limit. The ideal second-stage peak is $8.00(0.600)=4.80\,\text{V}$ . This exceeds the second amplifier's $4.50\,\text{V}$ limit, so its actual peak is clipped at $4.50\,\text{V}$ .	5
05.1	The gain is $7.50$ , the output peak is $5.40\,\text{V}$ and the smallest supply rails are $\pm6.25\,\text{V}$ .	The closed-loop gain is $G=1+R_f/R_1=1+91.0/14.0=7.50$ . The required output peak is therefore $7.50(0.720)=5.40\,\text{V}$ . Because the output must remain at least $0.850\,\text{V}$ inside each rail, the minimum rail magnitude is $5.40+0.850=6.25\,\text{V}$ . Hence the smallest symmetrical supply is $\pm6.25\,\text{V}$ .	3

3.13.4.3 · Operational amplifier: summing amplifier configuration

Tier 1 · Easy

Mark scheme for 3.13.4.3 Tier 1 · Easy
Question	Answers	Additional comments/Guidelines	Mark
01.1	$-1.6\,\text{V}$	Use $V_{\text{out}}=-R_f(V_1/R_1+V_2/R_2)$ . Thus $V_{\text{out}}=-20(0.30/10+0.50/10)=-1.6\,\text{V}$ when all resistances are in $\text{k}\Omega$ .	2
02.1	$2.7\,\text{V}$	Using the supplied relation, $V_{\text{out}}=(0.70-0.25)(6.0)=2.70\,\text{V}$ , or $2.7\,\text{V}$ .	1

Tier 2 · Standard

Mark scheme for 3.13.4.3 Tier 2 · Standard
Question	Answers	Additional comments/Guidelines	Mark
01.1	$-2.7\,\text{V}$	The input-current terms are $0.90/15=0.060$ , $-1.20/30=-0.040$ and $0.50/20=0.025$ in consistent units. Their algebraic sum is $0.045$ . Therefore $V_{\text{out}}=-60(0.045)=-2.70\,\text{V}$ , or $-2.7\,\text{V}$ .	3
02.1	$-0.30\,\text{V}$	Use $-0.60=-90(0.40/30+V_2/45)$ with resistances in $\text{k}\Omega$ . Dividing by $-90$ gives $0.006667=0.013333+V_2/45$ . Hence $V_2/45=-0.006667$ and $V_2=-0.300\,\text{V}$ .	2
03.1	$V_1=-1.0\,\text{V}$ , $V_2=0\,\text{V}$ and $V_3=-1.0\,\text{V}$ .	When active, the three inputs contribute $-40(-1.0)/10=+4.0\,\text{V}$ , $-40(-1.0)/20=+2.0\,\text{V}$ and $-40(-1.0)/40=+1.0\,\text{V}$ to the output. The unique combination giving $+5.0\,\text{V}$ is $4.0+1.0$ , so the first and third inputs are at $-1.0\,\text{V}$ and the second is at $0\,\text{V}$ .	3

Tier 3 · Hard

Mark scheme for 3.13.4.3 Tier 3 · Hard
Question	Answers	Additional comments/Guidelines	Mark
01.1	Ideal output $-15.5\,\text{V}$ ; actual output $-10.5\,\text{V}$ ; third input $0.50\,\text{V}$ .	The first two current terms are $1.20/12=0.100$ and $-0.60/30=-0.020$ in consistent units; the third is $1.50/20=0.075$ . Hence $V_{\text{out,ideal}}=-100(0.100-0.020+0.075)=-15.5\,\text{V}$ . The output therefore saturates at $-10.5\,\text{V}$ . At the boundary the current-term sum must be $10.5/100=0.105$ . If the reduced third input is $V_3$ , then $0.100-0.020+V_3/20=0.105$ , so $V_3/20=0.025$ and $V_3=0.50\,\text{V}$ .	6
02.1	$R_1=30\,\text{k}\Omega$ and $R_2=20\,\text{k}\Omega$	For channel 1 alone, $-2.00=-120(0.50/R_1)$ , so $R_1=120(0.50)/2.00=30\,\text{k}\Omega$ . The second channel changes the output by $+1.20\,\text{V}$ . Its contribution is $-R_fV_2/R_2=-120(-0.20)/R_2=24/R_2$ . Therefore $24/R_2=1.20$ , giving $R_2=20\,\text{k}\Omega$ .	3
03.1	$R_A=18.0\,\text{k}\Omega$ and $R_B=72.0\,\text{k}\Omega$	The two maximum-input contributions must add to $9.00\,\text{V}$ in the ratio $2:1$ , so sensor $A$ must contribute $6.00\,\text{V}$ and sensor $B$ must contribute $3.00\,\text{V}$ . For sensor $A$ , $R_fV_A/R_A=6.00$ , giving $R_A=180(0.600)/6.00=18.0\,\text{k}\Omega$ . For sensor $B$ , $R_fV_B/R_B=3.00$ , giving $R_B=180(1.20)/3.00=72.0\,\text{k}\Omega$ . Both inputs are positive, so the summed output at maximum input is $-(6.00+3.00)=-9.00\,\text{V}$ .	4
04.1	Input resistances $40.0\,\text{k}\Omega$ , $80.0\,\text{k}\Omega$ , $160\,\text{k}\Omega$ and $320\,\text{k}\Omega$ from MSB to LSB; first saturating code $1100$ .	The LSB must contribute $-0.400\,\text{V}$ when high, so its resistance is $R_{\text{LSB}}=R_f(0.800/0.400)=320\,\text{k}\Omega$ . Binary weighting requires contributions of $0.800\,\text{V}$ , $1.60\,\text{V}$ and $3.20\,\text{V}$ for the next bits, giving resistances $160\,\text{k}\Omega$ , $80.0\,\text{k}\Omega$ and $40.0\,\text{k}\Omega$ , respectively. A code of decimal value $n$ has ideal output $-0.400n\,\text{V}$ . Code $1011$ gives $-4.40\,\text{V}$ and remains unclipped, whereas the next code, $1100$ , requires $-4.80\,\text{V}$ and is the first to saturate.	5
05.1	Sensor input resistance $12.0\,\text{k}\Omega$ ; reference input resistance $15.0\,\text{k}\Omega$ .	For $V_{\text{out}}=-R_f(V_s/R_s+V_r/R_r)$ , the required output slope is $[-6.00-(+6.00)]/(1.00-0.200)=-15.0$ . Hence $-R_f/R_s=-15.0$ and $R_s=180/15.0=12.0\,\text{k}\Omega$ . At $V_s=0.200\,\text{V}$ , $6.00=-180[0.200/12.0-0.750/R_r]$ , with resistances in $\text{k}\Omega$ . Thus the bracket is $-0.033333\ldots$ , so $0.750/R_r=0.050000\ldots$ and $R_r=15.0\,\text{k}\Omega$ . Checking the other endpoint gives $-180[1.00/12.0-0.750/15.0]=-6.00\,\text{V}$ .	5

