# I don't understand negative feedback with operational amplifiers

#### wouter368

Joined Sep 22, 2022
20
Hey,

I stumbled upon the following paragraph with image but I couldn't figure out why this works:

I know that when Vin is higher than Vin', Vout will max out at negatively at -15V. I understand that otherwise, when Vin is lower than Vin', Vout will max out positively at +15V. However, when you connect Vout with Vin', I run into problems:

Let's suppose Vin is larger than Vin'. This means Vout will max out at -15V. That means that Vin' will decrease, nearing -15V. So, this means that Vin will become even larger than Vin', right? Because we first had Vin > Vin' and now Vin' has become even smaller.

So that means that the difference between Vin and Vin' increasing even more, instead of leveling out, what the text suggests. So where in my thought process do I make a mistake? Thanks for helping me out!

#### ronsimpson

Joined Oct 7, 2019
2,527
Let's suppose Vin is larger than Vin'. This means Vout will max out at -15V
Vin is larger than Vin' the output will head up. not down.

#### wouter368

Joined Sep 22, 2022
20
Vin is larger than Vin' the output will head up. not down.
To quote the source: " Suppose that Vin is larger than Vin‘. Then Vout will quickly become negative, approaching -15V." So would you say the source I am using made a mistake which should be ignored?

I don't think the source is wrong there though, and correct me if I'm wrong, because of the following formulas:

Vout = Adm (differential amplification factor) * (Vin-Vin')

Where Vin is the non-inverting input and Vin' is the inverting input.

Since Adm is always negative, when Vin > Vin', you will have the following result:

Vout = Adm * (Vin-Vin') = negative number * positive number = negative output

#### ronsimpson

Joined Oct 7, 2019
2,527
Input(+) causes the output to go positive.
Input (-) causes the output to go negative.

Vin=2V, Vin'=1V Vin-Vin'=1V 1V X open loop gain = Voutput (open loop)

in=2V, Vin'=1V Vin-Vin'=1V because 1V is positive the output will move up. Because Vout is connected to Vin' this causes Vin' to move up. Soon Vin=2V, Vout=2V and Vin'=2V now Vin-Vin'=0 and the output holds at that voltage.

#### BobTPH

Joined Jun 5, 2013
6,078
On the drawing, I see Vin on the + input and V’in on the - input. And nothing is labeled Vin’. I don’t think the drawing corresponds to the text.

#### WBahn

Joined Mar 31, 2012
27,880
To quote the source: " Suppose that Vin is larger than Vin‘. Then Vout will quickly become negative, approaching -15V." So would you say the source I am using made a mistake which should be ignored?

I don't think the source is wrong there though, and correct me if I'm wrong, because of the following formulas:

Vout = Adm (differential amplification factor) * (Vin-Vin')

Where Vin is the non-inverting input and Vin' is the inverting input.

Since Adm is always negative, when Vin > Vin', you will have the following result:

Vout = Adm * (Vin-Vin') = negative number * positive number = negative output

Adm is a POSITIVE number -- typically in the range of 100,000 to 10,000,000.

#### dl324

Joined Mar 30, 2015
15,453
So where in my thought process do I make a mistake?
The typical method for solving a circuit like this is to use the zero differential input theorem.

#### Papabravo

Joined Feb 24, 2006
19,580
Vin' and Vout are the same voltage because they are connected by a wire with a resistance so close to zero it is indistinguishable from zero. The negative feedback reduces the large open loop gain of the amplifier to unity. So, the amplifier makes Vout equal to a voltage that will minimize the quantity (Vin - Vout) to a number very close to zero. If Vin - Vout = 0, then it follows that Vout ≈ Vin.

#### crutschow

Joined Mar 14, 2008
31,123
The main thing to remember with negative feedback in an op amp, is that the op amp will move the output to whatever voltage is needed (as long as it's below saturation) to keep the two inputs at essentially the same voltage.
If you follow that, then you should be able to understand how any op amp circuit with negative feedback works.

From that is should be easy to see how the non-inverting follower circuit you posted, works.

#### wouter368

Joined Sep 22, 2022
20
Adm is a POSITIVE number -- typically in the range of 100,000 to 10,000,000.
If Adm is a positive number, that would clarify everything for me. However, I don't understand how Adm could be a positive number, because I thought that you calculate Adm like this (in a differential amplifier circuit):

This would give a negative value for Adm. So, is this formula for a differential amplifier wrong? Or might there be some additional component in op amps that inverts the output voltage?

