Class A Output Stage: perfect linearity, terrible efficiency
The transistor conducts through all 360° — and burns power even with no signal at all.
In a Class A stage the transistor is biased at the middle of its load line and never switches off. That guarantees the output follows the input faithfully, because the device never leaves its active region. But it also means full quiescent current flows continuously, whether you're playing music or silence — and that idle current is where the efficiency goes.
Why 25% is the ceiling
For a series-fed (resistive load) Class A stage:
Quiescent point at mid-supply for maximum swing:
V_CEQ = V_CC/2, I_CQ = V_CC/(2R_L)
DC power drawn from the supply (constant!):
P_dc = V_CC × I_CQ = V_CC²/(2R_L)
Maximum AC output power (peak swing = V_CC/2):
P_out(max) = (V_CC/2)²/(2R_L) = V_CC²/(8R_L)
Efficiency:
η = P_out/P_dc = [V_CC²/8R_L] / [V_CC²/2R_L]
= 2/8 = 0.25 = 25% ∎
Transformer-coupled version reaches 50%, because the
transformer's DC resistance is nearly zero so no supply
voltage is wasted across it.
The critical property: P_dc is constant and independent of the signal. With zero input the stage still draws V_CC × I_CQ from the supply and turns all of it into heat. That's why the transistor runs hottest when idle — the exact opposite of what intuition suggests, and a classic exam question.
Worked numerical 1 — full Class A analysis
A series-fed Class A stage has V_CC = 20 V, R_L = 50 Ω, biased at mid-point. Find I_CQ, P_dc, maximum P_out, efficiency, and the transistor dissipation at idle and at full output.
Q-point for maximum swing:
V_CEQ = V_CC/2 = 10 V
I_CQ = V_CC/(2R_L) = 20/100 = 0.2 A = 200 mA
DC power drawn (constant):
P_dc = V_CC × I_CQ = 20 × 0.2 = 4 W
Maximum output power:
peak output swing = 10 V, so V_rms = 10/√2 = 7.07 V
P_out = V_rms²/R_L = 50/50 = 1 W
(or V_CC²/8R_L = 400/400 = 1 W ✔)
Efficiency:
η = 1/4 = 25% ✔
Transistor dissipation:
At IDLE (no signal):
P_transistor = P_dc − P_out = 4 − 0 = 4 W ← maximum!
At FULL output:
P_transistor = 4 − 1 = 3 W
The transistor must be rated for 4 W, i.e. FOUR TIMES the
useful output power — sizing it from the 1 W output would
destroy it during silence.
Worked numerical 2 — transformer-coupled Class A
A transformer-coupled Class A stage runs from V_CC = 12 V with a reflected load of 8 Ω and I_CQ = 1 A. Find P_dc, maximum P_out and efficiency.
With a transformer, the DC drop across the primary ≈ 0,
so the collector sits at V_CC and can swing from
0 to 2V_CC (the transformer allows the "flyback" above V_CC):
V_CEQ = V_CC = 12 V
peak swing = V_CC = 12 V
P_dc = V_CC × I_CQ = 12 × 1 = 12 W
P_out(max) = (V_CC/√2)²/R_L' where R_L' is the reflected load
Actually with peak swing V_CC and peak current I_CQ:
P_out = (V_CC × I_CQ)/2 = (12 × 1)/2 = 6 W
η = 6/12 = 50% ✔ double the series-fed case
Transistor dissipation at idle = 12 W (still the worst case)
Required V_CE rating = 2V_CC = 24 V ← the transformer
flyback doubles the voltage stress. Overlooking this
destroys transistors in practice.
Worked numerical 3 — heatsink sizing
The 4 W Class A stage above uses a transistor with maximum junction temperature 150 °C and junction-to-case thermal resistance 2 °C/W, in a 40 °C ambient. What heatsink is needed?
Total allowable thermal resistance:
θ_total = (T_j(max) − T_ambient)/P
= (150 − 40)/4
= 27.5 °C/W
This budget splits between three resistances:
θ_total = θ_jc + θ_cs + θ_sa
θ_jc (junction–case) = 2 °C/W (given)
θ_cs (case–sink) ≈ 0.5 °C/W (with thermal paste)
θ_sa (sink–ambient) = ?
θ_sa = 27.5 − 2 − 0.5 = 25 °C/W
A small clip-on heatsink (~25 °C/W) suffices here.
Compare: the same 1 W output from a Class B stage would
dissipate only ~0.2 W, needing NO heatsink at all. That is
the practical consequence of the efficiency difference.
💡 Exam angle: the 25% derivation is a standard 5-mark question — set the Q-point at mid-supply, write P_dc and P_out(max), divide. The conceptual mark that most students miss: state that maximum transistor dissipation occurs at zero signal, and that the device must be rated for the full P_dc. Also remember the transformer-coupled version gives 50% but requires a 2V_CC voltage rating.
Syllabus points
Operation & efficiency (numerical)
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