Electronics - Bipolar Junction Transistors (BJT) - Discussion

Discussion Forum : Bipolar Junction Transistors (BJT) - General Questions (Q.No. 3)
3.
A transistor has a mca12_1001a1.gif of 250 and a base current, IB, of 20 mu.gif A. The collector current, IC, equals:
500 mu.gif A
5 mA
50 mA
5 A
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
17 comments Page 2 of 2.

Shreshthagupta said:   1 decade ago
Ic=Beta*Ib
=250*20*10^-6
=5*10^-6
=5mA

Jyothi said:   1 decade ago
Ic=BIb
=250*20*10^-6
5 mA

Micheal said:   1 decade ago
Bdc is 250 and current base Ib=20
so
they formula say that collector current(Ic)= Ib*Bdc
then IC= 20*250
=5000
sol: 5mA

Sashibhusan Panda said:   1 decade ago
For dc operation of bjd

Ic=Ie+(beta)Ib

But for Ib=20ua current value, Ie negligible so considered as zero.

=> Ic=250*20uA=5mA. (ans)

ARGHYA DAS said:   1 decade ago
Beta is 250,Ib=20
collector current Ic=(Ib*beta)/10^6
Ic=5000/10^6
=5mA

Rajnish Tiwari said:   1 decade ago
Beta=Ic/Ib
Ic=Beta*Ib
Ic=250*20/1000000

Ic=5mA

Laxman said:   1 decade ago
Ic=Beta*Ib

Ic=250*20/1000000

=5000/1000000
Ic=5mA


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