3.13.4.4 · Real operational amplifiers

Tier 1 · Easy

Mark scheme for 3.13.4.4 Tier 1 · Easy
Question	Answers	Additional comments/Guidelines	Mark
01.1	$1.0\times10^5\,\text{Hz}$	Using $Gf_B=3.0\times10^6\,\text{Hz}$ , the bandwidth is $f_B=(3.0\times10^6)/30=1.0\times10^5\,\text{Hz}$ .	2
02.1	Any two from: finite open-loop gain; finite input resistance; limited output voltage; limited output current; gain that falls at high frequency.	A real device does not have infinite gain or input resistance. Its output voltage and current are bounded by the device and supply, and its gain decreases as signal frequency rises. Any two distinct limitations score.	2

Tier 2 · Standard

Mark scheme for 3.13.4.4 Tier 2 · Standard
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01.1	No; the required slew rate is $2.0\,\text{V}\,\mu\text{s}^{-1}$ and the maximum frequency is $60\,\text{kHz}$ .	For a sine wave, the maximum gradient is $2\pi fV_p=2\pi(80\times10^3)(4.0)=2.01\times10^6\,\text{V s}^{-1}=2.01\,\text{V}\,\mu\text{s}^{-1}$ . This exceeds $1.5\,\text{V}\,\mu\text{s}^{-1}$ , so slew-rate distortion occurs. The limiting frequency is $f=\text{SR}/(2\pi V_p)=(1.5\times10^6)/(2\pi\times4.0)=5.97\times10^4\,\text{Hz}$ , or $60\,\text{kHz}$ to two significant figures.	4
02.1	The greatest usable gain is $40$ ; a gain of $60$ is unsuitable because its bandwidth would be only $30\,\text{kHz}$ .	Using $Gf_B=1.8\times10^6\,\text{Hz}$ gives $G=(1.8\times10^6)/(45\times10^3)=40$ . At gain $60$ , the bandwidth would be $(1.8\times10^6)/60=30\,\text{kHz}$ , which is below $45\,\text{kHz}$ , so gain $60$ is unsuitable.	2
03.1	Actual output $6.40\,\text{V}$ ; $20.0\%$ below the ideal output.	The finite input resistance loads the source, so the amplifier input is $0.800[800/(200+800)]=0.640\,\text{V}$ . The actual output is $10.0(0.640)=6.40\,\text{V}$ . With infinite input resistance the input would be $0.800\,\text{V}$ and the output $8.00\,\text{V}$ . The reduction is $(8.00-6.40)/8.00\times100\%=20.0\%$ .	3

Tier 3 · Hard

Mark scheme for 3.13.4.4 Tier 3 · Hard
Question	Answers	Additional comments/Guidelines	Mark
01.1	The $160\,\text{kHz}$ bandwidth is sufficient, but the required $1.1\,\text{V}\,\mu\text{s}^{-1}$ exceeds the slew rate; maximum undistorted peak $2.1\,\text{V}$ .	The closed-loop bandwidth is $(4.0\times10^6)/25=1.60\times10^5\,\text{Hz}=160\,\text{kHz}$ , so $60\,\text{kHz}$ is inside the small-signal bandwidth. The required slew rate is $2\pi fV_p=2\pi(60\times10^3)(3.0)=1.13\times10^6\,\text{V s}^{-1}=1.13\,\text{V}\,\mu\text{s}^{-1}$ , which exceeds the available rate. Slew rate is therefore the active limitation. Rearranging gives $V_{p,\max}=\text{SR}/(2\pi f)=(0.80\times10^6)/(2\pi\times60\times10^3)=2.12\,\text{V}$ , or $2.1\,\text{V}$ .	6
02.1	Op amp B only. A has bandwidth $60\,\text{kHz}$ , whereas B has bandwidth $150\,\text{kHz}$ and its required output peak is $7.20\,\text{V}$ , below the $8.0\,\text{V}$ limit.	At gain $40$ , op amp A has bandwidth $(2.4\times10^6)/40=60\,\text{kHz}$ , so it cannot amplify the $100\,\text{kHz}$ signal faithfully. Op amp B has bandwidth $(6.0\times10^6)/40=150\,\text{kHz}$ , which is sufficient. Its required output peak is $40(0.180)=7.20\,\text{V}$ , below the $8.0\,\text{V}$ output limit, so B is suitable.	3
03.1	Greatest gain $25$ ; the output peak is $3.75\,\text{V}$ , output current $7.98\,\text{mA}$ and bandwidth $144\,\text{kHz}$ .	The output-voltage limit gives $G\leq9.00/0.150=60.0$ . The bandwidth condition gives $G\leq(3.60\times10^6)/(80.0\times10^3)=45.0$ . The current limit requires $G(0.150)/470\leq8.00\times10^{-3}$ , so $G\leq25.066\ldots$ . The greatest permitted whole number is therefore $25$ . At this gain, $V_{\text{out,peak}}=25(0.150)=3.75\,\text{V}$ , $I_{\text{peak}}=3.75/470=7.98\,\text{mA}$ and the bandwidth is $(3.60\times10^6)/25=144\,\text{kHz}$ . A gain of $26$ would require $8.30\,\text{mA}$ , so it is not allowed.	4
04.1	The first two pairs both give $4.00\,\text{MHz}$ ; the gain- $125$ measurement is anomalous and its predicted bandwidth is $32.0\,\text{kHz}$ .	The first pair gives $25.0(160\,\text{kHz})=4.00\,\text{MHz}$ and the second gives $80.0(50.0\,\text{kHz})=4.00\,\text{MHz}$ , so they are mutually consistent. The third measured product is $125(28.0\,\text{kHz})=3.50\,\text{MHz}$ , so that pair is anomalous. Using the supported product, its predicted bandwidth is $(4.00\times10^6)/125=3.20\times10^4\,\text{Hz}=32.0\,\text{kHz}$ .	4
05.1	Minimum three stages; each has gain $10.0$ , bandwidth $210\,\text{kHz}$ and feedback resistance $135\,\text{k}\Omega$ .	The bandwidth requirement limits each stage gain to $(2.10\times10^6)/(75.0\times10^3)=28.0$ . With two identical stages, each would need gain $\sqrt{1000}=31.62\ldots$ , which exceeds this limit. Three identical stages each need gain $\sqrt[3]{1000}=10.0$ , so each has bandwidth $(2.10\times10^6)/10.0=210\,\text{kHz}$ and the signal is passed. For a non-inverting stage, $R_f=(G-1)R_1=(10.0-1)(15.0\,\text{k}\Omega)=135\,\text{k}\Omega$ .	5

