Electronic Devices - Transistor Bias Circuits - Discussion
Discussion Forum : Transistor Bias Circuits - General Questions (Q.No. 13)
13.
Refer to this figure. The value of 
DC is
DC is
Discussion:
7 comments Page 1 of 1.
Gangapriya said:
5 years ago
Thank you @Minitkumar.
Arnuel said:
8 years ago
Thank you so much @Minitkumar.
Minitkumar said:
1 decade ago
Ic = (Vcc-Vce)/Rc.
= (12-6)/1.2.
=5mA.
Ib = (Vcc-Vbe)/Rb.
= (12-0.7)/226.
= 0.05mA.
Bdc = Ic/Ib.
= 5/0.05.
= 100 ANSWER is D.
= (12-6)/1.2.
=5mA.
Ib = (Vcc-Vbe)/Rb.
= (12-0.7)/226.
= 0.05mA.
Bdc = Ic/Ib.
= 5/0.05.
= 100 ANSWER is D.
Amour Lesangwa Giabox said:
1 decade ago
No you guys ain't right lemme show you we work with these damn small things.
Take Ic=B (VCC-VBE/Rb) Making B the subject you get.
B=Ic/ (VCC-VBE/Rb).
Take Ic=B (VCC-VBE/Rb) Making B the subject you get.
B=Ic/ (VCC-VBE/Rb).
Arpitha said:
1 decade ago
A small correction here is:
Ib = (vcc-Vbe)/Rb.
Ib = (vcc-Vbe)/Rb.
Deepak said:
1 decade ago
Thank you baggy.
Baggy said:
1 decade ago
Bdc= ic/ib
ib=(Vcc-Vc)/Rb
ic=(Vcc-Vc)/Rc
ib=(Vcc-Vc)/Rb
ic=(Vcc-Vc)/Rc
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