# What happens on swapping the input terminals of an inverting opamp amplifier

#### silv3r.m00n

Joined Apr 15, 2010
51
Hi

The rules of the opamp say that the output will do all it can to bring the difference between the input terminals to zero.
So in the inverting amplifier configuration, if i swap the input terminals like this, why doesn't it still work like an inverting amplifier ?

As per the rule, since the inverting input is at ground (0), the output of opamp should bring the non-inverting input to ground (0) as well.
But the output is quite different.

#### BobaMosfet

Joined Jul 1, 2009
1,177
Try working with the OpAmp without the feedback loop in a DC environment first, then add feedback loop. Then AC signal.

Last edited:

#### LvW

Joined Jun 13, 2013
1,078
Hi
As per the rule, since the inverting input is at ground (0), the output of opamp should bring the non-inverting input to ground (0) as well.
But the output is quite different.
No - the mentioned rule is, of course, valid for negative feedback only .
Positive DC feedback brings the amplifier immediately into satration.

#### MrChips

Joined Oct 2, 2009
21,671
In normal op-amp circuit configuration, there is a resistor between the output pin and the inverting input pin of the op-amp.
This supplies something called negative feedback. Negative feedback in engineering systems is a good thing (unlike negative feedback in human relationships). This is what keeps the system self balancing and stable. This is why the voltage difference between the inverting and non-inverting inputs will tend towards zero volts.

If you swap the inputs, you now have a system with positive feedback. This is a run-away situation (self propelling system) and the output will saturate at either end of the supply rails.

#### crutschow

Joined Mar 14, 2008
25,269
The rules of the opamp say that the output will do all it can to bring the difference between the input terminals to zero.
That rule only applies for negative feedback (output to negative input), not for positive feedback (output to plus input).
You need to read about, and understand feedback.

Last edited:

#### ci139

Joined Jul 11, 2016
1,677
it depends the terminals swapped
the case of the signal input terminals of the dif.-amp.

the .txt goes to Falstad Circuit Simulator > Menu > File > Input From Text

#### Attachments

• 1.1 KB Views: 3

#### silv3r.m00n

Joined Apr 15, 2010
51
In normal op-amp circuit configuration, there is a resistor between the output pin and the inverting input pin of the op-amp.
This supplies something called negative feedback. Negative feedback in engineering systems is a good thing (unlike negative feedback in human relationships). This is what keeps the system self balancing and stable. This is why the voltage difference between the inverting and non-inverting inputs will tend towards zero volts.

If you swap the inputs, you now have a system with positive feedback. This is a run-away situation (self propelling system) and the output will saturate at either end of the supply rails.
thanks got it.

#### MrChips

Joined Oct 2, 2009
21,671
And another thing.

The "open loop" gain of an op-amp is very high, greater than 100,000.
That is, with no feedback, any voltage difference between the input pins gets amplified by a huge amount
(hence output saturation). This may be good if you are trying to build an analog comparator.

In most normal op-amp circuit configuration, one wants to limit the gain to a reasonable quantity, for example a gain of 20 using negative feedback. This has huge advantages such as very high linearity and very wide frequency band-width.

#### ci139

Joined Jul 11, 2016
1,677
i recently checked the formula for partially referencing +INP to OUTP . . .
i have some other past test that showed such is "stable" from Gain > 1 to Gain 4 (or 8)

Code:
.param gn = 2 ; can't be exactly 1
+ rg2 = 240k
+ rg3 = 270k
+ rg1 = 1/(gn-1)/(1/rg2 - 1/rg3)
+ rg0 = 1/(1/rg1 + 1/rg2 - 1/rg3)
+ gnChk = (1 + rg1/rg2) / (rg1/rg2 - rg0/rg3)
LM308.asy needs a "pointer" to the LM308.cir sub-dir = replace the path after the SYMATTR ModelFile ...\LM308.cir ← can be done with windows notepad (for example)
Code:
SYMATTR Prefix X
SYMATTR Value LM308
SYMATTR ModelFile D:\USER\Programs\LTC\LTspiceIV\lib\sct\LM308.cir

#### Attachments

• 3.6 KB Views: 0
• 920 bytes Views: 1
• 934 bytes Views: 0
Last edited:

#### LvW

Joined Jun 13, 2013
1,078
i recently checked the formula for partially referencing +INP to OUTP . . .
i have some other past test that showed such is "stable" from Gain > 1 to Gain 4 (or 8)
Yes, a positive feedback loop is allowed if we have - in addition - another negative feedback loop which dominates over the positive loop!
Hence, the net feedback must still be negative !!

