- 2082 Baisakh · 3+3 marks
Class A series-fed efficiency, 12 V supply
Short answer
Efficiency η = (VCE(p-p)²)/(4VCC²) = 25% × (VCEpp/VCC)². (a) For 12 V p-p output, η = 25%. (b) For 6 V p-p, η = 6.25%.
Calculate the efficiency of a class A series-fed amplifier for a supply of 12 V and outputs of: (a) V(P-P) = 12 V and (b) V(P-P) = 6 V.
Given: VCC = 12 V, series-fed class A stage.
Assumption: The Q point is set at the centre of the load line: VCEQ = VCC/2 = 6 V, ICQ = VCC/(2RC). The load resistance RC is not given, but it cancels in the efficiency.
Formulas
Pi = VCC · ICQ = VCC²/(2RC)
Po = VCE(p-p)² / (8 RC) (output power for a sine swing of peak-to-peak value VCE(p-p))
η = Po/Pi = VCE(p-p)² / (4 VCC²) = 25 % × (VCE(p-p)/VCC)²
(a) VCE(p-p) = 12 V
η = 25 × (12/12)² = 25 %
(Steps: Po/Pi = 144 / (4 × 144) = 0.25.)
This is the largest possible swing (full supply range 0 V to 12 V), so it is the maximum efficiency of a series-fed class A amplifier.
(b) VCE(p-p) = 6 V
η = 25 × (6/12)² = 25 × 0.25 = 6.25 %
(Steps: Po/Pi = 36 / (4 × 144) = 0.0625.)
Comparison
| VCE(p-p) | VCE(p-p)/VCC | η |
|---|---|---|
| 12 V | 1.0 | 25 % |
| 6 V | 0.5 | 6.25 % |
Halving the swing reduces the output power to one quarter, while the input power stays constant, so the efficiency falls to one quarter.
Answer: (a) 25 %, (b) 6.25 %.
✓ Answer checked by Claude Opus 5.5 (high effort)
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