Question 1 of 7: Two-Stage Diode Limiter with Resistive Divider

Question 1: Two-Stage Diode Limiter with Resistive Divider (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. $V_s(t)=10\sin(2\pi t)$ V feeds node $A$ through $R_1=5\text{k}\Omega$. $D_1$ (cathode to ground) shunts node $A$ to ground. $R_2=5\text{k}\Omega$ carries node $A$ to node $B=V_o$. At $B$, $D_2$ (cathode to ground) shunts to ground in parallel with a $10\text{k}\Omega$ resistor to ground.

Find. Diode reverse-voltage rating, the diode with the largest peak power (and a safe power rating), $V_s(t)$/$V_o(t)$ with peaks, and $I_{D2}(t)$.

Approach. Both diode cathodes are grounded, so each only conducts when its own anode node is pushed positive past 0.7V; walk the circuit from $V_s=-10\text{V}$ upward, tracking which diode turns on first and whether turning it on ever lets the downstream node reach 0.7V too.

1. Region 1 — both diodes off. With $D_1,D_2$ open, $R_1,R_2$ and the $10\text{k}\Omega$ resistor form a plain series divider from $V_s$ to ground: $$V_A=V_s\cdot\frac{R_2+10\text{k}}{R_1+R_2+10\text{k}}=0.75\,V_s,\qquad V_o=V_s\cdot\frac{10\text{k}}{R_1+R_2+10\text{k}}=0.5\,V_s$$ This holds while both $V_A<0.7\text{V}$ and $V_o<0.7\text{V}$. $V_A$ is the larger of the two (0.75 vs 0.5 of $V_s$), so $D_1$ reaches its threshold first: $$V_{s,\text{th}}=\frac{0.7}{0.75}=\boxed{0.933\text{ V}}\ \ (\text{giving }V_o=0.467\text{V at that instant — continuous with Region 2 below})$$ For the entire negative half-cycle both diodes stay off (their anodes go negative, i.e. reverse biased), so $V_o=0.5V_s$ all the way down to the trough.
2. Region 2 — $D_1$ on, $D_2$ still off ($V_s>0.933\text{V}$). $D_1$ clamps node $A$ at exactly 0.7V. With $V_A$ now fixed, the downstream network ($R_2$, $D_2$, $10\text{k}\Omega$) sees a constant 0.7V source, independent of how high $V_s$ climbs: $$V_o=0.7\text{V}\cdot\frac{10\text{k}}{R_2+10\text{k}}=0.7\cdot\frac{10}{15}=\boxed{0.4667\text{ V (flat — for all }V_s>0.933\text{V)}}$$ Because this value never reaches 0.7V, $D_2$ never turns on for any part of the cycle — once $D_1$ clamps node $A$, node $B$ is permanently safe below $D_2$'s threshold. This is easy to miss: a reader expecting a symmetric two-diode clipper would assume $D_2$ also clips, but the series resistor between the two clamp points prevents that here.
3. Part (c) — sketch of $V_s(t)$ and $V_o(t)$. Combining the two regions above gives the waveform pair plotted in Fig. Q1-c below: $V_o$ tracks $0.5V_s$ unclipped through the entire negative half-cycle down to its trough of $-5.0\text{V}$ at $V_s=-10\text{V}$, then flat-tops at $+0.4667\text{V}$ for the (small) portion of the positive half-cycle where $V_s>0.933\text{V}$.
4. Part (d) — sketch of $I_{D2}(t)$. Since $D_2$ never conducts at any point in the cycle (established above), its current sketch is trivial: $\boxed{I_{D2}(t)=0\text{ for all }t}$ — a flat line at zero, for every value of $V_s(t)$ from $-10\text{V}$ to $+10\text{V}$.
5. Part (a) — reverse voltage rating. Each diode's worst-case reverse bias occurs at the negative peak $V_s=-10\text{V}$ (both diodes off, plain-divider formulas apply): $$V_{A,\min}=0.75(-10)=-7.5\text{V}\ \Rightarrow\ V_{D1,\text{rev}}=0-(-7.5)=\boxed{7.5\text{ V}}$$ $$V_{o,\min}=0.5(-10)=-5.0\text{V}\ \Rightarrow\ V_{D2,\text{rev}}=0-(-5.0)=\boxed{5.0\text{ V}}$$ $D_1$ sees the larger reverse excursion. Choosing one rating for both diodes, a standard small-signal diode rated $\ge 15\text{V}$ PIV (e.g. a 1N4148 at 75V, or any $\ge 20\text{V}$ part) comfortably covers the 7.5V worst case with a safety margin.
6. Part (b) — peak power. $D_2$ never conducts, so its power is always zero. $D_1$'s peak current occurs at $V_s=+10\text{V}$, where $V_A=0.7\text{V}$ is fixed and the current continuing on to $R_2$ is set by the (constant) 0.7V-to-ground divider: $$I_{R1}=\frac{V_s-0.7}{R_1}=\frac{10-0.7}{5\text{k}}=1.860\text{ mA},\qquad I_{R2}=\frac{0.7-0.4667}{R_2}=\frac{0.2333}{5\text{k}}=0.0467\text{ mA}$$ $$I_{D1,\text{pk}}=I_{R1}-I_{R2}=1.860-0.047=\boxed{1.813\text{ mA}}$$ $$P_{D1,\text{pk}}=I_{D1,\text{pk}}\times V_D=(1.813\text{mA})(0.7\text{V})=\boxed{1.27\text{ mW}}$$ $D_1$ is the only diode that ever dissipates power. A $\tfrac{1}{8}\text{W}$ (125mW) small-signal diode gives nearly $100\times$ margin — any standard low-power diode is more than adequate.