Discussion :: Bipolar Junction Transistors (BJT) - General Questions (Q.No.3)
Koudinya said: (Oct 19, 2010) | |
Ic = Beta x Ib |
Laxman said: (Dec 3, 2010) | |
Ic=Beta*Ib Ic=250*20/1000000 =5000/1000000 Ic=5mA |
Rajnish Tiwari said: (Dec 11, 2010) | |
Beta=Ic/Ib Ic=Beta*Ib Ic=250*20/1000000 Ic=5mA |
Arghya Das said: (Jan 18, 2011) | |
Beta is 250,Ib=20 collector current Ic=(Ib*beta)/10^6 Ic=5000/10^6 =5mA |
Sashibhusan Panda said: (Sep 18, 2011) | |
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) |
Micheal said: (Oct 7, 2011) | |
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 |
Jyothi said: (Nov 10, 2011) | |
Ic=BIb =250*20*10^-6 5 mA |
Shreshthagupta said: (Feb 27, 2012) | |
Ic=Beta*Ib =250*20*10^-6 =5*10^-6 =5mA |
Bala.J said: (Apr 20, 2012) | |
THE FORMULA USED TO FIND OUT THE IC IS Ic = Ib*GAIN = (20*10^-6)*(250) =5*10^3* 10^-6 = 5mA |
Kalpana said: (Sep 29, 2012) | |
In common emiter configuration, amplification factor is given by BETE=Ic/Ib. Ic=BETA*Ib =250*(20*10^-6) =5000*10^-6 =5mA |
Sid said: (Jan 1, 2013) | |
From formula, Ic(collector current) = Beta*Ib(base current). =250*20*10^-6. =5*10^-3. =5mA. |
Tejaswini said: (Aug 4, 2013) | |
Ic = BIb. Ic = 250*20*10^-6. Ic = 5mA. |
Dieudonne Kubwimana said: (Apr 18, 2017) | |
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). |
Max Malick Electronic said: (Feb 7, 2018) | |
β = Ic/Ib. Ic=betta * Ib. Ic = 250 * 20micro Ampere. Ic = 5mA. |
Moin Ansari said: (Aug 25, 2018) | |
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. |
Deepen said: (Nov 12, 2018) | |
You are right, Agree @Kalpana. |
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