In an N-P-N transistor circuit,the collector | vedclass

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(B) Given that the collector current $I_C = 10 \ mA$. Since $90\%$ of the electrons emitted from the emitter reach the collector,we have the relation:

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$I_E = 11 \ mA; I_B = 1 \ mA$

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(B) Given that the collector current $I_C = 10 \ mA$. Since $90\%$ of the electrons emitted from the emitter reach the collector,we have the relation: $I_C = 0.90 	imes I_E$. Rearranging for $I_E$: $I_E = \frac{I_C}{0.90} = \frac{10 \ mA}{0.9} = 11.11 \ mA \approx 11 \ mA$. Using the Kirchhoff's current law for a transistor: $I_E = I_B + I_C$. Therefore,the base current is $I_B = I_E - I_C = 11 \ mA - 10 \ mA = 1 \ mA$. Thus,$I_E = 11 \ mA$ and $I_B = 1 \ mA$.

In an $N-P-N$ transistor circuit,the collector current is $10 \ mA$. If $90\%$ of the electrons emitted reach the collector,then the emitter current $(I_E)$ and base current $(I_B)$ are respectively:

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