Question 3 of 5: Series Voltage Regulator — Output, Ripple and Efficiency

Question 3: Series Voltage Regulator — Output, Ripple and Efficiency (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. A series regulator in which $R_1$ and the zener $D_1$ form the reference, the ideal op‑amp $A_1$ compares that reference with the output through a unity‑gain feedback path, and a Darlington pair $Q_1$–$Q_2$ passes the load current from $V_{DD}$.

Find. The nominal regulated output, the peak‑to‑peak ripple that survives at the output, and the power efficiency of the regulator.

Approach. Locate the zener on its piecewise‑linear characteristic to get the reference, propagate that through the unity‑gain loop for both the d.c. level and the ripple, then account for every ampere drawn from $V_{DD}$ to form the efficiency.

1. Part (a) — place the zener on its own load line. Model the diode as a battery in series with $R_Z$: from the quoted point, $V_{Z0}=V_Z-I_ZR_Z=6.7-(1\times10^{-3})(10\times10^{3})=-3.3\ \text{V}$, so $V_z=V_{Z0}+I R_Z$ with $I=(V_{DD}-V_z)/R_1$. Solving the two together, $$V_z=\frac{V_{Z0}+R_ZV_{DD}/R_1}{1+R_Z/R_1}=6.70\ \text{V},\qquad I=\frac{10-6.70}{3300}=1.00\ \text{mA}$$ The bias lands exactly on the quoted 1 mA test point, confirming that $R_1$ was chosen for it.
2. Close the loop. The op‑amp is ideal and the output is returned undivided to its inverting input, so the loop drives $V_-\to V_+$: $$V_{OUT}=V_z=\boxed{6.70\ \text{V}}$$ The Darlington sits inside the loop, so neither $V_{BE1}$ nor $V_{BE2}$ appears in the answer — the op‑amp simply raises its output by 1.4 V to cover them.
3. Part (b) — ask where ripple can enter. Ripple on $V_{DD}$ reaches the output by two routes: through the pass transistors, and through the reference. The first is killed by the loop — with an ideal op‑amp the loop gain is infinite, so the output is pinned to whatever the non‑inverting input says, independent of the collector supply. Only the reference route survives.
4. Divide the ripple between $R_1$ and the zener. To small signals the diode is just $R_Z$, so $R_1$ and $R_Z$ form a divider: $$\frac{v_z}{v_{dd}}=\frac{R_Z}{R_1+R_Z}=\frac{10}{3.3+10}=0.752$$ and since the output follows $v_z$ one for one, $$v_{OUT,\text{ripple}}=0.752\times1\ \text{V}_{pp}=\boxed{0.75\ \text{V}_{pp}}$$
5. Read what that means. Three quarters of the line ripple appears at the output; the regulator regulates against load changes but barely at all against the line. The culprit is the printed $R_Z=10\ \text{k}\Omega$, which is two to three orders of magnitude larger than a real zener’s dynamic resistance. Were $r_z$ a realistic $20\ \Omega$ the divider would give $20/3320=0.6\ \%$, i.e. about $6\ \text{mV}_{pp}$; the standard cure is to bootstrap the reference feed off the regulated output instead of off $V_{DD}$.
6. Part (c) — useful power out. The load draws $I_L=V_{OUT}/R_L=6.70/4=1.675\ \text{A}$, so $$P_{out}=V_{OUT}I_L=\frac{V_{OUT}^{2}}{R_L}=\frac{6.70^{2}}{4}=11.22\ \text{W}$$
7. Account for every path from the supply. The load current is the emitter current of $Q_2$, so $I_{B2}=I_L/(\beta+1)=1.675/101=16.58\ \text{mA}$ and $I_{C2}=1.658\ \text{A}$. That base current is in turn the emitter current of $Q_1$, giving $I_{C1}=16.58\times(100/101)=16.42\ \text{mA}$. Adding the 1.00 mA reference branch, $$I_{DD}=I_{C1}+I_{C2}+I_{R1}=0.01642+1.6584+0.001=1.676\ \text{A}$$
8. Form the efficiency. $$\eta=\frac{P_{out}}{P_{in}}=\frac{11.22}{(10)(1.676)}=\frac{11.22}{16.76}=\boxed{67.0\ \%}$$ The simple estimate $\eta\approx V_{OUT}/V_{DD}=67.0\ \%$ is indistinguishable, because the base and reference currents together amount to only 1 % of the load current. The missing 5.54 W is dropped across the Darlington as heat, which is the defining limitation of every linear series regulator and the reason a switching pre‑regulator is used when the raw supply sits far above the regulated rail.