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enter image description here

Let's consider an op-amp. In every book I've read till now, it is said that if we have:

  • negative feedback
  • finite output

then we have:

Vout = A * (v2 - v1) --> (v2 - v1) = Vout / A

If A is very very large, then v1 and v2 tend to be equal (although they are not equal)

Now the question: who says that at the very beginning (that is, when the circuit switches on), the difference between v2 and v1 is so small that we are allowed to use the linear region shown in image b (thus justifying the previous calculation)? What about if at the beginning (v2-v1) is not so small?

Thanks

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  • \$\begingroup\$ I don't know who says that, i don't. Actually it can be higher. What if? Nothing, feedback would fix it. \$\endgroup\$ – Gregory Kornblum Aug 24 '18 at 10:32
  • \$\begingroup\$ Yes, if we try to consider the transient behaviour immediately after power switch on, the output will be at Vout,max (one of the supply rails) before the negative feedback will bring back the output to the desired DC bias point (preferrable in the middle between both power rails). \$\endgroup\$ – LvW Aug 24 '18 at 11:01
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In a properly operating amplifier, the sense of the negative feedback will be correct, even if the magnitude of the output is limited as in the diagram you show. The feedback will drive the circuit back to balance, and eventually, to roughly equal inputs.

However, there are some opamps, usually older rail to tail types, where the sign of the gain reverses when the inputs exceed their input common mode range Vcm. These types of amplifier can therefore 'lock-up' as the intended negative feedback becomes positive. I guess this wasn't seen as a fault to the original amplifier designers, as inputs outside Vcm are not 'allowed'. But it's such a common fault condition to occur that it's tripped up scores of amplifier users.

It's now been fixed in newer opamps, that state proudly in their datasheets that the gain doesn't change sign, stays well-behaved, even when when the inputs are taken outside Vcm.

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Negative feedback works all the time, not just in the linear region. Outside that region, nothing changes other than the effective value of A is reduced.

Therefore, no other justification is required.

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