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The datasheet of this opamp says the input impedance is typically 100Meg. So I expect the voltage gain to be halved with a source resistance of 100Meg. But the following simulation doesn't reveal that and gain almost doesn't change:

enter image description here

Above the voltage gain is two. Applied input is 1V. So with a 100Meg source resistance I was expecting 1V output(half of the gain). But the plots show the gain is not affected from the source impedance and is still two.

Why 100Meg source impedance is not halving the output gain if the opamp input impedance is 100Meg?

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  • \$\begingroup\$ Why do you expect changing the input resistance to affect the gain in that circuit? What do you expect the gain to be and how did you calculate it? \$\endgroup\$ – pjc50 Feb 28 at 11:27
  • \$\begingroup\$ Voltage divider effect at the input stage. \$\endgroup\$ – user1245 Feb 28 at 11:34
  • \$\begingroup\$ See also electronicdesign.com/technologies/analog/article/21795474/… \$\endgroup\$ – Brian Drummond Feb 28 at 12:38
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There are other factors to consider such as input bias current (quoted at anything up to 4.5 nA). With 4.5 nA flowing from the input into 100 Mohm, that produces an offset error of anything up to 0.45 volts. That blows the effect of the input resistance out of the water.

But, realistically, with a simulation, it all depends on what is set up in the model parameters. Not all models accurately reflect all op-amp artifacts.

The datasheet of this opamp says the input impedance is typically 100Meg

That figure is the differential input impedance and not the input impedance of one input to ground so, it will have much less of an effect. If you look on page 6 of the data sheet you will see that the common mode input impedance is graphed at about 280 Gohm. But like I said, input bias currents will always blow things apart at extremes.

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  • \$\begingroup\$ Aren't "impedance of one input to ground" and "common mode input impedance" very different things? \$\endgroup\$ – user1245 Feb 28 at 11:59
  • \$\begingroup\$ @user1245 please look at page 1 here: analog.com/media/en/training-seminars/tutorials/MT-040.pdf \$\endgroup\$ – Andy aka Feb 28 at 12:01
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    \$\begingroup\$ The differential impedance is bootstrapped by the feedback loop so the effective input impedance is increased by the loop gain. (and no I'm not telling Andy this!) \$\endgroup\$ – Brian Drummond Feb 28 at 12:19
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    \$\begingroup\$ @BrianDrummond exactly !! \$\endgroup\$ – Andy aka Feb 28 at 12:22
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The circuit input resistance is different from the opamp input resistance. With negative feedback you need to multiply the opamp input resistance by the loop gain to calculate the resistance seen by the source at the non-inverting input.

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If you look at this circuit, the following can be understood:

  1. First understand, all Op-Amp cirçuits are designed so that the device operates to ensure almost same voltage at it's (+) &(-) inputs. The feed-back ensures that. Any wrong feed-back circuit configuration will not work.

2.The minimum input difference between (+) & (-) inputs required on open loop condition is very very low, more the open loop gain, less the difference voltage required to swing the output. The input current. required is is almost negligible , since all of Op-Amps have differential amplifiers at inputs, and many even have Mos-Fet at input differeintial amo stage.

  1. In the above mentioned non-inverting Op-Amp circuit, you assume both (+) & (-) inputs are at at same voltage. If you have V(i) at (+) input through the 100 meg ohm, then the same voltage is maintained at(-) input. In order to ensure that, the voltage divider between output ,(-) input and ground requires V(-)= V(O) × ( R÷(R+R)), i.e V(-) = (1/2)×V(O) Thus means, gain is (V(O) ÷V(I))= 2 , no matter what resistance you use at input. Of course if you try a 100 G ohm input resistor, it may not work. That may not ensure minimum input current required (pico amps) !
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