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Class B efficiency, 20 V peak into 8 Ω, 25 V supply

Short answer

Output power Po = 20²/(2×8) = 25 W; input power Pi = 2×25×20/(π×8) ≈ 39.79 W; efficiency η = (π/4)(20/25) = 62.8%.

A class B audio amplifier is providing 20V peak sine wave signal to 8Ω speaker with power supply of 25V (=VCC). At what efficiency is it operating?

Given: Class B push-pull, peak output voltage Vp = 20 V, load RL = 8 Ω, VCC = 25 V. The amplifier drives an 8 Ω speaker.

Formulas

  • Peak load current: Im = Vp/RL
  • Output power: Po = Vp²/(2RL)
  • Input (dc) power: Pi = VCC · Idc = 2 VCC Vp/(π RL), because each supply gives an average current Im/π per half-cycle, and the two together give 2Im/π
  • Efficiency: η = Po/Pi × 100 %

Calculation

  • Im = 20/8 = 2.500 A
  • Output power Po = 20² / (2 × 8) = 25.00 W
  • Input power Pi = 2 × 25 × 20 / (π × 8) = 39.79 W
  • Efficiency η = 25.00 / 39.79 = 62.8 %

Check with η = (π/4)(Vp/VCC) = 0.7854 × 20/25 = 62.8 %.

The power lost as heat in the two transistors is Pi − Po = 14.79 W.

Answer: Pi = 39.79 W, Po = 25.00 W, η = 62.8 %.

✓ Answer checked by Claude Opus 5.5 (high effort)

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