Skip to main content

Class B power and efficiency, 14 V peak, 16 Ω

Short answer

Output power Po = Vp²/(2RL) = 14²/(2×16) = 6.125 W; input power Pi = 2 Vcc Vp/(π RL) = 2×24×14/(π×16) ≈ 13.37 W; efficiency η = (π/4)(Vp/Vcc) = 45.8%.

For a class B amplifier providing a 14V peak signal to 16 Ω load and a power supply of Vcc = 24V, determine input power, output power and circuit efficiency.

Given: Class B push-pull, peak output voltage Vp = 14 V, load RL = 16 Ω, VCC = 24 V.

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 = 14/16 = 0.875 A
  • Output power Po = 14² / (2 × 16) = 6.125 W
  • Input power Pi = 2 × 24 × 14 / (π × 16) = 13.37 W
  • Efficiency η = 6.125 / 13.37 = 45.8 %

Check with η = (π/4)(Vp/VCC) = 0.7854 × 14/24 = 45.8 %.

The power lost as heat in the two transistors is Pi − Po = 7.24 W.

Answer: Pi = 13.37 W, Po = 6.125 W, η = 45.8 %.

✓ Answer checked by Claude Opus 5.5 (high effort)

Related questions

All Electronics Devices and Circuit chapter 4 questions →