Question 1 of 7: Two-Level Soft Limiter (Antiparallel Diode-Battery Branches)

Question 1: Two-Level Soft Limiter (Antiparallel Diode-Battery Branches) (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. $V_s(t)=10\sin(2\pi t)\text{V}$ (period $T=1\text{s}$); $R_s=10\text{k}\Omega$; $R_1=R_2=5\text{k}\Omega$; all diode drops $V_D=0.7\text{V}$; both branch batteries $3\text{V}$, wired with opposite polarity to each other so $D_1$ and $D_2$ conduct on opposite half-cycles.

Find. $V_o(V_s)$ while $D_1$ conducts; $V_s,V_o$ and $V_R$ waveforms with peaks; the resistor with the largest peak power and a suitable rating.

Approach. With $D_1,D_2$ off, no current flows in $R_s$ and $V_o=V_s$ exactly. $D_2$ (anode at the input node) turns on once $V_s$ exceeds $+(V_D+3)=+3.7\text{V}$, and $D_1$ (cathode at the input node) turns on once $V_s$ drops below $-(V_D+3)=-3.7\text{V}$; because $R_1,R_2$ are comparable to $R_s$ (not negligible), each clamp is a soft ($V_s$-dependent slope, not flat) limit, found from KCL at the output node with the conducting branch in the loop.

1. Part (a) — $V_o(V_s)$ with $D_1$ forward-biased. $D_1$ turns on for $V_s\le-3.7\text{V}$. With $D_1$ conducting (drop $V_D$) and its battery (net $3\text{V}$, oriented to add to the clamp) in series with $R_1$ to ground, KCL at the output node ($i$ from $V_s$ through $R_s$ equals $i$ down through $D_1$-battery-$R_1$) gives $$\frac{V_s-V_o}{R_s}=\frac{V_o+(V_D+3)}{R_1}\;\Longrightarrow\;\boxed{V_o=\frac{R_1V_s-R_s(V_D+3)}{R_s+R_1}}$$ Substituting numbers ($R_s=10\text{k}\Omega$, $R_1=5\text{k}\Omega$, $V_D+3=3.7\text{V}$): $$V_o=\frac{5000\,V_s-10000(3.7)}{15000}=\boxed{\frac{V_s}{3}-2.467\text{ V}}\qquad(V_s\le-3.7\text{V})$$ By the identical argument on $D_2$'s branch, the positive side is $V_o=\dfrac{V_s}{3}+2.467\text{V}$ for $V_s\ge+3.7\text{V}$; both meet the unity-slope middle region exactly at the $\pm3.7\text{V}$ breakpoints (continuity check: at $V_s=-3.7$, $V_o=-3.7/3-2.467=-3.7\text{V}$ ✓).
2. Part (b) — $V_s,V_o$ vs. time. For $|V_s|\le3.7\text{V}$ (no diode conducts), $V_o=V_s$ exactly. Beyond that, $V_o$ follows the soft-limited $\tfrac13$-slope lines from Part (a). At the sine's peaks $V_s=\pm10\text{V}$: $$V_{o,\text{pos,pk}}=\frac{10}{3}+2.467=\boxed{+5.8\text{ V}},\qquad V_{o,\text{neg,pk}}=-\frac{10}{3}-2.467=\boxed{-5.8\text{ V}}$$ So $V_o$ is a sine that tracks $V_s$ 1:1 through $\pm3.7\text{V}$ and then bends onto a shallow $1/3$-slope line up to $\pm5.8\text{V}$ at the peaks — a soft, symmetric limiter (Fig. Q1(b)).
3. Part (c) — $V_R$ vs. time. $V_R=V_s-V_o$ (the resistor's own $+/-$ reference in Figure 1), zero throughout the linear region and rising only once a diode conducts: $$V_{R,\text{pos,pk}}=10-5.8=\boxed{+4.2\text{ V}},\qquad V_{R,\text{neg,pk}}=-10-(-5.8)=\boxed{-4.2\text{ V}}$$ (Fig. Q1(c) — $V_R$ is flat at zero for roughly the middle third of each half-cycle, then bulges symmetrically to $\pm4.2\text{V}$ at the peaks.)
4. Part (d) — largest peak power, rating. The peak branch current is $I_{\text{pk}}=V_{R,\text{pk}}/R_s=4.2/10\text{k}=0.42\text{mA}$, and by KCL this same current flows through whichever of $R_1$/$R_2$ is conducting at that instant (series path). Peak dissipation in each: $$P_{R_s,\text{pk}}=I_{\text{pk}}^2R_s=(0.42\text{mA})^2(10\text{k}\Omega)=\boxed{1.764\text{ mW}}$$ $$P_{R_1,\text{pk}}=P_{R_2,\text{pk}}=I_{\text{pk}}^2R_1=(0.42\text{mA})^2(5\text{k}\Omega)=\boxed{0.882\text{ mW}}$$ Since the same current flows through both series elements, the larger resistor dissipates more — $R_s$ (double $R_1$'s or $R_2$'s value) has the largest peak power. Even $1.764\text{mW}$ is far below any standard resistor's rating, so a standard $\boxed{1/8\text{ W (}125\text{ mW)}}$ resistor already gives >70$\times$ margin; a $1/4\text{W}$ part is an equally safe, more commonly stocked choice.