Question

For a CE-transistor amplifier, the audio signal voltage across the collector resistance of $2{\text{ k}}\Omega$ is $2{\text{ V}}$. Suppose the current amplification factor of the transistor is $100$, find the input signal voltage and base current, if the base resistance is $1{\text{ k}}\Omega$ .

Answer 1

A transistor is an arrangement in which a thin layer of one type of semiconductor is placed between two thick layers of another type of semiconductor. A device used for increasing the amplitude of variation of alternating voltage or current or power is called an amplifier. A transistor can be used as an amplifier.
In a CE-transistor amplifier, the current amplification factor is given by $\beta = \dfrac{{\Delta {I_c}}}{{\Delta {I_b}}}$ where $\Delta {I_c}$ is change in collector current and $\Delta {I_b}$ is change in base current. In a CE-transistor, the input voltage is given in base and output is received at the collector.

Complete step by step answer:
Let us first discuss transistors and an amplifier.
A transistor is an arrangement in which a thin layer of one type of semiconductor is placed between two thick layers of another type of semiconductor. A device used for increasing the amplitude of variation of alternating voltage or current or power is called an amplifier. A transistor can be used as an amplifier.
Given:
Collector resistance, ${R_c} = 2{\text{ k}}\Omega$
Base resistance, ${R_b} = 1{\text{ k}}\Omega$
Audio signal voltage across collector resistance, ${V_c} = 2{\text{ V}}$
Current amplification factor of the transistor, $\beta = 100$
We know that in a CE-transistor amplifier, the current amplification factor is given by $\beta = \dfrac{{\Delta {I_c}}}{{\Delta {I_b}}}$ where $\Delta {I_c}$ is change in collector current and $\Delta {I_b}$ is change in base current. In a CE-transistor, the input voltage is given in base and output is received at the collector.
Therefore the current amplification factor can be written as
$\beta = \dfrac{{{V_c}/{R_c}}}{{{V_i}/{R_b}}}$ where ${V_i}$ is the input signal voltage.
On rearranging the expression we have
$\dfrac{{{V_c}}}{{{V_i}}} = \beta \dfrac{{{R_c}}}{{{R_b}}}$
On substituting the values we have
$\dfrac{2}{{{V_i}}} = 100 \times \dfrac{2}{1}$
So, ${V_i} = 0.01{\text{ V}}$
Now, we know that
${I_b} = \dfrac{{{V_i}}}{{{R_b}}}$
Substituting the values we have
${I_b} = \dfrac{{0.01}}{{1 \times {{10}^3}}} = 10 \times {10^{ - 6}}{\text{ A}} = 10{\text{ }}\mu {\text{A}}$
Hence, the input signal voltage is $0.01{\text{ V}}$ and the base current is $10{\text{ }}\mu {\text{A}}$ .

Note: A transistor acts as an amplifier as it raises the strength of a weak signal. Apart from common emitter transistors, common base transistors can also be used as an amplifier. The configuration remains in forward biased condition as the DC bias voltage is applied to the emitter base junction. But maximum amplification is done by the common emitter transistor amplifier.