--- title: "Design class B amplifier 20 W into 8 Ω" subject: "Electronics Devices and Circuit" chapter: 4 key: "semester_3/electronics_devices_and_circuit:class-b-design-20w-8ohm" source: https://www.bctengineer.com/q/semester_3/electronics_devices_and_circuit/class-b-design-20w-8ohm site: BCT Engineering --- # Design class B amplifier 20 W into 8 Ω Electronics Devices and Circuit · Chapter 4: Output Stages and Power Amplifiers - Asked in: 2079 Baisakh, 1+1+2+2 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/electronics_devices_and_circuit/class-b-design-20w-8ohm - Whole chapter: https://www.bctengineer.com/md/semester_3/electronics_devices_and_circuit/chapter-4 ## Question 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. ## 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. ## Answer **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.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the required supply voltage found for 20 W into 8 Ω with 5 V headroom? A. Vp = sqrt(2 Po RL) = 17.9 V, so VCC = Vp + 5 = about 23 V B. Vp = sqrt(Po RL) = 12.6 V, so VCC = 17.6 V C. Vp = 2 Po RL = 320 V D. Vp = Po/RL = 2.5 V, so VCC = 7.5 V Answer: A. Po = Vp^2/(2RL) gives Vp = sqrt(320) = 17.89 V; adding 5 V gives about 23 V. (Key point: Peak voltage and supply) 2. What is the total supply power? A. 20 W B. 32.7 W C. 45.8 W D. 26.2 W Answer: B. Pi = 2 VCC Im/π = 2 x 23 x 2.236/π = 32.7 W. (Key point: Total supply power) 3. Why is the efficiency 61.1% rather than 78.5%? A. The load is too small B. The transistors are class A C. Crossover distortion wastes power D. Vp is less than VCC because of the 5 V headroom Answer: D. Efficiency = (π/4)(Vp/VCC) = 0.785 x 17.89/23, below the ideal value reached only at Vp = VCC. (Key point: Efficiency reduced by headroom) 4. At what output peak voltage is transistor dissipation maximum, and what must each transistor handle? A. Vp = VCC, 6.7 W B. Vp = 2VCC/π = 14.6 V, 6.7 W per transistor C. Vp = 14.6 V, 13.4 W per transistor D. Vp = 0, 32.7 W Answer: B. Setting dP/dVp = 0 gives Vp = 2VCC/π; total loss is 2VCC^2/(π^2 RL) = 13.4 W, so each transistor handles 6.7 W. 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