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In studying the OpAmp amplifier, the derivations always take the voltage difference between the inverting- and non-inverting input to be approximately zero. But I can't find any explanation "why". This consideration is always taken for both the inverting- and non-inverting amplifier.

Why is this voltage difference always approximately zero? Any explanation?

I found a lecture of my senior professor, where he said that "as the ideal op-amp's input impedance is infinite, its positive- and negative terminal are virtually short. I mean very little voltage drop which can be neglected."

I can't understand why a very high (practical case) input impedance may cause a very little voltage drop.

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marked as duplicate by Ignacio Vazquez-Abrams, The Photon, Chetan Bhargava, Daniel Grillo, Matt Young Dec 1 '14 at 15:49

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  • \$\begingroup\$ Please see my answer to a similar question here \$\endgroup\$ – EM Fields Nov 30 '14 at 18:27
  • \$\begingroup\$ I understand that when output going to cross the limit of bias voltage it sends negative feedback and so voltage difference become very small . But when output voltage is less then bias voltage , like when input voltage is ac then output voltage varies across zero.How this output sends negative feedback as it is less then max output voltage an op amp can provide. Shouldn't it reach max value first ? \$\endgroup\$ – Anklon Nov 30 '14 at 18:44
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    \$\begingroup\$ If the professor actually said exactly what you wrote inside the quotes then he misspoke or there is some particular context that is missing. \$\endgroup\$ – Spehro Pefhany Nov 30 '14 at 18:50
  • \$\begingroup\$ I can only give negative feedback and that may (or may not) help you. :^) (Oh and don't believe everything your Prof's tell you.) \$\endgroup\$ – George Herold Dec 1 '14 at 1:22
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The main thing that is almost always mis-stated or mis-understood in all these examples is the fact that the inputs being close or even identical to each other is NOT a effect of the amplifier but is an effect of negative feedback. The amplifier is assumed to be ideal to simplify the discussion.

An ideal op-amp without negative feedback will NOT have the inputs being identical.

What your professor means with the input impedance statement is that if there is a measurable input impedance thane there will be a potential difference between the pins and the actual differential element inside the chip. this manifests itself as an offset voltage of one input vs. the other, which gets amplified and manifests itself as an offset on the output.

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The gain of an op-amp is usually very high (in the order of tens or hundreds of thousands at DC at moderate to low frequencies so, if the op-amp output is not end-stopped against the power rails, it's a reasonable assumption to say that the input voltage difference is Vout/100,000 for example.

If |Vout| is maybe 10 volts or less, the input voltage difference is going to be 0.1mV or less. In the whole scheme of things 0.1mV is not much and probably at least equal to the input offset voltage error most op-amps exhibit.

If the op-amp is behaving linearly, negative feedback keeps the inverting input terminal at substantially the same voltage as the non-inverting input. If the op-amp had moderate impedances to ground (at the input) these impedances can modify the circuit's gain from the ideal. However, these will not (on their own) alter the condition of the inputs being substantially the same so maybe dig into what your lecturer said or maybe I'm misinterpreting your meaning.

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  • \$\begingroup\$ Suppose the op amp works as a simple comparator with no feedback(saturation mode) and the voltage on inverter pin is 10V and the voltage on non inverter ping is 2V, based on that, I think that the voltage difference will be equal to 8V. How op amps will work to set the difference to ~zero? \$\endgroup\$ – hbak Jan 23 '16 at 8:58
  • \$\begingroup\$ @hbak in my opening paragraph I mention that the opamp is not end stopped. In your scenario it will be. A comparator is not what I'm talking about in my answer. \$\endgroup\$ – Andy aka Jan 23 '16 at 10:25
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One way to understand OPAMP in an intuitive way and simplistic is considering that it works as follows: If v+>v- then Vout increases otherwise Vout decreases.

Let see what happen if v+=4 V and v-=2 V. In this case Vout will increase up to Vcc voltage (positive power rail voltage, the maximum). This is what happens when we use an OPAMP as a comparator. You can see that the high impedance does NOT cause little voltage drop.

In contrast, let see what happens when there is negative feedback, for example when we use it as an amplifier. In this case v- is a fraction of vout. For example, let have a non-inverting amplifier where v-=0.2*vout. We set v+=1 V. Assuming OPAMP output is at 0 V at the beginning (and therefore v-=0.2*0=0), the output will then quickly increase (and so will v-) until v-=v+ (=1 V in this example) and at that point vout will settle (Vout will be 5 V).

Therefore the conditions that allows us to consider that v+=v- is negative feedback combined with a Vout below VCC and Vout over VEE (negative power supply).

This is a bit simplistic explanation of course, but it allows you to analyze circuits with ideal OPAMP. Technically an OPAMP is a very high gain differential amplifier (but not infinite), and thus v- wont settle exactly at v+ but really close.

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