Electronic Devices - BJT Amplifiers - Discussion
Discussion Forum : BJT Amplifiers - General Questions (Q.No. 40)
40.
Refer to this figure. If an emitter bypass capacitor was installed, determine the value of Rin(base).
416 
5 k
50 k
500
Discussion:
6 comments Page 1 of 1.
Prathmesh said:
10 years ago
Answer is wrong.
Rin(base) = Beta(r'e+Re).
Rin(base) = 100 (22.5+500) = 52.25 Kohm.
R'e = 22.5 it is come when we solve ac model.
Rin(base) = Beta(r'e+Re).
Rin(base) = 100 (22.5+500) = 52.25 Kohm.
R'e = 22.5 it is come when we solve ac model.
F fort said:
10 years ago
Answer is correct.
Vb = 20 (5 kohm/(5 kohm + 22 kohm)) = 3.7037.
Vb - 0.7 - IeRe = 0 --> Ie = 6mA.
R'e = 25 mV/Ie = 4.1667.
Rin base = Beta*R'e = (100)(4.1667) = 416.67.
Vb = 20 (5 kohm/(5 kohm + 22 kohm)) = 3.7037.
Vb - 0.7 - IeRe = 0 --> Ie = 6mA.
R'e = 25 mV/Ie = 4.1667.
Rin base = Beta*R'e = (100)(4.1667) = 416.67.
(2)
Krishna said:
8 years ago
How 25mV?
Annum Latif said:
8 years ago
@Krishna.
25mv is the internal volt drop across the Emitter junction layer provided in transistor data sheet.
25mv is the internal volt drop across the Emitter junction layer provided in transistor data sheet.
(1)
Sandy said:
5 years ago
How 25mV? Please explain it.
Irvin said:
4 years ago
25mv is constant when computing R'e.
Using formula of R'e=25mv/Ie @20C and r'e=26mv/Ie @25C.
But on this question, 25mv was used.
Using formula of R'e=25mv/Ie @20C and r'e=26mv/Ie @25C.
But on this question, 25mv was used.
Post your comments here:
Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers