Diodes & Applications

Question 1 (14 marks)

(a) Peak secondary voltage $V_{pk} = 12\sqrt{2} = 16.97\text{ V}$. Bridge has two diodes conducting in series each half-cycle, so $V_{peak,load} = 16.97 - 2(0.7) = 15.57\text{ V}$. (3) (1 mark $V_{pk}$, 1 for subtracting 2 drops, 1 answer)

(b) Ripple frequency for a full-wave/bridge rectifier is $2\times$ mains = $100\text{ Hz}$ (both half-cycles produce output pulses). (1 justify) $C = \frac{I_{load}}{f_{ripple}\,V_{ripple}} = \frac{0.25}{100 \times 0.5} = 5.0\times10^{-3}\text{ F} = 5000\ \mu\text{F}.$ (2 formula rearrange, 2 answer)

(c) Half-wave: ripple frequency = $50\text{ Hz}$ (only one pulse per cycle), so with same $C$: $V_{ripple} = \frac{I_{load}}{f\,C} = \frac{0.25}{50 \times 5000\times10^{-6}} = 1.0\text{ V (pp)}.$ This is double the target ripple — the half-wave design fails the $0.5\text{ V}$ spec and would need twice the capacitance. Full-wave charges the cap twice as often, halving ripple for the same $C$; hence the better choice. (2 calc, 2 comment)

(d) For a bridge, each diode sees a maximum reverse voltage of one peak = $16.97\text{ V}$ (≈ $V_{pk}$, since two diodes share the reverse blocking but each blocks $\approx V_{pk}$). Minimum PIV rating $\gtrsim 17\text{ V}$; choose with margin (e.g. 50 V). (2)

Question 2 (13 marks)

(a) Worst case for keeping the Zener in regulation is minimum input voltage (least available current) and maximum load current (most current diverted from Zener). At $V_{in}=9\text{ V}$, $I_L = 40\text{ mA}$, the Zener still needs $\ge 5\text{ mA}$. Total current through $R_S$ = $I_L + I_{Z(min)} = 40 + 5 = 45\text{ mA}$. $R_S = \frac{V_{in(min)} - V_Z}{I_L + I_{Z,min}} = \frac{9 - 5.1}{45\times10^{-3}} = \frac{3.9}{0.045} = 86.7\ \Omega.$ Choose $R_S \le 86.7\ \Omega$ (a smaller value gives more margin); take $R_S = 82\ \Omega$. (2 worst case, 2 formula, 2 answer)

(b) Max Zener power occurs at max input ($12\text{ V}$) and min load ($0\text{ mA}$) — all current flows through the Zener. With $R_S = 82\ \Omega$: $I_{S} = \frac{12 - 5.1}{82} = \frac{6.9}{82} = 84.1\text{ mA}, \quad I_Z = 84.1\text{ mA (all)}.$ $P_Z = V_Z I_Z = 5.1 \times 0.0841 = 0.429\text{ W}.$ So a Zener rated $\ge 0.5\text{ W}$ (with margin ~1 W) is required. (1 condition, 2 current, 2 power)

(If $R_S = 86.7\ \Omega$ used: $I_Z = 79.6\text{ mA}$, $P_Z = 0.406\text{ W}$ — accept.)

(c) Any of: max power dissipation $P_{Zmax}$ (must exceed 0.43 W); Zener dynamic impedance $Z_Z$ (affects regulation tightness); tolerance on $V_Z$. (2)

Question 3 (12 marks)

(a) Diode conducts when $v_{out}$ would exceed $+2\text{ V}$, clamping the top. Output = input for $v_{in} < 2\text{ V}$, and clipped flat at $+2\text{ V}$ for the portion where $v_{in} > 2\text{ V}$. Negative half-cycle (down to $-8\text{ V}$) passes unaltered. This is a positive clipper (limiter) at $+2\text{ V}$. (4)

(b) With silicon diode the branch conducts only when node voltage $= 2 + 0.7 = 2.7\text{ V}$. New clipping level $= +2.7\text{ V}$; the $0.7\text{ V}$ forward drop raises the clamp point. (3)

(c) Clamper (cathode to ground, ideal): the diode conducts on the positive peaks, charging the capacitor so the output cannot exceed $0\text{ V}$. The waveform is shifted down so its peak sits at $0\text{ V}$.

