Class A, B and AB Power Amplifiers

About this problem set

Power-stage calculations for Class A, B and AB operation: series-fed and transformer-coupled Class A efficiency, Class B push-pull power and the 78.5% ceiling, worst-case device dissipation, heat-sink specification from thermal resistances, and second-harmonic distortion.

Each problem below is solved in full — the given data is listed first, the governing relation is stated, and the arithmetic is carried through to a boxed answer. Work through the statement yourself before opening the solution. The underlying theory is covered in the Electronic Devices lecture series.

PROBLEM 01

Series-Fed Class A Amplifier Efficiency

Problem Statement

A series-fed Class A power stage works from \(V_{CC} = 20~\text{V}\) with the load \(R_C = 20~\Omega\) wired directly in the collector lead. Base bias comes from a \(R_B = 1~\text{k}\Omega\) resistor to the rail; \(\beta = 25\), \(V_{BE} = 0.7~\text{V}\).

Find the Q-point and the DC input power drawn from the supply.
An input signal swings the base current by \(I_{b(p)} = 10~\text{mA}\) peak. Find the AC output power and the efficiency.
Find the largest output power this stage can give and the efficiency at that point.
Show that no series-fed Class A stage can beat 25\%.

Solution

\(V_{CC} = 20~\text{V}\), \(R_B = 1~\text{k}\Omega\), \(R_C = 20~\Omega\), \(\beta = 25\), \(V_{BE} = 0.7~\text{V}\)
Class A: the device conducts for the full 360\(^\circ\), so \(I_{CQ}\) is drawn from the supply whether or not signal is present.

Part (a) — Q-point and DC input power

\[\begin{aligned} I_B &= \frac{V_{CC} - V_{BE}}{R_B} = \frac{19.3~\text{V}}{1~\text{k}\Omega} = 19.3~\text{mA} \\ I_{CQ} &= \beta I_B = 25 \times 19.3~\text{mA} = 482.5~\text{mA} \\ V_{CEQ} &= V_{CC} - I_{CQ}R_C = 20 - (482.5~\text{mA})(20~\Omega) = 10.35~\text{V} \\ P_{i(\text{dc})} &= V_{CC} I_{CQ} = (20~\text{V})(482.5~\text{mA}) = 9.650~\text{W} \end{aligned}\]

Q-point \((10.3~\text{V},\,483~\text{mA})\); \(P_{i(\text{dc})} = 9.65~\text{W}\), drawn constantly.

Part (b) — output power and efficiency for the given drive

\[\begin{aligned} I_{c(p)} &= \beta I_{b(p)} = 25 \times 10~\text{mA} = 250~\text{mA} \\ V_{ce(p)} &= I_{c(p)}R_C = (250~\text{mA})(20~\Omega) = 5.00~\text{V} \\ P_o(\text{ac}) &= \frac{I_{c(p)}^2 R_C}{2} = \frac{(0.2500~\text{A})^2(20~\Omega)}{2} = 0.6250~\text{W} \\ \eta &= \frac{P_o}{P_{i(\text{dc})}} \times 100\% = \frac{0.6250}{9.650} \times 100\% = 6.48\% \end{aligned}\]

\(P_o = 0.625~\text{W}\), \(\eta = 6.48\%\)

Part (c) — largest achievable output

The swing is clipped either by cutoff (\(I_c\) cannot exceed \(I_{CQ}\)) or by saturation (\(v_{ce}\) cannot exceed \(V_{CEQ}\), i.e. \(I_c \le V_{CEQ}/R_C = 517.5~\text{mA}\)). The cutoff limit bites first:

\[\begin{aligned} I_{c(p),\max} &= I_{CQ} = 482.5~\text{mA} \\ P_{o,\max} &= \frac{I_{CQ}^2 R_C}{2} = \frac{(0.48250)^2(20)}{2} = 2.328~\text{W} \\ \eta_{\max} &= \frac{2.328}{9.650} \times 100\% = 24.13\% \end{aligned}\]

\(P_{o,\max} = 2.33~\text{W}\) at \(\eta = 24.1\%\)

Part (d) — the 25\% ceiling

The best case is \(V_{CEQ} = V_{CC}/2\) with \(I_{CQ} = V_{CC}/2R_C\), giving a peak swing of \(V_{CC}/2\) in voltage and \(V_{CC}/2R_C\) in current:

\[\begin{aligned} P_{o,\max} &= \frac{(V_{CC}/2)^2}{2R_C} = \frac{V_{CC}^2}{8R_C} \\ P_{i(\text{dc})} &= V_{CC}\cdot\frac{V_{CC}}{2R_C} = \frac{V_{CC}^2}{2R_C} \\ \eta_{\max} &= \frac{V_{CC}^2/8R_C}{V_{CC}^2/2R_C} = \frac{1}{4} = 25\% \end{aligned}\]

For these numbers that would be \(P_{o,\max} = 2.500~\text{W}\) with \(P_{i} = 10.0~\text{W}\). The ceiling is structural: the supply current is constant, so at zero signal the whole 9.65 W is dissipated in the transistor. Class A wastes at least three watts for every watt delivered.

