Crossover Distortion and Thermal Runaway in a Class-B Amplifier
A student builds a 10 W class-B audio power amplifier for the school science exhibition. The output stage uses a complementary pair of 2N3055/MJ2955 transistors on a 100 mm × 75 mm heatsink, driven by a TIP31 driver. The design has no emitter resistors, and the bias network is a fixed 1 kΩ preset plus two 1N4148 diodes. On the bench, with ±18 V rails and an 8 Ω load, a 1 kHz sine input produces an output that is clean except for a flat step of about 0.65 V at every zero crossing. When the student adjusts the preset to remove the step, the output becomes clean for about 30 seconds, after which the quiescent current climbs from 25 mA to over 400 mA and one output transistor becomes very hot. The heatsink is shared, but the bias diodes are mounted on the PCB 40 mm away. The student has 0.22 Ω/5 W resistors, a BD139 transistor, and mica insulators available.
1. Identify the distortion seen at the zero crossing and explain why the adjustment that removed it caused the current to run away. Refer to the relevant device behaviour and the bias network.
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The flat step is crossover distortion. Near zero output both output transistors are essentially off until the input exceeds about 0.6–0.7 V, the base-emitter turn-on voltage of each device; the result is a dead band of roughly 1.2–1.4 V at the input, seen as a step at the output. The preset was increasing the forward bias so both devices conduct slightly at idle. That removes the dead band, but the fixed diode network does not track the output devices. As the output transistors heat, their Vbe falls by about 2 mV/°C per junction; for a 30 °C rise the bias needed falls by roughly 60 mV per device. The fixed bias voltage therefore overdrives them, and with no emitter resistors there is no local negative feedback to oppose the increase. Collector current rises exponentially with Vbe, so 25 mA can become 400 mA in seconds. The diodes being 40 mm away on the PCB also sense board temperature, not the heatsink, so the correction loop is slow and unstable.
2. Using the available parts, describe a modified bias and emitter arrangement, and calculate a safe quiescent setting. Explain how it prevents the same failure.
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Add 0.22 Ω/5 W emitter resistors in series with each output emitter and replace the two diodes with a Vbe multiplier using the BD139, mounted on the heatsink with mica insulation and thermal compound. The multiplier sets Vbias = Vbe(BD139)(1 + R1/R2). With R1 = 560 Ω and R2 = 1 kΩ, Vbias ≈ 0.65 × (1 + 0.56) ≈ 1.0 V; a preset should be included so the voltage across the two base-emitter junctions can be set to about 1.2 V. Set quiescent current by measuring the voltage across one 0.22 Ω resistor: 10 mV corresponds to 45 mA. At 45 mA the resistor drop is only 10 mV, but at 400 mA it is 88 mV, which reduces Vbe and provides immediate negative feedback. The BD139 on the heatsink senses the real temperature, so the bias voltage falls as the outputs heat. Check dissipation: with ±18 V and an 8 Ω load, each device may dissipate 8–12 W on programme peaks, so the 100 × 75 mm heatsink and mica insulators are needed; verify the tab is isolated from the heatsink before powering up.