# How to clasify if it's an inverting/non-inverting op amp

• xconwing
In summary, an inverting op amp is created when the (+) input node is more positive than the (-) input node, while a non-inverting op amp is created when the (−) input node is more positive than the (+) input node. This circuit is not an amplifier because it contains positive feedback which will immediately saturate one of the supply rails. After exchanging the input nodes, the circuit resembles an inverting amplifier with a gain of -9/3=-3.
xconwing
Hello,

Like the title said, how can you tell if it's an inverting or non-inverting op amp just by looking the the schematic like this one.

textbook said this is an inverting op amp. I thought whichever input node (+/-) the signal go through dictate the classification of an op amp if it's inverting or non-inverting.

Questions I still have regarding op amp:
• Op amp needed feedback (through '-' non-inverting input) to "stabilize" it. What does it means? I'm thinking that since the gains is most often in the magnitude of >100k, the feedback is there to take away some of the amplification so that the amplified voltage could be in the range of the supplied voltage of the op amp.
• what is it mean when the op amp is "saturated"?

X

Last edited:
xconwing said:
Like the title said, how can you tell if it's an inverting or non-inverting op amp just by looking the the schematic like this one.
There is no such thing as "an inverting or non-inverting op amp". An operational amplifier has two inputs, one inverting and one non-inverting. The circuit decides how you use it. The schematic shows it used in the non-inverting mode, which again means that it is not used as an amplifier (analyze the circuit yourself - assume that Vout max = 14V and Vout min = -14V).

hi there

I have seen that configuration before on here
The interesting things is, when you google images for inverting or non-inverting opamps
you never see that particular configuration for either mode

will wait for out resident opamp experts to chime in

Dave

The shown circuit cannot work as an amplifier at all (neither inverting or non-inverting) because it contains POSITIVE feedback. As a consequence, it will immediately latch/saturate at one of the supply rails. Each working opamp based circuit needs negative feedback for stabilizing a suitable operating point.
After exchanging both opamp input nodes the circuit resembles an inverting amplifier with a gain of -9/3=-3.

berkeman and davenn
LvW said:
The shown circuit cannot work as an amplifier at all (neither inverting or non-inverting) because it contains POSITIVE feedback. As a consequence, it will immediately latch/saturate at one of the supply rails. Each working opamp based circuit needs negative feedback for stabilizing a suitable operating point.
After exchanging both opamp input nodes the circuit resembles an inverting amplifier with a gain of -9/3=-3.

Okay, so conceptually. This doesn't work in practice, but the textbook stating that this configuration is "inverting" is wrong, correct?

Yes - it is wrong (just an unintended mixing of both input terminals).

So my initial thought was correct, whichever input is active will classify the op amp type? That is to say:
signals going through non-inverting input will make that op amp "non-inverting" and vice versa. What if they both active?

Just to show you a quick simulation of the shown circuit (caveat: I was using an op-amp with only ±6V supply voltage and an input signal of 100Hz AC). The green signal is input, the red is output.

Svein said:
Just to show you a quick simulation of the shown circuit (caveat: I was using an op-amp with only ±6V supply voltage and an input signal of 100Hz AC). The green signal is input, the red is output.
View attachment 78547

That data is purrty data man. If you don't mind, can you point out what is not right about it?

xconwing said:
So my initial thought was correct, whichever input is active will classify the op amp type? That is to say:
signals going through non-inverting input will make that op amp "non-inverting" and vice versa.
Providing it really is a proper amplifier, yes.
What if they both active?
One will dominate. But usually the circuit will involve frequency-dependent impedances (capacitors, etc), so the sign of the gain may change from + to - over some part of the usable range, and have a phase somewhere in between for most of the intermediate range.

xconwing said:
That data is purrty data man. If you don't mind, can you point out what is not right about it?
The output waveform doesn't resemble the input sinusoid! So not an "amplifier" in the ordinary sense of the word. The output is not linearly related to the input.

LvW said:
The shown circuit cannot work as an amplifier at all (neither inverting or non-inverting) because it contains POSITIVE feedback. As a consequence, it will immediately latch/saturate at one of the supply rails. Each working opamp based circuit needs negative feedback for stabilizing a suitable operating point.
After exchanging both opamp input nodes the circuit resembles an inverting amplifier with a gain of -9/3=-3.

and that was my whole point about never seeing this configuration, I couldn't see how it could be correct :)

davenn said:
and that was my whole point about never seeing this configuration, I couldn't see how it could be correct :)
The configuration is okay. It's a Schmitt comparator, and common enough. It's just not a linear amplifier.

davenn
NascentOxygen said:
It's a Schmitt comparator, and common enough.

Thankyou !
which is why it didn't show up in the inverting or non-inverting ccts
I really have to remember that one for next time I see it :)

Dave

oops i see many others posted while i was typing sorry if this is redundant... old jim
xconwing said:
how can you tell if it's an inverting or non-inverting op amp just by looking the the schematic
"Operational Amplifier" unfortunately has dual meaning.
It is the name given to an amplifier that could perform some math operation provided it has been surrounded by circuitry that let's it make its two inputs equal. That's how it "operates".
It has fallen into common usage as the name given to a circuit containing both such an amplifier and its surrounding circuitry when used with negative feedback to make its inputs equal. It performs a math operation, usually but not always a linear one.

The way to tell whether the latter inverts is to work it in your head. i talk my way through them.
Looking at yours
First question is can this operate ? That is, can it hold its + and - input terminals equal?

Let's see what we know:
we know input- is held firmly at 0.
we know input+ will be at some voltage that lies in between Vin and Vout, because the 3k and 9k resistors make a voltage divider.
we know that making input+ more positive makes Vout more positive because that's what the + means.
So at first glance it appears
applying a tiny positive voltage to Vin will make Vout go positive
so input+ will lie between Vin which is positive and Vout which is also positive

Now - the condition for operation was it must make its inputs equal, that is make input+ equal zero.
Can zero lie between two positive numbers? I think not.

So i think your textbook has thrown you a trick question - that circuit isn't a linear operational amplifier circuit, even though it has an operational amplifier in its heart.
Now there do exist the math operators for comparison, > for greater than and < for less than.
Your circuit can do that function.

Let's apply some arithmetic to your circuit.
Voltage between Vin and Vout is of course Vin-Vout
so voltage across the 3k is ¼(Vin-Vout)
and voltage across the 9k is ¾(Vin-Vout)
Write KVL around this closed loop::: Vin to common to input+ through 3k back to Vin
and you get
Vin - (input+) - ¼(Vin-Vout) = 0
re-arrange
input+ = Vin - ¼(Vin-Vout)

input+ = ¾Vin + ¼Vout

so in order for input+ to be zero, Vin and Vout must have opposite polarity. That's what is meant by inverting opamp circuit.
In the circuit above they don't, but if you swap the amplifier's input pins they will.
So as is, that circuit doesn't invert.

If this was a trick question, the circuit indeed doesn't invert so it must be noninverting. But i wouldn't call it an operational amplifier circuit

Since the opamp has supply of +/- 15 volts that's the highest its Vout can go.

What if Vout were at that limit, let's just say postive 15 volts?
could we drive input+ to zero by applying negative Vin ? If we went just a little further would we tip the balance and make Vout go to negative 15? Of course we could.

So let Vout be +15
what Vin will drive input+ to zero?
input+ = ¾Vin + ¼Vout
0 = ¾Vin + ¼(+15)
¾Vin = - ¼(+15)
Vin = -5

So that circuit flips its output state whenever input edit: reaches -Vout/3, as shown in svein's delightful post #8 scope trace . crosses 5 volts with polarity opposite Vout

which would be described as a comparator with hysteresis
also known as a "Schmitt Trigger"
and it does a logical math operation, the inequality "greater than" with memory of last state

and since output polarity follows input polarity
i'd call it a "non-inverting comparator with 5 volt hysteresis"

KVL is great for figuring out op-amps.

