Question

A transistor has a current amplification factor of 60. If a CE amplifier, input resistance is $1k\Omega $ and output voltage is 0.01V. The transconductance is (in SI unit)
$\text{A}\text{. }{{10}^{-5}}$
$\text{B}\text{. 6}\times {{10}^{-2}}$
$\text{C}\text{. 6}\times {{10}^{4}}$
$\text{D}\text{. }10$

Answer 1

Hint:
Trans-conductance is defined as the ratio of the change in collector to the change in emitter base voltage, i.e. ${{g}_{m}}=\dfrac{\Delta {{I}_{C}}}{\Delta {{V}_{BE}}}$. Use the formulas for current amplification factor $\beta =\dfrac{\Delta {{I}_{C}}}{\Delta {{I}_{B}}}$ and $\Delta {{V}_{BE}}=\Delta {{I}_{B}}{{R}_{i}}$, to find the value of trans-conductance.
Formula used:
${{g}_{m}}=\dfrac{\Delta {{I}_{C}}}{\Delta {{V}_{BE}}}$
$\beta =\dfrac{\Delta {{I}_{C}}}{\Delta {{I}_{B}}}$
$\Delta {{V}_{BE}}=\Delta {{I}_{B}}{{R}_{i}}$

Complete answer:
A transistor is a semiconductor device, which is formed by fusing two junction diodes. When the anodes of the two diodes are fused, the transistor is called a npn transistor. When the cathodes of the two diodes are fused, the transistor is called a pnp transistor.
A transistor consists of three parts – a collector, an emitter and a base. The middle part is base.
A transistor has many applications in the electronic sector. One of the applications is amplification. This means that a transistor can be used as an amplifier to amplify a current.
The amplification of the current is determined by the current amplification factor, also called as current gain factor. It is denoted by $\beta $.
And $\beta =\dfrac{\Delta {{I}_{C}}}{\Delta {{I}_{B}}}$ ….. (i),
where $\Delta {{I}_{C}}$ is the change in current in the collector and $\Delta {{I}_{B}}$ is change in current in the base.
The trans-conductance of an amplifier is defined as the ratio of the change in collector to the change in emitter base voltage.
i.e. ${{g}_{m}}=\dfrac{\Delta {{I}_{C}}}{\Delta {{V}_{BE}}}$ ….. (ii).
Divide (ii) by (i).
$\Rightarrow \dfrac{{{g}_{m}}}{\beta }=\dfrac{\dfrac{\Delta {{I}_{C}}}{\Delta {{V}_{BE}}}}{\dfrac{\Delta {{I}_{C}}}{\Delta {{I}_{B}}}}$
$\Rightarrow \dfrac{{{g}_{m}}}{\beta }=\dfrac{\Delta {{I}_{B}}}{\Delta {{V}_{BE}}}$ …. (iii).
And $\Delta {{V}_{BE}}=\Delta {{I}_{B}}{{R}_{i}}$, where ${{R}_{i}}$ is internal resistance.
Substitute this value in equation (iii).
$\Rightarrow \dfrac{{{g}_{m}}}{\beta }=\dfrac{\Delta {{I}_{B}}}{\Delta {{I}_{B}}{{R}_{i}}}$
$\Rightarrow {{g}_{m}}=\dfrac{\beta }{{{R}_{i}}}$ …. (iv)
It is given $\beta $=60 and ${{R}_{i}}=1k\Omega =1000\Omega $.
$\Rightarrow {{g}_{m}}=\dfrac{\beta }{{{R}_{i}}}=\dfrac{60}{1000}=6\times {{10}^{-2}}{{\Omega }^{-1}}$.
This means that the trans-conductance is $6\times {{10}^{-2}}{{\Omega }^{-1}}$.

Hence, the correct option is B.

Note:
From the above solution, we found that ${{g}_{m}}=\dfrac{\beta }{{{R}_{i}}}$.
Hence, we can also define the trans-conductance of an amplifier as the ratio of the current amplification factor to the internal resistance of the circuit.
Also note that $\beta $ is just a number and has no dimension. It is always greater than one.