Electronic Devices - BJT Amplifiers - Discussion
Discussion Forum : BJT Amplifiers - General Questions (Q.No. 4)
4.
Refer to this figure. Find the value of IE.
Discussion:
6 comments Page 1 of 1.
Ravi said:
7 years ago
Thank you all for the explanation
Raghav dixit said:
8 years ago
In it we start from Branch Vcc,R1,Vb,Re.
20-(R2/( R1+R2)*20)-0.7-Ie*Re=0,
then,
20(5/27)-0.7=Ie*500.
then Ie=6mA.
20-(R2/( R1+R2)*20)-0.7-Ie*Re=0,
then,
20(5/27)-0.7=Ie*500.
then Ie=6mA.
(1)
Muhammad Usman said:
1 decade ago
By Voltage divider rule we have:
Ie = Ve/Re, Re is given, we require Ve to get the value of Ie.
So,
At Base-Emitter Junction, Vbe = Vb-Ve, Implies that Ve = Vb-Vbe or
Ve = 3.07-0.7 = 3V. (since Vb = R2/R1+R2*20).
Ie = Ve/Re = 3V/500 ohm = 0.006 = 6mA.
Ie = Ve/Re, Re is given, we require Ve to get the value of Ie.
So,
At Base-Emitter Junction, Vbe = Vb-Ve, Implies that Ve = Vb-Vbe or
Ve = 3.07-0.7 = 3V. (since Vb = R2/R1+R2*20).
Ie = Ve/Re = 3V/500 ohm = 0.006 = 6mA.
(1)
Aravindhan said:
1 decade ago
Vb = 20(5K/(22K+5K)) = 3.70V.
Ve = Vb - 0.7 = 3V.
Ie = Ve/Re = 3V/500ohms = 0.006 = 6mA.
Ve = Vb - 0.7 = 3V.
Ie = Ve/Re = 3V/500ohms = 0.006 = 6mA.
Aalya said:
1 decade ago
Vb = 20(5*10^-3/(5+22)10^-3) =3.703
vbe = 0.7 Rb = 4.074
Ie= 3.703-0.7/(4.074/101 + 500) = 6 ma(appx)
vbe = 0.7 Rb = 4.074
Ie= 3.703-0.7/(4.074/101 + 500) = 6 ma(appx)
Katekani said:
1 decade ago
How this question is calculated? help please.
(1)
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