Diodes & Applications

Question 1 (10)

(a) Half-wave: diode conducts only for $0 \le \omega t \le \pi$. $V_{DC} = \frac{1}{2\pi}\int_0^{2\pi} v_o\, d(\omega t) = \frac{1}{2\pi}\int_0^{\pi} V_m\sin(\omega t)\, d(\omega t)$ (1 mark setup; 1 for limits — only half cycle passes) $= \frac{V_m}{2\pi}[-\cos\omega t]_0^{\pi} = \frac{V_m}{2\pi}(1-(-1)) = \frac{V_m}{2\pi}(2) = \frac{V_m}{\pi}$ $\boxed{V_{DC} = \frac{V_m}{\pi} \approx 0.318\,V_m}$ (2 marks evaluation/result)

(b) $V_{rms} = \sqrt{\frac{1}{2\pi}\int_0^{\pi} V_m^2\sin^2(\omega t)\,d(\omega t)}$ Using $\int_0^\pi \sin^2 = \pi/2$: $= \sqrt{\frac{V_m^2}{2\pi}\cdot\frac{\pi}{2}} = \sqrt{\frac{V_m^2}{4}} = \frac{V_m}{2}$. $\boxed{V_{rms} = \frac{V_m}{2}}$ (1 setup, 1 integral value, 1 result)

(c) Half-wave passes one pulse per input cycle → ripple fundamental = supply frequency (50 Hz). Full-wave inverts the negative half so two pulses per input cycle → ripple = 2× supply (100 Hz). Higher ripple frequency ⇒ smaller reservoir capacitor for same ripple. (3 marks for the two-pulses-per-cycle reasoning.)

Question 2 (10)

(a) $V_m = 12\sqrt{2} = 16.97\text{ V}$. (1) Bridge: two diodes conduct in series each half cycle, so $V_{peak,out} = 16.97 - 2(0.7) = 15.57\text{ V}$. (2)

(b) Ideal full-wave: $V_{DC} = \frac{2V_m}{\pi}$. Using loaded peak $15.57$: $V_{DC} = \frac{2(15.57)}{\pi} = 9.91\text{ V}$. (Using ideal $16.97$: $10.80$ V — accept either with stated assumption.) (2)

(c) PIV: in a bridge, the two non-conducting diodes see the peak secondary voltage across them (the conducting pair short the bridge to the rail). $\text{PIV} = V_m = 16.97\text{ V}$ (≈17 V). Choose ≥ 2× margin ⇒ e.g. 50 V diodes. (Award full even if student states PIV = $V_m$ for bridge vs $2V_m$ for centre-tapped.) (3)

(d) $f_{ripple} = 100\text{ Hz}$. $C = \frac{I}{f_{ripple}V_{ripple}} = \frac{0.5}{100 \times 1} = 5\times10^{-3} = 5000\,\mu\text{F}$. (2)

Question 3 (12)

(a) KCL: $I_{RS} = I_Z + I_L$. Zener stays regulated if $I_{Z,min} \le I_Z \le I_{Z,max}$. $R_S$ must be small enough that even at minimum input & maximum load, $I_Z \ge I_{Z,min}$: $R_S \le \frac{V_{in,min} - V_Z}{I_{Z,min} + I_{L,max}}$ Worst case for keeping Zener alive = lowest $V_{in}$ (least available current) with highest $I_L$ (load hogs current). (4)

(b) $V_{in,min} = 13\text{ V}$, $I_{L,max}=40\text{ mA}$: $R_S \le \frac{13 - 6.2}{(5+40)\text{ mA}} = \frac{6.8}{0.045} = 151\,\Omega$ Choose $R_S = 150\,\Omega$ (standard, satisfies ≤151). (4)

(c) Worst-case Zener power = maximum input, minimum load (max current diverted into Zener). $V_{in,max}=17\text{ V}$, $I_{L,min}=10\text{ mA}$: $I_{RS} = \frac{17-6.2}{150} = \frac{10.8}{150} = 72\text{ mA}$ $I_{Z,max} = 72 - 10 = 62\text{ mA}$ $P_Z = V_Z I_{Z,max} = 6.2 \times 0.062 = 0.384\text{ W} < 0.5\text{ W} \checkmark$ (2 for identifying worst case, 2 for computation & verdict.)

