Question 1 of 7: Two-Sided Diode Voltage Limiter

Question 1: Two-Sided Diode Voltage Limiter (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. Two series resistors $R_s=10\text{k}\Omega$ flank a middle node $M$ where the source $V_s(t)=10\sin(2\pi t)$ V is applied directly to ground. The left end (node $A$) feeds a shunt branch $D_1$–3V battery–$R_1=5\text{k}\Omega$ to ground; the right end (node $B=V_o$) feeds a mirror-image shunt branch $D_2$–3V battery–$R_2=5\text{k}\Omega$ to ground. $V_D=0.7\text{V}$ for both diodes.

Find. $V_o(t)$, $V_R(t)$, $I_s(t)$ with peak values, the resistor with the largest peak power, and a safe power rating for it.

Approach. Node $A$ and node $B=V_o$ are each driven from $M$ through their own $10\text{k}\Omega$ and clamped by their own diode-battery-resistor branch; because $R_1,R_2$ are not negligible next to $R_s$, each branch is a soft (voltage-divided) limiter, not an ideal clip — write one loop KVL per branch, find its turn-on threshold, then assemble $V_o$, $V_R=V_A-V_B$ and $I_s=I_1+I_2$ over the sine cycle.

1. Part (a)/(b)/(d) — Left branch ($D_1$): turn-on threshold and current. Diode $D_1$'s cathode sits at node $A$; its anode feeds through the 3V battery (oriented so the branch shunts current into $A$ once $A$ tries to rise) and $R_1$ to ground. Writing KVL around $M\text{-}R_s\text{-}A\text{-}D_1\text{-battery-}R_1\text{-}G$ with loop current $I_1$: $$I_1=\frac{V_M+2.3}{R_s+R_1}=\frac{V_M+2.3}{15\text{k}\Omega}\quad\text{(valid for }V_M\ge -2.3\text{V)}$$ Below $-2.3\text{V}$, $D_1$ is off and $I_1=0$, so $V_A=V_M$ (no drop across $R_s$). The node voltage while $D_1$ conducts is $$V_A=V_M-I_1R_s=\frac{R_1V_M-2.3R_s}{R_1+R_s}=\frac{5V_M-23}{15}\ \text{V}$$
2. Right branch ($D_2$): turn-on threshold and current. $D_2$ is the mirror image (battery flipped), giving the same form with a $+3.7\text{V}$ threshold: $$I_2=\frac{V_M-3.7}{R_s+R_2}=\frac{V_M-3.7}{15\text{k}\Omega}\quad\text{(valid for }V_M\ge 3.7\text{V)}$$ $$V_o=V_B=V_M-I_2R_s=\frac{R_2V_M+3.7R_s}{R_2+R_s}=\frac{5V_M+37}{15}\ \text{V, else }V_o=V_M$$
3. Assemble the three waveforms over one cycle ($V_M=V_s(t)=10\sin2\pi t$, amplitude $10\text{V}$): Since $D_2$ only conducts for $V_M>3.7\text{V}$ and $D_1$ only for $V_M>-2.3\text{V}$, $V_o$ tracks $V_s$ exactly for the entire negative half-cycle and up to $+3.7\text{V}$, then bends onto slope $R_2/(R_2+R_s)=1/3$ up to the peak. At the peak $V_M=+10\text{V}$: $$V_o\big|_{\text{pk}}=\frac{5(10)+37}{15}=\boxed{+5.8\text{ V}}$$ $$V_o\big|_{\text{trough}}=V_M=-10\text{ V (unclipped)}$$ $V_R=V_A-V_B$ is $0$ while both diodes are off, falls linearly to $-4\text{V}$ as $V_M$ rises through $[-2.3,3.7]\text{V}$, then stays flat at $-4\text{V}$ for the rest of the positive half (both formulas above give $V_A-V_B=-60/15=-4\text{V}$, independent of $V_M$, once $V_M>3.7\text{V}$): $$V_R\big|_{\text{peak magnitude}}=\boxed{-4.0\text{ V (flat-top for }V_s>3.7\text{V)}}$$ $I_s=I_1+I_2$ (both flow away from $M$) is zero below $-2.3\text{V}$, ramps up through the middle region on $I_1$ alone, then rises faster (both diodes conducting) to its peak at $V_M=+10\text{V}$: $$I_s\big|_{\text{pk}}=I_1+I_2=\frac{10+2.3}{15}+\frac{10-3.7}{15}=0.820+0.420=\boxed{1.24\text{ mA}}$$
4. Part (c) — peak resistor power. Peak current is monotonically increasing in $V_M$, so every resistor's peak power occurs at the same instant, $V_M=+10\text{V}$: $I_{1,\text{pk}}=0.820\text{ mA}$ flows through both $R_1$ and the LEFT $R_s$; $I_{2,\text{pk}}=0.420\text{ mA}$ flows through $R_2$ and the RIGHT $R_s$. $$P_{R_s,\text{left}}=I_{1,\text{pk}}^2R_s=(0.820\text{mA})^2(10\text{k}\Omega)=\boxed{6.72\text{ mW}}$$ $$P_{R_1}=3.36\text{ mW}\qquad P_{R_s,\text{right}}=1.76\text{ mW}\qquad P_{R_2}=0.88\text{ mW}$$ The left-hand $R_s$ (in series with $D_1$) dissipates the most, $\approx 6.7\text{mW}$ peak. A standard $\tfrac{1}{8}\text{W}$ (125 mW) resistor gives an $\approx18\times$ safety margin; a $\tfrac14\text{W}$ part is an equally common, more conservative choice.