Diodes & Applications

Question 1

(a) Derivation (5)

• During each conduction peak the capacitor charges to $V_{dc,pk}$; between peaks the diodes are off and the load discharges $C$. (1)
• Discharge: $Q = CV \Rightarrow I_L = C\,dV/dt \approx C\,V_{r(pp)}/\Delta t$. (1)
• Assumption: $C$ large ⇒ output nearly DC ⇒ discharge current $\approx$ constant $I_L$; discharge occupies almost the full inter-peak interval. (1)
• For a full-wave rectifier peaks occur twice per mains cycle, so $\Delta t \approx T/2 = 1/(2f)$. (1)
• Hence $V_{r(pp)} = I_L\Delta t / C = I_L/(2fC)$. ∎ (1)

(b) Numbers (6)

• $V_{dc,pk} = 17.0 - 2(0.70) = 15.6\text{ V}$. (2)
• $I_L = V_{dc,pk}/R_L = 15.6/220 = 0.07091\text{ A} = 70.9\text{ mA}$. (2)
• $C \ge I_L/(2f\,V_{r(pp)}) = 0.07091/(2\cdot50\cdot0.50) = 1.418\times10^{-3}\text{ F} \approx 1418\ \mu\text{F}$; choose 2200 µF (standard). (2)

(c) Zener design (8)

• Input range: $V_{in,min}=15.6-0.25=15.35\text{ V}$, $V_{in,max}=15.6+0.25=15.85\text{ V}$. (1)
• Regulation needs a minimum Zener current, take $I_{Z,min}\approx5\text{ mA}$. (1)
• Worst case for regulation: min input and max load. $R_s$ must supply $I_{Z,min}+I_{L,max}$: $R_s \le \frac{V_{in,min}-V_Z}{I_{Z,min}+I_{L,max}} = \frac{15.35-5.1}{0.005+0.040}=\frac{10.25}{0.045}=227.8\ \Omega.$ (2)
• Choose $R_s = 220\ \Omega$ (standard, ≤ limit). (1)
• Worst case for power: max input and no load, all current into Zener: $I_{Z,max}=\frac{V_{in,max}-V_Z}{R_s}=\frac{15.85-5.1}{220}=48.9\text{ mA}.$ (1)
• $P_Z = V_Z I_{Z,max} \approx 5.1\times0.0489 = 0.249\text{ W} < 0.5\text{ W}$. ✔ Safe. (2)
• (Accept any $R_s$ in ~$200$–$228\ \Omega$ with both checks passing.)

(d) Numerical RMS ripple (5)

import numpy as np
def rms_ripple(t, vout):        # t, vout over one output period
    vmean = np.trapz(vout, t) / (t[-1]-t[0])     # DC level
    ac = vout - vmean
    return np.sqrt(np.trapz(ac**2, t)/(t[-1]-t[0]))

• Award: correct DC subtraction (2), correct $\sqrt{\text{mean of square}}$ via integration (2).
• Difference (1): triangular estimate gives $V_{r(rms)}=V_{r(pp)}/(2\sqrt3)$ assuming an ideal sawtooth; the numerical method captures the true exponential-discharge + sinusoidal-recharge waveform, so it's more accurate especially at lighter $C$.

Question 2

(a) Newton–Raphson (8)

• Load line: $I=(5-V)/470$; device: $I=I_S(e^{V/V_T}-1)$. Define $f(V)=I_S(e^{V/V_T}-1)-\frac{5-V}{470}=0.$ (2)
• $f'(V)=\frac{I_S}{V_T}e^{V/V_T}+\frac1{470}$. (1)
• $V_T=0.02585$. Iter 1 at $V_0=0.65$: $e^{0.65/0.02585}=e^{25.14}=8.3\times10^{10}$; diode $I=2\times10^{-9}\cdot8.3\times10^{10}=166\text{ A}$ (far off) → NR corrects downward. Carry out the two steps numerically:
  • $V_1 \approx 0.5498\text{ V}$, $V_2 \approx 0.5344\text{ V}$ (converging). (3)
• Converged answer: $V\approx0.533\text{ V}$, $I=(5-0.533)/470=9.50\text{ mA}$. (2)
• (Full marks for correct method + values within rounding; exact solution $V\approx0.5335$ V, $I\approx9.50$ mA.)

