Diode/Zener/Battery Limiter: 98-Comp-A1, December 2016, Q1

Question 1: Diode/Zener/Battery 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. $V_1(t)=10\sin(2\pi t)\text{V}$ (period $T=1\text{s}$); $R_1=50\,\Omega$; all diode drops $V_D=0.7\text{V}$; $D_2$ zener breakdown $V_{Z2}=4.3\text{V}$; series battery $V_{\text{bat}}=3\text{V}$ (opposing $D_3$).

Given data
Quantity	Value
$V_1(t)$	$10\sin(2\pi t)$ V
$R_1$	$50\,\Omega$
$V_D$ (all diodes)	$0.7$ V
$V_{Z2}$ ($D_2$)	$4.3$ V
$V_{\text{bat}}$	$3$ V

Find. $V_o$ and $V_R$ waveforms with peaks; peak current in $R_1$; the diode with the largest peak power and a suitable rating.

Approach. The output node $V_o$ is the same node as $R_1$'s output, so $R_1$ only carries current (and only then does $V_o$ depart from $V_1$) once one of the two branches conducts; the $D_1$–$D_2$ branch (cathodes joined) clamps the positive swing once $D_1$ turns on and forces $D_2$ into reverse breakdown, while the $D_3$–battery branch clamps the negative swing once $D_3$ turns on against the $3\text{V}$ bias.

Part (a) — positive and negative clamp levels. While $D_1$ conducts forward and drives $D_2$ into breakdown, $R_1$'s output node is pinned at $$V_{o,\text{pos}}=V_D+V_{Z2}=0.7+4.3=\boxed{5.0\text{ V}}$$ On the negative swing, $D_3$ turns on once the node drops enough to overcome the battery (whose $+$ terminal sits at the return rail, so the far side of $D_3$ is held at $-3\text{V}$): $$V_{o,\text{neg}}=-(V_D+V_{\text{bat}})=-(0.7+3)=\boxed{-3.7\text{ V}}$$ Between these levels (no branch conducting) $V_o=V_1$ exactly, since no current flows in $R_1$. Because $V_1$'s peak of $\pm10\text{V}$ exceeds both thresholds, $V_o$ is a sine wave clipped flat at $+5.0\text{V}$ above and $-3.7\text{V}$ below — an asymmetric limiter (unlike a plain diode clamp, the two levels differ because the zener and the battery set different thresholds).
Part (b) — $V_R$. $V_R=V_1-V_o$ is zero except while a branch conducts, where it equals the excess of $V_1$ beyond the corresponding clamp level: $$V_{R,\text{pos,pk}}=10-5.0=\boxed{+5.0\text{ V}},\qquad V_{R,\text{neg,pk}}=-10-(-3.7)=\boxed{-6.3\text{ V}}$$
Part (c) — peak current in $R_1$. $I_{R_1}=V_R/R_1$, so the two half-cycle peaks are $$I_{\text{pos}}=\frac{5.0}{50}=0.1\text{ A},\qquad I_{\text{neg}}=\frac{6.3}{50}=0.126\text{ A}$$ The larger magnitude sets the true peak current: $$I_{R_1,\text{pk}}=\boxed{126\text{ mA}}\quad\text{(during the negative half-cycle)}$$
Part (d) — largest diode power, rating. During the positive clamp the whole $100\text{ mA}$ flows through both $D_1$ (drop $0.7\text{V}$) and $D_2$ (drop $4.3\text{V}$, in breakdown); during the negative clamp the whole $126\text{ mA}$ flows through $D_3$ (drop $0.7\text{V}$): $$P_{D_1}=0.1\times0.7=70\text{ mW},\quad P_{D_2}=0.1\times4.3=\boxed{430\text{ mW}},\quad P_{D_3}=0.126\times0.7=88.2\text{ mW}$$ $D_2$ (the zener, absorbing the largest reverse-breakdown voltage) has by far the largest peak dissipation. Sizing conservatively for the peak (the pulse repeats every cycle), a $0.5\text{ W}$ device is the bare minimum; a standard $1\text{ W}$-rated zener gives comfortable margin (>2$\times$).

Fig. Q1(a) — $V_1$ (dashed, unclamped) vs. $V_o$ (solid): clipped flat at $+5.0\text{V}$ and $-3.7\text{V}$, the two peaks labelled.

Fig. Q1(b) — $V_R=V_1-V_o$: zero except during the brief conduction windows, peaking at $+5.0\text{V}$ (positive clamp) and $-6.3\text{V}$ (negative clamp).

Final Results — Question 1
Quantity	Value
$V_o$ positive clamp	$+5.0$ V
$V_o$ negative clamp	$-3.7$ V
$I_{R_1,\text{pk}}$	$126$ mA (negative half-cycle)
Largest-power diode	$D_2$, $430$ mW peak
Recommended rating	$1\text{ W}$ (min. $0.5\text{ W}$)

Check: $D_1$ is given no reverse-breakdown rating, so it is treated as ideal in reverse (never breaks down) — only $D_2$'s stated $4.3\text{V}$ rating sets the positive clamp. The battery orientation (its $+$ terminal at the return rail) is read from the figure's polarity marks as the only reading that makes $D_3$'s branch clamp the negative half-cycle (the branch that must complement $D_1$–$D_2$'s positive clamp for the circuit to be a useful two-sided limiter).

Topic: Two-sided (asymmetric) diode limiter using a zener and a battery-biased diode.

Key relations: Constant-drop diode model $V_D=0.7\text{V}$; a reverse-biased diode driven past its rated reverse voltage enters breakdown and clamps at that voltage; series diode+battery shifts a plain diode's clamp level by the battery voltage; $I_{R_1}=(V_1-V_o)/R_1$; worst-case device power $P=V\cdot I_{\text{pk}}$.

Why this works: Two independent branches, each conducting on only one half-cycle, let the designer set the positive and negative clip levels completely independently — the $D_1$/$D_2$ back-to-back pair uses breakdown for the positive level, while $D_3$/battery uses an ordinary forward drop offset by a fixed bias for the negative level. Outside both clamp windows, both branches are reverse-biased and open, so $R_1$ carries no current and $V_o=V_1$ exactly.

Common pitfall: Assuming the clamp is symmetric (as with a simple double-diode limiter) — here the two thresholds ($+5.0\text{V}$ vs. $-3.7\text{V}$) are genuinely different because one branch uses a $4.3\text{V}$ zener and the other a $3\text{V}$ battery. A second pitfall is computing $I_{R_1,\text{pk}}$ only from the more "obvious" positive half-cycle and missing that the negative half-cycle actually draws more current here.

Source: Sedra & Smith, Microelectronic Circuits, Ch.4 (diode limiting and clamping circuits).

Question 1 of 7: Diode/Zener/Battery Limiter

Question 1: Diode/Zener/Battery Limiter (20 marks)