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Semiconductor Electronic: Material, Devices And Simple Circuits — Physics STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 SciencePhysicsSemiconductor Electronic: Material, Devices And Simple Circuits5 Marks
Question
Draw a simple circuit of a CE transistor amplifier. Explain its working. Show that the voltage gain, $A_V$, of the amplifier is given by $\text{A}_{V} = - \frac{\beta_{ac}\text{R}_{L}}{\text{r}_{i}},\text{ where} \beta_{ac}$is the current gain, $R_L$ is the load resistance and $r_i $ is the input resistance of the transistor. What is the significance of the negative sign in the expression for the voltage gain?
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Answer



When an ac input signal $v_i$ is superimposed on the bias $v_{BB}$, the output, which is measured between collector and ground, increases.
$\text{v}_{CC} = \text{v}_{CE} + \text{I}_{C}\text{R}_{L}$
$\text{v}_{BB} =\text{v}_{RE} + \text{I}_{B} \text{R}_{B}$
When $\text{v}_{i}$ is not zero, we have
$\text{v}_{BE} + \text{v}_{i} = \text{v}_{BE} + \text{I}_{B} \text{ R}_{B} +\Delta\text{I}_{B}(\text{R}_{B} + \text{R}_{i})$
$ = > \text{v}_{i} = \Delta\text{I}_{B}(\text{R}_{B} + \text{R}_{i})$
$\text{v}_{i} =\text{r}\Delta\text{I}_{B}$
Change in $\text{I}_{B}$ causes a change in $\text{I}_{C}$
Hence, $\beta_{ac} = \frac{\Delta\text{I}_{C}}{\Delta\text{I}_{B}} =\frac{\text{I}_{C}}{\text{I}_{B}}$
As $\Delta\text{V}_{CC} = \Delta\text{V}_{CE} + \text{R}_{L}\Delta\text{I}_{C} = 0 $
$ => \Delta\text{V}_{CE} = - \text{R}_{L}\Delta\text{I}_{C}$
$ = > \text{V}_{o} = -\text{R}_{L}\Delta\text{I}_{C}$
$\beta_{ac}\Delta\text{I}_{B}\text{R}_{L}$
$= > $ voltage gain of the amplifier
$\text{A}_{V} =\frac{\text{V}_{o}}{\text{V}_{i}} =\frac{\Delta\text{V}_{CE}}{\text{r}\Delta\text{I}_{B}} = \frac{-\beta_{ac}\Delta\text{I}_{B}\text{R}_{L}}{\text{r}\Delta\text{I}_{B}}$
Negative sign in the expression shows that output voltage and input voltage have phase difference of $\pi$.
Alternate Answer
(Also accept this derivation for voltage gain expression)
$\text{A}_{V}=\frac{\Delta\text{V}_{CE}}{\Delta\text{V}_{BE}} = \frac{-\Delta\text{I}_{C}\text{R}_{L}}{\Delta\text{I}_{B}\text{R}_{L}} $
But current gain
$\beta_{ac} = \frac{-\Delta\text{I}_{C}}{\Delta\text{I}_{B}}$
$ =\text{A}_{v} = - \beta_{ac}\times\frac{\text{R}_{L}}{\text{R}_{L}}\bigg].$

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