# Finding Vout on Op Amp

Discussion in 'Homework Help' started by wtrow, Oct 22, 2009.

1. ### wtrow Thread Starter New Member

Oct 21, 2009
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The picture is bad quality, I am sorry. The 2 nodes on the far left are v1(top) and v2(bottom). v1's resistor is R1 and v2's is R2. v1 is connected to the negative side of the op amp, v2 the positive side. The resistor above the op amp is R3 and the one below it is R4. The node on top of the op amp says Vcc=16V. The node on the far right is Vo (Vout).

Anyways, I normally approach these by doing KCL, however, there are no nodes here (unless I am allowed to count the currents that equal zero, as this is an ideal op amp). I haven't really been taught to solve op amps any other way. I am supposed to get Vo in terms of v1,v2, and the resistors. Any help would be appreciated. Thanks

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2. ### t_n_k AAC Fanatic!

Mar 6, 2009
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The "trick" to finding a solution is to realize that if the amplifier is ideal and the values are in the linear operating range then the voltage at the -ve and +ve inputs will be equal.

So denote the voltage at the negative input as Vx and the voltage at the positive input as Vy.

The voltage Vx will be given by Vx=V1-(R1/(R1+R2))*(V1-Vo)

You can prove this by writing the equation for the current flow from V1 via R1 and R2 to Vo and taking account of the voltage potential differences across R1 & R2. Since the amplifier is ideal, no current flows into either input terminal and hence the same current flows in R1 & R2.

You can also write an equation for Vy based on similar reasoning. You then determine the relationship between Vo and V1 & V2 by substituting the derived equations for Vx and Vy into the equality Vx=Vy.

3. ### wtrow Thread Starter New Member

Oct 21, 2009
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that seems to work all right, thanks.

4. ### Ratch New Member

Mar 20, 2007
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t_n_k,

I believe you really mean Vx=V1-(R1/(R1+R3))*(V1-Vo) .

Does anyone want to derive for The Electrician, the expression for Vo if the OP amp has finite gain?

Ratch

5. ### wtrow Thread Starter New Member

Oct 21, 2009
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Ratch - I caught that mistake already. Good eye though.

6. ### t_n_k AAC Fanatic!

Mar 6, 2009
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Thanks Ratch - it was my poor eyesight wrt the OP's fuzzy circuit. I transposed the resistors. Did you use optical enhancement to improve matters?

7. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
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I had a go and found.

Vo={[R4/(R2+R4)]V2-[R3/(R1+R3)]V1}/{[R1/(R1+R3)]-[R2/(R2+R4)]+[1/Av]}

Hopefully I had the resistors in the right configuration wrt the OP's circuit.

8. ### t_n_k AAC Fanatic!

Mar 6, 2009
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OK I see the OP's original details on the circuit layout in post#1. I didn't read it closely.....

I've a nagging feeling I've seen something like this circuit before in a long-forgotten post. It's the concurrent positive and negative feedback that "un-nerves" me a little.

In fact, to simulate this I had to decouple the +ve feedback by splitting the 1.2kΩ into two (equal) parts and bypassing the common point of the two R's (600Ω) via a capacitor (I used 10uF) to ground. Without the decoupling the output latches to one or other of the rails - depending on the relative values of V1 and V2. Presumably this was the positive feedback coming into play due to a simulation instability - I used a transient analysis.

With the decoupling applied the simulation seemed to follow my theoretical analysis.

So with Av=∞ (ideal op-amp) I have Vout=-3.2333..V with the values given in the original question.

Does that make sense to anyone?

9. ### wtrow Thread Starter New Member

Oct 21, 2009
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not really....but i got the same answer lol

10. ### t_n_k AAC Fanatic!

Mar 6, 2009
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If you have simulation software have you tried running a simulation of the circuit?

11. ### Ratch New Member

Mar 20, 2007
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To the Ineffable All,

I did the problem on Maple. As you can see, it saves a lot of work. All your solutions were correct.

Ratch

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12. ### Ratch New Member

Mar 20, 2007
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T N K,

No, I read the OP's description of the circuit.

Ratch

13. ### The Electrician AAC Fanatic!

Oct 9, 2007
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I would have expected this circuit to be unstable as regards its DC operating point. After you added the bypass cap, did you allow the simulation to run a long time? I would expect that it would eventually latch up to one of the rails.

If the resistors are swapped around:

R1 = 200
R2 = 100
R3 = 1200
R4 = 2400

then Vout = 3.83333, and it should be DC stable.

The problem of determining DC stability of circuits like this is still an unsolved problem as far as I know. I mentioned an interesting paper on the topic in post #34 of this thread: http://forum.allaboutcircuits.com/showthread.php?t=20636

14. ### hgmjr Retired Moderator

Jan 28, 2005
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t_n_k,

Perhaps you are thinking of the Howland Current Pump Circuit.

hgmjr

15. ### t_n_k AAC Fanatic!

Mar 6, 2009
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Aha! - Thanks hgmjr.

16. ### The Electrician AAC Fanatic!

Oct 9, 2007
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Could you run your simulation (with the bypass cap) for a long time, and see it the output eventually reaches the rail voltage?

17. ### t_n_k AAC Fanatic!

Mar 6, 2009
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Thanks Electrician - Curiously I have just re-run the same simulation and I couldn't get it to go unstable at all - even with the bypass removed. Looks like the app. software is puzzled as well. Thanks also for the pointer to your other post. I'll have a look.