• Peak: $0\text{ V}$
• Trough: $0 - (8-(-8)) = -16\text{ V}$
• DC level (average): $-8\text{ V}$. (2 peak/trough logic, 2 values, 1 DC level)

Question 4 (11 marks)

(a) $R = \frac{5 - 1.8}{15\times10^{-3}} = \frac{3.2}{0.015} = 213.3\ \Omega.$ Standard $E12 \ge 213.3\ \Omega$ is $220\ \Omega$. Resulting current $I = \dfrac{3.2}{220} = 14.5\text{ mA}$. Power in resistor $P = I^2 R = (0.01455)^2 \times 220 = 0.0466\text{ W} \approx 47\text{ mW}$ (or using computed $15$ mA: $P=3.2\times0.015=48\text{ mW}$). (2 R, 1 std value, 1 current, 1 power)

(b) Fill factor $FF = \frac{V_{mp} I_{mp}}{V_{oc} I_{sc}} = \frac{0.47 \times 0.110}{0.58 \times 0.120} = \frac{0.0517}{0.0696} = 0.743.$ Max power $P_{max} = V_{mp} I_{mp} = 0.47 \times 0.110 = 51.7\text{ mW}$. (2 FF, 2 Pmax)

(c) Reverse bias widens the depletion region → lower junction capacitance (faster response) and gives a linear, wide photocurrent-vs-illumination response with fast switching for sensing. A solar cell must deliver power, so it operates in the fourth quadrant (forward voltage, reverse current), the photovoltaic mode. (2)

Question 5 (10 marks)

(a) With pull-up to $+5\text{ V}$: any input at $0\text{ V}$ pulls $Y$ down via its diode to $0 + 0.7 = 0.7\text{ V}$. Only when both inputs are $5\text{ V}$ do both diodes turn off and $Y$ is pulled to $5\text{ V}$.

This is an AND gate (Y high only when A AND B high). (4 table, 1 identify)

(b) Each diode adds a $0.7\text{ V}$ offset (logic-0 is $0.7\text{ V}$, not $0\text{ V}$; logic-1 loses a drop in OR-type stages). Cascading accumulates these offsets, so after several stages the "0" and "1" levels drift toward each other and the logic margin collapses. Diode logic also cannot restore/amplify levels. (2)

(c) At $V_R = 0$: $C_j = 80/\sqrt{1+0} = 80\text{ pF}$. At $V_R = 6$: $C_j = 80/\sqrt{1 + 6/0.7} = 80/\sqrt{9.571} = 80/3.094 = 25.86\text{ pF}$. Tuning ratio $= 80/25.86 = 3.09:1$. (1 + 1 + 1)

[
  {"claim":"Bridge peak load voltage = 12*sqrt(2) - 1.4 = 15.57 V","code":"vpk=12*sqrt(2)-2*Rational(7,10); result=abs(float(vpk)-15.57)<0.02"},
  {"claim":"Smoothing C for full-wave = 5000 uF","code":"C=Rational(25,100)/(100*Rational(5,10)); result=abs(float(C)-0.005)<1e-9"},
  {"claim":"Half-wave ripple with same C = 1.0 V","code":"Vr=Rational(25,100)/(50*Rational(5,1000)); result=abs(float(Vr)-1.0)<1e-9"},
  {"claim":"Rs worst case = 86.67 ohm","code":"Rs=(9-Rational(51,10))/(Rational(45,1000)); result=abs(float(Rs)-86.6667)<0.01"},
  {"claim":"Max Zener power with Rs=82 is ~0.429 W","code":"Iz=(12-Rational(51,10))/82; Pz=Rational(51,10)*Iz; result=abs(float(Pz)-0.4293)<0.005"},
  {"claim":"Solar fill factor = 0.743","code":"FF=(Rational(47,100)*Rational(11,100))/(Rational(58,100)*Rational(12,100)); result=abs(float(FF)-0.743)<0.005"},
  {"claim":"Varactor Cj at 6V = 25.86 pF, ratio 3.09","code":"Cj=80/sqrt(1+Rational(6,1)/Rational(7,10)); ratio=80/Cj; result=abs(float(Cj)-25.86)<0.1 and abs(float(ratio)-3.094)<0.02"}
]