3.13.5.1 · Combinational logic

Tier 1 · Easy

Mark scheme for 3.13.5.1 Tier 1 · Easy
Question	Answers	Additional comments/Guidelines	Mark
01.1	$0,0,1,0$	When $A=0$ , the AND output is $0$ for both values of $B$ . For $AB=10$ , both $A$ and $\overline{B}$ are $1$ , so $Y=1$ . For $AB=11$ , $\overline{B}=0$ , so $Y=0$ . The ordered outputs are therefore $0,0,1,0$ .	2
02.1	NOR gate	$A+B$ is the output of an OR gate. The overline inverts that complete output, and an inverted OR function is a NOR gate.	1

Tier 2 · Standard

Mark scheme for 3.13.5.1 Tier 2 · Standard
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01.1	EOR gate; $Y=\overline{A}B+A\overline{B}$ .	The output is $1$ exactly when the inputs differ, which defines EOR. The row $01$ is selected by $\overline{A}B$ and the row $10$ by $A\overline{B}$ . OR combines the two mutually exclusive cases, giving $Y=\overline{A}B+A\overline{B}$ .	3
02.1	$Y=C(A+B)$ ; OR $A$ with $B$ , then AND the result with $C$ .	Every output- $1$ row has $C=1$ . Among the rows with $C=1$ , the output is $0$ only when both $A$ and $B$ are $0$ . Therefore $Y=C(A+B)$ . A minimal network uses one OR gate for $A+B$ followed by one AND gate with $C$ .	2
03.1	$Y=A\overline{B}\overline{C}+\overline{A}B\overline{C}+\overline{A}\overline{B}C$ ; form the three terms with NOT and AND gates, then OR them.	The three accepted input states are $100$ , $010$ and $001$ . They are selected by $A\overline{B}\overline{C}$ , $\overline{A}B\overline{C}$ and $\overline{A}\overline{B}C$ , respectively. OR combines these mutually exclusive terms, giving $Y=A\overline{B}\overline{C}+\overline{A}B\overline{C}+\overline{A}\overline{B}C$ . Invert each input as required, form the three products with AND gates, then feed them to an OR gate.	2

Tier 3 · Hard

Mark scheme for 3.13.5.1 Tier 3 · Hard
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01.1	Outputs $0,0,1,0,1,0,1,0$ . A NAND-only construction is $P=A\operatorname{\ NAND}A$ , $Q=B\operatorname{\ NAND}B$ , $R=P\operatorname{\ NAND}Q$ , $S=C\operatorname{\ NAND}C$ , $T=R\operatorname{\ NAND}S$ , $Y=T\operatorname{\ NAND}T$ .	The factor $\overline{C}$ makes every row with $C=1$ give $Y=0$ . When $C=0$ , the output equals $A+B$ , giving $0,1,1,1$ for $AB=00,01,10,11$ . In the stated row order the outputs are $0,0,1,0,1,0,1,0$ . For NAND-only logic, $P$ and $Q$ invert $A$ and $B$ . Then $R=\overline{\overline{A}\cdot\overline{B}}=A+B$ by De Morgan's law. $S=\overline{C}$ . The gate producing $T$ forms $\overline{R\cdot S}$ , and the final self-connected NAND inverts it, so $Y=R\cdot S=(A+B)\overline{C}$ .	6
02.1	$Q=\overline{\overline{AB}\cdot\overline{AC}}$ ; three NAND gates.	Use one NAND gate to produce $\overline{AB}$ and a second to produce $\overline{AC}$ . NANDing those outputs gives $\overline{\overline{AB}\cdot\overline{AC}}=AB+AC$ by De Morgan's law, so three NAND gates are required.	2
03.1	The $Y$ values are $0,1,1,0,1,0,0,1$ ; $Y=A\operatorname{\ EOR}B\operatorname{\ EOR}C$ .	First evaluate $X=A\operatorname{\ EOR}B$ , then compare $X$ with $C$ . This gives $Y=0,1,1,0,1,0,0,1$ for the stated row order. EOR is associative, so $(A\operatorname{\ EOR}B)\operatorname{\ EOR}C=A\operatorname{\ EOR}B\operatorname{\ EOR}C$ . The output is therefore $1$ exactly when an odd number of the three inputs is $1$ .	3
04.1	$Y=A\overline{C}+\overline{(A\operatorname{\ EOR}C)}B\overline{D}$ .	If the most significant bits differ, $X>Z$ exactly when $A=1$ and $C=0$ , giving $A\overline C$ . If $A=C$ , the lower bits decide the comparison and require $B=1$ , $D=0$ . Equality of $A$ and $C$ is $\overline{(A\operatorname{\ EOR}C)}$ , so this case is $\overline{(A\operatorname{\ EOR}C)}B\overline D$ . OR combines the mutually exclusive cases.	3
05.1	$Y=D_0\overline{S_1}\overline{S_0}+D_1\overline{S_1}S_0+D_2S_1\overline{S_0}+D_3S_1S_0$ ; with the stated data inputs, one EOR gate gives $Y=S_1\operatorname{\ EOR}S_0$ .	Each selector state enables one data input. The four selecting terms are therefore $D_0\overline{S_1}\overline{S_0}$ , $D_1\overline{S_1}S_0$ , $D_2S_1\overline{S_0}$ and $D_3S_1S_0$ , and OR combines them. With data pattern $0,1,1,0$ , the output is $1$ for selector states $01$ and $10$ only. Those are exactly the states in which $S_1$ and $S_0$ differ, so a single EOR gate is equivalent.	4

3.13.5.2 · Sequential logic

Tier 1 · Easy

Mark scheme for 3.13.5.2 Tier 1 · Easy
Question	Answers	Additional comments/Guidelines	Mark
01.1	$101$	The counter advances once per pulse: $000\to001\to010\to011\to100\to101$ . After five pulses its decimal count is $5$ , which is binary $101$ .	2
02.1	$0011$	BCD counts decimal digits and returns to $0000$ after $1001$ . Starting at decimal $7$ , six pulses give $8,9,0,1,2,3$ . Decimal $3$ is $0011$ in BCD.	2

Tier 2 · Standard

Mark scheme for 3.13.5.2 Tier 2 · Standard
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01.1	Johnson counter: $8$ states; BCD counter: $10$ states; BCD cycle rate $120\,\text{Hz}$ .	An $n$ -stage Johnson counter has $2n$ states, so four stages give $8$ . BCD represents decimal digits $0$ to $9$ , so it has $10$ states. One full BCD cycle requires $10$ clock pulses, hence the cycle rate is $1.20\times10^3/10=120\,\text{Hz}$ .	4
02.1	Count down; next output $0010$ .	The outputs represent decimal $5$ , $4$ and $3$ , so the control input is set to count down. The next state is decimal $2$ , written as four bits as $0010$ .	2
03.1	$1010$	The initial state $1101$ is decimal $13$ . Seven upward counts give $13+7=20$ , which wraps modulo $16$ to decimal $4$ , or $0100$ . Ten downward counts then give $4-10=-6$ , which wraps modulo $16$ to decimal $10$ . Decimal $10$ is $1010$ in four-bit binary.	2

