3.13 Electronics (A-level only) — coverage pack

Answer

• $V_{\mathrm{GS}}>V_{\mathrm{th}}$.

Method: A conducting channel begins to form when the gate-source potential difference exceeds the threshold value, so the condition is $V_{\mathrm{GS}}>V_{\mathrm{th}}$.

Answer

• $3.0\,\text{M}\Omega$

Method: 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 $R=3.0\,\text{M}\Omega$.

Answer

• $V_{\mathrm{DS}}=2.25\,\text{V}$, $P_{\mathrm{load}}=0.634\,\text{W}$ and $P_{\mathrm{MOSFET}}=0.146\,\text{W}$; it is not acting as an efficient fully-on switch.

Method: The load line is $V_{\mathrm{DS}}=12.0-I_{\mathrm{D}}(150)$. Using $I_{\mathrm{D}}=0.065\,\text{A}$ gives $V_{\mathrm{DS}}=12.0-(0.065)(150)=2.25\,\text{V}$. This is at least $2.0\,\text{V}$, so the assumed part of the output characteristic is self-consistent. The load power is $P_{\mathrm{load}}=I_{\mathrm{D}}^2R=(0.065)^2(150)=0.63375\,\text{W}=0.634\,\text{W}$. The MOSFET dissipates $P_{\mathrm{MOSFET}}=I_{\mathrm{D}}V_{\mathrm{DS}}=(0.065)(2.25)=0.146\,\text{W}$. 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.

Answer

• Reverse bias in the Zener breakdown region.

Method: 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.

Answer

• $400\,\Omega$

Method: 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 $R=V_R/I_R=6.4/0.016=400\,\Omega$.

Answer

• $330\,\Omega$; use at least a $0.25\,\text{W}$ resistor and a $0.25\,\text{W}$ Zener diode.

Method: 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 $0.25\,\text{W}$.

Answer

• Photoconductive mode.

Method: A reverse-biased photodiode is operating in photoconductive mode; photovoltaic mode uses no external bias.

Answer

• $0.60\,\text{V}$

Method: 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.

Answer

• 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.

Method: 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 $N=Q/e=(4.80\times10^{-14})/(1.60\times10^{-19})=3.0\times10^5$.

Answer

• The Hall voltage doubles.

Method: With current and geometry fixed, $V_{\mathrm{H}}\propto B$. Doubling $B$ therefore doubles $V_{\mathrm{H}}$.

Answer

• $2.492\,\text{V}$

Method: 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 $2.492\,\text{V}$.

Answer

• $40\,\text{mT}$ and $1.8\times10^3\,\text{rev min}^{-1}$

Method: 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 $30\times60=1.8\times10^3\,\text{rev min}^{-1}$.

Answer

• An analogue signal varies continuously, whereas a binary digital signal has two discrete levels.

Method: The defining contrast is continuous variation for analogue information and discrete allowed levels for digital information; binary uses two such levels.

Answer

• $18.75\,\text{mV}$ and $\pm9.38\,\text{mV}$

Method: 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 $\pm9.38\,\text{mV}$.

Answer

• 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.

Method: 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.

Answer

• $4.0\times10^4\,\text{Hz}$

Method: 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.

Answer

• $14\,\text{kHz}$ and $Q=11$

Method: 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.

Answer

• $C=88\,\text{pF}$ and $Q=28$; the $22\,\text{kHz}$ bandwidth passes the full $18\,\text{kHz}$ transmission while retaining some selectivity.

Method: 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.

Answer

• $0\,\text{A}$

Method: An ideal operational amplifier has infinite input impedance, so no current enters either input: $I_+=I_-=0$.

Answer

• 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.

Method: 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.

Answer

• $-4.0\,\text{V}$; this lies within the $\pm6.0\,\text{V}$ rails.

Method: 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.

Answer

• $-2.8\,\text{V}$

Method: 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.

Answer

• $+2.6\,\text{V}$

Method: 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.

Answer

• Ideal peak $11.2\,\text{V}$; the inverted output is clipped at $\pm8.0\,\text{V}$; maximum unclipped input peak $1.0\,\text{V}$.

Method: 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 $1.0\,\text{V}$.

Answer

• $3.9\,\text{V}$

Method: 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.

Answer

• $2.5\,\text{V}$

Method: 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.

Answer

• Maximum input $4.1\,\text{V peak-to-peak}$; output $23.0\,\text{V peak-to-peak}$.

Method: 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 $23.0\,\text{V peak-to-peak}$.

Answer

• $-1.6\,\text{V}$

Method: 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 $\text{k}\Omega$.

Answer

• $-2.7\,\text{V}$

Method: 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 $-2.7\,\text{V}$.

Answer

• Ideal output $-15.5\,\text{V}$; actual output $-10.5\,\text{V}$; third input $0.50\,\text{V}$.

Method: 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 $V_3=0.50\,\text{V}$.

Answer

• $1.0\times10^5\,\text{Hz}$

Method: Using $Gf_B=3.0\times10^6\,\text{Hz}$, the bandwidth is $f_B=(3.0\times10^6)/30=1.0\times10^5\,\text{Hz}$.

Answer

• No; the required slew rate is $2.0\,\text{V}\,\mu\text{s}^{-1}$ and the maximum frequency is $60\,\text{kHz}$.

Method: 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.

Answer

• 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 $2.1\,\text{V}$.

Method: 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 $2.1\,\text{V}$.

Answer

• $0,0,1,0$

Method: 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 $0,0,1,0$.

Answer

• EOR gate; $Y=\overline{A}B+A\overline{B}$.

Method: 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 $Y=\overline{A}B+A\overline{B}$.

Answer

• Outputs $0,0,1,0,1,0,1,0$. A NAND-only construction is $P=A\operatorname{\ NAND}A$, $Q=B\operatorname{\ NAND}B$, $R=P\operatorname{\ NAND}Q$, $S=C\operatorname{\ NAND}C$, $T=R\operatorname{\ NAND}S$, $Y=T\operatorname{\ NAND}T$.

Method: 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 $Y=R\cdot S=(A+B)\overline{C}$.

Answer

• $101$

Method: The counter advances once per pulse: $000\to001\to010\to011\to100\to101$. After five pulses its decimal count is $5$, which is binary $101$.

Answer

• Johnson counter: $8$ states; BCD counter: $10$ states; BCD cycle rate $120\,\text{Hz}$.

Method: 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 $1.20\times10^3/10=120\,\text{Hz}$.

Answer

• Reset on $110$; sequence $000,001,010,011,100,101$; after $53$ pulses the output is $101$; cycle rate $4.0\,\text{kHz}$.

Method: 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 $(24\times10^3)/6=4.0\times10^3\,\text{Hz}=4.0\,\text{kHz}$.