How is the required supply voltage found for 20 W into 8 Ω with 5 V headroom?

Vp = sqrt(2 Po RL) = 17.9 V, so VCC = Vp + 5 = about 23 V

Vp = sqrt(Po RL) = 12.6 V, so VCC = 17.6 V

Vp = Po/RL = 2.5 V, so VCC = 7.5 V

Why is the efficiency 61.1% rather than 78.5%?

Vp is less than VCC because of the 5 V headroom

Crossover distortion wastes power

At what output peak voltage is transistor dissipation maximum, and what must each transistor handle?

Vp = 2VCC/π = 14.6 V, 6.7 W per transistor

Vp = 14.6 V, 13.4 W per transistor

2079 Baisakh · 1+1+2+2 marks

Design class B amplifier 20 W into 8 Ω

Short answer

Peak output voltage Vp = √(2×20×8) = 17.89 V, supply VCC = 17.89+5 ≈ 23 V. Peak current Im = 17.89/8 = 2.24 A. DC input power Pi = 2×23×17.89/(π×8) ≈ 32.7 W. Efficiency η = 20/32.7 ≈ 61.1%. Maximum dissipation per transistor = VCC²/(π²RL) = 23²/(π²×8) ≈ 6.7 W.

It is required to design a class B power Amplifier to deliver an average power of 20 W to an 8 Ω load. The power supply is to be selected such that Vcc is about 5 V greater than the peak output voltage. This avoids transistor saturation and associated nonlinear distortion, and allows for including short circuit protection circuitry. Determine the supply voltage required, the peak current drawn from each supply, the total supply power, and the power conversion efficiency. Also determine the maximum power that each transistor must be able to dissipate safely.

Given: Class B push-pull, Po = 20 W average, RL = 8 Ω, VCC = Vp + 5 V.

1. Supply voltage

Po = Vp²/(2RL) → Vp = √(2 Po RL) = √(2 × 20 × 8) = √320 = 17.89 V

VCC = Vp + 5 = 22.89 V ≈ 23 V (that is, ±23 V supplies; 23 V is used in the calculations below)

2. Peak current from each supply

Im = Vp/RL = 17.89 / 8 = 2.24 A

3. Total supply power

Each supply gives an average current of Im/π = 0.712 A, so the total average current drawn from the two supplies is 1.424 A.

Pi = 2 VCC Im/π = 2 × 23 × 2.236 / π = 32.7 W

4. Efficiency

η = Po/Pi = 20 / 32.7 = 61.1 %

(An ideal class B stage would reach 78.5 % with Vp = VCC; the 5 V headroom reduces the efficiency.)

5. Maximum dissipation of each transistor

The total power dissipated by the two transistors is P2Q = Pi − Po = 2 VCC Vp/(π RL) − Vp²/(2RL). It is maximum when dP2Q/dVp = 0:

Vp = 2VCC/π = 2 × 23/π = 14.6 V

P2Q(max) = 2 VCC²/(π² RL) = 2 × 23² / (π² × 8) = 13.4 W

Each transistor dissipates half of this:

PDmax per transistor = VCC²/(π² RL) = 529 / (9.87 × 8) = 6.70 W

Since 14.6 V is below the available peak of 17.89 V, this worst case really occurs. Each transistor must be rated for at least about 6.7 W (in practice, with a safety margin, 10 W or more, mounted on a heat sink).

Answer: VCC ≈ 23 V; Im = 2.24 A; Pi = 32.7 W; η = 61.1 %; PD(max) per transistor = 6.7 W.

✓ Answer checked by Claude Opus 5.5 (high effort)

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