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I was wondering... I have read about ideal op amp, that we could change this circuit :

schematic

Into this one (with the ideal model), where the input resistance Ri is infinite, and the output resistance negligible (so Ro = 0 Ohm) :

schematic

simulate this circuit – Schematic created using CircuitLab

We then told me that, since the Ri is infinite, the voltage at node A is zero (since Vi = 0), so Va = 0 (node voltage).

But what happens if we have that?

schematic

simulate this circuit

Will the voltage at A still be zero? Does it changes absolutely nothing to have Vs2 there? I'm confused on this one. I've seen only problems with the + gate of the op amp at ground.

Thanks!

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    \$\begingroup\$ It is HIGHLY educational to set up and solve the equations for the first circuit, assuming that the input resistance is infinite and the gain is infinite. DO YOUR HOMEWORK, and you will experience enlightenment. Trust me on this one. (I speak from experience.) \$\endgroup\$ – John R. Strohm Mar 10 '14 at 5:24
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Will the voltage at A still be zero? Does it changes absolutely nothing to have Vs2 there?

To find the answer, write an equation for the node voltage \$v_A\$ using superposition. For simplicity, assume the input resistance \$R_I\$ is infinite.

$$v_A = v_S \frac{R_2}{R_1 + R_2} + v_O \frac{R_1}{R_1 + R_2}$$

We also have

$$v_I = v_{S2}-v_A $$

$$v_O = Av_I $$

Thus, the first equation becomes

$$v_A = v_S \frac{R_2}{R_1 + R_2} + A(v_{S2}-v_A) \frac{R_1}{R_1 + R_2}$$

Rearranging yields

$$v_A = \frac{v_SR_2 + Av_{S2}R_1}{(1 + A)R_1 + R_2}$$

This is the general expression for \$v_A\$. As the gain \$A\$ gets very large,

$$v_A \approx \frac{Av_{S2}R_1}{AR_1} = v_{S2}$$

In the ideal case where the gain is 'infinite', this expression is exact. This is why we say there is a virtual short between the two input terminals of an ideal op-amp with negative feedback - the negative feedback ensures the inverting input voltage is identical to the non-inverting input voltage.

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  • \$\begingroup\$ AHHHHHHhhhhh... I should have done by maths before assuming it was some uncommon and frightening case. Thanks a lot, exactly the kind of thing I was looking for! \$\endgroup\$ – Yannick Mar 10 '14 at 19:51
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Node A voltage, in the conventional perfect model of the opamp, always equals the voltage on the non inverting input. If it didn't, due to the opamp gain being infinite, the output would be infinite or, in a less than perfect model, the output would be hard against one of the power rails.

Try this. Think of the op-amp as a control system: -

enter image description here

You set the demand with one pot (on the left) and the motor rotates until the other pot reaches a position where the voltage it produces equals the demand voltage.

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  • \$\begingroup\$ But what does it changes exactly in the circuit that we put the Vs2 source on the +? Does it affect the circuit or not at all (i.e. could we completely ignore it in equations?). Sorry also, I don't study this in english, so what do you mean by "non inverting input"? The + gate, the - gate? Do you mean that the Va = Vs2 ? \$\endgroup\$ – Yannick Mar 10 '14 at 0:13
  • \$\begingroup\$ The non-inverting input is the + input. Since we have a linear system, superposition applies. So grounding Vs gives us an output of Vs2*(1+R2/R1). Grounding Vs2 gives us an output of -Vs*R2/R1. The final output is the sum of both. \$\endgroup\$ – John D Mar 10 '14 at 0:57
  • \$\begingroup\$ What I don't get is why voltage at node A would be equal to Vs2? Since we have an infinite Ri, it acts as an open circuit right? So the point under the open circuit (under Ri) should be Vs2 and the node at A should be zero, no? If someone could answer that with a simple schematic, I'd be very happy. \$\endgroup\$ – Yannick Mar 10 '14 at 1:44
  • \$\begingroup\$ @Yannick forget about Ri. Think of the op-amp as a control system; you put a demand voltage on +Vin (non inverting input) and the output voltage rises or falls to a level where the error is zero. The error is (+Vin minus -Vin). It's called negative feedback and an op-amp with very high gain means that the error voltage required to drive the output pin will be very small. As gain gets to infinite, the error voltage is zero. \$\endgroup\$ – Andy aka Mar 10 '14 at 8:14
  • \$\begingroup\$ @Andy, I see. So basically, we have to assume Vi is always zero (in the ideal op amp ofc), so that the infinite gain does not produce something infine at the output. Correct? \$\endgroup\$ – Yannick Mar 10 '14 at 9:46

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