Ask teacher if that was a trick question.

Have fun. Talk to your circuits, it'll amuse your roommates no end.

old jim

Last edited:
dlgoff and davenn
Svein's 'scope trace in post 8 shows Schmitt Trigger action quite nicely.

NIce Job !

NascentOxygen said:
The configuration is okay. It's a Schmitt comparator, and common enough. It's just not a linear amplifier.
Yes - it is a so called "Schmitt trigger". Watch the hysteresis effect (output goes high at another level as it goes down again).

Warning: One should know that some simulation runs do not reveal the unability of the circuit to amplify. For example, doing an ac analysis you get a result which looks quite "normal" (as expected).
However, you should not blame the program in this case. The result is quite correct.
Why? Because this analysis assumes (a) that the circuit was powered since infinite times (without switching transients) and (b) that there are absolutely no external influences (noise, temp. variations) which would disturb the quiescent conditions.
You must do a TRAN analysis (time domain) to show that the circuit cannot work in a linear mode.
This case is similar to the mechanical model of two balls - one riding upon another one. Under ideal conditions this arrangement could also be stable.

Last edited:
Just to show the basics:

Non-inverting circuit:
• Gain = 1 + R3/R2
• Input impedance: R1
Inverting circuit:
• Gain = -R5/R4
• Input impedance: R4

davenn and jim hardy
@xconwing haven't heard from you in a while.

Maybe in our enthusiasm we missed your original question:

...
xconwing said:
• Op amp needed feedback (through '-' non-inverting input) to "stabilize" it. What does it means? I'm thinking that since the gains is most often in the magnitude of >100k, the feedback is there to take away some of the amplification so that the amplified voltage could be in the range of the supplied voltage of the op amp.
That's exactly right.
We don't call it "taking away the amplification" though
we call it "negative feedback" because the output is "fed back" in such direction to subtract from the input signal. So it more takes away signal than amplification.
So, (input - feedback) X (gain of >100K) < supply. Maybe supply is 15 volts? If so, (input-feedback) must be less than 15/gain, around 150 microvolts. We usually round that off to zero for analysis.
Your 100K gain is called "open loop gain" and is designated on datasheets by Avol
When surrounded by a circuit including feedback, the gain of the entire circuit becomes called "closed loop gain"
• what is it mean when the op amp is "saturated"?
To use almost your exact words, it means the amplified voltage ISN'T within the range of the supplied voltage of the op amp.
that is,
try as it may the poor little amplifier just can't make its output voltage go all the way to (gain X input) because there's not that much voltage available to it.
Most amplifiers can drive output to about a volt shy of the supply voltage
so called "rail to rail" opamps can drive output to within millivolts of the supply voltage.Your questions suggest that you have an intuitive feel for these circuits.
Be rigorous in your thinking and vocabulary and you will do very well.

Good luck in your studies !

hmm where'd all those extra dots come from? see my signature...

old jim

@ jim hardy
I was just letting the thread chill for a bit so that I could soak up (or recalling) some of the information mentioned.

THANK YOU all for the replies.

xconwing said:
Op amp needed feedback ... to "stabilize" it.

xcongwing, may I add another comment - just to avoid misunderstandings.
For all linear amplification purposes using opamps we apply negative feedback - however, NOT to stabilize the circuit.

Just the opposite is true
: Negative feedback destabilizes the circuit and can cause unwanted oscillations.
The most critical case is 100% feedback (unity gain amplification). For this reason, most of the commercially available opamps are "unity-gain compensated", which means: They can be used down to unity gain without stability problems.
But note the consequence of such a compensation: The open-loop gain Aol starts to decrease already for rather low frequencies (10...1000 Hz).
In short: Negative feedback has many advantages (stable DC operating point, reduction of output impedance, THD improvement,...) , but it always causes a decrease of dynamic stability. For example, even if an amplifier with much feedback is stable, it will most probably show a step response with an (unwanted) overshoot or even ringing (decaying oscillations).

