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hi everyone so my question is how do i calculate the equivalent we see from Vo node (output of op-amp) in pspice beacause anything i do i get 0 value it will be appriciated if one of you help me to do this with pspice

and my circuit :

[![enter image description here][2]][2]

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  • \$\begingroup\$ For an ideal op-amp the output impedance is 0 ohm. For a real op-amp it will be higher but without knowing which one you are using we cant help more. \$\endgroup\$ – Warren Hill Jul 1 at 13:25
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    \$\begingroup\$ imagine im using ua741 and clearly its not ideal \$\endgroup\$ – mahyar jafari Jul 1 at 13:30
  • \$\begingroup\$ Show your complete PSpice circuit including the stimulus source, load, and simulation card. \$\endgroup\$ – The Photon Jul 1 at 13:46
  • \$\begingroup\$ added it you can see in post \$\endgroup\$ – mahyar jafari Jul 1 at 13:54
  • \$\begingroup\$ What is your op amp? Some simple op amps can't drive high and stay at 0 unless you pull output up to positive supply, which, I guess, could be the case. Put some 4.7k-20k between op amp output and vcc and see if it runs \$\endgroup\$ – Ilya Jul 1 at 14:00
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Because negative feedback is used in this op-amp circuit, the op-amp's massive open-loop gain ensures that the effective output impedance of the circuit is very close to zero ohms. As frequency rises, the op-amp open-loop gain falls at 6 dB/octave and the output impedance rises.

However, if you are just interested in DC or low-frequency conditions, you can assume that the open-loop gain is high enough to make the circuit's output impedance below 0.1 ohms (but don't assume a typical op-amp circuit can deliver amps of current because it cannot).

op-amp thevenin resistance calculation in pspice

If you wanted to determine this in PSPICE, then you could attach a load resistor to the output and see how much the voltage changes. That change in voltage divided by the change in output current (when you apply the load) is the dynamic output impedance of the circuit.

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  • \$\begingroup\$ i added my pspice circuit in my post too you can see it but when i add a load resistance i still get the save voltage i get when my load is infinity so it gives my zero can you please explain it more ? \$\endgroup\$ – mahyar jafari Jul 1 at 13:55
  • \$\begingroup\$ Show how you connected the load resistor and what value you chose. Ideally, you should use a DC level that maybe creates an offset of 10 mV at the output then add a 1 ohm resistor to 0 volts and see how the output changes. You need to examine it in fine detail. \$\endgroup\$ – Andy aka Jul 1 at 14:28
  • \$\begingroup\$ i edited to post with how i put r load in my output but no matter what value for it i choose it will always give me the offset value of 10mV which you mentioned i created it with 0.001 voltage source in the output \$\endgroup\$ – mahyar jafari Jul 1 at 14:55
  • \$\begingroup\$ How many decimal points of resolution does your simulator provide you? You could feed the output into a subtractor with the other input to the subtractor being 0.01 volts exactly and look at the output for microvolt levels. If you still get 0.00000000 volts then you have to conclude that the model for the 741 is not very representative of real life. \$\endgroup\$ – Andy aka Jul 1 at 15:03
  • \$\begingroup\$ thank you very much for your help \$\endgroup\$ – mahyar jafari Jul 2 at 11:14
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Here is one way to do it. I've used an LM358 because I don't have an LM741 model installed for obvious reasons.

enter image description here

The trace shows the op-amp output. AC signal is 1Hz 1V peak with an offset of 3V so 200uA p-p injected into the output (I also could have used a current source, the offset is to keep the output in class-A operation). The frequency was chosen to be well below the ~10Hz pole in the amplifier response.

End result is an amplitude of 21.6238mV - 21.6261mV = 2.3uV, for an output resistance of 11m\$\Omega\$. Double checking at 0.1Hz we get exactly the same number, but at 100Hz we get more than 10x worse - 145m\$\Omega\$ as the open-loop gain of the amplifier drops precipitously at such a high frequency.

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  • \$\begingroup\$ thanks a lot it was so helpful \$\endgroup\$ – mahyar jafari Jul 2 at 11:13
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Apply a small voltage to the input like 0.1V, then measure the output without any load connected to the opamp, my simulator shows -987.88072 mV, then apply a 1Kohm load to the output and measure the voltage, my simulator shows -987.87663 mV. Now, use the voltage divider formula and solve for the opamp resistance as follows:

$$ OutputR = (\frac{Unloaded Voltage}{Loaded Voltage}-1)Load$$

So it should be

$$ OutputR = (\frac{-987.88072}{-987.87663}-1)1000$$

This gives an output resistance of 0.00414 Ohm

To check for consistency, the closed loop output resistance can be roughly calculated from the open loop output resistance as

$$ ClosedLoopR = \frac{OpenLoopR}{1+a \beta}$$

Where \$ a \$ is the open loop gain and \$ \beta\$ is the feedback factor

In the data sheet of the 741 the open loop output resistance is 75 Ohm and the open loop gain is 200,000 V/V. In the opamp circuit you presented, \$ \beta = 0.0909091\$ , so plugin everything into the formula gives a closed loop output resistance of 0.00412 Ohm, which is pretty close to the simulation result.

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  • \$\begingroup\$ thank you very much sir \$\endgroup\$ – mahyar jafari Jul 2 at 11:13

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