Will a resistor fixed just in front of an op amp change its gain?

We have a circuit as shown in the image. Now if we didn't have the resistor Rp, the circuit is a simple negative feedback circuit with gain given by -(R2 / R1).

But the presence of Rp changes the circuit. However, what I think is that since no current flows via an ideal op amp, hence no current flows via Rp, therefore its presence changes nothing. So the circuit is the same without Rp, hence the gain of the circuit in the image it's still -(R2 / R1).

Will a resistor fixed just in front of an Op Amp change its gain?

RP is after the op-amp. (Note that op-amp is an abbreviation of operational-amplifier. It's not a proper noun so it doesn't get capital letters.)

However what I think is that since no current flows via an ideal Op Amp hence no current flows via Rp, therefore it's presence changes nothing.

No current flows into the input in an ideal op-amp. An op-amp with no output current would be useless.

RP does change the gain of the op-amp circuit (if you consider the left side of R1 as the input and the output of the op-amp itself as the output). To get the required VO the op-amp now has to "stretch" further because of the potential divider effect of RP and R2. The gain of the circuit is $$\ - \frac {R_2}{R_1} \$$ at VO but it's $$\ - \frac {R_2 + R_P}{R_1} \$$ at the output of the op-amp itself.

• Thank you , Do you mean that at the far right the circuit still has a Gain of -(R2/R1)? Commented Jun 24, 2020 at 14:47
• @Yasir Sadiq, read again Transistor's answer above; he has shown the same expression. Also, try to imagine what does "the op-amp now has to 'stretch' further" means. Yes, the answer is this... but it is more important to understand why... Commented Jun 24, 2020 at 16:45
• The signal at the op amp output will be sensitive to changes in loading. Commented Jun 24, 2020 at 22:48
• @Circuit, thanks for the links and I've looked at many of your answers but just found that they were way too much work to go through. I try to pitch the answers at a level that the OP can understand and using graphical / schematic answers as much as I can to bring them one step further. I'm an industrial automation engineer so I don't teach this subject as my job. Commented Jul 3, 2020 at 21:07
• @Transistor, Thanks for the clarification. It is a good idea for each of us to describe what his/her goal is, so that there is no misunderstanding. Your explanation is fine; it stands in the middle between two extremes - the formal approach that does not really explain anything, and my heuristic approach that reveals in depth the ideas behind circuits. I repeat, I explain basic circuit ideas and how they are implemented in the specific circuit solutions. For this purpose, I need words… sometimes, many words… and richly illustrated pictures. As I usually say, this is the price of understanding. Commented Jul 4, 2020 at 9:57

An op-amp can drive a load (and it certainly drives current through the feedback loop) so current does pass through $$\R_P\$$ but, $$\R_P\$$ might as well be inside the op-amp because the op-amp isn't perfect; it has output transistors that have finite resistance of many tens to hundreds of ohms hence: -

Yes, the load driving capability is reduced but that won't be a big deal for most op-amp application circuits.

EDIT for the fantasists who don't understand op-amps

If you look at the data sheet of the LF351 it shows the internal schematic: -

And I draw your attention to the output resistors - these are just as relevant as the resistor $$\R_P\$$ in the question but is anyone doubting that this op-amp delivers a gain of $$\-R_2/R_1\$$?

All op-amps have output resistances.

• No name-calling or personal attacks. Focus on the content, not the person. This includes terms that feel personal even when they're applied to content (e.g. “lazy”). Please see: electronics.stackexchange.com/conduct Commented Jun 24, 2020 at 19:57
• @Andy aka thank you, so we assume Rp is inside the op amp and hence the gain at VO remains the same. Commented Jun 24, 2020 at 23:53
• @YasirSadiq every op-amp output internally has resistance in series with the output transistors. This manifests itself as an equivalent resistor in series with the output but inside the chip package. But the actual output node is where we connect to and that is outside the package so, having another resistor that is outside the package and exclusively in series with the internal output resistor is of no consequence for most op-amp applications. So the gain remains the same at VO. Commented Jun 25, 2020 at 7:08
• @Yasir Sadiq...what would you anwer if somebody would ask you "WHY has a resistor inside the opamp" no influence"? To me, the answer is not yet given...just a claim that it would not matter. The answer is: Due to negative feedback the output resistance is reduced with a factor k=(1/loop gain). And the loop gain is Aol*R1/(R1+R2), which is a very large value (Aol=open-loop gain). Hence the reduction factor k is nearly zero.
– LvW
Commented Jun 25, 2020 at 15:13
• @LvW are you saying that my answer doesn't cover sufficient ground to explain this fairly straightforward problem? Commented Jun 25, 2020 at 15:29