Part B — Electronics: 23-Mechatronics-A2 Circuits and Electronics, December 2019, Q7

Part B, Question 1: Diode/zener bridge limiter — transfer characteristic [20]

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.

Quantity	Value
Series resistor $R_1$	1 kΩ
Diodes $D_1$–$D_4$	$V_{D0}=0.65$ V, $r_D=20\,\Omega$ each
Zener $D_Z$	$V_{Z0}=8.2$ V, $r_z=20\,\Omega$
Input range	$\pm20$ V

Find. The $v_o$–$v_I$ transfer characteristic: breakpoints and slopes in each region.

Transfer characteristic: unity slope for $|v_I|<9.5\,$V, then a shallow clipped slope beyond, symmetric about the origin.

Check. The diode-bridge network is analysed as the standard bidirectional zener limiter: whichever way current must flow to clip $v_o$, it is routed through two forward-biased regular diodes plus the zener (in breakdown for one polarity, essentially at its own forward drop for the other) — the textbook topology that gives symmetric clipping from a single zener.

Approach. Find the onset voltage where the diode/zener path first conducts (defining the breakpoints), then the reduced slope beyond it set by the network’s small-signal diode/zener resistance in the voltage divider with $R_1$.

Linear (unclipped) region. While $|v_o|$ is below the conduction threshold, no diode conducts, no current flows through $R_1$, and $$v_o=v_I,\quad \text{slope}=1.$$
Clipping onset (breakpoints). Conduction begins once $v_o$ reaches the series drop of two forward diodes plus the zener: $$V_{clip}=V_{Z0}+2V_{D0}=8.2+2(0.65)=\boxed{9.5\ \text{V}},$$ symmetric at $v_I=\pm9.5$ V (no drop across $R_1$ until conduction starts, so the breakpoint occurs exactly there).
Slope beyond the breakpoints. Once conducting, the total incremental diode-path resistance is $R_{path}=2r_D+r_z=2(20)+20=60\ \Omega$, dividing further input swing with $R_1$: $$m=\frac{R_{path}}{R_1+R_{path}}=\frac{60}{1000+60}=\boxed{0.0566}.$$
Endpoints at $v_I=\pm20$ V. $$v_o=V_{clip}+m\,(20-V_{clip})=9.5+0.0566\times(20-9.5)=\boxed{10.094\ \text{V}}\ (\text{and}\ -10.094\ \text{V at}\ v_I=-20).$$

Quantity	Result
Linear-region slope	1 (unity), $\|v_I\|<9.5$ V
Clip onset (breakpoints)	$\pm9.5$ V
Clipped-region slope	0.0566
$v_o$ at $v_I=\pm20$ V	$\pm10.094$ V

Concept & Theory

Topic. Piecewise-linear diode/zener limiter transfer characteristic.

Key relations. PWL diode model $v_D=V_{D0}+i_Dr_D$ (forward), zener $v_Z=V_{Z0}+i_Zr_z$ (breakdown); voltage divider between $R_1$ and the active clamp path once conducting.

Why this works. Below the conduction threshold the clamp path is an open circuit, so the network is a plain wire and $v_o$ tracks $v_I$ exactly; once the combined diode/zener drops are reached, the path turns on and forms a voltage divider with $R_1$, giving a much shallower (but non-zero, because of the finite $r_D,r_z$) slope rather than a hard clip.

Common pitfall. Treating the clipped region as perfectly flat (ignoring $r_D,r_z$) when the question explicitly asks for the slope there; using only one diode drop instead of two in the onset voltage.

Source. Sedra & Smith, Microelectronic Circuits §3.7–3.9 (diode limiting circuits, zener diodes, piecewise-linear models).

Question 7 of 11: Part B — Electronics

Part A — Circuits

Part B — Electronics

Part B, Question 1: Diode/zener bridge limiter — transfer characteristic [20]