Question

An npn transistor operates as a common emitter amplifier, with a power gain of 60 dB. The input circuit resistance Is 100 Ω and the output load resistance is 10 kΩ. the common emitter current gain β is:
(A) 6 × 10²
(B) 10²
(C) 60 (D) 10⁴

Answer 1

Answer:

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Answer 2

An npn transistor operates as a common emitter amplifier, with a power gain of 60 dB. The input circuit resistance Is 100 Ω and the output load resistance is 10 kΩ. the common emitter current gain β is (B) 10²

• Given:

Power gain in dB : Ap$_{dB}$ = 60 dB

Input ciruit resistance : Ri = 100 Ω

Output load resistance : Rout = 10 KΩ = 10 × 10³ Ω

Current gain : β = ?

• Power gain in dB can be converted using formula:

$Ap_{dB} = 10\ log_{10} (Ap)\\\\60 = 10\ log_{10} (Ap)\\\\6 = log_{10} (Ap)\\\\Ap = 10^6$

• Since Power is defined as product of current to the voltage and power gain is defined as the ratio of output power to the input power;

Power gain = Voltage Gain × Current Gain

• Ap = Av × β   ...................(1)

Voltage gain (Av) in terms of Current gain can be given as

• $Av = \beta (\frac{Rout}{Ri})$ ..................(2)

Substituting (2) in (1), we get

$Ap = \beta (\frac{Rout}{Ri}) \times \beta \\\\Ap = \beta ^2(\frac{Rout}{Ri})\\\\ \beta ^2 = \frac{Ap}{(\frac{Rout}{Ri})}\\\\\beta = \sqrt{ \frac{Ap\times Ri}{Rout}}}$

Substituting given values, we get

$\beta = \sqrt{ \frac{10^6\times 100}{10\times 10^3}}} = \sqrt{10^4} = 10^2 = 100$

Current Gain = 100 = 10²