PROBLEM 02

Transformer-Coupled Class A and the 50% Ceiling

Problem Statement

A Class A output stage drives an \(R_L = 8~\Omega\) loudspeaker through a transformer of turns ratio \(N_1:N_2 = 5:1\). The supply is \(V_{CC} = 12~\text{V}\) and the quiescent collector current is \(I_{CQ} = 60~\text{mA}\). The primary winding has negligible DC resistance.

Find the load reflected into the collector circuit.
Find the DC input power, the maximum AC output power and the maximum efficiency, and confirm that \(I_{CQ}\) is the correct value for maximum swing.
With a real drive the collector waveform measures \(V_{CE}\) between 4 V and 20 V and \(I_C\) between 20 mA and 100 mA. Find the delivered power, the efficiency and the transistor dissipation.

Solution

\(N_1/N_2 = 5\), \(R_L = 8~\Omega\), \(V_{CC} = 12~\text{V}\), \(I_{CQ} = 60~\text{mA}\)
Ideal, lossless transformer with zero primary DC resistance, so \(V_{CEQ} = V_{CC}\).

Part (a) — reflected load

\[R_L' = \left(\frac{N_1}{N_2}\right)^2 R_L = (5)^2(8~\Omega) = 200~\Omega\]

\(R_L' = 200~\Omega\) is what the collector actually works into.

Part (b) — power and the ceiling

Because the primary has no DC drop, \(V_{CEQ} = V_{CC} = 12~\text{V}\) and the collector can swing symmetrically from 0 to \(2V_{CC} = 24~\text{V}\) — the transformer's inductive kick is what allows \(V_{CE}\) to exceed the supply.

\[\begin{aligned} P_{i(\text{dc})} &= V_{CC}I_{CQ} = (12)(60~\text{mA}) = 0.720~\text{W} \\ P_o(\text{ac}) &= \frac{(V_{CE\max}-V_{CE\min})(I_{C\max}-I_{C\min})}{8} = \frac{(24-0)(120~\text{mA}-0)}{8} = 0.360~\text{W} \\ \eta_{\max} &= \frac{0.360}{0.720} \times 100\% = 50.0\% \end{aligned}\]

Check that the bias suits the load line: for a swing of \(V_{CC}\) across \(R_L'\) the peak current must be \(V_{CC}/R_L'\):

\[I_{CQ}\big|_{\text{required}} = \frac{V_{CC}}{R_L'} = \frac{12~\text{V}}{200~\Omega} = 60.0~\text{mA} \quad \checkmark\]

\(P_{i} = 0.720~\text{W}\), \(P_{o,\max} = 0.360~\text{W}\), \(\eta_{\max} = 50.0\%\), and \(I_{CQ} = 60~\text{mA}\) is exactly right for maximum swing.

Part (c) — the measured waveform

\[\begin{aligned} P_o &= \frac{(20-4)(100-20)~\text{mA}}{8} = \frac{(16~\text{V})(80~\text{mA})}{8} = 0.160~\text{W} \\ \eta &= \frac{0.160}{0.720} \times 100\% = 22.22\% \\ P_Q &= P_{i(\text{dc})} - P_o = 0.720 - 0.160 = 0.560~\text{W} \end{aligned}\]

\(P_o = 0.160~\text{W}\), \(\eta = 22.2\%\), transistor dissipation \(P_Q = 0.560~\text{W}\).

The transformer removes the DC loss in the load, which is what lifts the ceiling from 25\% to 50\%; but the supply current is still constant, so a half-amplitude drive still gives only 22.2\%, and the transistor absorbs the rest.

PROBLEM 03

Class B Push-Pull Power and Efficiency

Problem Statement

A complementary-symmetry Class B push-pull stage runs from \(\pm V_{CC} = \pm30~\text{V}\) into \(R_L = 16~\Omega\). Each transistor conducts for one half cycle. Crossover distortion is ignored and the transistors are taken as ideal switches in series with the load.