Thanks for helping me!

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#### WBahn

Joined Mar 31, 2012
27,880
I don't know why your differential pair has Vout on the same side as Vin. Every reference I have seen has Vout on the other side.

#### LvW

Joined Jun 13, 2013
1,586
I don't know why your differential pair has Vout on the same side as Vin. Every reference I have seen has Vout on the other side.
Yes - the nomenclature is rather "uncommon".
It should be clear that we have an inverted output signal at the most left output node with respect to the most left input signal - and vice versa.

#### BobaMosfet

Joined Jul 1, 2009
2,053
Hey,

I stumbled upon the following paragraph with image but I couldn't figure out why this works:
View attachment 280781
I know that when Vin is higher than Vin', Vout will max out at negatively at -15V. I understand that otherwise, when Vin is lower than Vin', Vout will max out positively at +15V. However, when you connect Vout with Vin', I run into problems:

Let's suppose Vin is larger than Vin'. This means Vout will max out at -15V. That means that Vin' will decrease, nearing -15V. So, this means that Vin will become even larger than Vin', right? Because we first had Vin > Vin' and now Vin' has become even smaller.

So that means that the difference between Vin and Vin' increasing even more, instead of leveling out, what the text suggests. So where in my thought process do I make a mistake? Thanks for helping me out!
I recommend you find a simulator- there are many available today- they are invaluable for letting you see the behavior of something like an OpAmp, and playing with it to see what happens. You can learn more in an afternoon this way, than a week in books.

I recommend this for learning:

https://everycircuit.com/app/

#### wouter368

Joined Sep 22, 2022
20
I don't know why your differential pair has Vout on the same side as Vin. Every reference I have seen has Vout on the other side.
Wow, thanks. That makes everything clear. One more question, is Acm (Common Mode Amplification) with differential amplifiers / op amps also supposed to be positive? Because right now, according to calculations from the same source, it should be negative. However, from testing with a circuit that I built my hypothesis was that it would be positive. Here's the calculations this source used:

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#### wouter368

Joined Sep 22, 2022
20
I have a question about differential amplifiers. Is Acm (Common Mode Amplification) with differential amplifiers / op amps supposed to be a positive number? Because right now, according to calculations from the same source, it should be negative. However, from testing with a circuit that I built my hypothesis was that it would be positive. Here's the calculations this source used:
View attachment 280851
View attachment 280852
View attachment 280854
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View attachment 280856
If someone could please explain why Acm would be either negative or positive, that would help me a lot.

#### WBahn

Joined Mar 31, 2012
27,880
If someone could please explain why Acm would be either negative or positive, that would help me a lot.
Conceptually it can be either and depends on the particulars of the implementation.

If the average output signal increases (becomes either more positive or less negative) when the average input signal increases, then the common-mode gain is positive.

In your simple BJT differential pair using a resistor as the current source, if you increase both of the input signals, the current in the common emitter resistor will increase. That will result in more current in each of the transistors and, hence, in each of the collector resistors. That will lower the voltage at both of the output ports. Thus, for that topology, the common-mode gain will be negative.

#### LvW

Joined Jun 13, 2013
1,586
I have a question about differential amplifiers. Is Acm (Common Mode Amplification) with differential amplifiers / op amps supposed to be a positive number? Because right now, according to calculations from the same source, it should be negative.
.......................
If someone could please explain why Acm would be either negative or positive, that would help me a lot.
The common mode gain for the classical transistor-based diff. amplifier (long-tailed pair) is always negative because of the phase inversion between base and collector.

#### wouter368

Joined Sep 22, 2022
20
Conceptually it can be either and depends on the particulars of the implementation.

If the average output signal increases (becomes either more positive or less negative) when the average input signal increases, then the common-mode gain is positive.

In your simple BJT differential pair using a resistor as the current source, if you increase both of the input signals, the current in the common emitter resistor will increase. That will result in more current in each of the transistors and, hence, in each of the collector resistors. That will lower the voltage at both of the output ports. Thus, for that topology, the common-mode gain will be negative.
Ah, that makes sense. Thanks. I have a follow up question to this topic regarding the circuit that I built but I will post that in another thread since I don't think it belongs to homework help.