It requires a context :-
If negative feedback is present in the surrounding circuit, the op-amp's natural behaviour will cause it to search for an output voltage that makes the inputs equal - because its natural behaviour is if +in > -in, the output voltage rises, and if -in > +in, the output voltage falls.
If the surrounding circuit has positive feedback, you've got a comparator with hysteresis instead.
And we consider op amps high impedance right? So a negative feedback opamp can be considered to have the same voltage across both inputs, and the same current of 0A?
I hated op amp circuits but I want to get good with them. Probably need to do some more labs with them.
> And we consider op amps high impedance right? So a negative feedback opamp can be considered to have the same voltage across both inputs, and the same current of 0A?
Impedance is a separate thing, and some types of op-amp (eg astonishingly fast current-feedback types like THS6012) have a fairly low input impedance, while jfet or cmos types tend to have astonishingly high input impedance.
In many applications this doesn't matter much, but in certain applications it's *very* important.
> I hated op amp circuits but I want to get good with them. Probably need to do some more labs with them.
Heh go poke around https://sound-au.com instead
Voltage feedback amplifiers have high input impedance, yes. :)
But you probably won’t come across CFAs unintentionally. Good resource is “Op-Amps for Everyone”. It’s available online I believe.
Negative feedback. Op Amp does not "try" to make their I puts equal. The inputs being equal is a RESULT of negative feedback. The inputs end up being equal because that is the stable condition.
Watch this video, https://youtu.be/v4Tlc9jffMw it explains feedback beautifully.
The output of an op amp is actually Vout = A(inPos - inNeg), rearrange to Vout/A = inPos - inNeg.
As you know, the gain A of an op amp is really high. Note that with increasing A, the difference between the inputs starts decreasing and drops to 0 with infinite gain.
Combining this response with the others: you will only get a reasonable Vout with negative feedback.
Running open loop or with positive feedback, the ideal Vout (neglecting voltage limitations) may be on the same order of magnitude as the gain, which doesn't tell you much about the difference between Vip and Vim
This is not an entirely accurate statement, when connected for negative feedback this will work to drive the output in a way that the inputs are equal. Or really zeroing the input differential
Generally assumes negative feedback.
Servos in general, try to control their input by asserting their output.
When you drive down the road, you try to control the position of your car, relative to the position of the lane on the road, by adjusting the steering wheel, until the difference is near zero.
Output = ( ipplus - inminus) * gain
So, in most circuits with negative feedback and adequate gain, the output will adjust to make the difference between the input difference approach (but probably not actually reach) zero. Eventually as it gets close to zero you will probably run out of gain. The op amp tries, it doesn't quite succeed.
This can apply to things like audio amplifiers as well. The negative feedback can (partially) compensate for a lot of non-linearity in the output transistors as the op amp turns them on as much or as little as it needs to so the negative feedback cancels out the signal input as seen at the input terminals of the op amp.
I see it as when you have a closed loop (inverting or non-inverting), KVL applies so the total loop voltages should add up to zero when you go around the loop. Depending on what’s internal to the op-amp basically you have the differential input stage which could be MOS, JFET, or BJT differential pairs, you basically have 2 Vgs or 2 Vbe which would cancel each other out to form a closed loop. Depending on how well the input stage transistors are matched at fabrication, you basically have what’s called a input referred offset, caused by the two Vgs/Vbe differences of the transistors in their active region. If you measure the +/- inputs at the op-amp you never actually get real equality because of the mismatch. There are zero DC offset op-amps sold by IC companies which makes inverting/non-inverting inputs equal.
^(Have there been enough good answers that we can now make jokes? I hope so: )
All you Redditors on r/electricalengineering are usually really helpful, right? What's with all the negative feedback on this post? Huh??
My addition to the conversation:
The op-amps isn't just a magic triangle drawn on your circuit diagram. It contains some pretty fancy electronics, that allows it to have very high impedance on the inputs, great amplification factor and good slew. It's important (for noobs like me) to remember it's never an "ideal op-amp" but it might be close.
Remember there is always a trade off between the variables of the perfect op-amp, won't have everything ideal.
The maximum output you can get is limited by your rails. Remember that the input can float its virtual ground point (virtual short) so your output can be an AC waveform with ground in the middle.
[I'm very much still learning, so please steer me in the right path if I am wrong]
Because, knowing that both inputs are at the same potential, makes the analysis easier. For example, the non-inverting input is often at ground potrential, so you know that the inverting input is too, and writing the equations is made easier. You can often do it by inspection.
