Question

The reverse leakage current of a transistor when connected in CB configuration is 0.2 macro ampere and it is 18macro ampere when the same transistor in connected in CE configuration calculate alpha and beta

Answer 1

Given:

reverse current  in CB configuration I$_B$ = 0.2 μA

reverse current  in CE configuration I$_E$ = 18 μA

To find:

1) α =?

2) β =?

Step-by-step explanation:

1)

• The CB configuration in which the base of the transistor is common between the emitter and collector circuit is called a common base configuration.
• When the transistor is connected in CB configuration, then the static current amplification factor which is denoted by α is given by:

$\displaystyle \alpha = 1-\frac{I_B}{I_E}$

    = $\displaystyle1- \frac{0.2}{18}$

α = 0.98

2)

• The CE configuration in which the emitter is connected between the collector and base is known as a common emitter configuration.
• When the transistor is connected in CE configuration,then the static current amplification factor which is denoted by β is given by:

$\displaystyle \beta = \frac{I_E}{I_B} -1$

   = $\displaystyle \frac{18}{0.2}-1$

β = 89