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The current gain of a transistor in common emitter mode is 35. The change in collector current is 140mV at constant collector to emitter voltage. The change in the base current will be.A. 10mAB. 2mAC. 0.1mAD. 4mA

Last updated date: 20th Jun 2024
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Hint: Current gain of a transistor in common emitter amplifier mode is given by,
$\beta = {\left( {\dfrac{{\vartriangle {I_C}}}{{\vartriangle {I_B}}}} \right)_{{V_{CE}}}}$

Complete step by step solution:
Current gain of a transistor in common emitter amplifier mode is given by,
$\beta = {\left( {\dfrac{{\vartriangle {I_C}}}{{\vartriangle {I_B}}}} \right)_{{V_{CE}}}}$

We can also write it as,
$\vartriangle {I_B} = \dfrac{{\vartriangle {I_C}}}{\beta }$

Now putting the values,
$= \dfrac{{140 \times {{10}^{ - 3}}}}{{35}} \\ = 4 \times {10^{ - 3}}A \\ = 4mA \\$

Therefore, the correct option is D.

This method of biasing the transistor greatly reduces the effects of varying Beta, ($\beta$) by holding the Base bias at a constant steady voltage level allowing for best stability. The quiescent Base voltage (Vb) is determined by the potential divider network formed by the two resistors, R1, R2 and the power supply voltage Vcc as shown with the current flowing through both resistors.
Note: Students need to be able to know this way of finding current gain. This method of biasing the transistor greatly reduces the effects of varying Beta, ($\beta$) by holding the Base bias at a constant steady voltage level allowing for best stability. The quiescent Base voltage (Vb) is determined by the potential divider network formed by the two resistors, R1, R2 and the power supply voltage Vcc as shown with the current flowing through both resistors.