Under what circumstances would the large internal input resistance of an actual op amp affect circuit operation?

I was able to find a lot about why the input resistance is high and basically infinite. I understand that the input resistance is high so that it doesn't become a load on the signal. I also know that it makes sense like a voltage divider, the high impedance means that all of the voltage drops on the op amp. But I can't find any research on a case where the high resistance might actually affect the circuit itself (despite how high it is).

• the high impedance means that all of the voltage drops on the op amp That's a weird sentence. You probably mean: the opamp's high input impedance means that it does not load the voltage divider so the value of the voltage is not affected Oct 9, 2018 at 9:26
• But I can't find any research on a case where the high resistance might actually affect the circuit itself (despite how high it is). Why would it be an issue? The high input impedance is there to not affect the circuit so why do you look for examples where there is an effect? Oct 9, 2018 at 9:31
• @Bimpelrekkie I want to know because despite the fact that it was built to not affect things, I imagine that there are scenarios where it could and want to know what these may be. Because I'm curious and maybe it could be an issue one day and I'd never know. Oct 9, 2018 at 9:34
• If you've designed an op-amp circuit and it is susceptible to problems due to high input impedance then you have designed it wrongly. Oct 9, 2018 at 9:37
• Instead of learning about what could be an issue, learn how to do things right. Suppose I make a list of things not to do and one with good practices. Which list will be longer? If you design an opamp circuit and the feedback resistors are in the same value as the opamp's input impedance (assuming that would be possible with a CMOS opamp with 1Tera ohm input resistance) then as Andy says: you're doing it wrong. So use reasonable value resistors. Oct 9, 2018 at 9:39

whether it's an inverting or non-inverting circuit, if the feedback resistance $$\R_F\$$ and the other resistor connected to the inverting input terminal (usually called $$\R_1\$$), gets as large as the internal input resistance, then the internal input resistance of the op-amp begins to affect the circuit.

• ....."begins to affect" ? According to my experineces, a remarkable influence of the finite opamp input impedance can be observed already when the effective external resistance at one of the input nodes is smaller by a factor of 5...10 than the opamps input resistance
– LvW
Oct 9, 2018 at 10:19
• you're right @LvW. i was understating it. Oct 9, 2018 at 21:05

Each opamp parameter which normally is neglected during calculation will affect the operation (the gain) of the amplifier (input and output impedances, finite and frequency-dependent open-loop gain,...).

However, in most cases we do not care about these effects because they will be either not too important (because the error is acceptable) or are overshadowed by external tolerances of the feedback network and/or parasitic influences caused by the hardware realisation (pin capacitances,...).

This is a typical example for the fact that in electronics no formnula is correct by 100%. It is simply not possible - and it makes no sense - to include all possible known physical effects in our expressions, functions and formulas. And it is one of the most challenging engineering tasks to decide if - for a specific application - the "simplified" expression may be applied with sufficient accuracy or not.

As far as the finite input impedance is concerned, we try to follow a general rule which requires that each of the external resistors should be small if compared with this input resistance. However, what means "small"? Factor 10 or 100 or 1000 ? The answer simply depends on the required accuracy of the gain value. But in most cases, the resistor tolerances are more important.

As an example: Here is the gain expression (non-inverting) for resistive feedback (ideal: 1+R2/R1) if input and output resistances are taken into account.

G=N/D with

N=(Eo R2 Rin + Eo R1 Rin + R1 Rout)

D=(Eo R1 Rin + Rin Rout + R2 Rin + R1 Rout + R1 Rin + R1 R2)

Note that the open-loop gain Eo is set to a fixed value. You can imagine how the expression would look like in case of a realistice frequency-dependent gain expression for a two pole model:

Eo(w)=Eoo/[(1+s/w1)(1+s/w2)]

If you have a low-gain opamp, perhaps in a high-frequency opamp where you are just happy as a clam to have a 1GHz UGBW opamp, and you are achieving that high UGBW with bipolar devices because the SiGe transistors are what you need to use, and you want high-gain-accuracy out where the gain-margin is poor (out near the closed loop F3dB) and the voltage across the "virtual ground" between (-) and (+) inputs becomes significant, that additional current thru the internal differential-mode resistance must be included in the modeling. {note: UGBW is Unity Gain Band Width}

Examine this comparison; The opamp model was edited to make UGBW=1GHz, keeping the high-value gain-set resistors. The RIn was edited: left graph is 2 MegOhms, middle is 2 KOhms; right graph is 2 KOHms, and openloopgain reduced from 100dB to 60dB.

• Your audience won't know what UGBW means Oct 9, 2018 at 11:04

You have it a little backwards. In an ideal op amp, there is no current entering the amplifier inputs. The behavior ddviates from ideal when this is not the case, meaning the equations are not accurate.

Thus, manufacturers make op amps with high input impedance so the behavior approaches ideal.