Question

The gain factor of an amplifier is increased from 10 to 12 as te load resistance si changed from $4k\Omega$ to $8k\Omega$ Calculate (a) the amplification factor and (b) the plate resistance.

Answer 1

Voltage gain $=\frac{\mu }{1+\frac{r_p}{R_L}}$
When A = 10
$R_L=4K\Omega 10=\frac{\mu }{1+\frac{r_p}{4\times {10}^3}}\Rightarrow 10=\frac{\mu \times 4\times {10}^3}{4\times {10}^3+r_p}40\times {10}^3+10r_p=4\times {10}^3\mu$
When A = 12
$R_L=8K\Omega 12=\frac{\mu }{1+\frac{r_p}{8\times {10}^3}}12=\frac{\mu \times 8\times {10}^3}{8\times {10}^3+r_p}=96\times {10}^3+12r_p=8\times {10}^3\mu$
Multiplying (2) in equation (1) and equating with equation (2)
$\Rightarrow 2(40\times {10}^8+10r_p)=96\times {10}^3+12r_p\Rightarrow 80\times {10}^3+20r_p=96\times {10}^3+12r_p\Rightarrow 8r_p^p=16\times {10}^3\Rightarrow r_p^p=2\times {10}^3\Omega =2K\Omega$
Putting the value in equation (1)
$40\times {10}^3+10(2\times {10}^3)=4\times {10}^8\mu \Rightarrow 40\times {10}^3+20\times {10}^3=4\times {10}^3\mu \Rightarrow 60=4\mu \Rightarrow \mu =\frac{60}{4}=15.$