For a peak output voltage \(V_m = 24~\text{V}\), find \(P_o\), the DC power drawn, the efficiency and the total device dissipation.
Find the same quantities at the maximum possible drive.
Show that the efficiency ceiling is \(\pi/4\).

Solution

\(V_{CC} = 30~\text{V}\) (each rail), \(R_L = 16~\Omega\), \(V_m = 24~\text{V}\) peak
Class B: each device conducts 180\(^\circ\); the supply current is a half-wave rectified sinusoid, whose average is \(I_m/\pi\) per device.

Part (a) — at the stated drive

\[\begin{aligned} I_m &= \frac{V_m}{R_L} = \frac{24~\text{V}}{16~\Omega} = 1.500~\text{A} \\ P_o &= \frac{V_m^2}{2R_L} = \frac{(24)^2}{2(16)} = 18.00~\text{W} \\ P_{i(\text{dc})} &= V_{CC}\,\frac{2I_m}{\pi} = (30)\frac{2(1.500)}{\pi} = 28.648~\text{W} \\ \eta &= \frac{18.00}{28.648} \times 100\% = 62.83\% \\ P_{2Q} &= P_{i(\text{dc})} - P_o = 28.648 - 18.00 = 10.65~\text{W} \end{aligned}\]

\(P_o = 18.0~\text{W}\), \(P_i = 28.65~\text{W}\), \(\eta = 62.8\%\), \(P_{2Q} = 10.6~\text{W}\) (i.e. 5.32 W per device).

Part (b) — full drive

The largest peak the output can reach is the rail itself, \(V_m = V_{CC} = 30~\text{V}\):

\[\begin{aligned} I_m &= \frac{30}{16} = 1.875~\text{A} \\ P_{o,\max} &= \frac{V_{CC}^2}{2R_L} = \frac{(30)^2}{2(16)} = 28.125~\text{W} \\ P_{i,\max} &= (30)\frac{2(1.875)}{\pi} = 35.810~\text{W} \\ \eta_{\max} &= \frac{28.125}{35.810} \times 100\% = 78.54\% \end{aligned}\]

\(P_{o,\max} = 28.125~\text{W}\) at \(\eta_{\max} = 78.54\%\)

Part (c) — the \(\pi/4\) result

\[\begin{aligned} \eta &= \frac{P_o}{P_{i(\text{dc})}} = \frac{V_m^2/2R_L}{V_{CC}(2V_m/\pi R_L)} = \frac{\pi}{4}\cdot\frac{V_m}{V_{CC}} \\ \eta_{\max} &= \frac{\pi}{4} = 0.7854 \quad\text{when } V_m = V_{CC} \end{aligned}\]

Efficiency rises linearly with drive because the supply current follows the signal — that is the essential difference from Class A. At the 24 V drive of part (a), \(\tfrac{\pi}{4}\times\tfrac{24}{30} = 62.83\%\), matching the arithmetic above.

PROBLEM 04

Maximum Dissipation per Transistor in Class B

Problem Statement

For the same push-pull stage (\(V_{CC} = 30~\text{V}\), \(R_L = 16~\Omega\)), the worst thermal case is not at maximum output.

Starting from \(P_{2Q} = P_{i(\text{dc})} - P_o\) as a function of \(V_m\), find the drive level at which the device dissipation peaks.
Evaluate the maximum dissipation, total and per transistor.
Verify the textbook rule \(P_{2Q,\max} \approx 0.4\,P_{o,\max}\) and quote the efficiency at that operating point.

Solution

\(V_{CC} = 30~\text{V}\), \(R_L = 16~\Omega\), \(P_{o,\max} = 28.125~\text{W}\) (from the previous problem)
Ideal Class B, no crossover region, both devices sharing the dissipation equally.

Part (a) — maximise \(P_{2Q}(V_m)\)

\[\begin{aligned} P_{2Q} &= P_{i(\text{dc})} - P_o = \frac{2V_{CC}V_m}{\pi R_L} - \frac{V_m^2}{2R_L} \\ \frac{dP_{2Q}}{dV_m} &= \frac{2V_{CC}}{\pi R_L} - \frac{V_m}{R_L} = 0 \\ \Rightarrow\; V_m &= \frac{2V_{CC}}{\pi} = 0.636\,V_{CC} \end{aligned}\]

\[V_m = \frac{2(30~\text{V})}{\pi} = 19.10~\text{V}, \qquad I_m = \frac{19.10}{16} = 1.194~\text{A}\]

Worst-case dissipation occurs at \(V_m = 19.1~\text{V}\), about 64\% of full drive.

