# How does an ideal op amp amplify a voltage input when the voltage difference is 0?

I am a first year Electrical Engineering student. Next week I have my exams on Op Amps and their applications and such.

While looking at the op amps, there was always this one this that was confusing to me: in a non inverting amplifier, the inputs of the op amp are both zero, since the one at the +-pole is grounded. But when they are both zero, how can there be any amplification? I would think that you can just draw the line of the circuit, without writing down the op amp, since no current flows in the inputs of the op amp.

I really hope anyone can help me, since my teachers only really gave me how to solve it. I know how to solve op-amps, but not the logic behind them.

• Think in terms of $lim$. start writing down the equations with a finite gane, then take it to infinity. Then the $lim$ of the voltage will be zero. – Eugene Sh. Oct 31 '19 at 20:54
• electronics.stackexchange.com/questions/441184/… – G36 Oct 31 '19 at 20:57
• You only stated one of the two critical facts about an ideal op amp. The voltage difference between the inputs is zero, and the input currents are also zero. Therefore, you can apply Kirchhoff's current and voltage laws to solve the rest of circuit. For practical purposes, just think of an ideal op amp as a magical device which makes the conditions at its inputs always true - just like an ideal BJT is a magical device that makes the collector current $\beta$ times the base current. – alephzero Nov 1 '19 at 21:35
• @alephzero, a good summary. I'll add that it's important to remember that those rules of thumb (or "facts") are true only when the opamp has negative feedback. – spuck Nov 1 '19 at 22:02
• The differential input is not zero, but very close. The gain of an ideal op-amp is infinity. The gain of a real, physical op-amp is more like 10,000; 100,000; or more. Let's say your 10,000 gain real op-amp is producing a output voltage of 1 volt. Then it must have a differential voltage of 100 microvolts. That's close enough for a rough analysis to call it 0. – penguin359 Nov 2 '19 at 0:42

Uin = Uout / A, so if A is big, Uin is small. That's it.

Just like it took ages to "invent" zero, the concept of "infinite" is also not that easy to understand. For now, just consider that an (ideal) opamp has an amplification that is "very large", so that Uin is "very small". You may consider Uin "zero" as long as you understand that it is not really zero, but small enough so that you are having difficulty measuring it, and big enough so that it gets amplified to get Uout.

• This. It answers exactly the question asked, without confusing the answer with showing off. – Neil_UK Nov 1 '19 at 17:13
• @Neil_UK Thanks. Indeed, other answers are about how to tame an infinitely strong amplifier. But I am cheating, long ago, I actually studied this kind of stuff at university.... – Roland Nov 3 '19 at 23:39

A transfer function of closed loop system is:

$$\H(s) = \dfrac{F(s)}{1+F(s)}\$$ simulate this circuit – Schematic created using CircuitLab

Wiki

A transfer function of closed loop system with a feedback is:

$$\H(s) = \dfrac{F(s)}{1+F(s)\cdot G(s)}\$$ simulate this circuit

Now let's suppose you have a opamp buffer - the output is feedback to inverting input, as case 1 in this answer. F(s) = G (Gain), you can notice that output is never equal to the voltage on non-inverting input, but very close. simulate this circuit

The error is :

$$\\varepsilon = \dfrac{1}{1+G}\$$

So, the voltages on both inputs are never equal, but very close, and higher gain makes them closer. Since the opamp's gain is insane high, for the sake of simplicity we can say that voltage of inverting input is equal to the voltage of the non-inverting input. If you ground the non-inverting input, then you get a virtual ground on the inverting input, but it is never zero voltage for real. And as you already discovered it also can't be zero, else no signal would come out.

• I suggested an edit to improve a sentence that didn't quite make sense, please adjust if that's not what you meant. – tolos Nov 1 '19 at 12:53

It's a Platonic ideal of an op-amp; it's not a thing that could ever exist in the real world.

So you say to yourself "I'll just assume that my system is stable" (real op-amp circuits aren't, always), and "I'll just assume that my output does whatever it needs to hold the input voltage difference at zero" (real op-amps don't, quite), and "I'll just assume my op-amp is infinitely fast" (this is impossible). And then you work really hard to just stop worrying about it because it's all pretend, to make the math easier, and because for a lot of real-world problems it's Good EnoughTM.

Later on, if you pursue analog circuit design, they'll start expecting you to solve problems like "how accurate is the circuit if the gain is only $$\10^5\$$?" and "how do the op-amp's very real bandwidth limitations affect the circuit performance?" But those are much harder questions to answer if you can't go back to that ideal op-amp touchstone and figure out what the ideal circuit would do.

The difference between the inputs is not exactly zero, but it is arbitrarily close to zero (i.e., infinitesimal), and the arbitrarily high gain of the opamp turns that into a finite output value.

The ideal opamp is an abstraction, based on the concept of limits. Start with an amplifier that has a finite gain value. As you increase the gain to arbitrarily high values, other circuit parameters, including the input difference, approach but do not quite reach some definite limiting value.

When we use ideal opamps in our analysis, we use the limit value rather than the actual value.

The two inputs are at the same voltage because the op amp is ideal, and because there is negative feedback. If the gain is infinite and the output voltage is finite, then the difference voltage must be zero.

To see how there is amplification you should follow the current from the input side to the output. Although there is no current into or out of the op amp inputs, there is likely to be current flowing in the other parts of the circuit.

Virtual ground--the potential at the negative op amp input of your inverting amplifier--is actually at ground potential only if the gain of the op amp is infinite. If it's an ideal op amp, you can use limits (as in calculus) to work it out, but in real life the amplifier has a large but finite gain and there will be a small but extant voltage difference between the inputs.

However, infinite gain and virtual ground are good enough approximations to use for initial design in most cases.

I think you are not grasping the concept of feedback from output to input. That is the thing that keeps the inverting and non-inverting inputs ideally at same voltage.

An ideal op-amp is at equilibrium when inverting and non-inverting inputs are at the same voltage. If you just connect output directly to inverting input to be measured as feedback, indeed, there is no voltage gain, output voltage is equal to inverting input, and inverting input is equal to non-inverting input. It is a unity gain buffer, it takes no current from the input, and can drive infinite current on output, but no voltage gain.

If you now add a voltage divider on the output, so that the inverting input measures only smaller proportion of the output voltage, then op-amp output must still do everything it can to make inverting and non-inverting inputs equal. It means that the output voltage has to be higher and there is voltage gain in the system.

in a non inverting amplifier, the inputs of the op amp are both zero, since the one at the +-pole is grounded.

... doesn't quite make sense to me.

Op-amps amplify the difference in voltage between their inputs. Their (open-loop) gain is the change in output voltage resulting from a given change in the difference between the input terminals. That is to say that if we start with both inputs at 0V, and then change the + input to +1mV (keeping the - input at 0V), and if the output voltage changes by 1V, then the op-amp's open-loop gain is 1000.

Ideal op-amps have infinite open-loop gain; real-world op-amps have very large but finite open-loop gain. We assume ideal behavior because:

1. Real-world gain is high enough that for practical purposes, it can be satisfactorily modeled as infinite, and
2. It simplifies the design process.

If we consider the ideal behavior (infinite gain), whatever the op-amp's output is at any given time, the input difference signal must be infinitely smaller i.e. zero difference. For the open-loop case, we still have infinite amplification, and for the feedback case, we have our feedback-defined gain.