Consider a Darlington amplifier. In the self bias network, the biasing resistances are 220k? and 400 k?. What can be the correct value of input resistance if hfe=50 and emitter resistance = 10k?.
A)141 k?
B)15 m?
C)20 m?
D)200 k?

Question

Quizplus · Accepted Answer

The input resistance of a Darlington amplifier can be approximated as Rin = (2 * hfe^2 * Re) + (R1 || R2), where Re is the emitter resistance, R1 and R2 are the biasing resistances, and || denotes parallel resistance. Given hfe = 50 and Re = 10kΩ, the first term 2 * (50^2) * 10kΩ = 2 * 2500 * 10kΩ = 50MΩ. The parallel resistance of R1 = 220kΩ and R2 = 400kΩ is approximately 146.67kΩ. Adding these, we get a value much larger than the options provided, indicating a mistake in the calculation. The correct approach for a Darlington configuration is to consider the effective input resistance seen at the base of the first transistor, which is significantly increased by the hfe of both transistors in the pair. The correct formula should consider the base-emitter resistance and the effect of hfe on the emitter resistance, leading to a significantly lower value than calculated by the incorrect formula. Given the options, the closest realistic value for a Darlington amplifier's input resistance, considering the high current gain and the provided resistances, would be in the range of the options given, with 141kΩ being a plausible approximation based on typical Darlington input resistance characteristics.

Consider a Darlington Amplifier