Tier 3 · Hard

Mark scheme for 3.13.5.2 Tier 3 · Hard
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01.1	Reset on $110$ ; sequence $000,001,010,011,100,101$ ; after $53$ pulses the output is $101$ ; cycle rate $4.0\,\text{kHz}$ .	Modulo $6$ requires six stable states representing decimal $0$ to $5$ . Decode the next state, decimal $6=110$ , and use it to reset immediately to $000$ . The stable sequence is therefore $000,001,010,011,100,101$ and then back to $000$ . Since $53=8\times6+5$ , the state after $53$ pulses is the state five steps after reset, $101$ . Each cycle uses six pulses, so its repetition rate is $(24\times10^3)/6=4.0\times10^3\,\text{Hz}=4.0\,\text{kHz}$ .	6
02.1	$0111$	Applying the stated complemented feedback and shift rule gives $0000\to1000\to1100\to1110\to1111\to0111\to0011\to0001\to0000$ . The cycle has eight states. Since $37=4\times8+5$ , the required state is five steps after $0000$ , which is $0111$ .	4
03.1	States $00000,10000,11000,11100,11110,11111,01111,00111,00011,00001$ ; $Y$ repeats as $0,1,1,0,0$ ; frequency $6.00\,\text{kHz}$ .	The register advances through $00000\to10000\to11000\to11100\to11110\to11111\to01111\to00111\to00011\to00001\to00000$ . Evaluating $Q_4\operatorname{\ EOR}Q_2$ gives $0,1,1,0,0,0,1,1,0,0$ , which is two repetitions of $0,1,1,0,0$ . Thus one output cycle lasts five clock periods and $f_Y=30.0\,\text{kHz}/5=6.00\,\text{kHz}$ .	3
04.1	$f_{Q_0}=800\,\text{kHz}$ , $f_{Q_3}=100\,\text{kHz}$ ; counter output $1011$ .	$Q_0$ changes state on every clock pulse, so one complete $Q_0$ cycle takes two pulses and $f_{Q_0}=1.60\,\text{MHz}/2=800\,\text{kHz}$ . The most significant output $Q_3$ completes one cycle every $16$ pulses, so $f_{Q_3}=1.60\,\text{MHz}/16=100\,\text{kHz}$ . A four-bit counter repeats every $16$ pulses. Since $2027=126(16)+11$ , its state is decimal $11$ , or $1011$ .	4
05.1	The tens counter receives $14$ pulses; final display $05$ ; tens output $0000$ and units output $0101$ ; pattern rate $25.0\,\text{Hz}$ .	The units digit starts at $2$ and passes from $0$ to $9$ on pulses $3,13,23,\ldots,133$ , so the tens counter receives $14$ pulses. Equivalently, $42-137=-95$ , which is $5$ modulo $100$ , giving display $05$ . Decimal $0$ and $5$ are $0000$ and $0101$ in BCD. The complete two-counter sequence contains $100$ states, so its repetition rate is $(2.50\times10^3)/100=25.0\,\text{Hz}$ .	5

3.13.5.3 · Astables

Tier 1 · Easy

Mark scheme for 3.13.5.3 Tier 1 · Easy
Question	Answers	Additional comments/Guidelines	Mark
01.1	$400\,\text{Hz}$ ; $40\%$	The frequency is $f=1/T=1/(2.5\times10^{-3})=400\,\text{Hz}$ . The duty cycle is $(1.0/2.5)\times100\%=40\%$ .	2
02.1	$0.32\,\text{ms}$	The period is $T=1/f=1/(2.0\times10^3)=5.0\times10^{-4}\,\text{s}=0.50\,\text{ms}$ . Therefore $t_{\text{low}}=0.50-0.18=0.32\,\text{ms}$ .	2

Tier 2 · Standard

Mark scheme for 3.13.5.3 Tier 2 · Standard
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01.1	$1.43\,\text{ms}$ ; $698\,\text{Hz}$	During charging, $V=V_S-(V_S-V_0)e^{-t/RC}$ . With $V_0=V_S/3$ and $V=2V_S/3$ , $1/3=(2/3)e^{-t/RC}$ , so $e^{-t/RC}=1/2$ and $t=RC\ln2$ . The discharge over the reverse interval has the same duration, so $T=2RC\ln2=2(47\times10^3)(22\times10^{-9})\ln2=1.433\times10^{-3}\,\text{s}$ . Hence $f=1/T=697.8\,\text{Hz}$ , giving $1.43\,\text{ms}$ and $698\,\text{Hz}$ .	4
02.1	$42\,\text{nF}$	Since $T=1/f$ , $C=1/(1.4Rf)$ . Thus $C=1/[1.4(68\times10^3)(250)]=4.20\times10^{-8}\,\text{F}=42\,\text{nF}$ to two significant figures.	2
03.1	New resistance $60.9\,\text{k}\Omega$ ; increase $8.70\%$ .	Because $f=1/(1.4RC)$ and $C$ is unchanged, resistance is inversely proportional to frequency. The new frequency is $0.9200$ of its initial value, so $R_{\text{new}}=56.0/0.9200=60.87\,\text{k}\Omega$ , or $60.9\,\text{k}\Omega$ . The percentage increase is $(60.87-56.0)/56.0\times100\%=8.70\%$ .	2

