Electronics and Communication Engineering - Analog Electronics - Discussion

We will apply for normal kvl.

First, understand that here feedback resistor is placed between collector and base to keep the transistor in active region irrespective of beta value

So current through 5.4k will be Ic+Ib. Current through 400k will be Ib.
Now apply kvl.
20 = 5.4k *(Ic+Ib) + 400k * Ib + Vbe.
Substitute Ic with beta*Ib.
We will get Ib = .02mA.
Now Ic =beta*Ib = 1.98mA.
Vc = Vcc- Rc( Ib + Ic)= 9.2V.

It should be 400/99 not 400/9.

This question 400/9 how come 9?

What is 400/9?

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.

I am Not clear with the solution. Please explain.

Operating Vbe should be 0.7.

@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.

In question, they mention it as Veb but it is Vbe.

Use these formulas:

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

Ic = BIb.

Vc = Vcc-IcRc.