How does an op amp "see" the feedback voltage?

[pacman22]

Consider an inverting op amp with a feedback resistor connected to the inverting input from the output. How does the op amp "see" this feedback voltage when an input voltage is already being fed to the inverting terminal of the op amp?

 

[danadakk]

Because the inputs to an OpAmp are close to ideal, that is they draw no current.
And as long as Aol is high enough at the frequency of interest this holds true.

Therefore the only relationship left is the voltage divider actions of Vo and Vin applied
to the feedback R's.

https://www.electronics-tutorials.ws/opamp/opamp_2.html

T(s) as a f() of Aol considerations found here :

https://www.ti.com/lit/an/slyt367/slyt367.pdf?ts=1685554757882&ref_url=https%253A%252F%252Fwww.google.com%252F

Regards, Dana.

 

[LvW]

Consider an inverting op amp with a feedback resistor connected to the inverting input from the output. How does the op amp "see" this feedback voltage when an input voltage is already being fed to the inverting terminal of the op amp?

The answer is simple:
There are two currents which meet in the node which belongs to the inverting opamp node: (1) The current driven by the input signal voltage and (2) the current driven by the opamps output voltage Vout (voltage source due to the low output impedance).
Both currents together create a voltage v_n at this node wich is in the range of some µVolts only (Vout/Aol with Aol=open-loop voltage gain).
Therefore, in this case we have voltage driven current feedback.
In practice and for calculation purposes, this voltage v_n is mostly set equal to the voltage v_p at the non-inv. opamp input (which is zero for the inverting amplifier).

 

[D.A.(Tony)Stewart]

Consider an inverting op amp with a feedback resistor connected to the inverting input from the output. How does the op amp "see" this feedback voltage when an input voltage is already being fed to the inverting terminal of the op amp?

Using an differential input amplifier with negative feedback always amplifies the error between the applied input and the stable reference which is normally DC in the mid-supply range or the desired input average. When the op. amp. has a gain of 10e6 or 1 million the input differential signal is the output divided by this gain. Although we consider the input differential voltage as a virtual null, the input offset and signal is predictable. The output is then a function of the resistor ratios.

If the output ever "hits the supply rail" or saturated the feedback stops and thusthe gain stops and becomes ZERO. This is important to remember this that when you have a complex control system that depends on gain for nice stable performance. So saturating leads to abnormal recovery.

 

[LvW]

Consider an inverting op amp with a feedback resistor connected to the inverting input from the output. How does the op amp "see" this feedback voltage when an input voltage is already being fed to the inverting terminal of the op amp?

Just a short comment:
* In case of an inverting gain stage the opamp does not "see" the feedback voltage, but the CURRENT driven by the feedback voltage (feedback signal is a current).

* Only for non-inverting operation (input signal at the pos. opamp input node) the feedback signal is "seen" by the opamp in form of a voltage (at the neg. opamp input node).

 

[KlausST]

Hi,

Consider an inverting op amp with a feedback resistor connected to the inverting input from the output. How does the op amp "see" this feedback voltage when an input voltage is already being fed to the inverting terminal of the op amp?

A little correction:
* the input voltage (V_in) is not fed to the inverting terminal (IN-). Instead it is fed to a resistor. Let´s call it "R_i". You need this resistor.
(if you´d fed the input voltage directly to the OPAMP then it won´t work properly)
Since there is (considered) no current into the OPAMP .. the current through R_i must be the same as through R_f.

Ther´s just one other "must" for a properly working OPAMP: (idealized) the voltage at IN- is the same as at IN+.
The rest is just Ohm´s law.

****
I´m pretty sure if you do an internet search on this topic you find millions of good explanations with diagrams and math. Even thousands of videos.
Please watch a couple of them. Then if there still are questions, then post a link to the video/document and ask a detailed question.
****

@LvW :

In case of an inverting gain stage the opamp does not "see" the feedback voltage, but the CURRENT driven by the feedback voltage (feedback signal is a current).

Can we really say the OPAMP "sees" the current?
How can it "see" the current while we consider the input current to be zero.
And if we consider it not zero, then we have maybe a million times higher current at an BJT input OPAMP than on a CMOS input OPAMP .. but still the output for both OPAMPs will be (almost) the same.

Klaus

 

[danadakk]

(if you´d fed the input voltage directly to the OPAMP then it won´t work properly)

Of course we do this quite often, connect an input V source directly to an OpAmp,
either terminal, Vin + for a follower, and Vin - for feedback V from other network
and sensor sources, like classic current sink with a bipolar output....

I would add we "see" current and V with OpAmp because they are interdependent
and determinate / analytic as such.


Regards, Dana.

