Electronic Devices - Power Amplifiers - Discussion

Discussion Forum : Power Amplifiers - General Questions (Q.No. 2)
2.
Calculate VB2.

0 V
0.7 V
9.3 V
10.7 V
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
11 comments Page 1 of 2.

JJ T said:   1 decade ago
Hello, I am curious as to why 9.3V is the answer.

If we have 20 Vcc across a branch of resistance totaling R1 + D1 + D2 + R2, which is 1200 + 0.7 +0.7 + 1200 = 2401.4 Ohm.

If I run a voltage divider on this at VB2, I am getting 1200/2401.4(20) = which is 9.99V.

I tried to do it using voltage drops as well and I am still getting about 10V at VB2.

Could you please let me know if I am doing something wrong?

Rathi said:   1 decade ago
It is simple.

Voltage at the point between two diode is 10v(Vcc/2).

i.e. R1 + D1 + D2 + R2 voltage drops is 9.3 + 0.7 + 0.7 + 9.3 = 20v.

So VB2 = 9.3V.

Digvijay said:   5 years ago
See do dc analysis i.e open ckt all capacitors.

Then by kvl in loop of r1 d1 d2 r2.
20=i(1.2k+1.2k)+.7+.7,
So, i=7.75mA.
So drop at base= 1.2k * 7.75mA= 9.3V.
(5)

Abdul naweed said:   9 years ago
v/g across each diode is 10 v a considering diode as a diode as a silicon the vk of silicon is 0.7 therefore 10 - 0.7 give 9.3 v.

Saila said:   10 years ago
One confusion is how could you consider R1 as 9.3. Please answer.

A.Manoj Kumar said:   8 years ago
I am not getting this, Please explain the answer.

Anomi said:   7 years ago
Vcc-.7-.7 = 18.6.
(R2/(r1+r2))(18.6) = 9.3.

Harsh said:   4 years ago
@Digvijay.

Thank you for explaining.

Biradar Mayuri said:   5 years ago
Nice explanation. Thanks all.

Pragati said:   6 years ago
Please explain in detail.


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