In a p-n-p transistor working as common base amplifier, current gain is 0.96 and emitter current is 7.2 mA. Then base current will be:

Question

In a p-n-p transistor working as common base amplifier, current gain is 0.96 and emitter current is 7.2 mA. Then base current will be: (A) 0.2 mA (B) 0.29 mA (C) 0.35 mA (D) 0.4 mA

Tardigrade Admin · Accepted Answer

Here, current gain α =0.96 and emitter current ie=7.2 mA from the relation of current gain α =0.96=(Ic/Ie)=(Ic/7.2), So Ic=0.96× 7.2=6.91 mA Hence the base current Ib=Ie-Ic=7.2-6.91=0.29 mA

Q. In a p-n-p transistor working as common base amplifier, current gain is 0.96 and emitter current is 7.2 mA. Then base current will be:

0.2 mA

0.29 mA

0.35 mA

0.4 mA

Here, current gain $α = 0.96$ and emitter current $i_{e} = 7.2 m A$ from the relation of current gain $α = 0.96 = \frac{I _{c}}{I _{e}} = \frac{I _{c}}{7.2},$ So $I_{c} = 0.96 \times 7.2 = 6.91 m A$ Hence the base current $I_{b} = I_{e} - I_{c} = 7.2 - 6.91 = 0.29 m A$

Q. In a p-n-p transistor working as common base amplifier, current gain is 0.96 and emitter current is 7.2 mA. Then base current will be:

0.2 mA

0.29 mA

0.35 mA

0.4 mA

Here, current gain α=0.96 and emitter current ie​=7.2mA from the relation of current gain α=0.96=Ie​Ic​​=7.2Ic​​, So Ic​=0.96×7.2=6.91mA Hence the base current Ib​=Ie​−Ic​=7.2−6.91=0.29mA

Here, current gain $α = 0.96$ and emitter current $i_{e} = 7.2 m A$ from the relation of current gain $α = 0.96 = \frac{I _{c}}{I _{e}} = \frac{I _{c}}{7.2},$ So $I_{c} = 0.96 \times 7.2 = 6.91 m A$ Hence the base current $I_{b} = I_{e} - I_{c} = 7.2 - 6.91 = 0.29 m A$