 

[KlausST]

Hi,

(if you´d fed the input voltage directly to the OPAMP then it won´t work properly)

The OP wrote "inverting OPAMP" and "input voltage is already being fed to the inverting terminal of the op amp".
Also my context referred to "IN-" and "direct"

I was referring to the OP's topic. And explained what's the meaning of V_in and IN-.

*******
You are free to discuss many other Opamp circuits, but - respectfully - please don't refer them to my posts.

Klaus

 

[danadakk]

I will preface all future posts with @OP, that was the intended person I
wanted to make sure was not confused by prior posts.

Knight.

 

[LvW]

@LvW :

Can we really say the OPAMP "sees" the current?
How can it "see" the current while we consider the input current to be zero.
And if we consider it not zero, then we have maybe a million times higher current at an BJT input OPAMP than on a CMOS input OPAMP .. but still the output for both OPAMPs will be (almost) the same.

Klaus

Well, in my answer I have tried to use the same term ("see") as it is used in post#1.
Of course, the opamp does not "see" anything - however, I think we can say that the opamp "reacts" upon the current that is fed back from the output because this current - when combined with the input current - creates the tiny voltage at the inverting input node which we normally set equal to zero.

 

[KlausST]

Hi,

A misunderstandig here. I want to explain.

I was focussed on current vs voltage. Because I treat the OPAMP inputs functionally as voltage_input .. while the current at the input functionally has no meaning. And to explain "functionally": I mean that OPAMP_gain is specified as V/V and not I/I.

I guess you focussed on "see" because I´ve put it in quotation marks. I used the quotation marks as so called "scare quotes", because the OPAMP obviously can´t see.

***

English isn´t my mother tongue, but if it was, there was still room for misunderstandings.
So no complaints from my side.

Klaus

 

[D.A.(Tony)Stewart]

- Lots of ways to say the same thing for a virtual input null in an OA (Op Amp).

Negative feedback always drives to make both inputs match.,

For inverting input the feedback inverts and cancels the input to match the Vin+ reference.
For non-inverting input, the reference moved to inverting input resistor end.

This is normally set to (Vdd+Vss)/2 or (Vcc+Vee)/2 and that makes the common mode input virtual null in the middle of "most" OA's operating range.,

For a differential amplifier there is a 0V reference for bipolar or mid-Vcc reference for single supply, this time using the non-inverting input.

Here are the differential inputs with a differential configuration. https://tinyurl.com/2lmtomhb

 

[danadakk]

@OP

There are two key types of OpAmp, ones that operate in voltage mode
and ones that more closely approximate current mode, called Nortyon OpAmps.
Each with common design concerns and individual attributes.

https://www.ti.com/lit/an/snoa653/snoa653.pdf?ts=1685714565192&ref_url=https%253A%252F%252Fwww.google.com%252F

http://www.bitsavers.org/components/national/_appNotes/AN-0278.pdf

Regards, Dana.

 

[LvW]

@OP

There are two key types of OpAmp, ones that operate in voltage mode
and ones that more closely approximate current mode, called Nortyon OpAmps.
Each with common design concerns and individual attributes.

To complete the picture - the second-generation Current Conveyor (CCII) is used much more often used than the Norton type. The CCII has one high-impedance non-inverting and a second low-impedance inverting input node (and a current output).

 

[D.A.(Tony)Stewart]

I "see" some interesting feedback to @pacman22 on atypical Op Amps. I hope this does not confuse him. FWIW, Falstad's Op Amps are ideal but also has two real models that run slower and also has voltage and current controlled voltage and current source "black boxes.

Consider an inverting op amp with a feedback resistor connected to the inverting input from the output. How does the op amp "see" this feedback voltage when an input voltage is already being fed to the inverting terminal of the op amp?

Wikiwand - Error amplifier (electronics)

An error amplifier is most commonly encountered in feedback unidirectional voltage control circuits, where the sampled output voltage of the circuit under control, is fed back and compared to a stable reference voltage. Any difference between the two generates a compensating error voltage which...

www.wikiwand.com

Essentially an Op Amp is an error amplifier with an integrating compensation capacitor built-in.

For "Control Systems" reasons of minimum phase shift with negative feedback in order to avoid unwanted oscillations for stable operation, minimal overshoot and minimize the "virtual null" error on the input.

Here the word virtual or virtually is like the english word "literally" where it is used for emphasis (meaning almost the same, but not exactly) .Again this is because the input signal "error" is just the output signal divided by the gain , or a virtual null input, but if you shorted the diff. inputs it wouldn't work and tiny offsets would drive the output to one extreme limit or the other.

I hope you provide some feedback, negative or positive on the answers you have received.