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:


Discussion:
17 comments Page 1 of 2.
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
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).
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)
Ic=Ie+(beta)Ib
But for Ib=20ua current value, Ie negligible so considered as zero.
=> Ic=250*20uA=5mA. (ans)
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.
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
so
they formula say that collector current(Ic)= Ib*Bdc
then IC= 20*250
=5000
sol: 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)
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.
ARGHYA DAS said:
1 decade ago
Beta is 250,Ib=20
collector current Ic=(Ib*beta)/10^6
Ic=5000/10^6
=5mA
collector current Ic=(Ib*beta)/10^6
Ic=5000/10^6
=5mA
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.
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)
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