#### BobaMosfet

Joined Jul 1, 2009
1,177
Hi

The rules of the opamp say that the output will do all it can to bring the difference between the input terminals to zero.
So in the inverting amplifier configuration, if i swap the input terminals like this, why doesn't it still work like an inverting amplifier ?

As per the rule, since the inverting input is at ground (0), the output of opamp should bring the non-inverting input to ground (0) as well.
But the output is quite different.
The rules of the OpAmp are written in a way that is confusing to you. What an OpAmp actually does is it throws the output to the rail (as close as it's designed to get to it) in a direction based on which input pin is reference, and which is signal. What you think you understand, and what you actually understand are 2 different things. Understand what I'm telling you, and you'll grasp what the 'rules' are actually telling you (or rather what they imply without saying).

That is _ALL_ an OpAmp does. Period. End of Story.

Therefore, if you apply ground to the positive input (a reference), and you apply +5V to the negative input, then the output is going to be ground because the output is driven from the power and ground pins, not from the inputs, and ground is as far in the opposite direction of +5 (because the input is inverting) as it can get.

And this has nothing to do with the feedback loop. A negative feedback loop has just one purpose- make the output oscillate around a set point. That's it. That's how it keeps it around that 'set point' created by a resistor divider- an oscillating output.

#### MrChips

Joined Oct 2, 2009
21,671
The rules of the OpAmp are written in a way that is confusing to you. What an OpAmp actually does is it throws the output to the rail (as close as it's designed to get to it) in a direction based on which input pin is reference, and which is signal. What you think you understand, and what you actually understand are 2 different things. Understand what I'm telling you, and you'll grasp what the 'rules' are actually telling you (or rather what they imply without saying).

That is _ALL_ an OpAmp does. Period. End of Story.

Therefore, if you apply ground to the positive input (a reference), and you apply +5V to the negative input, then the output is going to be ground because the output is driven from the power and ground pins, not from the inputs, and ground is as far in the opposite direction of +5 (because the input is inverting) as it can get.

And this has nothing to do with the feedback loop. A negative feedback loop has just one purpose- make the output oscillate around a set point. That's it. That's how it keeps it around that 'set point' created by a resistor divider- an oscillating output.
That is categorically incorrect.
An op-amp is not an oscillator.
The output of an op-amp is given by:

Vout = AOL (V+ - V-)

Reference: https://en.wikipedia.org/wiki/Operational_amplifier

#### WBahn

Joined Mar 31, 2012
26,073
The rules of the OpAmp are written in a way that is confusing to you. What an OpAmp actually does is it throws the output to the rail (as close as it's designed to get to it) in a direction based on which input pin is reference, and which is signal. What you think you understand, and what you actually understand are 2 different things. Understand what I'm telling you, and you'll grasp what the 'rules' are actually telling you (or rather what they imply without saying).

That is _ALL_ an OpAmp does. Period. End of Story.

Therefore, if you apply ground to the positive input (a reference), and you apply +5V to the negative input, then the output is going to be ground because the output is driven from the power and ground pins, not from the inputs, and ground is as far in the opposite direction of +5 (because the input is inverting) as it can get.

And this has nothing to do with the feedback loop. A negative feedback loop has just one purpose- make the output oscillate around a set point. That's it. That's how it keeps it around that 'set point' created by a resistor divider- an oscillating output.
Big HUH???

An op amp is nothing more than a very high gain differential amplifier.

Ignoring issues such as input offset voltages and currents, the output of an opamp is given by

Vout = Av·(Vp - Vn)

where Av is the voltage gain, Vp is the voltage applied to the non-inverting input, and Vn is the voltage applied to the inverting input.

If you construct a circuit that allows you to apply microvolt-scale differential signals to the input and run the opamp (or most opamps, anyway) open loop, you will find that it behaves like pretty nonlinear amplifier over a very small input range. It will not oscillate beyond that due to the noise on the input signals (it's hard to get clean signals at that level without feedback).

If your circuit is configured with negative feedback, then an increase in the magnitude of the output voltage (from it's equilibrium point) will serve to reduce the differential voltage at the input pins, which will cause the output voltage to go down. Similarly, a reduction in the output voltage will result in the differential input voltage going up which will result in the output voltage increasing back toward the equilibrium point. The equilibrium point is where the differential input signal produces an output at the opamp that is exactly sufficient to produce that same differential signal at the input.