Standby Power, Swing, Drive, Output Power and Efficiency of a Class AB Stage: 22-Elec-B5 Advanced Electronics, December 2016, Q4

Question 4: Standby Power, Swing, Drive, Output Power and Efficiency of a Class AB Stage (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 complementary Darlington output stage biased by a four-diode string. \(R_1\) and \(R_2\) feed the string from the rails, the driver \(A\) injects \(v_S\) at the centre of the string, and the emitter resistors \(R_3,R_4\) set the quiescent current and stabilise it thermally.

Quantity	Symbol	Value
Supplies	\(\pm V_{CC}\)	\(\pm 12\) V
Bias-string feed resistors	\(R_1=R_2\)	1 kΩ
Emitter resistors	\(R_3=R_4\)	5 Ω
Load	\(R_L\)	8 Ω
Diode drop (each of four)	\(V_D\)	0.7 V
Darlington base–emitter drop	\(\|V_{BE}\|\)	1.2 V
Darlington current gain	\(\beta\)	500
Saturation drop	\(\|V_{CE(\mathrm{sat})}\|\)	0.3 V

Find. Standby power; maximum peak-to-peak output swing; the drive amplitude needed to reach it; the maximum rms output power; and the efficiency at that operating point.

Question 4. Class AB stage. The four-diode string spans both Darlington base-emitter drops plus the drops across R3 and R4, which is what sets the quiescent current.

Approach. Work the quiescent state first — the diode string fixes both base voltages, the Darlington drops fix both emitter voltages, and the difference across \(R_3+R_4\) fixes the standing current. Then take the positive peak, where the conducting device bottoms out at \(V_{CE(\mathrm{sat})}\) and its emitter resistor carries the whole load current, and finish with the standard half-sine power integrals.

Part (a) — node voltages of the bias string at \(v_S=0\). Two diodes sit above the injection point and two below it, so \[V_{B1}=v_S+2V_D=+1.4\ \text{V},\qquad V_{B2}=v_S-2V_D=-1.4\ \text{V}.\] The feed resistors therefore carry \[I_{R1}=\frac{12-1.4}{1\ \text{k}\Omega}=10.6\ \text{mA},\qquad I_{R2}=\frac{-1.4+12}{1\ \text{k}\Omega}=10.6\ \text{mA},\] equal, as symmetry demands and as the ideal (zero-power) driver requires.
Quiescent current in the output devices. Each Darlington drops 1.2 V from base to emitter, so with the output at 0 V \[V_{E1}=1.4-1.2=+0.2\ \text{V},\qquad V_{E2}=-1.4+1.2=-0.2\ \text{V},\] and the standing current flows from \(+V_{CC}\) down through \(Q_1,R_3,R_4,Q_2\) to \(-V_{CC}\): \[\boxed{I_Q=\frac{V_{E1}-0}{R_3}=\frac{0.2\ \text{V}}{5\ \Omega}=40\ \text{mA}}\] The load carries nothing because both ends of it are at 0 V. This is the class-AB compromise: 40 mA of standing current is enough to keep both devices conducting through the crossover region.
Total supply current and standby power. The Darlington base current is \(I_B=I_Q/(\beta+1)=40/501=79.8\ \mu\text{A}\), so the diode string actually carries 10.52 mA and each rail delivers \[I_{supply}=I_{R1}+I_{C1}=10.6+39.92=50.52\ \text{mA}.\] Both rails contribute, so \[P_{standby}=(V_{CC}-(-V_{CC}))I_{supply}=24\ \text{V}\times 50.52\ \text{mA}=\boxed{1.21\ \text{W}}\] Ignoring the base current altogether gives 1.214 W, so the "estimate" the question asks for is insensitive to that refinement — but it is worth showing that it was checked.
Part (b) — positive peak output. At the top of the swing \(Q_2\) is cut off, so \(R_4\) carries nothing and the whole load current flows through \(R_3\). The limit is reached when \(Q_1\) reaches saturation: \[V_{CC}-|V_{CE(\mathrm{sat})}|=\hat{V}_O+\hat{I}_OR_3=\hat{V}_O\left(1+\frac{R_3}{R_L}\right).\] Solving with \(R_3/R_L=5/8\), \[\hat{V}_O=\frac{12-0.3}{1.625}=7.20\ \text{V},\qquad\boxed{V_{O(p\text{-}p)}=2\hat{V}_O=14.4\ \text{V}}\] Without the emitter resistors the answer would have been 23.4 V p-p; the 5 Ω resistors cost nearly 40 % of the available swing, which is exactly what the hint is pointing at.
Part (c) — drive required at \(v_S\). At the peak the load current is \(\hat{I}_O=7.20/8=0.90\ \text{A}\), so the emitter of \(Q_1\) stands at \(7.20+0.90(5)=11.7\ \text{V}\) and its base one Darlington drop higher, at 12.9 V. Two diode drops below that is the driver node: \[\hat{v}_S=\hat{V}_O+\hat{I}_OR_3+|V_{BE}|-2V_D=11.7+1.2-1.4,\] \[\boxed{\hat{v}_S=11.5\ \text{V peak}\ \ (23\ \text{V peak-to-peak})}\] Note that the base of \(Q_1\) must be driven 0.9 V above the positive rail, so \(R_1\) can no longer supply it and the driver \(A\) must both source that current and swing beyond the rail — a bootstrapped or separately supplied driver is implied. That is why the question specifies an ideal amplifier.
Part (d) — maximum output power. For a sinusoid of peak \(\hat{V}_O\) the rms value is \(\hat{V}_O/\sqrt{2}\), so \[P_{out(max)}=\frac{\hat{V}_O^{2}}{2R_L}=\frac{(7.20)^{2}}{2(8)}=\boxed{3.24\ \text{W}}\]
Part (e) — efficiency. Each device conducts a half sine of peak 0.9 A, whose average is \(\hat{I}_O/\pi\), so the two rails together deliver \[P_S=2V_{CC}\frac{\hat{I}_O}{\pi}=2(12)\frac{0.90}{\pi}=6.88\ \text{W},\qquad \eta=\frac{P_{out}}{P_S}=\frac{3.24}{6.88}.\] \[\boxed{\eta=47.1\ \%=\frac{\pi}{4}\cdot\frac{\hat{V}_O}{V_{CC}}}\] The theoretical class-B ceiling of 78.5 % is only reached when \(\hat{V}_O\to V_{CC}\); here the saturation voltage and the emitter resistors hold the peak to 60 % of the rail, and efficiency scales directly with that ratio. Including the 10.6 mA bias string in the supply current lowers the figure slightly, to 45.4 %.

