Calculated output is greater than the op amp can produce?

Question

I had to analyze a differentiator and integrator op amp. I am now asked the question of what would happen if the calculated output is greater than the op amp can produce?

I am given a hint that it is usually a little less than the magnitude of the power supply voltage.

I think the answer is that the op amp would damped on suppress the output, so that the output is within the range that the op amp can produce, but I can't find anything that can confirm this for me.

Can someone tell me if I'm on the right track or not? Or if I'm going in the completely wrong direction.

Answer 1

Take the integrator. Suppose you apply a voltage to the resistor, and the capacitor starts off with 0V across it.

^{simulate this circuit – Schematic created using CircuitLab}

The voltage at the output will rise from 0V towards +10V at +1 volt/second, so after 10 seconds the output voltage will (ideally) reach the power supply voltage. At that point the op-amp will saturate or "rail" at the (ideally) positive supply voltage.

The op-amp can no longer operate in its active region so the output looks like its shorted to the +10V supply. At the moment of saturation, the voltage on the inverting input is 0V, but after that it looks like a classic RC circuit and you'll see the kind of RC response you'd expect on the inverting input (while the non-inverting input obviously will remain at 0V since it is grounded).

After a very long time the voltage at the inverting input will thus be V1. Since it's driving the op-amp output towards the positive rail (negative input on the inverting input wrt the positive input) the op-amp stays saturated at the positive rail.

You can do a similar analysis for a positive input to the integrator and similar inputs to the differentiator that cause the op-amp output to saturate.

Answer 2

Yes, you are absolutely right. Op amp cannot produce voltage greater than the supply voltage. It will give it's maximum possible voltage on the output that is a little less than the supply voltage.

But there is one important point. - After the input differential voltage multiplied on the op amp gain coefficient (Uin_diff*k) exceeds the maximum output voltage, the op amp enters so called over-saturation mode. And to enter its' normal working mode then it takes some time to leave over-saturation mode, and thus for this time it can not give correct output voltage corresponding to the input signal.

(If the op amp belongs to "rail-to-rail" class, its' maximum output voltage is more close to supply voltage or it is actually the supply voltage itself.)

Answer 3

There are mainly 4 blocks of op amp 1)input stage which is a dual input balanced out put differential amplifier.This differential amplifier provides the voltage gain of the op amp.Then intermediate stage which is directly coupled to the differential amplifier.This intermediate stage is a dual input unbalanced output differential amplifier.Because of direct coupling the output of this stage is above ground potential.A level shifter is used to bring the output to ground level.The fourth stage is a push pull amplifier which increases the output voltage swing.If the level shifter is not making the DC level exactly to zero and other non linearities associated with these stages causes the output voltage to be above our expected values I believe .