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The value of emitter resistance in Emitter Biased circuit are RE1=25kΩ & RE2=16kΩ. Find RE

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in Linear Integrated Circuits by (38.2k points)
The value of emitter resistance in Emitter Biased circuit are RE1=25kΩ & RE2=16kΩ. Find RE

(a) 9.756kΩ

(b) 41kΩ

(c) 9.723kΩ

(d) 10kΩ

This question was posed to me in homework.

Question is taken from Differential Amplifier and Circuit Configuration topic in portion Operational Amplifier of Linear Integrated Circuits

1 Answer

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by (39.8k points)
Right choice is (a) 9.756kΩ

Explore the finest explanation: In emitter biased circuit, RE1 & RE2 is connected in parallel combination.

⇒ RE = RE1 II RE2 = (RE1× RE2)/(RE1+RE2) = (25kΩ×16kΩ)/(25kΩ+16kΩ) = 9.7561kΩ.

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