Question
The gain factor of an amplifier in increased from 10 to 12 as the load resistance is changed from $4\text{k}\Omega\text{ to }8\text{k}\Omega$ Calculate (a) the amplification factor and (b) the plate resistance.
✓
Answer
Voltage gain $=\frac{\mu}{1+\frac{\text{r}_\text{p}}{\text{R}_\text{L}}}$ When $\text{A}=10,\ \text{R}_\text{L}=4\text{K}\Omega$$10=\frac{\mu}{1+\frac{\text{r}_\text{p}}{4\times10^{3}}}$
$10=\frac{\mu\times4\times\times10^{3}}{4\times10^{3}+\text{r}_\text{p}}$
$40\times10^{3}\times10\text{r}_\text{p}=4\times10^{3}\mu\ ....(\text{i})$
When $\text{A}=12,\text{R}_\text{L}=8\text{K}\Omega$$12=\frac{\mu}{1+\frac{\text{r}_\text{p}}{8\times10^{3}}}$
$12=\frac{\mu\times8\times10^{3}}{8\times10^{3}+\text{r}_\text{p}}$
$96\times10^{3}+12\text{r}_\text{p}=8\times10^{3}\mu\ ...(\text{ii})$
Multiplying (ii) in equation (i) and equating with equation (ii)$2(40\times10^{3}+10\text{r}_\text{p})=96\times10+3+12\text{r}_\text{p}$
$\text{r}_\text{p}=2\times10^{3}\Omega=2\text{K}\Omega$
Putting the value in equation (i)$40\times10^{3}+10(2\times10^{3})=4\times10^{3}\mu$
$40\times10^{3}+20\times10^{3}=4\times10^{3}\mu$
$\mu=\frac{60}{4}=15$
$10=\frac{\mu\times4\times\times10^{3}}{4\times10^{3}+\text{r}_\text{p}}$
$40\times10^{3}\times10\text{r}_\text{p}=4\times10^{3}\mu\ ....(\text{i})$
When $\text{A}=12,\text{R}_\text{L}=8\text{K}\Omega$$12=\frac{\mu}{1+\frac{\text{r}_\text{p}}{8\times10^{3}}}$
$12=\frac{\mu\times8\times10^{3}}{8\times10^{3}+\text{r}_\text{p}}$
$96\times10^{3}+12\text{r}_\text{p}=8\times10^{3}\mu\ ...(\text{ii})$
Multiplying (ii) in equation (i) and equating with equation (ii)$2(40\times10^{3}+10\text{r}_\text{p})=96\times10+3+12\text{r}_\text{p}$
$\text{r}_\text{p}=2\times10^{3}\Omega=2\text{K}\Omega$
Putting the value in equation (i)$40\times10^{3}+10(2\times10^{3})=4\times10^{3}\mu$
$40\times10^{3}+20\times10^{3}=4\times10^{3}\mu$
$\mu=\frac{60}{4}=15$
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