Tier 3 · Hard

Mark scheme for 3.13.5.3 Tier 3 · Hard
Question	Answers	Additional comments/Guidelines	Mark
01.1	$t_{\text{high}}=0.85\,\text{ms}$ , $t_{\text{low}}=0.34\,\text{ms}$ , $f=8.4\times10^2\,\text{Hz}$ , duty cycle $71\%$ , mark-to-space ratio $2.5:1$ .	The high interval is the charge time: $t_{\text{high}}=(82\times10^3)(15\times10^{-9})\ln2=8.53\times10^{-4}\,\text{s}=0.853\,\text{ms}$ . The low interval is $t_{\text{low}}=(33\times10^3)(15\times10^{-9})\ln2=3.43\times10^{-4}\,\text{s}=0.343\,\text{ms}$ . Thus $T=1.196\,\text{ms}$ and $f=1/T=836\,\text{Hz}$ , or $8.4\times10^2\,\text{Hz}$ . The duty cycle is $0.853/1.196\times100\%=71.3\%$ , or $71\%$ . The mark-to-space ratio is $0.853:0.343=2.48:1$ , or $2.5:1$ .	6
02.1	$R_1=6.80\,\text{k}\Omega$ and $R_2=13.6\,\text{k}\Omega$	The period is $T=1/(3.50\times10^3)=285.714\ldots\,\mu\text{s}$ . Hence $t_{\text{high}}=0.600T=171.429\ldots\,\mu\text{s}$ and $t_{\text{low}}=0.400T=114.286\ldots\,\mu\text{s}$ . Therefore $R_2=t_{\text{low}}/(0.7C)=13.605\ldots\,\text{k}\Omega$ . Also $R_1+R_2=t_{\text{high}}/(0.7C)=20.408\ldots\,\text{k}\Omega$ , so $R_1=6.8027\ldots\,\text{k}\Omega$ . To three significant figures, $R_1=6.80\,\text{k}\Omega$ and $R_2=13.6\,\text{k}\Omega$ .	2
03.1	Frequency $847.7\,\text{Hz}$ to $1145.2\,\text{Hz}$ ( $848\,\text{Hz}$ to $1.15\,\text{kHz}$ , $3$ s.f.); duty cycle $84.0\%$ to $86.1\%$ (both to $3$ s.f.).	The lowest frequency uses all maximum component values: $T_{\max}=0.7[(48.846+2(10.395))\times10^3](24.2\times10^{-9})=1.17963\,\text{ms}$ , so $f_{\min}=847.7\,\text{Hz}$ . The highest frequency uses all minimum values: $T_{\min}=0.7[(44.194+2(9.405))\times10^3](19.8\times10^{-9})=0.87324\,\text{ms}$ , so $f_{\max}=1145.2\,\text{Hz}$ . Duty cycle is $(R_1+R_2)/(R_1+2R_2)$ , independent of $C$ . It is smallest for $R_1=44.194\,\text{k}\Omega$ and $R_2=10.395\,\text{k}\Omega$ , giving $84.0038\%$ , and largest for $R_1=48.846\,\text{k}\Omega$ and $R_2=9.405\,\text{k}\Omega$ , giving $86.0988\%$ .	4
04.1	Lower temperature: $2.04\,\text{kHz}$ and $71.4\%$ ; higher temperature: $2.86\,\text{kHz}$ and $60.0\%$ . The lower thermistor resistance shortens only the high interval, so frequency rises and duty cycle falls.	At the lower temperature, $t_{\text{high}}=0.7(50.0\times10^3)(10.0\times10^{-9})=350\,\mu\text{s}$ and $t_{\text{low}}=0.7(20.0\times10^3)(10.0\times10^{-9})=140\,\mu\text{s}$ . Thus $T=490\,\mu\text{s}$ , $f=1/T=2.041\ldots\,\text{kHz}$ and duty cycle $=350/490\times100\%=71.43\ldots\%$ . At the higher temperature, $t_{\text{high}}=0.7(30.0\times10^3)(10.0\times10^{-9})=210\,\mu\text{s}$ while $t_{\text{low}}$ remains $140\,\mu\text{s}$ . Therefore $T=350\,\mu\text{s}$ , $f=2.857\ldots\,\text{kHz}$ and duty cycle $=210/350\times100\%=60.0\%$ . Only the high-time resistance contains $R_T$ , so its decrease produces both trends.	5
05.1	Initial frequency $2.04\,\text{kHz}$ and average power $0.257\,\text{W}$ , so the limit is exceeded; greatest $R_1=10.0\,\text{k}\Omega$ ; resulting frequency $2.38\,\text{kHz}$ .	Initially $t_{\text{high}}=0.7(25.0\times10^3)(20.0\times10^{-9})=350\,\mu\text{s}$ and $t_{\text{low}}=0.7(10.0\times10^3)(20.0\times10^{-9})=140\,\mu\text{s}$ . Hence $T=490\,\mu\text{s}$ , $f=2040.8\ldots\,\text{Hz}$ and duty cycle $=350/490=0.714286\ldots$ . The average power is $0.360(0.714286\ldots)=0.25714\ldots\,\text{W}$ . The power limit requires duty cycle at most $0.240/0.360=2/3$ . Since duty cycle is $(R_1+R_2)/(R_1+2R_2)$ , setting it to $2/3$ gives $3(R_1+R_2)=2(R_1+2R_2)$ and $R_1=R_2=10.0\,\text{k}\Omega$ . The new period is $0.7[10.0+2(10.0)]\times10^3(20.0\times10^{-9})=420\,\mu\text{s}$ , so $f=2.381\ldots\,\text{kHz}$ .	5

3.13.6.1 · Principles of communication systems

Tier 1 · Easy

Mark scheme for 3.13.6.1 Tier 1 · Easy
Question	Answers	Additional comments/Guidelines	Mark
01.1	Input transducer, transmitter, transmission channel, receiver, output transducer.	The message is first converted into an electrical signal by the input transducer. The transmitter sends a suitable signal through the channel. The receiver recovers the information and the output transducer converts it to the required output, so the stated order follows.	2
02.1	The microphone converts sound energy into an electrical signal; the loudspeaker converts electrical energy into sound.	The microphone is the input transducer, so it changes the original sound into an electrical information signal. The loudspeaker is the output transducer and performs the reverse conversion to reproduce sound.	2

Tier 2 · Standard

Mark scheme for 3.13.6.1 Tier 2 · Standard
Question	Answers	Additional comments/Guidelines	Mark
01.1	The transmitter converts the information into a suitable signal and supplies it to the channel; the channel carries the signal; the receiver selects the wanted signal and recovers its information.	Award one linked purpose per stage: the transmitter prepares the message for transmission, for example by using it to modulate a carrier and amplifying the result; the channel is the physical medium through which the signal travels; the receiver selects the required transmission from other signals and noise, then demodulates it to recover the information.	3
02.1	The output transducer is faulty. The correct signal at the receiver output shows that the transmitter, channel and receiver have already conveyed and recovered the information.	A correct electrical waveform at the receiver output proves that information passed through the transmitter and channel and was recovered by the receiver. The remaining conversion from electrical energy to sound is performed by the output transducer, the loudspeaker, so that is the faulty block.	3
03.1	The input transducer converts the water-level information into an electrical signal for the transmitter. The output transducer converts the recovered electrical signal into a visible indication of water level.	A water-level or pressure sensor is the input transducer: it changes a non-electrical physical quantity carrying the information into a corresponding electrical signal. After transmission and recovery, the output transducer changes that electrical signal into the required visible form, for example a meter movement or display reading. The two transducers therefore perform opposite conversions at the ends of the communication system.	3

