Consider an NPN transistor amplifier in commo | vedclass

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(B) Given: Current gain $\beta = 100$,Change in collector current $\Delta I_c = 1\, mA$.We know that the current gain in common-emitter configuration

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$1.01$

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(B) Given: Current gain $\beta = 100$,Change in collector current $\Delta I_c = 1\, mA$.We know that the current gain in common-emitter configuration is defined as $\beta = \frac{\Delta I_c}{\Delta I_b}$.Therefore,the change in base current is $\Delta I_b = \frac{\Delta I_c}{\beta} = \frac{1\, mA}{100} = 0.01\, mA$.The relationship between emitter,collector,and base currents is $\Delta I_e = \Delta I_c + \Delta I_b$.Substituting the values,we get $\Delta I_e = 1\, mA + 0.01\, mA = 1.01\, mA$.Thus,the change in emitter current is $1.01\, mA$.

Consider an $NPN$ transistor amplifier in common-emitter configuration. The current gain of the transistor is $100$. If the collector current changes by $1\, mA$,what will be the change in emitter current in $mA$?

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