How Do Op Amps Respond to Inverting and Non-Inverting Inputs?

[Guineafowl]

I've read quite a few guides on op amps, and most are very similar. I still don't quite get them, though.

Rule: The output always acts to make the input voltages the same.

But there are inverting and non-inverting inputs. Which does the op amp act on? For example, if it's open loop, and the inverting input is lower than the non, does the ouput go high to raise the inverting, or low to lower the non?

 

[NascentOxygen]

The OP-AMP amplifies the difference voltage:

v_o = A_o(v₊ – v_–)

The OP-AMP is unaware of the source or origin of the voltages it sees on its input terminals, so even if the signal on the inverting input is taken as a fraction of what's at its output, the OP-AMP equation still holds and it amplifies the difference voltage as before. This means that with fixed feedback forming the v_– signal we have

v_o = A_o(v₊ – k.v_o)

You should be able to rearrange this to have v_o appear only on the left.

Should it happen that the equation leads to the output voltage needing to exceed the voltage supply that is powering the OP-AMP chip, typically ± 9 volts, then linear amplification cannot be maintained and at these times the equation doesn't hold.

 

[Guineafowl]

The OP-AMP amplifies the difference voltage:

v_o = A_o(v₊ – v_–)

The OP-AMP is unaware of the source or origin of the voltages it sees on its input terminals, so even if the signal on the inverting input is taken as a fraction of what's at its output, the OP-AMP equation still holds and it amplifies the difference voltage as before. This means that with fixed feedback forming the v_– signal we have

v_o = A_o(v₊ – k.v_o)

You should be able to rearrange this to have v_o appear only on the left.

Should it happen that the equation leads to the output voltage needing to exceed the voltage supply that is powering the OP-AMP chip, typically ± 9 volts, then linear amplification cannot be maintained and at these times the equation doesn't hold.

Thanks.

So the Vo is proportional to (V+ - V-); let's call that dV, and assume open loop.

If dV is positive, Vo will swing to the +ve rail.
If dV is negative, Vo will swing to the -ve rail.

I get that the op amp sees a difference, rather than discrete voltages, but it could be said that the output's response only makes sense if you think of it as acting on the inverting input - eg if dV is positive, that means V+ is higher, and the response (swing high) is best directed at the V- input, to raise it.

 

[berkeman]

Rule: The output always acts to make the input voltages the same.

Only when configured with negative feedback.

If dV is positive, Vo will swing to the +ve rail.
If dV is negative, Vo will swing to the -ve rail.

Sort of. Most practical opamps cannot swing the output voltage to the rails. They will usually get to within about a volt or so of the rails. There are some nice (slightly more expensive) CMOS opamps that do have nearly rail-to-rail output drive capability, though.

I get that the op amp sees a difference, rather than discrete voltages, but it could be said that the output's response only makes sense if you think of it as acting on the inverting input

That's not a great way to put it. Just start thinking in terms of opamp circuits using negative feedback to stabilize the gain. There are a number of different topologies of components around the opamp that give you inverting, non-inverting, summing, filtering, etc.

Can you post a link to the reading you have been doing about those uses of opamps with the different configurations of parts around them? They should all be uising negative feedback.

The exception would be if you use an opamp as a "comparator", where you actually use a type of positive feedback to get more of a discrete output as a result of the input difference voltage comparison. Have you looked at comparators as well?

 

[Averagesupernova]

I think a search on this forum would turn up some good results. I know it has been discussed many times. I know I personally have participated. Lots of resources on this forum if you want to dig.

 

[jim hardy]

Rule: The output always acts to make the input voltages the same.

It is the duty of the circuit designer to surround the amplifier with circuitry that allows it to make them equal.
Otherwise it can't "operate"

But there are inverting and non-inverting inputs. Which does the op amp act on? For example, if it's open loop, and the inverting input is lower than the non, does the ouput go high to raise the inverting, or low to lower the non?

It acts on whichever one the circuit designer gave it control over. That's almost always(but not always always) the inverting one.

Find AN31 there's a nice copy at https://www.ti.com/ww/en/bobpease/assets/AN-31.pdf
Here's the first three circuits

Left circuit:
Noninverting input is locked at zero by tying it to circuit common.
Input signal is applied to inverting through R1.
Amplifier holds Inverting at zero through R2.

Right circuit:
Input signal is applied to Noninverting input.
Amplifier drives inverting input to match noninverting via voltage divider R2&R1.

........................................

This one has two Input signals, V1 applied to inverting input and V2 to noninveting . .
TO understand it, first assume all four resistors are equal. Value doesn't matter, use 1K if you like.
Noninverting input is at V2/2.
By voltage divider action, inverting input is at (V1 + Vout)/2
Equating them,
V2/2 = (V1 + Vout)/2
V2 = (V1 + Vout)
Vout = V2 - V1
Now you're ready to try giving it gain.

old jim

 

[Tom.G]

Here is the image @jim hardy posted:

Let's try an approach that some people may be more comfortable with. We will be using the Inverting Amplifier circuit on the left.