This prooves the general rule in electronics: Each modification to improve one or more circuit parameters will always have a negative effect to some other parameters.
Hence, every design will be a trade-off between conflicting effects.

LvW said:
Just the opposite is true: Negative feedback destabilizes the circuit and can cause unwanted oscillations.

That is carrying it a bit far. Of course you have to be aware of phase margins and all that, but if you use a modern op-amp (designed after 1990) and read the data sheet carefully, there should be no problems.

I suspect you are thinking about early op-amps like 709 and 741, where compensation was something you needed to very aware of. But modern operational amplifiers are better in all respects (advances in semiconductor design and fabrication are not reserved for digital circuits). Check out operational amplifiers from Linear Technology and Texas Instruments.

Svein said:
That is carrying it a bit far. Of course you have to be aware of phase margins and all that, but if you use a modern op-amp (designed after 1990) and read the data sheet carefully, there should be no problems.

I suspect you are thinking about early op-amps like 709 and 741, where compensation was something you needed to very aware of. But modern operational amplifiers are better in all respects (advances in semiconductor design and fabrication are not reserved for digital circuits). Check out operational amplifiers from Linear Technology and Texas Instruments.

No, he is not carrying it a bit far.

Negative feedback is not applied to "stabilize" an amplifier unless you are trying to compensate for existing positive feedback. That happens, but it isn't the common situation.

Negative feedback is generally applied to linearize an amplifier, control its gain, control its frequency response, etc, etc. Your comments relate to its use for those reasons.

If there is no positive feedback, the circuit will be stable.

meBigGuy said:
If there is no positive feedback, the circuit will be stable.
Yes. But when you use an operational amplifier with a sizable gain above 50MHz, positive feedback can be hard to avoid...

Svein said:
That is carrying it a bit far. Of course you have to be aware of phase margins and all that, but if you use a modern op-amp (designed after 1990) and read the data sheet carefully, there should be no problems.
.
I didnt mention any "problems". It was my only intention to point o the fact that negative feedback always degrades the stability margin. If - for a specific application - this could cause problems (or not) is another question. But one should know the consequences. By the way - these consequences of neg. feedback gave reason for developing new concepts like the principle of "current feedback".

Svein said:
Yes. But when you use an operational amplifier with a sizable gain above 50MHz, positive feedback can be hard to avoid...
Yes - with increasing frequencies negative feedback will ALWAYS turn into positive feedback.
However, in most cases, we are lucky and the loop gain already is too small to cause oscillations at these frequencies..

LvW said:
Negative feedback destabilizes the circuit and can cause unwanted oscillations.

I think that statement suffers from lack of introduction...
Negative feedback calms things so long as it remains negative.
That is demonstrated via algebra in Mark's Mechanical Engineer's Handbook chapter on automatic controls.
Here it is also from TI online, see page 7 of the pdf, section 5.3
edit forgot to include link , here it is
http://www.ti.com/lit/ml/sloa077/sloa077.pdf

An amplifier with gain G surrounded by negative feedback H is called a "closed loop".
It creates a circuit (or physical process) having "closed loop gain" of
G/(1+GH)
That's true both for DC analysis and for AC analysis with sinewaves.

Now - there always exists some finite tme delay propagating a signal through an amplifier and back around through the feedback network to close the loop.
So for AC analysis we should regard both G and H as complex numbers, capable of shifting the phase of a sinewave.Again the closed loop response is

G/(1+GH)

For stability, what nails you is that GH term in the denominator.
GH has the name "Loop Gain"
and is the product of two complex numbers , so it'll affect the phase of a sinewave.

