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So I have started learning about Operational Amplifiers. In case of idea Op-Amps, Vd=0 as amplification is very large and Vo is limited by Vcc(the supply voltage).

Here in this inverting Op-Amp configuration ,enter image description here

Vout= - (Rf/Rin) * Vin which I understand perfectly.

But I have 2 doubts :

1.What happens physically if we remove Rin i.e Rin tends to zero ? By the formula , Vout should become infinite . Does it happen really?

2.In an ideal op-amp , voltage at positive terminal=voltage at negative terminal but if we remove Rin , voltage at + terminal=Vin and voltage at negative is 0 which is kind of paradoxical.

What is the explanation here?

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    \$\begingroup\$ The question has some confusing wording. "if we remove Rin", implies that \$R_{in}\$ is replaced by an open circuit (\$R_{in}\to+\infty\$), while the rest of the question implies that \$R_{in}\to 0\$. Please pick one, or make a distinction between the two. \$\endgroup\$ – Sven B Jul 5 '18 at 10:00
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    \$\begingroup\$ I think, you are asking what happens if Rin becomes smaller and smaller until we have Rin=0. In this case, the signal source Vin is connected directly at the inv. terminal. Now, we will have no feedback at all (the feedback signal is shorted) and we also can remove Rf. That means: We have an opamp without negative feedback and the actual voltage gain will be 1E5....1E6. Hence, if the signal voltage is larger than app. 10µV....100µV the output will go into saturation. This consideration assumes that we will have no input offset voltage at all. In reality, the open-loop amplifier will not work. \$\endgroup\$ – LvW Jul 5 '18 at 15:58
  • \$\begingroup\$ Sorry about that. I meant to say Rin -> 0. \$\endgroup\$ – Souhardya Mondal Jul 6 '18 at 5:13
  • \$\begingroup\$ Sorry about that. I meant to say Rin -> 0. \$\endgroup\$ – Souhardya Mondal Jul 6 '18 at 5:13
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When you replace the resistor with a short you will eliminate the feedback. So you effectively have the situation of the input voltage directly across the amplifier input terminals with a 10K load on the amplifier output. The output will saturate near one of the supply rails unless the input voltage is very near to zero.

Here's an LTspice simulation:

enter image description here

X-axis is the input voltage, Y axis is the output voltage.

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  • \$\begingroup\$ I am new to Op-Amps , can you please explain why removing Rin means removing the feedback? \$\endgroup\$ – Souhardya Mondal Jul 6 '18 at 5:12
  • \$\begingroup\$ Speaking about "feedback" we refer to the case when a part of the output voltage is coupled back to the input. For this purpose, we are using a simple voltage divider Rin/(Rin + Rf). However, if Rin>>0 there is no resistor which can develop such a feedback voltage. This assumes, of course, that the connected signal source is an ideal source (internal resistance zero). Otherwise, the internal source resustance can act as a feedback resistor Rf. \$\endgroup\$ – LvW Jul 6 '18 at 8:40
  • \$\begingroup\$ What he said... \$\endgroup\$ – Spehro Pefhany Jul 6 '18 at 12:13
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General explanation:

In electronics, there is no formula which is correct by 100%. All formulas and equations are approximations because some (minor) effects are always neglected. But this situation is not a bad one because (a) it is quite impossible to consider all effects which might have a small influence on the result (temperature, parasitic capacitances, non-linearities which are always present) and (b) it would make no sense to include such effects as long as there are other uncertainties causing larger deviations (in particular tolerances of all parts).

In the present case (inverting amplifier), the given gain expression (-Rf/Rin) is a (good) approximation only as long as the following conditions are met:

  • Open-loop gain of the opamp much larger than the closed loop gain (this is equivalent to require a very large loop gain);

  • This requirement can be fulfilled within a certain and limited frequency range only;

  • The large-signal bandwidth and the associated slewing capabilities of the opamp have no influence

  • All resistors are much larger than the (neglected) finite output impedance of the opamp;

  • All resistors are much smaller than the (neglected) finite input impedance of the opamp;

  • The noise gain of the closed-loop amplifier (1+Rf/Rin) is small enough not to disturb the desired DC operational point more than acceptable. (The unavoidable input offset voltage appears at the opamp output multiplied with the noise gain).

    Comment: This requirement is violated if Rin becomes to small.

*......(other minor conditions)

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I am not talking with experience to about what happens, but I can give my guess as to what happens.

Theoretically what you say is correct, but you have to remember that there is still a circuit inside the chip trying to replicate the theoretical behaviour. And if you push such a circuit out of it's limits (e.g. by removing Rin and pushing the gain to infinity) it will not be able to handle it.

