Homework Help: What kind of op-amp is that?

1. Dec 2, 2011

Femme_physics

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2. Dec 2, 2011

technician

I have never seen anything like this !!! Are you certain you have the correct diagram.
There are applications where Vout is connected back to the - input (this gives a gain of x1) but this circuit seems to me to show some inconsistencies.
I cannot see how Va can be equal to Vb.
I look forward to see the response of others

3. Dec 2, 2011

Femme_physics

Oh, that reminds me! I forgot to ask my teacher about what you confirmed was correct! I have a class with him sunday, I will definitely remember.

Yes, I'm looking forward to other replies as well-- even though you seem to have great knowledge at the field yourself.

4. Dec 2, 2011

technician

The first thing I did was re-draw your circuit.... I kept R1, R3 and R4 exactly as they are but I then drew R2 just above R3 connecting to the - input.
The connection from Vout to point b remains as it is.
This just did not 'look right'....
I love op amps.... they look very complicated but the principles are straight forward and I like sorting them out.

5. Dec 2, 2011

Staff: Mentor

It's a tricky divide by 4 circuit. That is, Vout = Vin/4. To analyze, recognize that the feedback from Vout to b will make a = b = Vout. Then forget about the op-amp and concentrate on the resistor network, putting in a controlled source "Vout" to stand in for the the op-amp:

All resistors are 5KΩ.

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6. Dec 2, 2011

technician

Thanks gneill I tried something like that to make Vout = Va =Vb and got tied in Knots !!!!
It is not really a normal question about op amps.... difficult to see this as a practical application.

7. Dec 2, 2011

Staff: Mentor

Yeah, you might say that it's more of an academic exercise

As for its practicality, I suppose one would have to check to see how accurate the division operation is for the given component tolerances as compared to other configurations.

Last edited: Dec 2, 2011
8. Dec 2, 2011

Femme_physics

I forgot to add that the supply voltages are +-10

Is there a name for this op-amp, gneill?

REally? That's it? That's the answer?

9. Dec 2, 2011

technician

I can think of a name for it (if it doesn't already have one)

10. Dec 2, 2011

Staff: Mentor

I don't know if there's any specific name, but it certainly falls under the general heading of non-inverting amplifier. I'll admit that the positive feedback loop in addition to the negative feedback loop makes it a tad unusual. And yes, the amplification factor is 1/4. You should see if you can derive it from the hints given.

11. Dec 2, 2011

Femme_physics

Can you elaborate more please? I can't see the logic of just dividing by 4.....is it purely based on KVL and KCL?

12. Dec 2, 2011

Staff: Mentor

Yes, KVL, KCL, the usual analysis tools. If you can see how the diagram I gave arises (here it is again):

Then you can start from there. Suppose that all resistors are R. Apply nodal analysis at the node labelled Vx. You'll get an equation involving Vin, Vout, and Vx. Make Vx vanish by noting that i3 = Vx/(2R), and that Vout = i3*R. Re-arrange what's left to find Vout in terms of Vin.

13. Dec 2, 2011

Staff: Mentor

Just in case it's not obvious where the "reduced" circuit came from, here it is again but with the op-amp shown in place. Note that all I really did was to make the op-amp into a controlled source.

If you think about it, the op-amp is in a voltage follower configuration with the output fed back directly to its "-" input. So it really is a controlled source: it's output is 1x the voltage across the resistor at its + input.

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14. Dec 3, 2011

Femme_physics

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15. Dec 3, 2011

I like Serena

Yes, it is like that.

But your Sum Va is not quite right.
You're forgetting the voltage of the controlled power supply, which is the same as Vout.
I'm afraid gneill left that out of his (otherwise neat and insightful) drawing.

16. Dec 3, 2011

Staff: Mentor

It's the diamond shape at the bottom left, labelled with the carefully concealed clue "Vout".

17. Dec 3, 2011

I like Serena

Sorry!
Yes it is there.

@Fp: so you should include Vout in your equation.

18. Dec 3, 2011

Staff: Mentor

@FP: also note that your path V(b) should include the voltage drop across R1 due to I1.

19. Dec 3, 2011

Femme_physics

OK, rather surprising! but ok..
:shy:
-------------------------------
So I3R4 = Vout?

Ok, think I got it.

http://img72.imageshack.us/img72/6636/pitaron.jpg [Broken]

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20. Dec 3, 2011

I like Serena

How so?

Yes.

Looks good!
Don't forget to use Vout=I3R4.

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