Here's the key to intuitively grasping this concept
---
When GH affects that sinewave's phase by half a cycle, 180 degrees, it has changed the sign of that sinewave.
So at that frequency , closed loop response is now
G/(1 - GH) , observe negative sign. What was negative feedback has become positive because of the delays around the loop.

AHA ! Should GH ever become equal to -1 the denominator goes to zero !
So the closed loop suddenly acquires infinite gain. It'll oscillate, usually making beautiful sinewaves. Mother Nature just loves a sine !
That's what lvW meant by "negative feedback destabilizes..."

Prudent design requires that product GH be less than 1 at the particular frequency that gives 180 degree phase shift around the GH loop,
because that keeps the denominator from disappearing(going to zero)
and that is what was meant above by that undefined term "phase margin".

Where i worked 1 hz was high frequency. So we didn't worry about phase shift in opamps , only in the whole loop including some physical process like flow or pressure..
G might be a PID controller where we can dial in integral and derivative terms to balance those frequency sensitive terms in the H part of a closed loop. A flow control valve might have a time constant measured in seconds, so Svein's 50 mhz example also holds true in the sub-hz range.
The math is identical.
Mother nature loves feedback loops, she uses them everywhere.

I hope this conveyed at least a clue for a beginner of what is meant by these mysterious control theory terms "Closed Loop", Phase Margin", "Stability", etc. It's a fascinating world.

Modern Control Theory is as the name suggests a relative newcomer to engineering.
I've read that Descartes stumbled across the algebra of closed loops, but since there were no machines to control in the early 1600's
he set it aside as just another math curiosity no more useful than Hero's steam engine.
German rocket scientists found and revived his work in WW2 for their flying machines.
Their textbooks were among war prizes we brought home in 1945 - see "Operation Paperclip"
old jim

Last edited by a moderator:
davenn
jim hardy said:
...and that is what was meant above by that undefined term "phase margin".
Hi Old Jim - just one short comment to your long and detailed explanation.
Since many decades the term "phase margin" is (a) defined and (b) extensively in use - in control theory as well as for all amplifier configurations with feedback.
I was not sure if it make sense to repeat the definition of such well-known parameters here in the forum.
And when I say "negative feedback destabilizes" I simply mean that the stability properties are not as good as without feedback (I have mentioned overshoot and ringing).
It was my only intention to correct and comment the (false) statement: Feedback stabilizes.
Regards
LvW

I think it needs to be established what exactly 'stable' is. Whoever said that negative feedback is seldom used to compensate positive feedback is quite wrong. This is very common although it may be subtle. Multi-stage amplifiers with 3 stages each having a gain of 50 instead of a single stage with a gain of 125000 is a perfect example. There is enough negative feedback in each individual stage to easily keep the stage from oscillating. Each stage can be isolated from the other enough to minimize feedback from stage to stage. I totally get that negative feedback can turn into positive feedback at certain frequencies. When this scenario is unavoidable the fix is to guarantee negative feedback at the pertinent frequency.

LvW said:
Hi Old Jim - just one short comment to your long and detailed explanation.
Since many decades the term "phase margin" is (a) defined and (b) extensively in use - in control theory as well as for all amplifier configurations with feedback.
I was not sure if it make sense to repeat the definition of such well-known parameters here in the forum.

Thanks LvW

sorry i got so long winded.
I estimated xconwing to be not yet an "Old Hand"
so took the risk of going back to the basic of basics ,
introducing those well known( to old hands) definitions.

It's important for beginners
to distinguish instability caused by running out of phase margin
from instability caused by noise, friction, loose connections, thermal runaway and the like.

For me the "light came on" when i grasped that bit about denominator going to zero because GH=1 with phase inversion; the math and my intuition suddenly agreed.
I wanted to offer that to xconwing

please take no offense at my remark about 'lack of introduction' for none was meant.
I just wanted to describe the concept clearly.

