# Using the ideal OP-Amp model

Discussion in 'Homework Help' started by Brutalmouse, Mar 21, 2014.

1. ### Brutalmouse Thread Starter New Member

Mar 21, 2014
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I'm stuck on this problem, and I feel like if I could understand this one I'd be able to understand a lot more of my homework. I don't even know where to start..

Mar 31, 2012
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3. ### Brutalmouse Thread Starter New Member

Mar 21, 2014
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OK!

Well the currents entering and leaving the + and - leads are zero, I know that much for an ideal op amp. Now, my book states that negative feedback causes the voltage Vo to assume the value required to make the error voltage ΔV 0. I know what the error voltage is, or I'm pretty sure it's the difference between V+ and V-. But I'm not sure what the mean by negative feedback.

4. ### WBahn Moderator

Mar 31, 2012
18,093
4,920
Negative feedback simply means that the output voltage affects the voltage at one or both of the inputs such that, all other things being equal, if the output voltage goes one way the resulting differential input voltage will be such as to drive the output voltage back in the direction it came from. Positive feedback would do the opposite and drive it further in the direction it was going.

5. ### crutschow Expert

Mar 14, 2008
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Negative feedback is voltage fed from the output back to an input of the opposite polarity. In the case of an opamp it goes from the output to the (-) input. Due to the high open loop gain of an opamp the negative feedback thus will always try to keep the (-) input at essentially the same voltage as the (+) input. If there is a difference in voltage between the two inputs then the opamp gain will change the output so as to minimize this difference.

Looking at your circuit if the input source voltage raises the (+) input voltage, then the output voltage will rise until the current through the "Unknown device" is sufficient to raise the voltage across R1 so the (-) input voltage equals the (+) input voltage. Thus the current through R equals Vin / R.

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6. ### LvW Active Member

Jun 13, 2013
674
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Yes - good explanation.

@Brutalmouse, now you should consider if the wanted output current i has the same value or not.

7. ### Brutalmouse Thread Starter New Member

Mar 21, 2014
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Very awesome. Okay, so I'm thinking the line connecting the resistance node and the - input doesn't even need to be there. Would that then make i -Vin/R? And where does a dependent source come in?

8. ### LvW Active Member

Jun 13, 2013
674
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Yes, you are right - if the task would be to create a current through a single resistor R.
But the task is another one. Therefore, my question in post#6.

9. ### Brutalmouse Thread Starter New Member

Mar 21, 2014
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Shouldn't it be the same but negative, since there is no current coming out of the top of the resistor?

10. ### LvW Active Member

Jun 13, 2013
674
100
That´s the question. Remember Ohm`s law.

11. ### Brutalmouse Thread Starter New Member

Mar 21, 2014
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OK, so would the equivalent circuit consist of a voltage controlled current source then, so that it depends on Vs? So it would still have Vs, R, the unknown device, and then a Vs/R current source pointing into the negative terminal of Vs?

12. ### Brutalmouse Thread Starter New Member

Mar 21, 2014
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or would the current i just be Vo / R ?

13. ### WBahn Moderator

Mar 31, 2012
18,093
4,920
The line connecting the resistor to the -input is where your negative feedback comes from. Disconnect it and you don't know what will happen because the opamp is running without feedback and an undefined input signal on one input. It will probably either rail or flail.

The opamp IS the dependent voltage source.

Vout = Av*(Vd)

Where Vd = (Vp-Vn)

14. ### WBahn Moderator

Mar 31, 2012
18,093
4,920
You should have in your text the dependent-voltage source model of an opamp. Try to find that (or look around on the web).

15. ### Brutalmouse Thread Starter New Member

Mar 21, 2014
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Well, they mean an equivalent circuit with one of these When they say dependent source.

16. ### Brutalmouse Thread Starter New Member

Mar 21, 2014
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OK, I am still having trouble understanding these. It makes absolutely no sense to me. Here is a picture of the configuration you mentioned:

Why is there no connection between the input and the output? Is that what the arrow indicates?

17. ### LvW Active Member

Jun 13, 2013
674
100
The figure shows an "equivalent circuit diagram" (not a "classical" circuit diagram with BJT´s or opamps) containing a controlled source.
There is no need to show a wired connection. The control function is included in written form (read the function which controls the source).
However, coming back to your original question, you do not need at all an equivalent circuit diagram to find an answer to your question (first posting). Ohm`s law and/or KCL is sufficient.

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