Electronics and Communication Engineering - Electronic Devices and Circuits - Discussion
Discussion Forum : Electronic Devices and Circuits - Section 4 (Q.No. 48)
48.
In a reverse biased p-n junction, the reverse bias is 4V. The junction capacitance is about
Discussion:
7 comments Page 1 of 1.
Syam krishnan b said:
10 years ago
How it comes please explain?
Hari charan said:
9 years ago
Explanation please.
Mukund said:
9 years ago
Anyone please explain it.
Miles Carlson said:
8 years ago
How did this happen? please explain.
Nivedha said:
7 years ago
Actually, we can guess the answer. Junction capacitance is inversely proportional to reverse voltage. Here reverse voltage is more so capacitance has to be less.
(2)
Nevada said:
4 years ago
Please explain it in detail.
Kyle said:
9 months ago
Here’s the equation in a simple copy-paste format:
Csubj = Csub0/sqrt(1 + Vsubr/Vsub0).
Given:
Vr = 4V (reverse bias).
V0 ≈ 0.7V (built-in potential for silicon).
Since junction capacitance decreases with increasing reverse bias, the expected value is 20 pF.
Final Answer: 20 pF.
Csubj = Csub0/sqrt(1 + Vsubr/Vsub0).
Given:
Vr = 4V (reverse bias).
V0 ≈ 0.7V (built-in potential for silicon).
Since junction capacitance decreases with increasing reverse bias, the expected value is 20 pF.
Final Answer: 20 pF.
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