Question

Why is the base region of the transistor thinly doped?
The input resistance of a transistor is $1000\Omega$. On changing the base current by $10\mu A$, the collector current increases by $2mA$. If a load resistance of $5k\Omega$ is used in the circuit, calculate (i) the current gain, (ii) voltage gain of the amplifier.

Answer 1

Base region of the transistor is thinly doped, so, the number of majority carriers is low.so, less number of electrons holes recombination take place and whole current emitted from the emitter with a slight difference is passed to the collector.Since, base provides the proper junction for interconnection between emitter and collector.

Given,

Input Resistance,$R_i=1000\Omega =1K\Omega$

Load resistance,$R_L=5K\Omega$

Change in base current,$△I_B=10\mu A$

Change in collector current,$△I_C=2mA$

(i) The current gain $\left(\beta \right)$ is given by,

$\beta ={\left(\frac{△I_C}{△I_B}\right)}_{V_{CE}}$

(ii) The voltage gain $\left(Av\right)$ is given by

$A_V=\beta \times \frac{R_L}{R_i}=200\times \frac{5}{1}=1000$