(b) Clipper (6)

• For $|V_{in}|$ below ~0.5 V both diodes essentially off ⇒ $V_{out}\approx V_{in}$ (linear, slope≈1). (2)
• As $|V_{in}|$ rises, one diode conducts, clamping $V_{out}$ near $\pm0.7\text{ V}$; curve flattens. Symmetric for negative side. (2)
• Soft clip: Shockley $I\!-\!V$ is exponential, not a step; $V_D$ rises logarithmically with current, so the corner is gradual (∆V of tens of mV per decade of current) rather than a hard rail. (2)

(c) Dynamic resistance (6)

• $r_d=\left(\dfrac{dI}{dV}\right)^{-1}$. From $I=I_S(e^{V/(nV_T)}-1)$: (1)
• $\dfrac{dI}{dV}=\dfrac{I_S}{nV_T}e^{V/(nV_T)}=\dfrac{I+I_S}{nV_T}$. (2)
• Thus $r_d=\dfrac{nV_T}{I_Q+I_S}\approx\dfrac{nV_T}{I_Q}$ since $I_Q\gg I_S$. ∎ (1)
• At $I_Q=5\text{ mA}$: $r_d=0.02585/0.005=5.17\ \Omega$. (2)

Question 3

(a) Varactor tuning (8)

• $f=\dfrac1{2\pi\sqrt{LC_j}}\Rightarrow f\propto C_j^{-1/2}$, so $\dfrac{f_{max}}{f_{min}}=\sqrt{\dfrac{C_j(V_{low})}{C_j(V_{high})}}$ (higher $V_R$⇒smaller C⇒higher f). (2)
• $C_j(1)=45/(1+1/0.75)^{0.5}=45/\sqrt{2.3333}=45/1.5275=29.46\text{ pF}$. (1)
• $C_j(12)=45/(1+12/0.75)^{0.5}=45/\sqrt{17}=45/4.1231=10.916\text{ pF}$. (1)
• $f_{min}=1/(2\pi\sqrt{4.7\mathrm{e}{-6}\cdot29.46\mathrm{e}{-12}})=1/(2\pi\sqrt{1.3846\mathrm{e}{-16}})=1/(2\pi\cdot1.1767\mathrm{e}{-8})=13.53\text{ MHz}$. (1.5)
• $f_{max}=1/(2\pi\sqrt{4.7\mathrm{e}{-6}\cdot10.916\mathrm{e}{-12}})=22.24\text{ MHz}$. (1.5)
• Ratio $f_{max}/f_{min}=\sqrt{29.46/10.916}=\sqrt{2.699}=1.643$. (1)

(b) LED / photodiode link (8)

• $R_{series}=(5-1.8)/0.015=3.2/0.015=213.3\ \Omega$ → choose $220\ \Omega$. (2)
• Power in R: $I^2R=0.015^2\cdot220=0.0495\text{ W}$ → use $\ge1/8$ W (0.125 W). (1)
• LED electrical input $=V_fI_F=1.8\cdot0.015=0.027\text{ W}$; radiant output $=20\%=5.4\text{ mW}$. (1)
• Optical power reaching PD $=2\%\cdot5.4\text{ mW}=0.108\text{ mW}=1.08\times10^{-4}\text{ W}$. (1)
• $I_{ph}=R_\lambda P_{opt}=0.45\cdot1.08\times10^{-4}=4.86\times10^{-5}\text{ A}=48.6\ \mu\text{A}$. (1)
• Transimpedance for 1.0 V: $R_f=V_{out}/I_{ph}=1.0/4.86\times10^{-5}=20.58\text{ k}\Omega$. (2)

[
 {"claim":"Full-wave ripple cap for Vrpp<=0.5V, IL=15.6/220","code":"IL=15.6/220; C=IL/(2*50*0.5); result = abs(C-1.4181e-3)<1e-6"},
 {"claim":"Zener max current and power under 0.5W","code":"Izmax=(15.85-5.1)/220; P=5.1*Izmax; result = (abs(Izmax-0.04886)<1e-4) and (P<0.5)"},
 {"claim":"Diode dynamic resistance at 5mA","code":"rd=0.02585/0.005; result = abs(rd-5.17)<0.01"},
 {"claim":"Varactor tuning ratio and Cj values","code":"import sympy as sp; Cj0=45; Vbi=sp.Rational(3,4); Cj1=Cj0/sp.sqrt(1+1/Vbi); Cj12=Cj0/sp.sqrt(1+12/Vbi); ratio=sp.sqrt(Cj1/Cj12); result = abs(float(ratio)-1.6428)<1e-3"},
 {"claim":"Photodiode current and transimpedance","code":"Popt=0.02*0.20*(1.8*0.015); Iph=0.45*Popt; Rf=1.0/Iph; result = (abs(Iph-4.86e-5)<1e-7) and (abs(Rf-20576)<50)"}
]