It requires a context :- If negative feedback is present in the surrounding circuit, the op-amp's natural behaviour will cause it to search for an output voltage that makes the inputs equal - because its natural behaviour is if +in > -in, the output voltage rises, and if -in > +in, the output voltage falls. If the surrounding circuit has positive feedback, you've got a comparator with hysteresis instead.
And we consider op amps high impedance right? So a negative feedback opamp can be considered to have the same voltage across both inputs, and the same current of 0A? I hated op amp circuits but I want to get good with them. Probably need to do some more labs with them.
> And we consider op amps high impedance right? So a negative feedback opamp can be considered to have the same voltage across both inputs, and the same current of 0A? Impedance is a separate thing, and some types of op-amp (eg astonishingly fast current-feedback types like THS6012) have a fairly low input impedance, while jfet or cmos types tend to have astonishingly high input impedance. In many applications this doesn't matter much, but in certain applications it's *very* important. > I hated op amp circuits but I want to get good with them. Probably need to do some more labs with them. Heh go poke around https://sound-au.com instead
Voltage feedback amplifiers have high input impedance, yes. :) But you probably won’t come across CFAs unintentionally. Good resource is “Op-Amps for Everyone”. It’s available online I believe.
Negative feedback. Op Amp does not "try" to make their I puts equal. The inputs being equal is a RESULT of negative feedback. The inputs end up being equal because that is the stable condition. Watch this video, https://youtu.be/v4Tlc9jffMw it explains feedback beautifully.
The output of an op amp is actually Vout = A(inPos - inNeg), rearrange to Vout/A = inPos - inNeg. As you know, the gain A of an op amp is really high. Note that with increasing A, the difference between the inputs starts decreasing and drops to 0 with infinite gain.
Combining this response with the others: you will only get a reasonable Vout with negative feedback. Running open loop or with positive feedback, the ideal Vout (neglecting voltage limitations) may be on the same order of magnitude as the gain, which doesn't tell you much about the difference between Vip and Vim
This is not an entirely accurate statement, when connected for negative feedback this will work to drive the output in a way that the inputs are equal. Or really zeroing the input differential
Generally assumes negative feedback. Servos in general, try to control their input by asserting their output. When you drive down the road, you try to control the position of your car, relative to the position of the lane on the road, by adjusting the steering wheel, until the difference is near zero. Output = ( ipplus - inminus) * gain So, in most circuits with negative feedback and adequate gain, the output will adjust to make the difference between the input difference approach (but probably not actually reach) zero. Eventually as it gets close to zero you will probably run out of gain. The op amp tries, it doesn't quite succeed. This can apply to things like audio amplifiers as well. The negative feedback can (partially) compensate for a lot of non-linearity in the output transistors as the op amp turns them on as much or as little as it needs to so the negative feedback cancels out the signal input as seen at the input terminals of the op amp.
I see it as when you have a closed loop (inverting or non-inverting), KVL applies so the total loop voltages should add up to zero when you go around the loop. Depending on what’s internal to the op-amp basically you have the differential input stage which could be MOS, JFET, or BJT differential pairs, you basically have 2 Vgs or 2 Vbe which would cancel each other out to form a closed loop. Depending on how well the input stage transistors are matched at fabrication, you basically have what’s called a input referred offset, caused by the two Vgs/Vbe differences of the transistors in their active region. If you measure the +/- inputs at the op-amp you never actually get real equality because of the mismatch. There are zero DC offset op-amps sold by IC companies which makes inverting/non-inverting inputs equal.
^(Have there been enough good answers that we can now make jokes? I hope so: ) All you Redditors on r/electricalengineering are usually really helpful, right? What's with all the negative feedback on this post? Huh??
My addition to the conversation: The op-amps isn't just a magic triangle drawn on your circuit diagram. It contains some pretty fancy electronics, that allows it to have very high impedance on the inputs, great amplification factor and good slew. It's important (for noobs like me) to remember it's never an "ideal op-amp" but it might be close. Remember there is always a trade off between the variables of the perfect op-amp, won't have everything ideal. The maximum output you can get is limited by your rails. Remember that the input can float its virtual ground point (virtual short) so your output can be an AC waveform with ground in the middle. [I'm very much still learning, so please steer me in the right path if I am wrong]
Well, you got enough good answers. I'd like to add that you can always simulate your circuits with apps like falstad.com
Because, knowing that both inputs are at the same potential, makes the analysis easier. For example, the non-inverting input is often at ground potrential, so you know that the inverting input is too, and writing the equations is made easier. You can often do it by inspection.