3.13 Electronics (A-level only) — AQA 7408 revision notes

3.13.1.1

MOSFET (metal-oxide semiconducting field-effect transistor)

Notes

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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

ORIGINAL

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

Tier 2 · Standard

ORIGINAL

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

Tier 3 · Hard

ORIGINAL

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

3.13.1.2

Zener diode

Notes

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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

ORIGINAL

State the bias direction and operating region used when a Zener diode provides a constant-voltage reference.

[1 mark]

Total for this question: 1

Tier 2 · Standard

ORIGINAL

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

Tier 3 · Hard

ORIGINAL

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

3.13.1.3

Photodiode

Notes

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Your confidence:

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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

ORIGINAL

Name the operating mode of a photodiode that has an external reverse-bias voltage.

[1 mark]

Total for this question: 1

Tier 2 · Standard

ORIGINAL

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

Tier 3 · Hard

ORIGINAL

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

3.13.1.4

Hall effect sensor

Notes

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Your confidence:

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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

ORIGINAL

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

Tier 2 · Standard

ORIGINAL

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

Tier 3 · Hard

ORIGINAL

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

3.13.2.1

Difference between analogue and digital signals

Notes

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Your confidence:

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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

ORIGINAL

State one difference between an analogue signal and a binary digital signal.

[1 mark]

Total for this question: 1

Tier 2 · Standard

ORIGINAL

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

Tier 3 · Hard

ORIGINAL

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

3.13.3.1

LC resonance filters

Notes

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Your confidence:

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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

ORIGINAL

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

Tier 2 · Standard

ORIGINAL

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

Tier 3 · Hard

ORIGINAL

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

3.13.3.2

The ideal operational amplifier

Notes

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Your confidence:

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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

ORIGINAL

State the current entering either input terminal of an ideal operational amplifier.

[1 mark]

Total for this question: 1

Evidence from answers you checked

Checked automatically against the model answer once you submit.

Tier 2 · Standard

ORIGINAL

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

Tier 3 · Hard

ORIGINAL

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

3.13.4.1

Operational amplifier: inverting amplifier configuration

Notes

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Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

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

ORIGINAL

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

Tier 2 · Standard

ORIGINAL

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

Tier 3 · Hard

ORIGINAL

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

3.13.4.2

Operational amplifier: non-inverting amplifier configuration

Notes

Evidence from your answers: none yet

Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

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

ORIGINAL

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

Tier 2 · Standard

ORIGINAL

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

Tier 3 · Hard

ORIGINAL

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

3.13.4.3

Operational amplifier: summing amplifier configuration

Notes

Evidence from your answers: none yet

Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

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

ORIGINAL

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

Tier 2 · Standard

ORIGINAL

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

Tier 3 · Hard

ORIGINAL

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

3.13.4.4

Real operational amplifiers

Notes

Evidence from your answers: none yet

Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

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

ORIGINAL

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

Tier 2 · Standard

ORIGINAL

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

Tier 3 · Hard

ORIGINAL

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

3.13.5.1

Combinational logic

Notes

Evidence from your answers: none yet

Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

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

ORIGINAL

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

Tier 2 · Standard

ORIGINAL

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

Tier 3 · Hard

ORIGINAL

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

3.13.5.2

Sequential logic

Notes

Evidence from your answers: none yet

Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

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

ORIGINAL

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

Tier 2 · Standard

ORIGINAL

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

Tier 3 · Hard

ORIGINAL

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

3.13.5.3

Astables

Notes

Evidence from your answers: none yet

Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

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

ORIGINAL

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

Tier 2 · Standard

ORIGINAL

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

Tier 3 · Hard

ORIGINAL

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

3.13.6.1

Principles of communication systems

Notes

Evidence from your answers: none yet

Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

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

ORIGINAL

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

Tier 2 · Standard

ORIGINAL

In a live radio link, explain the different purposes of the transmitter, transmission channel and receiver.

[3 marks]

Total for this question: 3

Tier 3 · Hard

ORIGINAL

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

3.13.6.2

Transmission media

Notes

Evidence from your answers: none yet

Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

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

ORIGINAL

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

Tier 2 · Standard

ORIGINAL

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

Tier 3 · Hard

ORIGINAL

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

3.13.6.3

Time-division multiplexing

Notes

Evidence from your answers: none yet

Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

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

ORIGINAL

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

Tier 2 · Standard

ORIGINAL

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

Tier 3 · Hard

ORIGINAL

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

3.13.6.4

Amplitude (AM) and frequency modulation (FM) techniques

Notes

Evidence from your answers: none yet

Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

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

ORIGINAL

An AM transmission carries a highest information frequency of

6.0\,\text{kHz}

. Calculate its minimum bandwidth.

[2 marks]

Total for this question: 2

Tier 2 · Standard

ORIGINAL

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

Tier 3 · Hard

ORIGINAL

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

MOSFET (metal-oxide semiconducting field-effect transistor)

Tier 1 · Easy

Tier 2 · Standard

Tier 3 · Hard

Zener diode

Tier 1 · Easy

Tier 2 · Standard

Tier 3 · Hard

Photodiode

Tier 1 · Easy

Tier 2 · Standard

Tier 3 · Hard

Hall effect sensor

Tier 1 · Easy

Tier 2 · Standard

Tier 3 · Hard

Difference between analogue and digital signals

Tier 1 · Easy

Tier 2 · Standard

Tier 3 · Hard

LC resonance filters

Tier 1 · Easy

Tier 2 · Standard

Tier 3 · Hard

The ideal operational amplifier

Tier 1 · Easy

Tier 2 · Standard

Tier 3 · Hard

Operational amplifier: inverting amplifier configuration

Tier 1 · Easy

Tier 2 · Standard

Tier 3 · Hard

Operational amplifier: non-inverting amplifier configuration

Tier 1 · Easy

Tier 2 · Standard

Tier 3 · Hard

Operational amplifier: summing amplifier configuration

Tier 1 · Easy

Tier 2 · Standard

Tier 3 · Hard

Real operational amplifiers

Tier 1 · Easy

Tier 2 · Standard

Tier 3 · Hard

Combinational logic

Tier 1 · Easy

Tier 2 · Standard

Tier 3 · Hard

Sequential logic

Tier 1 · Easy

Tier 2 · Standard

Tier 3 · Hard

Astables

Tier 1 · Easy

Tier 2 · Standard

Tier 3 · Hard

Principles of communication systems

Tier 1 · Easy

Tier 2 · Standard

Tier 3 · Hard

Transmission media

Tier 1 · Easy

Tier 2 · Standard

Tier 3 · Hard

Time-division multiplexing

Tier 1 · Easy

Tier 2 · Standard

Tier 3 · Hard

Amplitude (AM) and frequency modulation (FM) techniques

Tier 1 · Easy

Tier 2 · Standard

Tier 3 · Hard

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