Electronics - Bipolar Junction Transistors (BJT) - Discussion
Discussion Forum : Bipolar Junction Transistors (BJT) - General Questions (Q.No. 3)
3.
A transistor has a 
of 250 and a base current, IB, of 20 
A. The collector current, IC, equals:
of 250 and a base current, IB, of 20 A. The collector current, IC, equals:500 A
A5 mA
50 mA
5 A
Discussion:
17 comments Page 1 of 2.
Syed Shakeeb said:
4 years ago
Ic=β*Ib.
Ic = 250 * 20 * 10^-6,
Ic = 0.005.
Ic = 5mA.
Ic = 250 * 20 * 10^-6,
Ic = 0.005.
Ic = 5mA.
(1)
Deepen said:
7 years ago
You are right, Agree @Kalpana.
Moin ansari said:
7 years ago
As we know that the amplification factor in ce mode β=Ic/Ib.
and Ib is in micro range so 20*1000000,
250*20/1000000=5ma answer.
and Ib is in micro range so 20*1000000,
250*20/1000000=5ma answer.
Max Malick Electronic said:
8 years ago
β = Ic/Ib.
Ic=betta * Ib.
Ic = 250 * 20micro Ampere.
Ic = 5mA.
Ic=betta * Ib.
Ic = 250 * 20micro Ampere.
Ic = 5mA.
Dieudonne kubwimana said:
8 years ago
Given datas:bette = 250,current base = 20micro ampere.
Then IC = bette*current base.
IC = 250 * 20 = 5000micro ampere.
Convetion of 5000MA into mA will be 5mA(Ans).
Then IC = bette*current base.
IC = 250 * 20 = 5000micro ampere.
Convetion of 5000MA into mA will be 5mA(Ans).
Tejaswini said:
1 decade ago
Ic = BIb.
Ic = 250*20*10^-6.
Ic = 5mA.
Ic = 250*20*10^-6.
Ic = 5mA.
Sid said:
1 decade ago
From formula,
Ic(collector current) = Beta*Ib(base current).
=250*20*10^-6.
=5*10^-3.
=5mA.
Ic(collector current) = Beta*Ib(base current).
=250*20*10^-6.
=5*10^-3.
=5mA.
KALPANA said:
1 decade ago
In common emiter configuration, amplification factor is given by BETE=Ic/Ib.
Ic=BETA*Ib
=250*(20*10^-6)
=5000*10^-6
=5mA
Ic=BETA*Ib
=250*(20*10^-6)
=5000*10^-6
=5mA
(1)
Bala.j said:
1 decade ago
THE FORMULA USED TO FIND OUT THE IC IS Ic = Ib*GAIN
= (20*10^-6)*(250)
=5*10^3* 10^-6
= 5mA
= (20*10^-6)*(250)
=5*10^3* 10^-6
= 5mA
Shreshthagupta said:
1 decade ago
Ic=Beta*Ib
=250*20*10^-6
=5*10^-6
=5mA
=250*20*10^-6
=5*10^-6
=5mA
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