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Below is a basic inverting op amp. I'm having trouble seeing the negative feedback operation.

schematic

simulate this circuit – Schematic created using CircuitLab

If I understand the "golden rules" of op amps correctly, the voltage at the negative terminal should be at ground (at least in equilibrium). Let's say for some reason that the negative terminal is held at a small positive voltage then because of the differential amplifier, vout should be a large positive number. How does this generate negative feedback (aka, how does this force the negative terminal to go down)? I'm also having trouble seeing the same situation but if the negative terminal is slightly below ground.

Hopefully there is a "intuitive" reasoning behind this. Thanks!

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  • \$\begingroup\$ The amp drives Vin- until the differential input is null. Since it is inverted it also matches the input current. such that you can neglect the signal current to OP input. So iR1=iR2 electronics.stackexchange.com/questions/50460/… \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 May 8 at 5:00
  • \$\begingroup\$ If you hold the inverting input at a certain voltage you are preventing feedback, so the output will simply rail (in your example, at the negative rail, or negative infinity for an ideal op-amp). \$\endgroup\$ – Spehro Pefhany May 8 at 8:49
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It's better if you were to examine the behavior of a long-tailed pair. The long-tailed pair has two outputs, not one. So it's a little bit different in that sense. But suppose you selected only one of the outputs for your opamp? Then one of the inputs would have a (-) behavior relative to that output and the other input would have a (+) behavior relative to that output. An opamp is basically a long-tailed pair with only one of the possible outputs selected, with an added system to provide lots of gain and some output current compliance. So if you understand the long-tailed pair well enough, you've pretty much got what you need for understanding an opamp. (It's basically a long-tailed pair without only one of the two symmetrical outputs selected, lots of added gain, and an output driver section to boost the current compliance.)

If you get a chance to do so, study the long-tailed pair. For a long time. It's worth it.


The essential idea here is to recognize how the two different inputs affect the output. If the (+) input rises above the (-) input then the output will be driven higher in order to correct for the problem. If the (-) input rises above the (+) input then the output will be driven lower in order to correct for the problem. That's about it. What this means is that if the (+) input moves upward in voltage, then the output will also move upwards to "compensate." (The output has no idea if that will work, or fail. But it tries.) Also, if the (-) input moves upward in voltage, then the output will move downwards in voltage to "compensate." (Again, the output has no idea if that will work, or fail. But it tries and keeps on trying.)

So to understand why the circuit works, all you need to do is hold the above details in mind.


The (+) input is grounded. So it is NOT going to change. We only need to look at the (-) input and worry about that.

In your circuit, if the input falls (goes downward) then the output will do the opposite (go positive) in order to "fix the problem." In this case, that's a good thing. Because if the input goes in the negative direction then it is exactly that we want the output to go in a more positive direction in order to pull more current out of the (-) input node. (Which is what you want, because the net current at that node [or any node] must be 0.)

Suppose \$V_{\text{IN}}=-1\:\text{V}\$. Then this means that \$V_{\text{IN}}\$ is sinking \$I_\text{SINK}=\frac{\left(V_{\text{IN}}=-1 \:\text{V}\right)}{R_1}\$ current. In order that \$V_{\left(-\right)}\$ continues to match \$V_{\left(+\right)}=0\:\text{V}\$, the output must be \$V_{\text{OUT}}=-I_\text{SINK}\cdot R_2\$. If you solve, you find that \$\frac{V_\text{OUT}}{\text{IN}}=-\frac{R_2}{R_1}\$. As you already know, I'm sure.

Now suppose there is a very tiny change at \$V_\text{IN}\$ in the negative direction. Then this implies that there will be a tiny change at \$V_\text{OUT}\$ in the positive direction. The tiny negative change in \$V_\text{IN}\$ increases the sinking current, via \$R_1\$. But the tiny positive change in \$V_\text{OUT}\$ which occurs in reaction to the input's change means more current supplied via \$R_2\$ to compensate. This is exactly the right direction for \$V_\text{OUT}\$ to go in order to fix things up.

The opamp will very quickly apply its very high gain in order to quickly adjust \$V_\text{OUT}\$ in the right direction in order to get change in the sinking current in \$R_1\$ compensated by a change in the sourcing current in \$R_2\$. (The reverse is also true, because of the use of the (-) input to the opamp.)

At some point, the output voltage will be "just right" and the two inputs will be equal to each other (with the ability of the opamp to decide it.) At this point, the two inputs are basically "equal" and the opamp stops changing its output.


The "negative feedback" idea is essentially the idea that a downward change in the (-) input node will be counted by an upward change in the output voltage. That's "negative feedback." Upward-going at the input is countered by downward-going at the output; and downward-going at the input is countered up upward-going at the output. The very essence of this is the opposing direction one takes in response to the other.

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What you call the negative terminal is actually the inverting terminal. Because of the inverting behavior, the output will actually be a large negative voltage. That should explain the rest of the operation, that is, forcing the original terminal's voltage back down.

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No - the inverting terminal is NOT at ground potential.

This would be true for an IDEAL opamp only (infinite open loop gain).

Perhaps it helps to realize that your assumption (small pos. voltage at the inv. terminal) is realistic. Yes, for a pos. input signal Vin, there will be always a positive voltage at the inv. terminal - because otherwise the output cannot show any value (without offset considerations).

Buth this voltage Vn is very, very small(µV range): Vn=Vout/Ao with Ao being the open loop gain which is very large (1E5 or more).

Because this voltage Vn is so small, we always NEGLECT it during calculations, but it always does exist. This approach is equivalent to the classical assumption that the opamp gain would be infinite. In most cases, this simplification is acceptable.

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