# Multistage amplifier problem

Discussion in 'Homework Help' started by epsilonjon, Mar 5, 2012.

1. ### epsilonjon Thread Starter Member

Feb 15, 2011
65
1
Question: Calculate the dc voltage at the collector of Q1 if the capacitor C3 becomes shorted.

Attempt:

Assuming a stiff voltage divider at the base of Q1 I get VE1=10k*10/(10K+47k) - 0.7 = 1.05v. Therefore IE1 = 1.05mA. If I assume that the dc collector and emitter currents for Q1 are roughly the same, then I think we have this situation:

Applying KCL at node VC1 and doing the maths yields VC1=3.77v.

Could someone tell me if this is a correct method of solving the problem? The solutions manual I found for the book does it in some weird way which doesn't make sense to me, and gets a different answer too

Thanks!
Jon.

Last edited: Mar 5, 2012
2. ### Jony130 AAC Fanatic!

Feb 17, 2009
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My answer will look like this

Vc1 = 10V * (R6||((Hfe+1)*R8)/ ( (R3||R5) + (R6||((Hfe+1)*R8) ) - 1mA * R3||R5||(R6||((Hfe+1)*R8)

(R6||((Hfe+1)*R8) = 9.375K

R3||R5 = 4.27K

Vc1 = 10V * 9.375K/(4.27K + 9.375K) - 1mA * 4.27K||9.375K = 6.870V - 2.933V = 3.937V

Or if we simplify the circuit

Vc1 = 10V * R6/(R3 + R6) - 1mA * R6||R3 = 6.8V - 3.2V = 3.6V

But now Q2 is in saturation so our calculations are incorrect.

3. ### Jony130 AAC Fanatic!

Feb 17, 2009
4,901
1,378
So we now that for Vc1 3.9V Q2 is in saturation. So to find new Vc1 value we need use Thévenin's theorem.

Vth = 10V * R6/(R6 + R3||R5) - 1mA * R3||R5||R6 = 7V - 2.99V = 4.01V ( I use superposition)

Rth = R3||R5||R6 = 2.99K

And now situation look like this

So now we need to solve for Vb.
But it will be easier to solve for Ve first because Vb = Ve + Vbe
In saturation the only equations which holds for the BJT is
Ie = Ic + Ib

Ie = Ve / Re

Ib = ( Vth - (Vbe + Ve) /Rth

Ic = (Vcc - (Vce_sat + Ve) / Rc

Solve this for Ve I get this result

And if I assume Vbe = 0.65V and Vce_sat = 0.1V

Ve = 2.08V
and Vb = 2.08 + Vbe = 2.73V

So my final answer to you question

Calculate the dc voltage at the collector of Q1 if the capacitor C3 becomes shorted

is Vc1 ≈ 2.73V

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Last edited: Mar 5, 2012
4. ### epsilonjon Thread Starter Member

Feb 15, 2011
65
1

I see now where I went wrong in my initial attempt. I took the input resistance at the base of Q2 to be βR8 which is completely wrong (I was getting confused with the AC case). I corrected for this and get the same answer as you in your first post, which takes Q2 into saturation.

I've just finished Thevenizing the circuit and I get the same as you again .

Interestingly the solutions manual gives an answer of VC1=4.03v with some very strange workings out! :s

5. ### Jony130 AAC Fanatic!

Feb 17, 2009
4,901
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can you post the answer (picture) from the book?

Feb 15, 2011
65
1
Sure thing:

7. ### Jony130 AAC Fanatic!

Feb 17, 2009
4,901
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For sure there must be a error in the book.

8. ### epsilonjon Thread Starter Member

Feb 15, 2011
65
1
Yeah i'm really not sure what is going on with that one. It's a good job we have kind people like you to help out

9. ### Brownout Well-Known Member

Jan 10, 2012
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1,000
The book answer is for C2 shorted. The orginal question was for C3 shorted. Sure you read the question correctly?

10. ### epsilonjon Thread Starter Member

Feb 15, 2011
65
1
Ah yeah you're quite right. My book definitely says C3 shorted, but the solution does say C2 shorted. However, the answer they give wouldn't make sense for C2 shorted either. And all the resistor values they refer to are the same as in my diagram. I can't work out what they're doing! But nevermind, i'm going onto the next question anyway

Thanks again!

11. ### Brownout Well-Known Member

Jan 10, 2012
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1,000
You got me there. I see what they are doing. In the first part, they are calculating an equivilant resistance for the transistor r'ce as vce/ic. In the second part, they are using that equililant resistance in a voltage divider equation with (r'ce+r4)||rin(q2) at the bottom leg, and R3 at the top leg.

12. ### Brownout Well-Known Member

Jan 10, 2012
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However, the top leg of the divider should be R3||R5.

13. ### epsilonjon Thread Starter Member

Feb 15, 2011
65
1
Can I just ask one more quick related question?

If I have this circuit:

The question is, if R1 opens what happens to Q1? And what will be the Q1 dc collector voltage?

I am sure that it will go into saturation, but my book (the same book as before) says that it will go into cut-off and VC1 = VEE. (The answer for this question is at the back of the book, but with no working).

I really think this is incorrect, but could someone just confirm for me? I have only been learning about transistors for a short time so i'm not yet confident, and i'm just reading from books so don't have teachers to help.

Thanks again!!!

14. ### Jony130 AAC Fanatic!

Feb 17, 2009
4,901
1,378
If we assume Hfe = 150 and Vbe = 0.65V

Ib = ( 30V - 0.65V) / ( 330K + 151*34K) = 5.4uA

And Ic = 806uA

But

Ic_max = 30V/(22K + 1K + 33K ) = 536uA

So yes BJT is in saturation

15. ### epsilonjon Thread Starter Member

Feb 15, 2011
65
1
Thanks, that's what I got

Stupid book!

16. ### Brownout Well-Known Member

Jan 10, 2012
2,364
1,000
It probably gave the answer for R2 open.

17. ### epsilonjon Thread Starter Member

Feb 15, 2011
65
1
Yeah must be. This book is so filled with errors it's criminal