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The gain factor of an amplifier is incre...

The gain factor of an amplifier is increased from 10 to 12 as the load resistance is changed from `4 kOmega` to `8 kOmega`. Calculate (a) the amplification factor and (b) the plate resistance.

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Voltage gain = ` (mu/(1+r_p/R_L))`
` When A = 10 `
` R_L = 4K Omega `
10 = `(mu /(1+ r_p/4 xx (10^3)))`
` rArr 10 = ((mu xx 4 xx (10^3))/(4 xx (10^3) xx r_p))`
` 40 xx (10^3) + 10r_p = 4 xx (10^3) mu`
` When A = 12 ……….(1) `
` R_L = 8K Omega `
` 12= (mu/(1+(r_p/8 xx(10^3))))`
` 12 = ((mu xx 8 xx (10^3))/(8 xx (10^3) xx r_p))`
` = 96 xx (10^3) xx 12r_p`
` = 8 xx (10^3)mu` .........(2)
Multiplying (2) in equation (1) and
equating with equation (2)
`rArr 2(40 x(10^3)+10r_p) = 96 xx (10^3) + 12r_p`
`rArr 80 xx (10^3) + 20r_p = 96 xx (10^3) + 12r_p `
` rArr 8r_p = 16 xx (10^3) `
` rArr r_p = 2 xx (10^3) Omega = 2K Omega `
Putting the value in equation (1)
` 40 x (10^3) + 10(2 xx(10^3)) = 4 xx (10^3)mu `
` rArr (40 xx (10^3) + 20 xx (10^3)) = 4 xx (10^3)mu `
` rArr 60 = 4 mu `
` rArr mu = 60/4 = 15.`
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