Part (b) — the peak dissipation

Substituting \(V_m = 2V_{CC}/\pi\) back:

\[\begin{aligned} P_{2Q,\max} &= \frac{2V_{CC}}{\pi R_L}\cdot\frac{2V_{CC}}{\pi} - \frac{1}{2R_L}\left(\frac{2V_{CC}}{\pi}\right)^2 = \frac{2V_{CC}^2}{\pi^2 R_L} \\ &= \frac{2(30)^2}{\pi^2(16)} = 11.399~\text{W} \\ P_{D,\max}\big|_{\text{per device}} &= \frac{V_{CC}^2}{\pi^2 R_L} = \frac{900}{157.91} = 5.699~\text{W} \end{aligned}\]

\(P_{2Q,\max} = 11.40~\text{W}\) total, i.e. \(P_{D,\max} = 5.699~\text{W}\) in each transistor.

Part (c) — the 0.4 rule and the efficiency there

\[\begin{aligned} \frac{P_{2Q,\max}}{P_{o,\max}} &= \frac{2V_{CC}^2/\pi^2R_L}{V_{CC}^2/2R_L} = \frac{4}{\pi^2} = 0.40528 \approx 0.4 \\ 0.4\,P_{o,\max} &= 0.4 \times 28.125 = 11.25~\text{W} \quad\text{versus}\quad 11.399~\text{W} \end{aligned}\]

\[\begin{aligned} P_o\big|_{V_m = 2V_{CC}/\pi} &= \frac{(19.10)^2}{2(16)} = 11.399~\text{W} \\ P_{i(\text{dc})} &= (30)\frac{2(1.194)}{\pi} = 22.797~\text{W}, \qquad \eta = 50.00\% \end{aligned}\]

The rule is exact at \(4/\pi^2 = 0.4053\); efficiency at the thermal worst case is exactly 50.0\%.

Size the heat sink for 5.70 W per device, not for the 14.1 W each device handles at full output. A stage tested only at full power and at idle will pass and then fail in the field at half volume.

PROBLEM 05

Heat-Sink Design from Thermal Resistances

Problem Statement

Design. Each output transistor of the Class B stage above must dissipate its worst-case \(P_D = 5.699~\text{W}\). The device is rated \(T_{J,\max} = 150^\circ\)C, has \(\theta_{JC} = 1.5^\circ\)C/W, and is mounted with a mica washer and grease giving \(\theta_{CS} = 0.6^\circ\)C/W. The ambient is \(T_A = 40^\circ\)C.

In free air the device has \(\theta_{JA} = 40^\circ\)C/W. Find \(T_J\) and the maximum power it could dissipate unaided.
Specify the largest sink-to-ambient thermal resistance \(\theta_{SA}\) that will keep the junction inside its rating.
A stock extrusion with \(\theta_{SA} = 8^\circ\)C/W is fitted. Find the actual junction temperature, the margin, and the power the assembly could take.

Solution

\(P_D = 5.699~\text{W}\), \(T_{J,\max} = 150^\circ\)C, \(T_A = 40^\circ\)C
\(\theta_{JC} = 1.5^\circ\)C/W, \(\theta_{CS} = 0.6^\circ\)C/W
Thermal resistances in series add, exactly like a resistive divider driven by a current source \(P_D\): \(T_J = T_A + P_D\theta_{JA}\).

Part (a) — no heat sink

\[\begin{aligned} T_J &= T_A + P_D\theta_{JA} = 40 + (5.699~\text{W})(40^\circ\text{C/W}) = 268.0^\circ\text{C} \\ P_{\max}\big|_{\text{no sink}} &= \frac{T_{J,\max} - T_A}{\theta_{JA}} = \frac{110^\circ\text{C}}{40^\circ\text{C/W}} = 2.75~\text{W} \end{aligned}\]

\(T_J = 268^\circ\)C — far beyond the 150\(^\circ\)C rating. Unaided, the device can take only 2.75 W, well under the 5.70 W required. A sink is mandatory.

Part (b) — specifying \(\theta_{SA}\)

\[\begin{aligned} \theta_{JA}\big|_{\text{allowed}} &= \frac{T_{J,\max} - T_A}{P_D} = \frac{110}{5.699} = 19.30^\circ\text{C/W} \\ \theta_{JA} &= \theta_{JC} + \theta_{CS} + \theta_{SA} \\ \theta_{SA} &\le 19.30 - 1.5 - 0.6 = 17.20^\circ\text{C/W} \end{aligned}\]

Specify a heat sink with \(\theta_{SA} \le 17.2^\circ\)C/W.