Tier 3 · Hard

Mark scheme for 3.13.6.1 Tier 3 · Hard
Question	Answers	Additional comments/Guidelines	Mark
01.1	A temperature sensor produces an electrical information signal; a transmitter encodes or modulates and launches it into a channel; a receiver selects and recovers it; a display converts it to a readable output, while channel noise can alter the received signal.	The input transducer is a temperature sensor that converts temperature into an electrical information signal. The transmitter prepares that signal for the chosen channel, for example by encoding it or modulating a carrier, and supplies sufficient power. The wire, fibre or radio path is the transmission channel. Noise or interference added mainly in the channel changes the received waveform and can reduce signal-to-noise ratio. The receiver selects the wanted signal and decodes or demodulates it. Finally, the output stage drives a display so the recovered information is presented as a temperature. Continuous repetition keeps the result real time.	5
02.1	The first fault is in the transmission channel; the second fault is in the receiver or demodulator.	A correct transmitter output followed by no receiver input places the first fault between those test points, in the transmission channel. For the second symptom, the correct receiver input shows that the transmitter and channel are working, but distortion introduced by the receiver output places the fault in the receiver's decoding or demodulation stage.	3
03.1	Greater transmitter power makes the signal entering the channel larger, so it remains larger relative to noise added by the channel after attenuation. Receiver gain acts only after that noise has been added and amplifies signal and noise together, leaving their ratio unchanged; it cannot recover detail already masked by noise.	The transmitter supplies signal power before the transmission channel. Raising that power increases the signal that remains at the receiver input after channel attenuation, but it does not increase noise generated later in the channel, so the signal-to-noise ratio can improve. Receiver amplification occurs after signal and channel noise have arrived together. It multiplies both by the same factor, so their ratio and the information quality do not improve. A receiver can raise the output level but cannot reconstruct variations already hidden by noise.	4
04.1	Microphone $\to$ transmitter $\to$ first channel $\to$ relay receiver $\to$ relay transmitter $\to$ second channel $\to$ final receiver $\to$ loudspeaker. The relay receiver recovers the information; the relay transmitter prepares and launches a new signal into the second channel.	The signal passes through the blocks in this order: microphone, transmitter, first channel, relay receiver, relay transmitter, second channel, final receiver and loudspeaker. At the relay, the receiver selects the wanted signal and recovers its information. The transmitter then prepares and launches a new signal carrying that information through the second channel.	3
05.1	Outward: control transducer $\to$ transmitter $\to$ channel $\to$ receiver $\to$ actuator. Return: position sensor $\to$ transmitter $\to$ channel $\to$ receiver $\to$ display. Two chains are needed because the control and position information travel in opposite directions.	The outward chain is control transducer $\to$ transmitter $\to$ channel $\to$ receiver $\to$ actuator, carrying the operator's command to the machine. The return chain is position sensor $\to$ transmitter $\to$ channel $\to$ receiver $\to$ display, carrying the measured position back. The information travels in opposite directions, so one one-way chain cannot perform both tasks.	3

3.13.6.2 · Transmission media

Tier 1 · Easy

Mark scheme for 3.13.6.2 Tier 1 · Easy
Question	Answers	Additional comments/Guidelines	Mark
01.1	Optical fibre, because it has high bandwidth and does not radiate a signal that can easily be intercepted.	Optical fibre supports a high data rate because of its large available bandwidth. Light remains guided within the fibre, so interception generally requires physical access and is easier to detect than receiving a broadcast radio signal.	2
02.1	Microwaves have short wavelengths and diffract very little, so Earth or other obstacles can block the beam.	The short wavelength produces little diffraction around large obstacles. The transmitting and receiving aerials must therefore see each other above terrain and Earth's curvature.	2

Tier 2 · Standard

Mark scheme for 3.13.6.2 Tier 2 · Standard
Question	Answers	Additional comments/Guidelines	Mark
01.1	A long-wavelength ground wave diffracts around Earth's surface; a sky wave is refracted or reflected by the ionosphere back towards Earth.	A ground wave of sufficiently long wavelength undergoes appreciable diffraction, so it follows the curvature of Earth's surface rather than being blocked at the geometric horizon. Alternatively, a sky wave is directed upwards and is refracted through, or described as reflected by, ionised layers in the atmosphere so that it returns to Earth at a distant point. Each route therefore avoids a purely line-of-sight path by a different wave process.	4
02.1	Radio is more suitable for rapid, low-cost installation, but it is less secure because its broadcast signal is easier to intercept.	A radio link needs no trench or long physical cable, so it can be installed and moved quickly at lower initial cost. Its signal spreads through free space, however, so an unauthorised receiver can intercept it more easily. Buried copper provides a confined path and can be more physically secure, but excavation and cable installation make it slower and more costly for a temporary event.	3
03.1	Use $200\,\text{kHz}$ : its wavelength is $1.50\,\text{km}$ and it can diffract as a ground wave, whereas the $5.00\,\text{GHz}$ wavelength is $0.0600\,\text{m}$ and its microwave path is approximately line of sight.	For $200\,\text{kHz}$ , $\lambda=c/f=(3.00\times10^8)/(2.00\times10^5)=1.50\times10^3\,\text{m}$ . This long wavelength diffracts appreciably around terrain and Earth's surface, allowing ground-wave reception beyond the horizon. For $5.00\,\text{GHz}$ , $\lambda=(3.00\times10^8)/(5.00\times10^9)=0.0600\,\text{m}$ . Such microwaves diffract very little around hills and normally require line of sight, so the $200\,\text{kHz}$ carrier is suitable.	3

