# Why does the opamp output voltage saturate?

simulate this circuit – Schematic created using CircuitLab

Vout = Vin (1 + R2/R1) According to this positive feedback topology - made an error here, it's Vout = Vin(-R2/R1) (edit)

Now I read in a tutorial - http://www.electronics-tutorials.ws/systems/feedback-systems.html , that

If the input voltage Vin is positive, the op-amp amplifies this positive signal and the output becomes more positive. Some of this output voltage is returned back to the input by the feedback network.

Thus the input voltage becomes more positive, causing an even larger output voltage and so on. Eventually the output becomes saturated at its positive supply rail.

How does this happen? Shouldn't Vout be constant at a particular value, because we aren't changing Vin or the resisitors right? Shouldn't the equation Vout= Vin ( 1 +r2/r1) always hold? Error Vout = Vin(-R2/R1) (edit)

So how is the Vout increasing iteratively?

Vout= Vin (1 +r2/r1) is the equation for a non-inverting op-amp with negative feedback.

simulate this circuit – Schematic created using CircuitLab

Figure 1. Non-inverting amplifier configuration.

Your configuration is non-inverting but has positive feedback. This will give a Schmitt trigger effect.

Remember that the op-amp output will be $(V_+ - V_-)A$ where A is the open loop gain and typically > 1M. If we apply V1 = 1 V as shown in your schematic when the output is zero then $V_+$ goes to 0.5 V. The output will then try to go to $(0.5 - 0)1M$ $= 500,000~V$! It will get as far as the positive supply rail and stop with $V_+$ half-way between 1 V and the supply voltage. This will hold the opamp in that condition indefinitely.

Schmitt trigger, ʃmɪt/, noun, ELECTRONICS: a bistable circuit in which the output increases to a steady maximum when the input rises above a certain threshold, and decreases almost to zero when the input voltage falls below another threshold.

• "If we apply V1 = 1 V as shown in your schematic when the output is zero then V+ goes to 0.5 V. The output will then try to go to (0.5−0)1M or 500,000 V. It will get as far as the positive supply rail" Can't the same thing happen in negative feedback also? Aug 12, 2016 at 9:36
• Also if the amplifier is ideal, wouldn't V+ = V- ? Or in this case we relax the ideality assumption? Aug 12, 2016 at 9:42
• No, an ideal opamp will give the same result - faster! Negative feedback will work to bring the two inputs together. In my Figure 1 if $V_{IN}$ = 1 V the output will go to 2 V which brings $V_-$ to 1 V also. If the output goes any higher then $V_- > V_+$ and output will decrease. It settles at the desired value. Aug 12, 2016 at 9:54
• Great. Thank you, that was a very clear example to clear my doubt. I have one more, a generic one. Does, positive feedback mean infinite gain then? Always? Because my schematic has a gain(finite) of -R2/R1 on first look, but effectively the gain is coming out to be infinite right?(as it saturates) Aug 12, 2016 at 10:36
• The gain is +R2/R1. (I think your edit is incorrect.) The saturation is because the feedback is positive, not because of the gain. Aug 12, 2016 at 11:06

Just look at what you've done - you are mixing up the circuit of an op-amp with positive feedback with the theory of an inverting op-amp with negative feedback.

Look at the two scenarios in the link you gave - in the negative feedback scenario the gain is -Rf/Ri (the article says Rf ÷ Rin but there should be a negative sign).

In the positive feedback scenario they say: -

Eventually the output becomes saturated at its positive supply rail

And this of course is true when the input signal is positive enough to exceed the schmitt threshold imposed by the positive feedback resistor and input resistor (otherwise it will be saturated negative).

"So how is the Vout increasing iteratively?"

It is not possible to demonstrate the saturation effect as a sequence of several steps ("iteratively"). The input voltage of 1V (at first, without taking the feedback path into consideration) will appear at the non-inv. opamp node and will bring the output immediately into saturation (pos, supply voltage Vcc). Now - we have two voltage sources at both ends of the resistor chain: At the left Vin and at the right Vout. That means: The voltage at the non-inv. node will even be larger than 1V and there is no reason that Vout changes its value - it remains at Vcc.

• But Vout = -R2/R1 still has to hold right? According to KVL? Aug 12, 2016 at 9:32
• " The input voltage of 1V (at first, without taking the feedback path into consideration) will appear at the non-inv. opamp node and will bring the output immediately into saturation " Shouldn't this happen with negative feedback also then? Aug 12, 2016 at 9:34
• First question: No - this would be the case for operation within the LINEAR transfer function of the opamp only. Second question: Yes - this would be the start of the sequence for negative feedback also. However, in this case, a positive Vin would cause a NEGATIV Vout. Therefore, the voltage at the inv. opamp node would change its sign and, therefore, the output voltage would cross the linear opamp transfer region where it will find a stable operational point.
– LvW
Aug 12, 2016 at 9:41
• Second part- Positive Vin needn't cause a negative Vout. What if I'm using negative feedback,but non inverting topology ? In that case, Vout is still positive, but there seems to be no saturation. Aug 12, 2016 at 9:46
• Positive Vin causes pos. Vout - however, feedback to the inv. input makes the negative voltage at the inv. input larger than the pos. voltage at the non-inv. input. Thus, the output again changes sign and crosses the linear region. Same effect as for inv. amplifier.,
– LvW
Aug 12, 2016 at 10:33