Question 1 of 5: Series Voltage Regulator

Question 1: Series Voltage Regulator (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.

Find. The nominal regulated output voltage, the peak-to-peak ripple that survives at the output when the raw supply carries 1 V p-p, and the power efficiency of the regulator delivering that output into 4 Ω.

Approach. Fix the zener operating point first (it is not simply the rated 6.7 V until the bias current is checked), read the output straight off the unity-gain feedback loop, treat the ripple as a small-signal divider between $R_1$ and the zener's incremental resistance, and finish with a power balance that counts every current drawn from $V_{DD}$ — collector, base and reference alike.

1. Part (a) — locate the zener's actual operating point. The op-amp input draws no current, so the whole of the current through $R_1$ flows in the diode, and the diode's own model must agree with it: $$I_Z=\frac{V_{DD}-V_Z}{R_1},\qquad V_Z=V_{Z0}+r_z\,(I_Z-I_{Z0}).$$ Substituting the first into the second and solving the single linear equation, $$V_Z=\frac{V_{Z0}-r_z I_{Z0}+r_z V_{DD}/R_1}{1+r_z/R_1} =\frac{6.7-0.01+0.03030}{1.003030}=6.700\ \text{V},$$ whence $I_Z=(10-6.700)/3300=1.000$ mA. The design lands exactly on the rated test point, which is why the incremental term vanishes — a deliberate choice by the examiner and a useful check that the reading of the circuit is right.
2. Read the output from the feedback connection. The output node is wired straight back to the inverting input, so the loop is a unity-gain follower around the reference. An ideal op amp forces $v_-=v_+$, and here $v_+=V_Z$ and $v_-=V_{OUT}$: $$\boxed{V_{OUT}=V_Z=6.70\ \text{V}}$$ There is no feedback divider in this circuit, so the output is the reference itself; the Darlington pair contributes only current gain, and its two $V_{BE}$ drops are absorbed by the op amp's output swing rather than appearing at the load.
3. Part (b) — reduce the ripple problem to a divider. For small perturbations the diode is just its incremental resistance $r_z$, and $R_1$ feeds it from the rippling supply. Because the op-amp input draws no signal current either, the reference node sees a plain resistive divider: $$\Delta v_Z=\Delta v_{DD}\,\frac{r_z}{R_1+r_z} =1.0\times\frac{10}{3300+10}=3.02\times10^{-3}\ \text{V}.$$
4. Carry that ripple through the loop to the output. With infinite loop gain the error voltage is driven to zero, so whatever appears on the reference appears on the output and nothing else does — the ripple that couples through the pass transistors is suppressed by the loop: $$\boxed{\Delta V_{OUT}\approx 3.02\ \text{mV peak-to-peak}}$$ Expressed as line rejection this is $20\log_{10}(1/3.02\times10^{-3})=50.4$ dB. Notice that the answer is set entirely by the reference, not by the pass device: improving this regulator means a lower r_z or a bigger $R_1$ (or a current source in place of $R_1$), not a better transistor.
5. Part (c) — account for every current the supply must deliver. The load takes $$I_L=\frac{V_{OUT}}{R_L}=\frac{6.70}{4}=1.675\ \text{A},$$ which is the emitter current of $Q_2$. Working back up the Darlington, $$I_{B2}=\frac{I_{E2}}{\beta+1}=\frac{1.675}{101}=16.58\ \text{mA},\qquad I_{C2}=I_{E2}-I_{B2}=1.6584\ \text{A},$$ and since $I_{E1}=I_{B2}$, $$I_{B1}=\frac{16.58\ \text{mA}}{101}=0.164\ \text{mA},\qquad I_{C1}=16.42\ \text{mA}.$$ As a check on the bookkeeping, $I_{C1}+I_{C2}+I_{B1}=1.675$ A, which is exactly the emitter current leaving the pair.
6. Form the power balance and the efficiency. Both collectors and the op-amp output stage (which sources $I_{B1}$) are fed from $V_{DD}$, and so is the reference string: $$I_{DD}=I_{C1}+I_{C2}+I_{B1}+I_Z=1.675+0.001=1.676\ \text{A}.$$ $$P_{in}=V_{DD}I_{DD}=10\times1.676=16.76\ \text{W},$$ $$P_{L}=V_{OUT}I_L=6.70\times1.675=11.22\ \text{W},$$ $$\boxed{\eta=\frac{P_L}{P_{in}}=\frac{11.22}{16.76}=0.670\;\;\text{i.e.}\;\;67.0\ \%}$$
7. Sanity-check the number against the physics. A series regulator can never beat the ratio of the voltages it stands between, and here $V_{OUT}/V_{DD}=67.00$ percent. The computed 66.96 percent sits just below that ceiling, the small shortfall being the 1 mA reference current and the base currents; the missing 5.5 W is dissipated in the pass pair. This is the standing argument for a switching regulator whenever the drop is large.

Final results.