Add 1.4 V of bias to a Class B stage and the notch disappears. That's the whole idea.
Class B is efficient but has a dead band. Class A has no dead band but wastes power. The fix is almost embarrassingly simple. Bias both output transistors just barely into conduction. Now at the zero crossing both are slightly on, so one takes over smoothly as the other fades. You give up a little efficiency for a small quiescent current, and crossover distortion vanishes.
The bias requirement
Need ≈ 1.4 V between the two bases
(2 × V_BE for a silicon complementary pair)
Conduction angle: slightly more than 180°
Quiescent current: small but non-zero (typically 1–50 mA)
Efficiency: ~50–70% (between Class A's 25% and B's 78.5%)
Three ways to generate that bias
🔧 Biasing methods, worst to best
Resistor dividerSimplest — two resistors set ~1.4 V. But V_BE falls about 2 mV/°C as the transistors warm, while the resistor bias stays fixed. The quiescent current therefore rises with temperature, causing more heating: thermal runaway.
Diode biasTwo forward-biased diodes give 1.4 V. Crucially, the diodes' forward voltage also falls ~2 mV/°C, so if they are thermally bonded to the heatsink they track the transistors and cancel the drift. This is the classic, reliable solution.
V_BE multiplierA transistor with a resistor divider across its base–emitter produces V = V_BE(1 + R₂/R₁) — an adjustable bias, trimmed with a potentiometer. Used in quality audio amplifiers where quiescent current must be set precisely.
The reason diode bias works is matched temperature coefficients. Both a silicon diode and a silicon base–emitter junction lose about 2 mV/°C. Mount the diodes on the same heatsink as the output devices and the bias voltage falls exactly as fast as the required V_BE — the quiescent current stays put. A resistor divider cannot do this, because a resistor's value doesn't fall with temperature.
Worked numerical 1 — V_BE multiplier design
Design a V_BE multiplier to give 1.4 V bias, given V_BE = 0.7 V. Then find the bias if the transistor warms by 30 °C.
V_bias = V_BE(1 + R₂/R₁)
1.4 = 0.7(1 + R₂/R₁)
2.0 = 1 + R₂/R₁
R₂/R₁ = 1.0 → R₂ = R₁
Choose R₁ = R₂ = 1 kΩ, giving V_bias = 1.4 V ✔
After a 30 °C rise, V_BE falls by 2 mV/°C × 30 = 60 mV:
V_BE = 0.7 − 0.06 = 0.64 V
V_bias = 0.64 × 2 = 1.28 V
Meanwhile the output transistors also need 60 mV less each,
so their requirement drops from 1.4 V to 1.28 V.
Bias supplied: 1.28 V. Bias needed: 1.28 V. ✔ Tracked.
With a fixed resistor divider instead, V_bias would have
stayed at 1.4 V while only 1.28 V was needed — the extra
120 mV of overdrive would raise quiescent current sharply
(V_BE is exponential), heating the devices further.
Worked numerical 2 — Class AB efficiency
A Class AB stage on ±20 V into 8 Ω has a quiescent current of 25 mA. Find the idle dissipation, maximum output power, and efficiency at full output.
Idle dissipation (both rails supply 25 mA):
P_idle = 2 × V_CC × I_Q = 2 × 20 × 0.025 = 1.0 W
(Class B would be 0 W; Class A at this output level
would be tens of watts)
Maximum output:
P_out = V_p²/(2R_L) = 20²/16 = 25 W
P_dc at maximum ≈ 2V_CC I_p/π + P_idle
I_p = 20/8 = 2.5 A
P_dc = 2 × 20 × 2.5/π + 1.0
= 31.83 + 1.0 = 32.83 W
Efficiency at full output:
η = 25/32.83 = 76.2%
Nearly Class B's 78.5%, but with NO crossover distortion.
The 1 W idle cost buys a completely clean waveform — which
is why essentially every audio power amplifier built since
the 1970s is Class AB.
Worked numerical 3 — thermal runaway check
A Class AB output transistor has θ_ja = 5 °C/W and dissipates 3 W at idle in a 35 °C ambient. If quiescent current rises 8% per 10 °C of junction temperature, does it run away?
Initial junction temperature:
T_j = T_a + P × θ_ja = 35 + 3 × 5 = 50 °C
→ 15 °C above ambient
That 15 °C rise increases I_Q by:
15/10 × 8% = 12%
New dissipation = 3 × 1.12 = 3.36 W
Second iteration:
T_j = 35 + 3.36 × 5 = 51.8 °C (+16.8 °C)
I_Q rise = 13.4% → P = 3.40 W
Third iteration:
T_j = 35 + 3.40 × 5 = 52.0 °C → P = 3.41 W
The sequence CONVERGES (3.00 → 3.36 → 3.40 → 3.41 W).
Stable — no runaway.
If instead I_Q rose 40% per 10 °C (poor bias tracking):
T_j = 50 °C → P = 3 × 1.6 = 4.8 W
T_j = 59 °C → P = 3 × 1.96 = 5.88 W
T_j = 64 °C → P = 3 × 2.16 = 6.5 W … DIVERGES
Thermal runaway — the device destroys itself. Bias tracking
is what keeps the loop gain of this thermal feedback below 1.
💡 Exam angle: two questions come up repeatedly. "How is crossover distortion eliminated?" — bias both devices slightly on, which is Class AB. "Why use diodes rather than resistors?" — their matched −2 mV/°C tracking prevents thermal runaway. The V_BE multiplier formula V_BE(1 + R₂/R₁) is worth memorising for design questions.
Syllabus points
Class AB operation
Biasing the Class AB stage
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