Tier 3 · Hard

Mark scheme for 3.13.6.2 Tier 3 · Hard
Question	Answers	Additional comments/Guidelines	Mark
01.1	$0.240\,\text{s}$ ; different frequencies prevent receiver de-sensing; fibre generally offers high data rate and greater physical security than a broadcast satellite link.	The minimum path is ground to satellite and back to ground, so $d=2(3.60\times10^7)=7.20\times10^7\,\text{m}$ . The delay is $t=d/c=(7.20\times10^7)/(3.00\times10^8)=0.240\,\text{s}$ . A satellite transmits the downlink while receiving the much weaker uplink; separating their frequencies prevents its own strong transmission from overwhelming, or de-sensing, the receiver. Satellite microwave links cover large areas without laying cable but broadcast through free space and have limited allocated bandwidth. Optical fibre commonly provides a higher data rate and confines the signal to a physical cable, making interception harder, although laying that cable can be costly.	6
02.1	Optical fibre: transfer time $3.0\,\text{s}$ . Radio would take $15\,\text{s}$ and copper $30\,\text{s}$ ; fibre also confines the signal and is harder to intercept.	$750\,\text{MB}=750\times8=6000\,\text{Mbit}$ . Fibre takes $6000/2000=3.0\,\text{s}$ , radio takes $6000/400=15\,\text{s}$ and copper takes $6000/200=30\,\text{s}$ . Only fibre meets the $5.0\,\text{s}$ limit. Its guided light remains within the installed cable, so interception normally requires physical access and is more difficult than receiving a radio broadcast.	4
03.1	The minimum reduction factor is $2.5\times10^6$ ; select the $1.0\times10^7$ filter, which leaves $2.0\times10^{-13}\,\text{W}$ of leakage.	The greatest permitted leakage is $0.10(8.0\times10^{-12})=8.0\times10^{-13}\,\text{W}$ . The required reduction factor is therefore $(2.0\times10^{-6})/(8.0\times10^{-13})=2.5\times10^6$ . A factor of $1.0\times10^6$ is insufficient. The $1.0\times10^7$ filter leaves $(2.0\times10^{-6})/(1.0\times10^7)=2.0\times10^{-13}\,\text{W}$ , which is below the permitted value.	4
04.1	The direct horizon range is $55.9\,\text{km}$ , so the stations are not in line of sight. One midpoint relay is sufficient: the two section limits are $41.8\,\text{km}$ and $49.8\,\text{km}$ , each exceeding $35.0\,\text{km}$ .	The horizon distances are $\sqrt{2(6.37\times10^6)(45.0)}=23.94\ldots\,\text{km}$ and $\sqrt{2(6.37\times10^6)(80.0)}=31.92\ldots\,\text{km}$ . Their sum is $55.87\ldots\,\text{km}$ , less than $70.0\,\text{km}$ . For the relay, $d_r=\sqrt{2(6.37\times10^6)(25.0)}=17.85\ldots\,\text{km}$ . The line-of-sight limits are therefore $23.94\ldots+17.85\ldots=41.79\ldots\,\text{km}$ and $31.92\ldots+17.85\ldots=49.77\ldots\,\text{km}$ . Each exceeds the corresponding $35.0\,\text{km}$ section, so one relay is sufficient.	5
05.1	Greatest one-hop ground distance $846\,\text{km}$ ; minimum $4$ hops; equal-hop angle $56.8^\circ$ to the vertical (accept $56.7^\circ$ to $56.8^\circ$ ).	For half a symmetrical hop, the vertical side is $225\,\text{km}$ and the greatest permitted horizontal side is $225\tan62.0^\circ$ . The greatest full ground distance is therefore $2(225)\tan62.0^\circ=846.326\ldots\,\text{km}$ . Since $2750/846.326\ldots=3.249\ldots$ , three hops are insufficient and at least four are required. Four equal hops each cover $2750/4=687.5\,\text{km}$ , so each half-hop has horizontal length $343.75\,\text{km}$ . Hence $\tan\theta=343.75/225$ and $\theta=56.793\ldots^\circ$ , which is within the aerial's radiation limit.	5

3.13.6.3 · Time-division multiplexing

Tier 1 · Easy

Mark scheme for 3.13.6.3 Tier 1 · Easy
Question	Answers	Additional comments/Guidelines	Mark
01.1	$4$ slots; $32$ data bits	There is one slot per channel, so each frame has $4$ slots. Each slot contains $8$ bits, giving $4\times8=32$ data bits per frame.	2
02.1	$50\,\mu\text{s}$	The frame duration is the reciprocal of frame rate: $T=1/(2.0\times10^4)=5.0\times10^{-5}\,\text{s}=50\,\mu\text{s}$ .	1

Tier 2 · Standard

Mark scheme for 3.13.6.3 Tier 2 · Standard
Question	Answers	Additional comments/Guidelines	Mark
01.1	Frame rate $8.0\,\text{kHz}$ ; slot rate $48\,\text{kslot s}^{-1}$ ; bit rate $480\,\text{kbit s}^{-1}$ .	Every frame must carry one new sample from each channel, so the frame rate equals the sampling rate, $8.0\,\text{kframe s}^{-1}$ . There are six slots per frame, hence the slot rate is $6(8.0\times10^3)=4.8\times10^4\,\text{slot s}^{-1}$ . Each slot contains $10$ bits, so the bit rate is $(4.8\times10^4)(10)=4.8\times10^5\,\text{bit s}^{-1}=480\,\text{kbit s}^{-1}$ .	4
02.1	$12.0\,\text{kHz}$	Each frame contains $5(12)+5=65$ bits. The greatest frame rate is $(780\times10^3)/65=12.0\times10^3\,\text{frame s}^{-1}$ . One sample from each channel is carried per frame, so each channel can be sampled at $12.0\,\text{kHz}$ .	2
03.1	The receiver loses the frame boundary, so it groups the following bits into the wrong slot positions. Samples are corrupted or routed to the wrong channels until a later synchronisation marker restores the correct $A$ , $B$ , $C$ slot order.	The synchronisation bit marks where the repeating frame starts. If it is lost, the receiver's bit groups no longer coincide with the transmitted slot boundaries, so the demultiplexer cannot reliably assign the next groups to $A$ , $B$ and $C$ . The displaced groups either combine bits from adjacent samples or are sent to the wrong output. Correct separation resumes only when the receiver recognises a later synchronisation marker and restores the frame boundary and slot order.	3

Tier 3 · Hard

Mark scheme for 3.13.6.3 Tier 3 · Hard
Question	Answers	Additional comments/Guidelines	Mark
01.1	Frame rate $12.0\,\text{kHz}$ ; data-slot rate $288\,\text{kslot s}^{-1}$ ; $296$ bits per frame; bit rate $3.55\,\text{Mbit s}^{-1}$ ; the link is sufficient.	One sample per channel is sent in each frame, so the frame rate is $12.0\times10^3\,\text{frame s}^{-1}$ . With $24$ data slots per frame, the data-slot rate is $24(12.0\times10^3)=2.88\times10^5\,\text{slot s}^{-1}$ . Data occupy $24\times12=288$ bits and synchronisation adds $8$ , giving $296$ bits per frame. The bit rate is $296(12.0\times10^3)=3.552\times10^6\,\text{bit s}^{-1}=3.55\,\text{Mbit s}^{-1}$ . Since $3.55<4.00$ , the link is sufficient, with about $0.45\,\text{Mbit s}^{-1}$ spare capacity.	6
02.1	$352\,\text{kbit s}^{-1}$ ; synchronisation identifies each frame so the receiver assigns successive time slots to the correct channels.	Each frame contains $4(10)+4=44$ bits. The bit rate is $44(8.0\times10^3)=3.52\times10^5\,\text{bit s}^{-1}=352\,\text{kbit s}^{-1}$ . The synchronisation bits mark the frame boundary, allowing the receiver to demultiplex the repeating time slots in order and route each sample to its channel.	3
03.1	$10$ channels; yes, because $4.00\,\text{kHz}>2(1.50\,\text{kHz})$ .	One sample per channel per frame means that the frame rate is the common sampling rate, $4.00\times10^3\,\text{s}^{-1}$ . The number of bits in each frame is $(512\times10^3)/(4.00\times10^3)=128$ . Removing the $8$ synchronisation bits leaves $120$ sample bits, so the number of channels is $120/12=10$ . Nyquist requires a sampling rate of at least $2f_{\max}=2(1.50)=3.00\,\text{kHz}$ . The actual $4.00\,\text{kHz}$ rate is greater than this, so it satisfies the condition.	3
04.1	Slots per master frame: $A:4$ , $B:2$ , $C:1$ ; $86$ bits per master frame; bit rate $258\,\text{kbit s}^{-1}$ .	At $3.00\,\text{kframe s}^{-1}$ , channel $A$ needs $12.0/3.00=4$ samples per frame, $B$ needs $6.00/3.00=2$ and $C$ needs $3.00/3.00=1$ . Their data occupy $4(10)+2(12)+1(16)=80$ bits. Adding $6$ synchronisation bits gives $86$ bits per master frame. The bit rate is $86(3.00\times10^3)=2.58\times10^5\,\text{bit s}^{-1}=258\,\text{kbit s}^{-1}$ .	5
05.1	$60$ spare bits per frame; at most $6$ sensors.	The link permits $(1.12\times10^6)/8000=140$ bits per frame. The six audio samples and synchronisation use $6(12)+8=80$ bits, leaving $140-80=60$ spare bits. Each sensor needs $10$ bits in every frame, so the number that fit is $60/10=6$ .	2