Question 4 (10)

(a) Positive clipper with series diode + 3 V battery: output follows input until $v_{in}$ exceeds the clip level. Clipping level = $V_{battery} + V_f = 3 + 0.7 = 3.7\text{ V}$. Output = $v_{in}$ for $v_{in} < 3.7$ V, clamped flat at $3.7$ V for the portion above. Negative half ($-10$ V peak) passes unaffected. (2 for level, 2 for waveform description.)

(b) Negative clamper (diode conducts on positive half, charging C to $+V_m = 5$ V; diode then off, capacitor voltage in series shifts waveform down). Steady state: $v_{out} = v_{in} - V_m = 5\sin(\omega t) - 5$ Peaks: max $= 0$ V, min $= -10$ V, DC level $= -5$ V. (2 derivation, 2 values.)

(c) RC ≫ T so the capacitor barely discharges between cycles, holding its charge ⇒ the DC shift stays constant and waveform shape is preserved (no droop/distortion). (2)

Question 5 (10)

(a) Diode OR (anodes to inputs, cathodes common to output, R to ground). Output HIGH when any input HIGH; $V_{OH} = 5 - 0.7 = 4.3\text{ V}$.

(2 truth table, 2 for $V_{OH}=4.3$ V/HIGH level.)

(b) (i) Each stage drops $V_f$ (~0.7 V), so logic levels degrade after every stage. (ii) No signal gain / no active restoration ⇒ noise margin shrinks and levels drift; loading of later stages pulls levels further. (2)

(c) (1) Lower $V_f$ (0.3 V) ⇒ less forward loss / better efficiency in low-voltage rectifiers. (2) Negligible reverse recovery ⇒ works at high switching frequencies (fast switch-mode rectifiers, RF). (2)

(d) $R = \frac{5 - 2.0}{0.015} = \frac{3.0}{0.015} = 200\,\Omega$. (2)

Question 6 (8)

(a) Photovoltaic mode: zero bias, the diode generates voltage/current from light (solar-cell-like), lower noise but slower/less linear. Photoconductive mode: reverse-biased, wider depletion region ⇒ lower capacitance, faster response, current linear with illumination — suits fast, linear sensing. (3)

(b) A varactor is a reverse-biased diode whose depletion-layer width (and hence junction capacitance $C_j$) varies with reverse voltage. Placed in the LC tank, changing the DC reverse bias changes $C$ and thus the resonant frequency $f = 1/(2\pi\sqrt{LC})$ — enabling voltage-controlled tuning (VCO/PLL). Reverse bias is essential to avoid conduction and to set $C_j$. (3)

(c) $I_R$ (reverse leakage) constrains reverse/off-state losses & sensor noise floor; $V_f @ I$ constrains forward power dissipation/heat; $V_{RRM}$ constrains the maximum reverse voltage (PIV) the diode can block. (2)

[
  {"claim":"Half-wave ideal Vdc = Vm/pi = 0.3183 Vm","code":"wt=symbols('wt'); Vm=symbols('Vm',positive=True); Vdc=integrate(Vm*sin(wt),(wt,0,pi))/(2*pi); result=simplify(Vdc-Vm/pi)==0"},
  {"claim":"HW rms output = Vm/2","code":"wt=symbols('wt'); Vm=symbols('Vm',positive=True); Vrms=sqrt(integrate((Vm*sin(wt))**2,(wt,0,pi))/(2*pi)); result=simplify(Vrms-Vm/2)==0"},
  {"claim":"Zener RS<=151 ohm at worst case (13-6.2)/0.045","code":"R=(13-6.2)/0.045; result=abs(R-151.11)<1"},
  {"claim":"Zener max power 0.384 W under Vin=17,IL=10mA,RS=150","code":"Irs=(17-6.2)/150; Iz=Irs-0.010; Pz=6.2*Iz; result=abs(Pz-0.384)<0.01 and Pz<0.5"},
  {"claim":"LED resistor (5-2)/0.015 = 200 ohm","code":"R=(5-2.0)/0.015; result=abs(R-200)<1e-6"},
  {"claim":"Bridge Vdc loaded = 2*(12*sqrt(2)-1.4)/pi approx 9.91","code":"Vdc=2*(12*sqrt(2)-1.4)/pi; result=abs(float(Vdc)-9.91)<0.05"}
]