And i see i left the TI link out of post, will edit it in...
http://www.ti.com/lit/ml/sloa077/sloa077.pdf old jim

Last edited by a moderator:
jim hardy said:
It's important for beginners to distinguish instability caused by running out of phase margin
from instability caused by noise, friction, loose connections, thermal runaway and the like.
old jim

Jim - thanks for clarification.
And yes - the above cited problem is an important one. Thats the reason we should strictly distinguish between "dynamic instability" and other forms of unwanted fluctuations ("static" instability - although not a very descriptive term).
LvW

meBigGuy said:
No, he is not carrying it a bit far.
Negative feedback is not applied to "stabilize" an amplifier unless you are trying to compensate for existing positive feedback. That happens, but it isn't the common situation. Negative feedback is generally applied to linearize an amplifier, control its gain, control its frequency response, etc, etc. Your comments relate to its use for those reasons. If there is no positive feedback, the circuit will be stable.

Averagesupernova said:
Whoever said that negative feedback is seldom used to compensate positive feedback is quite wrong. This is very common although it may be subtle. .

When I apply negative feedback to control the gain, I naturally reduce the gain at high frequencies where positive feedback may be occurring. Did I apply feedback to control the gain? Or to reduce the effects of positive feedback? It may be used often to compensate for positive feedback, but it is used "oftener" for other reasons (that's a judgement call).

Maybe if the OP had just said that negative feedback CAN be applied to stabilize an amplifier ...

Well I DID say it was subtle.

## 1. How do I determine if an op amp is inverting or non-inverting?

To determine if an op amp is inverting or non-inverting, you can look at the circuit diagram. In an inverting op amp, the input signal is connected to the inverting input (-) and the feedback is connected to the non-inverting input (+). In a non-inverting op amp, the input signal is connected to the non-inverting input (+) and the feedback is connected to the inverting input (-).

## 2. What is the difference between an inverting and non-inverting op amp?

The main difference between an inverting and non-inverting op amp is the polarity of the output signal. In an inverting op amp, the output signal is inverted (180 degrees out of phase) with respect to the input signal. In a non-inverting op amp, the output signal is not inverted (in phase) with respect to the input signal.

## 3. How does an inverting op amp work?

In an inverting op amp, the input signal is connected to the inverting input (-) and the feedback is connected to the non-inverting input (+). The op amp amplifies the difference between the two inputs and produces an inverted output signal. The gain of the op amp is determined by the feedback resistor and the input resistor.

## 4. What is the purpose of an op amp inverting/non-inverting configuration?

The purpose of an op amp inverting/non-inverting configuration is to amplify and/or modify an input signal. In an inverting configuration, the input signal can be inverted, amplified, or both. In a non-inverting configuration, the input signal can be amplified without being inverted. These configurations are commonly used in electronic circuits to perform mathematical operations, such as addition, subtraction, and integration.

## 5. How can I calculate the gain of an inverting/non-inverting op amp circuit?

To calculate the gain of an inverting/non-inverting op amp circuit, you can use the following formulas:

In an inverting configuration: Gain = - (Rf/Rin)

In a non-inverting configuration: Gain = 1 + (Rf/Rin)

Where Rf is the feedback resistor and Rin is the input resistor. The negative sign in the inverting configuration indicates the inverted output signal.

• Electrical Engineering
Replies
3
Views
796
• Electrical Engineering
Replies
6
Views
1K
• Electrical Engineering
Replies
9
Views
4K
• Electrical Engineering
Replies
9
Views
3K
• Electrical Engineering
Replies
4
Views
1K
• Electrical Engineering
Replies
6
Views
1K
• Electrical Engineering
Replies
6
Views
1K
• Electrical Engineering
Replies
5
Views
2K
• Engineering and Comp Sci Homework Help
Replies
34
Views
2K
• Electrical Engineering
Replies
11
Views
2K