Class B Output Stage: two transistors sharing the work
78.5% efficient — but there's a notch in the middle of the waveform.
Class A wastes three-quarters of its supply power because the transistor never switches off. So switch it off: bias it exactly at cutoff, and it conducts only for the positive half-cycle. Now use a second transistor for the negative half. Together they reconstruct the whole waveform, and neither draws any current at idle. The efficiency jumps to 78.5% — but a new problem appears exactly where the two hand over.
Push-pull operation
η = (π/4)(V_p / V_CC) × 100% With Vp = 14 V, Vcc = 15 V, RL = 8 Ω, eta = 73.3 %, Pout = 12.25 W, Pd = 4.46 W.
The famous 78.5% happens only when the output swings to the full supply. Back the drive off to half and the efficiency halves too — so a Class B stage run at ordinary listening levels is far less efficient than the headline figure suggests.
Complementary pair (one NPN, one PNP):
Positive half-cycle → NPN conducts, PNP off
Negative half-cycle → PNP conducts, NPN off
At zero signal → BOTH off, zero current drawn
Each device handles 180°; together they cover 360°.
Deriving the 78.5%
For a complementary push-pull stage on ±V_CC:
Peak output voltage: V_p (up to V_CC)
Peak load current: I_p = V_p/R_L
Output power:
P_out = V_p²/(2R_L) (RMS² / R)
DC power drawn — the average of a half-sine of current
from each supply:
I_dc(avg) = I_p/π per supply
P_dc = 2 × V_CC × I_p/π = 2V_CC·V_p/(πR_L)
Efficiency:
η = P_out/P_dc = [V_p²/2R_L] × [πR_L/(2V_CC V_p)]
= πV_p/(4V_CC)
At maximum swing V_p = V_CC:
η_max = π/4 = 0.785 = 78.5% ∎
Notice η = πV_p/(4V_CC) depends on the signal level. Unlike Class A, a Class B stage's efficiency rises with output. At half swing (V_p = V_CC/2), η = π/8 = 39% — so Class B is only efficient when driven hard. This is also why its worst-case transistor dissipation occurs at about 63% of maximum output, not at full output.
Crossover distortion — the price
Silicon needs about 0.7 V of base-emitter voltage before it conducts at all. So around the zero-crossing, when the input is between −0.7 V and +0.7 V, neither transistor is on. The output sits at zero, producing a flat notch — a "dead band" of about 1.4 V total.
⚔️ Class A vs Class B
EfficiencyA: 25% fixed. B: up to 78.5%, rising with signal level. B draws no idle current, which is decisive for battery equipment.
DistortionA: very low, smooth. B: crossover distortion at every zero crossing — worst for quiet signals, since the 1.4 V dead band is a larger fraction of a small waveform.
Worst-case heatA: at idle. B: at about 63% of full output (not at maximum). Design the heatsink for that point.
DevicesA: one transistor. B: a matched complementary pair, which must be well matched or the two half-cycles differ.
Worked numerical 1 — complete Class B analysis
A complementary Class B stage runs on ±15 V with R_L = 8 Ω. Find maximum output power, P_dc, efficiency, and the maximum dissipation per transistor.
Maximum output (peak swing = V_CC = 15 V):
P_out(max) = V_p²/(2R_L) = 15²/(2 × 8)
= 225/16 = 14.06 W
Peak current:
I_p = V_p/R_L = 15/8 = 1.875 A
DC power at maximum output:
P_dc = 2V_CC·I_p/π = 2 × 15 × 1.875/3.1416
= 56.25/3.1416 = 17.90 W
Efficiency:
η = 14.06/17.90 = 0.785 = 78.5% ✔
Maximum dissipation per transistor — occurs at
V_p = 2V_CC/π = 0.637 × 15 = 9.55 V (about 63% of max):
P_d(max per device) = V_CC²/(π²R_L)
= 225/(9.87 × 8) = 2.85 W
Total device dissipation at that point = 5.70 W
Compare Class A delivering the same 14 W:
P_dc = 14/0.25 = 56 W, heat = 42 W
Class B: heat = 5.7 W → 7× less heat
Worked numerical 2 — efficiency at partial output
For the same stage, find the efficiency at 1 W output (typical listening level).
From P_out = V_p²/(2R_L):
1 = V_p²/16 → V_p = 4 V
η = πV_p/(4V_CC) = π × 4/(4 × 15)
= 12.566/60 = 0.209 = 20.9%
P_dc = 1/0.209 = 4.78 W, so heat = 3.78 W
At 1 W output the stage is only 21% efficient — barely
better than Class A. Class B's advantage only materialises
near full power. (Class A at 1 W from this supply would
still draw its full quiescent power, so it remains worse.)
Worked numerical 3 — crossover distortion magnitude
Estimate the distortion introduced by the 1.4 V dead band for a 10 V peak signal and a 0.5 V peak signal.
Dead band = ±0.7 V, total 1.4 V
For a 10 V peak signal:
affected fraction = 1.4/(2 × 10) = 7%
Audible, but modest — and negative feedback around the
stage reduces it substantially.
For a 0.5 V peak signal:
the entire signal (±0.5 V) lies INSIDE the ±0.7 V dead
band → output is ZERO. The signal vanishes completely.
This is why Class B is unusable for audio without
modification: quiet passages disappear entirely. The fix
is Class AB, biasing both devices just into conduction so
there is no dead band at all.
💡 Exam angle: the η = 78.5% derivation is a standard 6–8 mark question — the step students get wrong is the average current I_p/π per supply. Then define crossover distortion and explain the 0.7 V cause. The high-value detail: maximum transistor dissipation is V_CC²/(π²R_L) occurring at 63% of full output, not at maximum output. Examiners specifically probe that.