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Question: In a triode,\(g_{m} = 2 \times 10^{- 3}ohm^{- 1};\mu = 42\), resistance load, \(R = 50\) kilo ohm. T...

In a triode,gm=2×10−3ohm−1;μ=42g_{m} = 2 \times 10^{- 3}ohm^{- 1};\mu = 42, resistance load, R=50R = 50 kilo ohm. The voltage amplification obtained from this triode will be

A

30.42

B

29.57

C

28.18

D

27.15

Answer

29.57

Explanation

Solution

Voltage gain

Av=μ1+rpRLA_{v} = \frac{\mu}{1 + \frac{r_{p}}{R_{L}}} and μ=rp×gm\mu = r_{p} \times g_{m}

⇒rp=422×10−3=21000Ωr_{p} = \frac{42}{2 \times 10^{- 3}} = 21000\Omega⇒Av=421+2100050×103=29.57A_{v} = \frac{42}{1 + \frac{21000}{50 \times 10^{3}}} = 29.57