Electronic – Op amp analysis: when are the “negative feedback rules” applicable

constraintsfeedbacknegativeoperational-amplifier

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?

EDIT:

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.

Best Answer

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.

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