Result	Symbol	Value
Bias string current	\(I_{R1}=I_{R2}\)	10.6 mA
Quiescent output-device current	\(I_Q\)	40 mA
Standby power	\(P_{standby}\)	1.21 W
Maximum peak / peak-to-peak output	\(\hat{V}_O,\ V_{O(p\text{-}p)}\)	7.20 V, 14.4 V
Peak load current	\(\hat{I}_O\)	0.90 A
Minimum drive amplitude	\(\hat{v}_S\)	11.5 V peak (23 V p-p)
Maximum rms output power	\(P_{out(max)}\)	3.24 W
Supply power at full drive	\(P_S\)	6.88 W
Efficiency (output devices / including bias)	\(\eta\)	47.1 % / 45.4 %

Topic. Class AB output stage: diode biasing, swing limits set by saturation and emitter degeneration, and half-sine power and efficiency.

Key relations. Bias string \(V_{B1}-V_{B2}=4V_D\) spans \(2|V_{BE}|+I_Q(R_3+R_4)\); \(\hat{V}_O=(V_{CC}-V_{CE(\mathrm{sat})})/(1+R_3/R_L)\); \(P_L=\hat{V}_O^2/2R_L\); \(P_S=2V_{CC}\hat{I}_O/\pi\); \(\eta=(\pi/4)(\hat{V}_O/V_{CC})\); per-device dissipation \(P_D=V_{CC}\hat{V}_O/(\pi R_L)-\hat{V}_O^2/(4R_L)\) with the free check \(P_S-P_L=2P_D\).

Why this works. The four diodes are chosen so that their total drop exceeds the two \(|V_{BE}|\) drops by exactly the amount needed across \(R_3+R_4\) to set the standing current; because the diodes track the transistor junctions with temperature, the quiescent current is stabilised without trimming. The emitter resistors add local series feedback that prevents thermal runaway, at the cost of swing and efficiency, which is the trade the numbers above quantify. On the power side, each device conducts for half a cycle only, so the average rail current is \(\hat{I}_O/\pi\) rather than \(\hat{I}_O/2\), and the \(\pi/4\) factor in the efficiency expression is a direct consequence.

Common pitfall. Quoting the textbook 78.5 % for part (e) without scaling it by \(\hat{V}_O/V_{CC}\) — the maximum applies only to a stage swinging to the rail. The other trap is forgetting \(R_3\) when finding the swing (giving 11.7 V peak instead of 7.20 V, and hence 8.6 W instead of 3.24 W), or including \(R_4\) as well: on the positive half-cycle \(Q_2\) is off, so \(R_4\) carries no current and contributes no drop.

Source. A. S. Sedra & K. C. Smith, Microelectronic Circuits, 7th/8th ed., Ch. 12 §12.4–12.6 (class AB operation, biasing with diodes, and the Darlington output stage).

Question 4 of 5: Standby Power, Swing, Drive, Output Power and Efficiency of a Class AB Stage

Question 4: Standby Power, Swing, Drive, Output Power and Efficiency of a Class AB Stage (20 marks)