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Recently we have been working with opamps in the lab and I never can understand how to determin the output and input impedance of a circuit with op-amps (not the input and output inpedance of the circuit itself)

We have been measuring it by placing a voltage signal in the output of an inverting amplifier with its imputs grounded. Measuring the voltage over an impedance we have found the current flowing in

schematic

simulate this circuit – Schematic created using CircuitLab

Then the output impedance is found to be the voltage at the output (Vout) divided by the current.

I would like to know what the input and output resistances in opamp circuits represent and if possible how to obtain them from equations

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  • \$\begingroup\$ Good question. You would not believe now many new users post a terse, "this doesn't work" question like they assume that everyone's been watching over their shoulder. \$\endgroup\$ – AaronD Feb 11 '15 at 20:50
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The input resistance is the ratio between the change in the input voltage and the input current (or just between the input voltage and current in case of linear systems). In your case, assuming the input is on the negative side of the opamp, the input current is \$V_{in}/R_2\$ (because of the virtual ground on the negative terminal). So the input resistance would be \$R_2\$. If \$V_{in}\$ is applied to the positive side, the input resistance will be close to infinity, since there is no input current.
As for the output resistance, it can be obtained by connecting a known load \$R_L\$, measuring the voltage on it \$V_L\$, and then calculate the simple voltage divider problem: \$V_L=V_{out}R_L/(R_L+R_o)\$, where \$R_o\$ is the output resistance, and \$V_{out}\$ is calculated as for ideal opamp.

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  • \$\begingroup\$ Is there anyway that R0 can be obtained without experimentation? it is for an exam. Thanks anyway \$\endgroup\$ – gorilon Feb 11 '15 at 22:05
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    \$\begingroup\$ Without additional data it is not possible. Either the open-loop resistance should be given or the output currents and/or voltages when connected to a load \$\endgroup\$ – Eugene Sh. Feb 11 '15 at 22:12
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I would like to know what the input and output resistances in opamp circuits represent and if possible how to obtain them from equations

I assume you are asking for "theortical" formulas, right? OK - from system theory we derive the following expressions for the whole circuit; all opamp resistances without feedback will be drastically altered due to feedback (Loop Gain LG):

1.) Non-inv. input: r,p=rp,o*(1+LG)

2.) inv. input: r,i=rn,o/(1+LG)

3.) input R2: r,2=R2+r,i (very close to R2)

4.) output: r,out=r,o/(1+LG)

rp,o and rn,o: dynamic input resistances without feedback;

r,o: dyn. output resistance without feedback;

Loop Gain: LG=Aol*feedback factor=AoR2/(R1+R2).

PS: Your measurement of r,out is OK (in principle). However, I suggest (a) to use a smaller external resitor (R3=1...5 Ohms) to realize a suitable voltage divider (together with r,out) or (b) to use a much larger resistor for realizing a "good" current source (current practically determined by R3 only).

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As you may have covered in class (or maybe not yet), the three rules for (ideal) opamps are:

  1. The inputs take no current.
  2. The output voltage travels in the direction of (+in minus -in). If +in is bigger, the output increases; if -in is bigger, the output decreases.
  3. The output will go as far as it needs to to make the two inputs equal. (Note: for real opamps, it cannot exceed the power supply and so will stop there)

Now, you've tied +in to ground, so it's going to do whatever is necessary to hold -in also at ground. The two resistors R1 and R2 make a sort of lever, if you will, because their center tap takes no current. By tying one end of that to ground, the opamp trying to keep the center tap also at ground must therefore keep its output at ground. So now you can replace the entire opamp circuit with ground and see what you get.

If you're studying non-idealities, like the finite output impedance of a real opamp, then R3 is physically inside of the opamp chip. Not completely literally, but it behaves that way, because of actual resistance in the silicon die and the wires that connect it to the pins, and because of the overload protection that's built into even the dirt cheap ones. However, by including that non-ideality into the feedback (consider what would happen if you moved your right-most vertical wire to the right of R3), you can all but eliminate that effect at the expense of some headroom. (the output will clip easier because it's swinging farther to compensate for the weirdness)

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For the model of an ideal opamp with negative feedback, the output impedance of the circuit is zero.

The next more realistic model (for the inverting amplifier) would be $$R_{out\_of\_circuit} = R_{o\_of\_opamp} \frac{A_{CL\_CloseLoopGain}}{A_{O\_OpAmpOpenLoopGain}}$$
Which in the ideal case, \$A_O=\infty\$, it reverts back to the model on top.
I hope this helps.

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