Electronics and Communication Engineering - Analog Electronics - Discussion
Discussion Forum : Analog Electronics - Section 1 (Q.No. 20)
20.
The transistor of following figure in Si diode with a base current of 40 μA and ICBO = 0, if VBB = 6V, RE = 2 kΩ and β = 90, IBQ = 20 μA then RB =
Answer: Option
Explanation:
.
Discussion:
13 comments Page 2 of 2.
Artigo said:
7 years ago
ICBO = 0 implying IC = 0 => IE.
So emitter voltage was not considered.
Hence, the equation has only VBB - VRBQ - VBEQ = 0.
So emitter voltage was not considered.
Hence, the equation has only VBB - VRBQ - VBEQ = 0.
Arthi said:
5 years ago
@All.
We need only resistance at RB.
So there is no need of considering resistance across emitter.
Its just Ohm's law.
R = V/I.
We need only resistance at RB.
So there is no need of considering resistance across emitter.
Its just Ohm's law.
R = V/I.
C.THAMILARASI said:
3 years ago
Ohm law cannot be applied. We have to apply Kirchhoff's voltage law then;
RB = VBB - VBE - IERE/IB = 85 Kohm.
RB = VBB - VBE - IERE/IB = 85 Kohm.
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