The basic starting assumptions are that the op-amp has very high gain, perhaps 100 000, an input impedance high enough to be ignored, and an output impedance low enough that it too can be ignored.
It is supplied with equal positive and negative supply voltages whos common point is circuit common ("ground").

Just as an example set both R1 and R2 to 10kΩ and turn on the power. Vin is not connected to anything. Since the op-amp has gain, any voltage difference between the input pins will show up at the output.

Now assume that the op-amp is NOT ideal and has a small input offset voltage, perhaps making the Inverting input 1mV above the Non-Inverting input. With the op-amp gain being 100 000, this would try to drive the output to -100V. But keep in mind that the output is connected to Inverting input through R2. When the output gets to -1mV it has canceled the +1mV of offset at the input. That's also why the inputs are often referred to as being "at Virtual Ground", because the Inverting input is driven to match (very closely) the Non-Inverting input.

That's the long explanation of your quote in the OP: "Rule: The output always acts to make the input voltages the same." And it holds when there is a net negative feedback in the circuit, i.e when the output is fed back the Inverting input.

Now it's time to complicate matters by considering what happens when there is a voltage applied at Vin. Here it is easier to explain if we consider the currents through R1 and R2. As we saw above, due to the negative feedback via R2, the Inverting input is driven to match the Non-Inverting Input.

Lets start with the power off. Apply a voltage, let's say +1V, to Vin, which applies a voltage to Inverting input.

Now turn on the power. As described above, the feedback from the output through R2 to the Inverting input, will tend to match the voltage at the Inverting input to the voltage at the Non-Inverting input. To do this, there must be the same amount of current flowing thru R2 as there is flowing thru R1, but of the opposite polarity. You now have an Op-Amp circuit with a gain of minus one (-1). The output is at -1V.

If you want a higher gain, you increase the ratio of R2/R1. For instance for a gain of three, R2/R1 would be 3; you could change R2 to 30kΩ or change R1 to 3 333Ω. This means the output voltage must be 3 times Vin to drive a matching, compensating current thru R2.

So there is a version of OP-AMPS 101. Hope it helped.

Cheers,
Tom

 

[Guineafowl]

Only when configured with negative feedback

Aha. That makes more sense. I've been reading mainly published books, such as Electronics for Dummies (yes, I know), and Practical Electronics for Inventors. Also some online guides/Youtube. In every one, that rule is stated without qualifying the negative feedback criterion. That's where my confusion lay. I wonder if this is a case of teaching being handed down without questioning it, or perhaps experts writing a guide to something they understand too well, so they forget the little details that might confuse a newbie.

So it's the configuration of the op amp, not the op amp itself, that the rule applies to. Got it - thanks!

 

[Guineafowl]

And it holds when there is a net negative feedback in the circuit, i.e when the output is fed back the Inverting input.

Bingo! The rule is stated to apply to op amps, when really it's the configuration. Trouble is, you can't ask an author questions like that so you end up re-reading the text until it makes sense. In this case, it didn't.

Thanks to you and Jim also. I see that the first stage to analysing an op amp circuit is to identify the configuration, then work out what voltages should be present. Your posts will certainly help me do that.

 

[jim hardy]

Bingo! The rule is stated to apply to op amps, when really it's the configuration.

Yes. That's the key point, somebody has to wrap that amplifier in a circuit that gives feedback.

Word "Operational" is significant. remember these came out of the days of analog computing before digital computers were fast enough to solve differential equations real time.
When configured with feedback the amplifier circuit performs some mathematical operation. The ones we discussed above just add or multiply by a constant, very simple operations. If you take a course in analog computing solving differential equations you'll use them to do integration .

You might enjoy this historic little book
http://www.waynekirkwood.com/images/pdf/Applications_Manual_for_Operational_Amplifiers_Part_1.pdf

and this archive
http://www.philbrickarchive.org/

 

[berkeman]

Bingo! The rule is stated to apply to op amps, when really it's the configuration. Trouble is, you can't ask an author questions like that so you end up re-reading the text until it makes sense. In this case, it didn't.

Here is a thread that I think you will enjoy. It's about bad circuits and why they are bad. it follows the theme you mention of instruction not always giving the best info/advice.

https://www.physicsforums.com/threads/bad-circuits-test-your-knowledge.178516/

 

[dlgoff]

Word "Operational" is significant. remember these came out of the days of analog computing before digital computers were fast enough to solve differential equations real time.

You might enjoy this historic little book
http://www.waynekirkwood.com/images/pdf/Applications_Manual_for_Operational_Amplifiers_Part_1.pdf

and this archive
http://www.philbrickarchive.org/

Wikipedia had a good Historical timeline for Operational Amplifiers.

 

[Guineafowl]

Here is a thread that I think you will enjoy. It's about bad circuits and why they are bad. it follows the theme you mention of instruction not always giving the best info/advice.

https://www.physicsforums.com/threads/bad-circuits-test-your-knowledge.178516/

Quite a fun exercise - a little 'above my pay grade' at the moment but I'm sure I'll come back to it now and again.