When we build op amp circuits that use negative feedback, like so:

inverting amplifier connection

... we can analyze the circuit very easily, by assuming that $$v^- = v^+$$ due to negative feedback (when also assuming the op amp is ideal, of course).

Besides the obvious high-precision cases where these simplified models break down, when is this and when is this not valid?
For example, if we replace the feedback resistor with some other element - perhaps a capacitor, inductor, diode (regular silicon diode, zener diode, etc.), or some combination of them and other common circuit elements - how do we know where this simplification is valid?
Also, even if we stay with a resistor as the feedback element, as the resistance becomes very, very high, at some point we can pretty much consider it an open circuit, and so clearly this model breaks down somewhere along the way.

So, the question is: under what constraints is this approximation "true enough" to give useful results?


For another example, consider the basic inverting log amplifier circuit:

log amp circuit

If we solve the Shockley diode equation

$$i_D = I_S(e^{vD/VT} - 1)$$

for vD, we get $$v_D = VT \ln{\left(\frac{i_D}{I_S} \right)}$$ (ignoring the 1, which is mostly irrelevant as the exponential will be rather huge)

If we then use the virtual short method to see that $$i_D = \frac{v_{in} - 0}{R_{in}}$$ we get the correct expression for the output:

$$v_{out} = -VT \cdot \ln{\left( \frac{v_{in}}{I_S R_{in}} \right)}$$

So, the virtual short method works here. But since this diode will be an open circuit when $$v_{out} > v^-$$ I'm not sure how to figure out beforehand that the analysis will be valid.

  • \$\begingroup\$ With an ideal op-amp, the + and - terminals will be equal independent of the op-amp's use in a circuit. \$\endgroup\$
    – kevlar1818
    Jun 4, 2012 at 12:33
  • 2
    \$\begingroup\$ @kevlar1818 How would that work? If there is no connection between the output and the inputs, how could it possibly change the inputs? \$\endgroup\$
    – exscape
    Jun 4, 2012 at 12:44
  • \$\begingroup\$ See my answer for clarification. \$\endgroup\$
    – kevlar1818
    Jun 4, 2012 at 13:36
  • \$\begingroup\$ @kevlar1818: The assumption that op amp inputs will be equal relies in some measure not only upon the op amp being ideal, but also the other components in the circuit. If other components in the circuit cause the first derivative of the feedback path voltage with respect to the output voltage to be zero (as could happen if there's any uncompensated RC delay), the op amp would not be able to instantaneously balance the inputs in response to a step stimulus. \$\endgroup\$
    – supercat
    Jun 4, 2012 at 18:28

3 Answers 3


Like you said, the fact that the two opamp inputs will be nearly equal is a simplification and depends on parameters often not explicitly stated. This is a good question in that it is essential to know the limits of any shortcuts or rules of thumb you use.

As clabacchio already said, one place the assumption are violated is if the opamp output is clipped, or would need to exceed its available range to make the desired signal. Other reasons that make the assumption invalid include:

  1. The feedback isn't negative. This may sound stupid, but I've actually shown someone a simple opamp hysteresis circuit in a interview and asked them to draw a plot of output voltage as a function of input voltage. More than one candidate started off saying the opamp will try to keep its two inputs the same, and then dug himself into a deeper hole from there. Needless to say, those were short interviews.

  2. The gain isn't sufficient. Note that the rule of keeping the inputs equal assumes infinite gain. Likewise the rule that Gain = -Rf / Rin assumes infinite gain. Normally opamp open loop gains are around 100k or more and we don't ask for more than 100 or maybe 1000 at most from a single stage, so it would seem this is a small issue.

    However, that forgets about the effect of frequency on gain. A 1 MHz opamp may be specified for 100k open loop voltage gain at DC, but if you use it for audio and want to pass 20 kHz, then you only have a open loop gain of 50 worst case. If you set the feedback resistors for a gain of 25, that only leaves 2x headroom at the high end, which will seriously reduce closed loop gain at high frequencies.

  3. Slew rate limitation. Even with enough gain and proper feedback, the opamp can only change its output so fast. That is what the slew rate spec is for. The gain*bandwidth product is for small signals. Large amplitude signals may run into slew rate issues. For most opamps, the full swing output signal is rather lower frequency that what the gain*bandwidth product implied.

  • \$\begingroup\$ Nice answer. I supposed the opamp to be ideal, because otherwise the hyphothesis is always false :) \$\endgroup\$
    – clabacchio
    Jun 3, 2012 at 17:06

As long as the Op-amp can set the inputs equal driving the output at a certain voltage, it will.

That assumption falls when it cannot, such as if it has an open circuit in the feedback (positive or negative). Then, it'll saturate to one of the rails, depending on which input is driven higher. Note that the open circuit feedback can be also a reversed diode.

Another case may be if the voltage that allows the equilibrium at the inputs is beyond the saturation voltages. Again, the op-amp will saturate and the input will unbalance.

But why the inputs have to be equal?

The op-amp has three operating regions, one called high-gain region, and two saturation regions. The rule that the inputs must be equal holds only for the high gain region, and comes from the fact that for the ideal op-amp:

$$V_{out} = \infty (V_d) = \infty (V_+ - V_-)$$

which means that the output voltage is finite only if the input voltages are equal, so the op-amp will force the output voltage to the value that zeroes the difference.

When the op-amp saturates, though, the output voltage is just given by

$$V_{out} = V_{sat}$$

which means that the op-amp is doing its best to set the inputs equal, but it's clashing against an amovable wall. So the inputs can unbalance to satisfy the output voltage.

In your example, you can find that the op-amp saturates when the input is equal or greater than:

$$ V_{in}^{SAT-} = - V_{SAT-}\frac{R_{in}}{R_{f}} $$

In your example circuit, when Vin is negative, V+ will be higher and then the output will saturate. There is no way then for the feedback to restore the equilibrium, because the diode will be reversed, so for every negative input the output will be the saturation voltage.

  • \$\begingroup\$ Thanks, but I knew most of that already (I've analyzed tons of different op amp circuits, but they all had one thing in common: it was usually obvious whether this method would apply or not). I guess what I'm confused about what counts as an open circuit - for example, a diode can be one (an ideal one, at least), but the method appears to still work there. I added the example of the log amp. \$\endgroup\$
    – exscape
    Jun 3, 2012 at 15:30
  • \$\begingroup\$ I have just remember you in those old days! I'm curiose about the formula of saturation(last one. Could you give me a reference to that formula, Instead asking you to talk more about it. \$\endgroup\$
    – hbak
    Jan 24, 2016 at 21:11

In this answer I make the derivation of the transfer function and conclude with why we can suppose both inputs are equal.

There's a minor simplification in the calculation, which is forgivable if the open loop gain is very high. This is true for most opamps, I used the figure 100 000.

What I did not talk about is offset error. This is a voltage difference between the input pins which will be amplified by -Rf/Rin. A 0.25mV offset may show as a 250mV error at the output if you choose for a \$\times\$1000 amplification. Some opamps have an offset null option, which allows you to trim the offset away by means of a potentiometer.
Anyway, limit the amplification of your opamp.


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