Electronics and Communication Engineering - Analog Electronics - Discussion

12. 

In figure, VEB = 0.6 V, β = 99. Then VC and IC are

[A]. 9.3 V and 1.98 mA respectively
[B]. 4.6 V and 1.98 mA respectively
[C]. 9.3 V and 0.02 mA respectively
[D]. 4.6 V and 0.02 mA respectively

Answer: Option A

Explanation:

VC = 20 - 1.98 x 10-3 x 5.4 x 103 9.3 V.


Sandeep said: (Jan 26, 2015)  
Why 400/9 is taken?

Arif said: (Feb 28, 2015)  
Use these formulas:

Ib = (Vcc-Vbe)/(Rb+BRc).

Ic = BIb.

Vc = Vcc-IcRc.

Sivaiah said: (Jul 28, 2016)  
In question, they mention it as Veb but it is Vbe.

Chenelyn Cedron said: (Aug 3, 2016)  
@Sandeep It's 400/99, just a typo error.

In fact, if you are bookish enough and you hate approximations, your first equation might look like this:

Vcc = V{5.4k} + V{400k} + V{be}.
Vcc = ((B)(Ib) + 1)(5.4k)(Ic)/B + (400k)(Ic)/B + V{be}.

Substituting the given gives:
20 = (99 + 1)(5.4k)(Ic)/(99) + (400k)(Ic)/(99) + 0.6.
This would give you an Ic of 2.043 mA.

Then since Vce = Vc = Vcc - IcRc,
Vc = 20 - (2.043mA)(5.4k) = 8.97 V. Pretty close to letter A.

Neeraj said: (Mar 21, 2017)  
Operating Vbe should be 0.7.

Prutha said: (Oct 19, 2017)  
I am Not clear with the solution. Please explain.

Aman said: (Jan 10, 2019)  
Apply kvl from 20-Ic * 5.4-Ib * 400-0.6 = 0.
You will get Ib=. 02ma.then Ic=BIb=1.98ma.
Now you can easily find Vc.

Karthik said: (Jan 28, 2019)  
What is 400/9?

Vasanthakumara D said: (Oct 6, 2019)  
This question 400/9 how come 9?

Rajani said: (Oct 10, 2019)  
It should be 400/99 not 400/9.

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