Part (c) — with the 8 \(^\circ\)C/W extrusion

\[\begin{aligned} \theta_{JA} &= 1.5 + 0.6 + 8 = 10.1^\circ\text{C/W} \\ T_J &= 40 + (5.699)(10.1) = 97.56^\circ\text{C} \\ \text{margin} &= 150 - 97.56 = 52.44^\circ\text{C} \\ P_{\max} &= \frac{110}{10.1} = 10.89~\text{W} \end{aligned}\]

\(T_J = 97.6^\circ\)C, a margin of 52.4\(^\circ\)C; the mounting could handle 10.9 W.

Note where the resistance sits: \(\theta_{JC} + \theta_{CS} = 2.1\)\(^\circ\)C/W is fixed by the package and the washer, so the sink dominates. Also note that ambient enters directly — the same assembly inside a 65\(^\circ\)C enclosure would lose 2.48 W of capability.

PROBLEM 06

Second-Harmonic Distortion from the Output Waveform

Problem Statement

An oscilloscope trace of a single-ended power stage shows the collector current swinging between \(I_{\max} = 140~\text{mA}\) and \(I_{\min} = 20~\text{mA}\); the quiescent value measured with the signal removed is \(I_{CQ} = 75~\text{mA}\). The load is \(R_L = 8~\Omega\).

Find the fundamental and second-harmonic current amplitudes and the percentage second-harmonic distortion \(D_2\).
Find the fundamental-frequency output power.
Find the total output power including the harmonic and comment.

Solution

\(I_{\max} = 140~\text{mA}\), \(I_{\min} = 20~\text{mA}\), \(I_{CQ} = 75~\text{mA}\), \(R_L = 8~\Omega\)
Only the second harmonic is significant; the transfer characteristic is modelled as \(i_C = I_{CQ} + I_0 + I_1\cos\omega t + I_2\cos 2\omega t\).

Part (a) — harmonic amplitudes

Evaluating the assumed waveform at \(\omega t = 0,\,\pi/2,\,\pi\) and solving the three equations gives

\[\begin{aligned} I_1 &= \frac{I_{\max} - I_{\min}}{2}, \qquad I_2 = I_0 = \frac{I_{\max} + I_{\min} - 2I_{CQ}}{4} \\ D_2 &= \left|\frac{I_2}{I_1}\right| = \frac{\left|\tfrac{1}{2}(I_{\max}+I_{\min}) - I_{CQ}\right|}{I_{\max}-I_{\min}} \end{aligned}\]

\[\begin{aligned} I_1 &= \frac{140 - 20}{2} = 60.0~\text{mA} \\ \tfrac{1}{2}(I_{\max}+I_{\min}) &= \frac{140 + 20}{2} = 80.0~\text{mA} \\ D_2 &= \frac{|80.0 - 75|}{120} = \frac{5.00}{120} = 0.041667 = 4.167\% \\ I_2 &= D_2 I_1 = 0.041667 \times 60.0~\text{mA} = 2.50~\text{mA} \end{aligned}\]

\(I_1 = 60.0~\text{mA}\), \(I_2 = 2.50~\text{mA}\), \(D_2 = 4.17\%\)

Part (b) — fundamental power

\[P_1 = \frac{I_1^2 R_L}{2} = \frac{(0.06000~\text{A})^2(8~\Omega)}{2} = 0.01440~\text{W} = 14.4~\text{mW}\]

\(P_1 = 14.4~\text{mW}\)

Part (c) — total power

\[\begin{aligned} P &= (1 + D_2^2)P_1 = \left(1 + (0.041667)^2\right)(14.4~\text{mW}) \\ &= (1.00174)(14.4) = 14.425~\text{mW} \end{aligned}\]

\(P = 14.43~\text{mW}\), only 0.174\% above the fundamental power.

The distortion enters the power as \(D_2^2\), so a 4.17\% distortion adds a negligible 0.174\% to the delivered power — yet 4.17\% is plainly audible. Never judge distortion by a power measurement. The physical cause here is that \(\tfrac{1}{2}(I_{\max}+I_{\min}) \ne I_{CQ}\): the signal has shifted the average current, which is the signature of an asymmetric, second-order nonlinearity. A push-pull pair cancels exactly this term.