# Operational amplifier

Discussion in 'Homework Help' started by regexp, Dec 28, 2010.

1. ### regexp Thread Starter New Member

Nov 20, 2010
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Hello,

I have the following operational amplifier:

My task is to calculate the expression that yields the quotient $~~\frac{u_0}{i_s}$

Hm, i don't really know how to start, any advice?

2. ### hgmjr Moderator

Jan 28, 2005
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Consider the features of an ideal opamp.

1. Infinite open loop gain.
2. Infinite input impedance on both input terminals
3. Output impedance is 0 ohms.

Also consider that an opamp's mission is to drives its output to whatever voltage it takes to make the two input terminals equal in voltage. In this opamp configuration, the positive terminal is at 0v. That means that the opamp is going to drive its output to a voltage that will force its negative terminal to be 0V.

hgmjr

3. ### shteii01 AAC Fanatic!

Feb 19, 2010
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$\frac{U_0}{I_s}=R_3(\frac{R_1}{R_2}+\frac{R_1}{R_3}-1)$
Can someone check me?

4. ### Ron H AAC Fanatic!

Apr 14, 2005
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That's not correct.

5. ### thatoneguy AAC Fanatic!

Feb 19, 2009
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Think about it as "What current will flow through R3?"

Since R1 and R2 are voltage dividers across the current source, and R3 is the tap to the output.....

6. ### shteii01 AAC Fanatic!

Feb 19, 2010
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Ok. So. There is no current in R3? Then U0=R2(Is)?

7. ### Ron H AAC Fanatic!

Apr 14, 2005
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No one said there is no current in R3. How did you conclude that?

8. ### shteii01 AAC Fanatic!

Feb 19, 2010
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I am just stumbling in the dark. When I solved it the first time, I thought R2 and R3 would act as current divider, that was shut down. Now I am out of ideas.

9. ### Jony130 AAC Fanatic!

Feb 17, 2009
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Here's a tip:
You can convert the T-pad network to a PI-pad network using the Y-Δ transformation:
http://en.wikipedia.org/wiki/Wye-delta

Or you could just write out your node equations

Is = ( Va - Vb) /R1 (1)

Is = Vb/R2 + (Vb - Vout)/R3 (2)

Va = 0V ( thanks to op amp "action" )

Last edited: Dec 29, 2010
10. ### Ron H AAC Fanatic!

Apr 14, 2005
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Are you the same person as regexp?

11. ### hgmjr Moderator

Jan 28, 2005
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Greetings Shteii01,

Like RonH, I too am confused. Your replies are phrased in such a manner that it sounds like you are the originator of this thread.

Can you please confirm that you "are" or "are not" regexp?

hgmjr

12. ### shteii01 AAC Fanatic!

Feb 19, 2010
3,297
482
This is not my thread. I am trying to solve it because I am not very good at op-amp circuits, I thought it would be good practice for me. Note that I only posted my answers, I still want regex do his own homework.

13. ### shteii01 AAC Fanatic!

Feb 19, 2010
3,297
482
I see what you mean. I think I can do node equations on the original circuit.

Ok, I redid the problem using node voltage method on the original circuit. I got following: $\frac{U_0}{I_s}=-R_3(\frac{R_1}{R_2}+\frac{R_1}{R_3}+1)$

Last edited: Dec 29, 2010
14. ### Jony130 AAC Fanatic!

Feb 17, 2009
3,939
1,089
Your equation look good for me

15. ### Ron H AAC Fanatic!

Apr 14, 2005
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Yeah, that's what I got too.

16. ### BillO Well-Known Member

Nov 24, 2008
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The minus sign looks wrong. Everything else looks right.

17. ### hgmjr Moderator

Jan 28, 2005
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I obtained the same answer as shteii01 using Millman's Theorem.

hgmjr

18. ### Ron H AAC Fanatic!

Apr 14, 2005
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The output voltage is bound to be negative, due to the direction of the current and the feedback.

19. ### BillO Well-Known Member

Nov 24, 2008
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Hmm, suit your self, but there are valid ways to work this out so that you do not end up with negaitve ohms as your answer.

Apr 14, 2005
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