Question

𝞪=0.99 for CE transistor amplifier circuit. The input resistance is equal to $1K\Omega$ and load resistance is equal to $10K\Omega$. The voltage gain of the circuit is-
(A). $99$
(B). $990$
(C). $9900$
(D). $~99000$

Answer 1

Voltage gain is the ratio of voltage in the output circuit to the voltage in the input circuit. It is also equal to the product of current gain and resistance gain. Calculating the value of current gain and substituting corresponding values of resistances voltage gain is calculated.

Formula used:
$G=\dfrac{{{I}_{L}}\cdot {{R}_{L}}}{{{I}_{B}}\cdot {{R}_{B}}}$
$G=\dfrac{\beta \cdot {{R}_{l}}}{{{R}_{B}}}$

Complete step by step solution:
A transistor is a semiconductor used to amplify, regulate or control electrical signals. When a transistor is used as an amplifier then the circuit is known as a transistor amplifier circuit.
For an amplifier circuit, the input is taken across the emitter-base junction which is in forward-bias and we get the output across the load connected to the collector-base junction which is in reverse-bias.
The voltage gain is the ratio of input voltage to the ratio of voltage difference produced in the output circuit.
$Voltage\,gain(G)=\dfrac{{{V}_{L}}}{{{V}_{B}}}=\dfrac{{{R}_{B}}}{{{R}_{L}}}$ ………….. (1)
Here, ${{V}_{L}}$ is the voltage drop in the output circuit
${{V}_{B}}$ is the voltage in the input circuit

We know,
$V=IR$ using this relation in eq (1) we get,

$G=\dfrac{{{I}_{L}}\cdot {{R}_{L}}}{{{I}_{B}}\cdot {{R}_{B}}}$ ……………. (2)
Here,
${{I}_{L}}$ is the current in the output circuit
${{I}_{B}}$ is the current in the input circuit
${{R}_{L}}$ is resistance across the output load resistor
${{R}_{B}}$ is resistance in the input circuit
We know that,
$\beta =\dfrac{{{I}_{C}}}{{{I}_{B}}}$
$\beta$ is also called the current gain. It is the ratio of current in the output circuit to the current in the input circuit.
We can rewrite eq (2) as-
$G=\dfrac{\beta \cdot {{R}_{l}}}{{{R}_{B}}}$ …………...(3)
To calculate the value of $\beta$ from $\alpha$,
$\beta =\dfrac{\alpha }{1-\alpha }$ ……………. (4)
Here,
$\alpha$ is also another form of representation of current gain. It is the ratio of output or collector current to the emitter current.
$\alpha =\dfrac{{{I}_{C}}}{{{I}_{E}}}$
Substituting the value of $\alpha$ in eq (4) we get,

& \Rightarrow \beta =\dfrac{0.99}{1-0.99} \\\ & \therefore \beta =99 \\\ \end{aligned}$$ Substituting the value of $${{R}_{B}},\,{{R}_{C}},\,\beta $$ in eq (3), we, get, $$\begin{aligned} & G=\dfrac{99\times 10\times {{10}^{3}}}{{{10}^{3}}} \\\ & \Rightarrow G=990 \\\ \end{aligned}$$ The value of voltage gain is $$990$$ . **So, the correct answer is “Option B”.** **Note:** Basically, a transistor works as a switch. It is made up of three terminals namely, emitter, base, collector. The Base is the smallest part while the collector is the largest part. These three terminals give rise to two junctions. The emitter-base junction is in forward-bias where the P-side of a diode is attached to the positive terminal while the N-side is attached to the negative terminal.