I think there are two things that might happen:

  • The output voltage will hit one of the input voltage rails (the voltages you supply to power the opamp. Typically Vcc and Ground). This is the most likely result.
  • The opamp will break. It's trying to reach an unreachable voltage, which could cause large currents inside the chip, which can fry it. I think this is unlikely though, because opamps are very robust and tough. I don't think I have ever broken a opamp.
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Firstly, no the output will not become infinite. The output will just go as high (for vin <0) or as low (for vin>0) as it can. This will be close to the supply voltages. Exactly how close will depend on the opamp and whatever its output is connected to.

Secondly, yes, in this configuration, the opamp may be damaged. Many opamps have a limit on the maximum difference allowable between their inputs. The AD8021, for example, has a maximum differential input voltage of 0.8V. If you exceed this, internal protection diodes will start to conduct, clamping the voltage at 0.8V. At this point, the inputs are no longer high impedance. If enough current flows through the diodes, it will break the chip. Many opamps do not have this differential voltage limitation.

Other than this failure mode, however, you should be fine. Also, just to add, the "ideal opamp" assumptions are only valid if the opamp is in a negative feedback configuration. If you remove Rin and drive the input with an ideal voltage source, you're removing this feedback, and so the ideal opamp assumptions are no longer valid.

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For the case of removing Rin, theoretically the gain will be infinity. But actually typical real values range from about 20,000 to 200,000.

On the other hand, An “ideal” or perfect operational amplifier is a device with certain special characteristics such as infinite open-loop gain AO, infinite input resistance RIN, zero output resistance ROUT, infinite bandwidth 0 to ∞ and zero offset (the output is exactly zero when the input is zero).

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With Rin at 10 kohm, the input impedance is 10 kohm. This is because the inverting input must be at the same voltage level as the non-inverting input (due to feedback and infinite gain) and, the non-inverting input is connected to 0 volts. This is called a virtual earth.

As you lower Rin the input impedance to Vin gets smaller and smaller and gain rises accordingly. As Rin approaches zero ohms, the input impedance also approaches 0 ohms and any voltage at Vin produces a massive input current.

All the above is for an ideal op-amp with infinite gain and clearly you can only ever attain an output voltage level that is limited by the supply rails so, in reality, as Rin gets really low in value, the current supplied from Vin cannot be driven into the op-amp output pin and you lose linearity i.e. the virtual earth situation is lost and you should not analyse this as a linear amplifier circuit any more.

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  • \$\begingroup\$ Sorry but for an ideal op-amp , input impedance is infinite . So even if we remove the Rin , impedance is still infinite. \$\endgroup\$ – Souhardya Mondal Jul 6 '18 at 5:18
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    \$\begingroup\$ @SouhardyaMondal the actual input pins are infinite impedance but the effect of negative feedback from the output means the inverting input node is forced to have the same voltage as the non-inverting node hence a zero ohms is "manufactured" through the use of feedback. Go read about virtual earths on op-amps. \$\endgroup\$ – Andy aka Jul 6 '18 at 9:44
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Question 1

No, if you remove Rin physically, output will not be infinite, not even close to VDD, but zero.

As you said, the opamp will try to keep the voltages at both the input nodes same. It will achieve it in this case by outputting a zero.

After removing Rin the formula is no longer valid. You have to derive it again for the new configuration again, this time without Rin.

The new configuration is voltage follower with positive input grounded. Whatever is there at the positive pin (since grounded, zero Volts) will be present at the output of the opamp.

enter image description here

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    \$\begingroup\$ It will not turn into a voltage follower, as the negative input will still be connected to ground. \$\endgroup\$ – Hein Wessels Jul 5 '18 at 8:49
  • \$\begingroup\$ @HeinWessels please look at the OPs picture. The positive pin is connected to ground. Not the negative terminal. \$\endgroup\$ – Umar Jul 5 '18 at 8:51
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    \$\begingroup\$ I do not understand why this got downvoted. "Removing Rin" will lead to a voltage follower... \$\endgroup\$ – Sven B Jul 5 '18 at 9:48
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    \$\begingroup\$ It's a voltage follower with a few uV voltage offset and lower bandwidth. \$\endgroup\$ – Jason Han Jul 5 '18 at 10:09
  • \$\begingroup\$ You are assuming "removing resistor" means open circuit, but it's clear from the question that the OP means replacing it with a short. \$\endgroup\$ – Spehro Pefhany Jul 5 '18 at 17:41

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