Question 1 of 5: Series voltage regulator — output voltage, line ripple and efficiency

Question 1: Series voltage regulator — output voltage, line 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.

Find. The nominal regulated output V_OUT, the peak-to-peak ripple that a 1 V p-p disturbance on V_DD produces at the output, and the power efficiency η = P_L/P_supply.

Approach. Fix the reference by finding the zener current, use the ideal op amp with 100 % negative feedback to transfer that reference to the output, treat the ripple as a small signal divided between R₁ and the zener's incremental resistance, then compare load power with the total power drawn from V_DD.

1. Part (a) — Check that the zener sits at its specified operating point. The reference branch is R₁ from V_DD into the zener; the op-amp input draws no current, so all of that current is zener current: $$I_Z=\frac{V_{DD}-V_Z}{R_1}=\frac{10-6.7}{3.3\ \text{k}\Omega}=1.00\ \text{mA}$$ which is exactly the current at which the data sheet quotes V_Z = 6.7 V. The reference is therefore 6.7 V with no correction needed.
2. Close the feedback loop to obtain the output. The inverting input is tied directly to the output node, so the loop is a unity-gain follower wrapped around the Darlington pair. An ideal op amp in negative feedback forces $v_- = v_+$, i.e. $V_{OUT}=V_Z$: $$\boxed{V_{OUT}=V_Z=6.70\ \text{V}}$$ The two base-emitter drops of Q₁ and Q₂ (1.4 V in total) are absorbed inside the loop — the op amp simply raises its output to $V_{OUT}+2V_{BE}\approx 8.1\ \text{V}$ to supply them, which is well within its ±15 V rails.
3. Part (b) — Convert the supply ripple into a reference ripple. For small signals the zener is replaced by its incremental resistance $r_z = R_Z = 10\ \text{k}\Omega$, which forms a divider with R₁ across the ripple source: $$\frac{v_z}{v_{dd}}=\frac{r_z}{R_1+r_z}=\frac{10}{3.3+10}=0.752$$ The op amp and pass pair reproduce that reference ripple one-for-one at the output, so $$\boxed{v_{out(\text{p-p})}=0.752\times 1\ \text{V}=0.75\ \text{V p-p}}$$
4. State the line regulation that this implies. Expressed as a ratio the line regulation is $0.752\ \text{V/V}$, i.e. $20\log_{10}(0.752)=-2.5\ \text{dB}$ of ripple rejection. This is very poor for a regulator, and it is entirely the fault of the 10 kΩ incremental zener resistance quoted for D₁: a real 6.7 V zener biased at 1 mA would show tens of ohms, which would give a rejection near 80 dB. The circuit topology is sound; the reference device is the weak link, and the calculation is worth doing precisely because it exposes that.
5. Part (c) — Compute the delivered load power. With the output regulated at 6.70 V across 4 Ω, $$I_L=\frac{V_{OUT}}{R_L}=\frac{6.70}{4}=1.675\ \text{A},\qquad P_L=V_{OUT}I_L=6.70\times1.675=11.22\ \text{W}$$
6. Account for every ampere drawn from the supply. The load current is the emitter current of Q₂, so $I_{B2}=I_L/(\beta+1)=1.675/101=16.58\ \text{mA}$, and that base current is in turn the emitter current of Q₁, giving $I_{B1}=I_{B2}/(\beta+1)=0.164\ \text{mA}$. Only the two collector currents and the reference branch are drawn from V_DD (the op amp runs from its own rails), and by KCL these total $$I_{DD}=I_Z+I_{C1}+I_{C2}=I_Z+I_L-I_{B1}=1.00\ \text{mA}+1.675\ \text{A}-0.164\ \text{mA}=1.676\ \text{A}$$ $$P_{supply}=V_{DD}I_{DD}=10\times1.6758=16.76\ \text{W}$$
7. Form the efficiency. $$\eta=\frac{P_L}{P_{supply}}=\frac{11.22}{16.76}=0.6697 \quad\Rightarrow\quad \boxed{\eta=67.0\ \%}$$ The dissipation is dominated by the 3.3 V that the pass pair drops at 1.675 A (5.53 W), which is the unavoidable price of a linear series regulator: the efficiency ceiling is $V_{OUT}/V_{DD}=67.0\ \%$ and this design essentially reaches it.