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Chapter 20: Problem 54

A common emitter amplifier is designed with NPN transistor \((\alpha=0.99)\). The input impedance is \(1 \mathrm{k} \Omega\) and load is \(10 \mathrm{k} \Omega\). The voltage gain will be (A) \(9.9\) (B) 99 (C) 990 (D) 9900

Short Answer

Expert verified
The voltage gain of the common emitter amplifier with given parameters is -9.9. Therefore, the correct answer is (A) 9.9.

Step by step solution

01

Write the expression for the voltage gain

The voltage gain (Av) of a common emitter amplifier is given by: \(Av = \frac{- \alpha RL}{Ri}\)
02

Plug in the given parameters into the expression

Now, let's plug in the given parameters into the expression for the voltage gain: 伪 = 0.99 (current gain) Ri = 1k惟 (input impedance) RL = 10k惟 (load) \(Av = \frac{- (0.99)(10k\Omega)}{1k\Omega}\)
03

Calculate the voltage gain

Now, we will perform the calculation to find the voltage gain: \(Av = \frac{- (9.9 k\Omega)}{1 k\Omega}\) Av = -9.9
04

Select the correct option

Since the magnitude of the calculated voltage gain is 9.9, we can infer that the correct answer is (A) \(9.9\) (with a negative sign to denote an inverted output).

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