Question 1 of 5

Question 1

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. A diode bridge (D₁–D₄) with a single zener D_Z across the output diagonal, driven from v_I through R₁; the output v_O is the node at the top of the bridge.

Find. The complete v_O–v_I transfer characteristic over ±20 V: every break point and the slope of every segment.

Approach. The four pn diodes form a full-wave bridge whose output diagonal (L–R) always presents the same polarity to the zener, so for either sign of v_I the conducting path is two forward diodes in series with the zener in breakdown; below the turn-on level nothing conducts and the output simply follows the input.

1. Trace the conduction path for each polarity. For v_I > 0 the current flows $T\!\rightarrow\!D_1\!\rightarrow\!R\!\rightarrow\!D_Z\!\rightarrow\!L\!\rightarrow\!D_3\!\rightarrow\!B$; for v_I < 0 it flows $B\!\rightarrow\!D_2\!\rightarrow\!R\!\rightarrow\!D_Z\!\rightarrow\!L\!\rightarrow\!D_4\!\rightarrow\!T$. Either way the path is two forward pn diodes in series with the zener operating in reverse (breakdown).
2. Off state — central segment. While the total voltage across the bridge is below the string’s turn-on value, the zener is not yet in breakdown, no current flows, and there is no drop across $R_1$: $$v_O = v_I \qquad (\text{slope} = 1).$$
3. Turn-on (break-point) level. Conduction begins when the branch voltage reaches the sum of the two forward diode offsets and the zener knee: $$V_{\text{th}} = 2V_{D0} + V_{Z0} = 2(0.65) + 8.2 = \boxed{9.5\ \text{V}}.$$ Because no current flows up to this point, $v_O = v_I$ right up to it, so the break points sit at $v_I = \pm 9.5\ \text{V}$, $v_O = \pm 9.5\ \text{V}$.
4. Incremental (bulk) resistance of the conducting string. Two diode bulk resistances plus the zener bulk resistance appear in series: $$r_s = 2r_D + r_z = 2(20) + 20 = 60\ \Omega.$$ The conducting bridge therefore behaves as a $9.5\ \text{V}$ battery in series with $60\ \Omega$.
5. Limiting-region slope. That Thévenin branch and $R_1$ form a divider for increments of $v_I$: $$\frac{dv_O}{dv_I} = \frac{r_s}{R_1 + r_s} = \frac{60}{1000+60} = \boxed{0.0566}\quad(\approx \tfrac{1}{17.7}).$$
6. Solve the limiting-region line and evaluate the endpoints. With $v_O = 9.5 + r_s I$ and $v_I = v_O + R_1 I$, $$v_O = \frac{9.5 + 0.06\,v_I}{1.06}.$$ At $v_I = 20\ \text{V}$: $v_O = (9.5 + 1.2)/1.06 = \boxed{10.09\ \text{V}}$; by symmetry $v_O(-20) = -10.09\ \text{V}$. The curve is odd-symmetric.