3.13.6.4 · Amplitude (AM) and frequency modulation (FM) techniques

Tier 1 · Easy

Mark scheme for 3.13.6.4 Tier 1 · Easy
Question	Answers	Additional comments/Guidelines	Mark
01.1	$12\,\text{kHz}$	Simple AM has two sidebands, so its bandwidth is $2f_m=2(6.0\,\text{kHz})=12\,\text{kHz}$ .	2
02.1	Frequency modulation	Changing cycle spacing means that the instantaneous carrier frequency varies. Since the amplitude stays constant, the description is FM rather than AM.	1

Tier 2 · Standard

Mark scheme for 3.13.6.4 Tier 2 · Standard
Question	Answers	Additional comments/Guidelines	Mark
01.1	FM $66\,\text{kHz}$ ; AM $16\,\text{kHz}$	For FM, $B=2(\Delta f+f_m)=2(25+8.0)\,\text{kHz}=66\,\text{kHz}$ . For AM, $B=2f_m=2(8.0\,\text{kHz})=16\,\text{kHz}$ .	4
02.1	Carrier: $1.438\,\text{MHz}$ ; lower and upper sidebands: $1.431\,\text{MHz}$ and $1.445\,\text{MHz}$ ; highest information frequency: $7\,\text{kHz}$ ; occupied bandwidth: $14\,\text{kHz}$ .	The carrier is the central line, $1.438\,\text{MHz}$ , and the equally spaced outer lines are the lower and upper sidebands. Their separation from the carrier is $(1.438-1.431)\,\text{MHz}=0.007\,\text{MHz}=7\,\text{kHz}$ , so this is the highest information frequency. The occupied bandwidth is the full sideband span, $(1.445-1.431)\,\text{MHz}=0.014\,\text{MHz}=14\,\text{kHz}$ .	3
03.1	Carrier frequency $96.000\,\text{MHz}$ ; maximum deviation $75.0\,\text{kHz}$ ; bandwidth $174\,\text{kHz}$ .	The unmodulated carrier is the midpoint, $(95.925+96.075)/2=96.000\,\text{MHz}$ . The maximum deviation is half the range, $(96.075-95.925)/2=0.0750\,\text{MHz}=75.0\,\text{kHz}$ . Hence $B=2(\Delta f+f_M)=2(75.0+12.0)\,\text{kHz}=174\,\text{kHz}$ .	3

Tier 3 · Hard

Mark scheme for 3.13.6.4 Tier 3 · Hard
Question	Answers	Additional comments/Guidelines	Mark
01.1	FM bandwidth $180\,\text{kHz}$ and $5$ channels; AM bandwidth $30\,\text{kHz}$ and $30$ channels. FM better rejects amplitude noise but accommodates fewer channels.	For FM, $B=2(\Delta f+f_m)=2(75+15)\,\text{kHz}=180\,\text{kHz}$ . The ideal channel count is $900/180=5$ . For AM, $B=2f_m=2(15)=30\,\text{kHz}$ , so $900/30=30$ ideal channels. In FM the information is carried by frequency changes, so amplitude-limiting can remove much amplitude noise without removing the information, giving a better signal-to-noise performance. Its larger bandwidth, however, means far fewer stations fit the same allocation. These ideal counts ignore guard bands.	6
02.1	Carrier frequency $250\,\text{kHz}$ ; information frequency $2.0\,\text{kHz}$ .	The carrier frequency is the number of carrier cycles divided by the elapsed time: $f_c=125/(0.50\times10^{-3})=2.5\times10^5\,\text{Hz}=250\,\text{kHz}$ . The cycle-spacing pattern repeats every $0.50\,\text{ms}$ , so $f_m=1/(0.50\times10^{-3})=2.0\times10^3\,\text{Hz}=2.0\,\text{kHz}$ .	2
03.1	Greatest information frequency $17.0\,\text{kHz}$ ; minimum frequency deviation $51.0\,\text{kHz}$ .	At the greatest possible information frequency the deviation takes its smallest allowed value, $\Delta f=3f_M$ . Substitution into $B=2(\Delta f+f_M)$ gives $136=2(3f_M+f_M)=8f_M$ . Hence $f_M=136/8=17.0\,\text{kHz}$ and $\Delta f=3(17.0)=51.0\,\text{kHz}$ .	2
04.1	The first transmission occupies $711$ to $729\,\text{kHz}$ and the second occupies $733$ to $759\,\text{kHz}$ ; the unused interval is $4.00\,\text{kHz}$ .	For AM, the occupied frequencies extend by the highest information frequency either side of the carrier. The first range is $720\pm9.00\,\text{kHz}$ , or $711$ to $729\,\text{kHz}$ . The second is $746\pm13.0\,\text{kHz}$ , or $733$ to $759\,\text{kHz}$ . The interval from $729$ to $733\,\text{kHz}$ is $4.00\,\text{kHz}$ .	3
05.1	Worst-case greatest information frequency $36.0\,\text{kHz}$ and bandwidth $176\,\text{kHz}$ .	In the worst case, carrier drift uses $7.00\,\text{kHz}$ of the $95.0\,\text{kHz}$ allowance on one side. The remaining spectral half-width is $95.0-7.00=88.0\,\text{kHz}$ . Since the half-width from the actual carrier is $\Delta f+f_M$ , $52.0+f_M\leq88.0$ , so $f_M=36.0\,\text{kHz}$ at the limit. The bandwidth is then $2(52.0+36.0)=176\,\text{kHz}$ .	2