Question

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A CE amplifier is drawn by a voltage source of internal resistance of 500Ω and load impedance of 800 Ω .The h parameters hie = 2kΩ, hre = 2 X 10-4, hfe = 50, hoe = 25μA/V and compute Ai, Ri, AV and RO using exact analysis.​

Answer 1

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Answer 2

Ai = - 49.02

Ri = 2007.84 Ω

Av = - 19.53

Ro = 47.62 KΩ

Given: internal resistance = 500 Ω, load impedance = 800 Ω .

hie = 2 kΩ, hre = 2 X 10-4, hfe = 50, hoe = 25 μA / V.

To Find: Ai, Ri, AV, and RO using exact analysis.​

Solution:

From exact analysis, we know that Ai is the current gain, given by the formula,

             Ai = - hfe / ( 1 + hoe × RL )

Putting respective values, we get,

            Ai  = - 50 / ( 1 + 25 × 10^-6 × 800 )

                  = - 49.02

Again, the input resistance is said to be Ri and the formula is

            Ri = hie - (( - hfe × hre ) / ( hoe + 1 / RL ))

Putting respective values in the formula, we get,

            Ri = 2000 - (( - 50 × 2 × 10^-4 ) / ( 25 × 10^-6 + 1 / 800 ))

                = 2000 + 7.84 Ω

                = 2007.84 Ω

Again, voltage gain is termed as Av which is given by the formula,

            Av = Ai × ( RL / Ri )

Putting respective values in the formula, we get,

            Av = ( - 49.02 ) × ( 800 / 2007.84 )

                 = - 19.53

Lastly, output resistance is denoted by Ro, given by the formula,

            Ro = hoe - ( hfe × hre / ( hie + Rs )

Putting respective values in the formula, we get,

            Yo = 25 × 10^-6 - ( 50 × 2 × 10^-4 / ( 2000 + 500 )

                  = 2.1 × 10^-5 mho

now,    Ro = 1/Yo

                  = 1 / 2.1 × 10^-5

                  = 47.62 KΩ

compiling all the answers we get,

Ai = - 49.02

Ri = 2007.84 Ω

Av = - 19.53

Ro = 47.62 KΩ

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