# Op Amp, trying to approach it again.

#### Xenon02

Joined Feb 24, 2021
355
Hello !

I know I've had this post about Op Amp, but I wanted to start this again with fresh mind (also there is a lot of pages and it's hard to find the answears sorry).

I guess there are only 2 questions? So comming with the first one : Op Amp with only negative feedback simple inverting OpAmp.

It is known that giving Vin negative input will result in reverting the input. So Vin = 1V will result in Vout = -1V.
Also V- is equal = 0V so that V- is equal V+. But according to this : Vout = Aol(V+ - V-), the output should result in something different ? Like Vout= 0V ? More logical would be if V- = 1V so Vout = Aol(0 - 1V). It is probably wrong but it is important for next question !

Second question is for both feedbacks :

I've had from the last post all calculations I will post them here

Here there are calculations for Vx (which is here V+) that it is equal to V-. But for the first one the Vx is positive and for the other one the Vx is negative while Vin is constant 1V positive.
I've heard that it depends on the which feedback is "stronger". I've seen those equations for stronger feedbacks, but I don't understand it when I look att this equation : Vout = Aol(V+ - V-). Let's take aside this "stronger" feedback equations. Why V- and V+ can't be negative while Vin is positive ? the results mentions that they can be equal so it should work.
Also the next answear was also because I have connected Vin which was positive to positive input, and ?I mean why connecting Vin to positive input must result in positive output when V- and V+ is not defined ? Unknown values. For the first question it is also a good question. Why Vin defines the Output ? when Vout = Aol(V+ - V-) and V- = 0V ?

If something is not clear I'll try to write it better.

#### MrChips

Joined Oct 2, 2009
27,114
For question #1 you are confusing Vin and V-.
They are not the same thing.

Let us assume for now that R1 = R2.
V- = V+ = 0V.

No current flows in the V- pin.
Calculate the current in R1.
Calculate the current in R2.

#### Xenon02

Joined Feb 24, 2021
355
Let us assume for now that R1 = R2.
V- = V+ = 0V.

No current flows in the V- pin.
Calculate the current in R1.
Calculate the current in R2.
Current in R1 is Vin/R1.
Current in R2 = Vout/R2.

I'm aware that V- isn't the same as Vin. I just mean that why Vin determines the Vout when in this equation Vout = Aol(V+ - V-) most likely V- determines. I just read that if I plug Vin to negative input then the output must be reversed for example 1V Vin results in Vout = -1V. Or somethine like that When I saw that equation I was like hmmm. Then why we say that Vin determines the Vout when V- determines it.

#### MrChips

Joined Oct 2, 2009
27,114

Now, no current flows into V- pin.
What does this tell you about the currents in R1 and R2?

#### Xenon02

Joined Feb 24, 2021
355
Which equation ?

Now, no current flows into V- pin.
What does this tell you about the currents in R1 and R2?
Same current flows through R1 and R2.

#### MrChips

Joined Oct 2, 2009
27,114

#### Xenon02

Joined Feb 24, 2021
355
Vout = Aol(V+ - V-)
This is a wrong equation ? I've seen it in many websites and tutorials.

So write this in the form of an equation.
Current for R1 is equal Vin/R1 and it must be equal to Vout/R2.
Vout =Vin*R2/R1.

#### ericgibbs

Joined Jan 29, 2010
16,364
hi X02,
I would suggest that you first focus on the Inverting OPA circuit.

Gain for, -Vout = Vin( Rfb/Rin)

E

#### Xenon02

Joined Feb 24, 2021
355
hi X02,
I would suggest that you first focus on the Inverting OPA circuit.

Gain for, -Vout = Vin( Rfb/Rin)

E
I believe I do. Using the assumptions that V+ = V-.

#### ericgibbs

Joined Jan 29, 2010
16,364
Hi,
Exactly what do you mean by: I believe I do. Using the assumptions that V+ = V-.
E

#### Xenon02

Joined Feb 24, 2021
355
Hi,
Exactly what do you mean by: I believe I do. Using the assumptions that V+ = V-.
E
Hi

I've calculated Vout using the assumption that V+is equal V-. So V- is = 0V it is grounded so current in R1 is equal the current in R2.
I am trying to grasp the idea how this : Vout = Aol(V+ - V-) works as well.
I just remembered that for positive feedback it was also possible to calculate V+ = V-. But it was incorrect because Vin if it was connected for example in positive input then the output shouldn't be reversed.
Wierd.

#### ericgibbs

Joined Jan 29, 2010
16,364
hi X,
You must define more clearly what you mean by V+ and V-.

Is your V+ the Vin Voltage on the Inverting Input and the V- the -Vin voltage on the Non Inverting Input of the OPA.?

E

#### Xenon02

Joined Feb 24, 2021
355
hi X,
You must define more clearly what you mean by V+ and V-.

Is your V+ the Vin Voltage on the Inverting Input and the V- the -Vin voltage on the Non Inverting Input of the OPA.?

E
I'm sorry
V+ and V- is the potential on the Op Amp inputs.
Vin is just the voltage that is connected to R1.

check the first image of this post

#### ericgibbs

Joined Jan 29, 2010
16,364
hi,
Ref your Vout = Aol(V+ - V-),
The equation is intended to be read this way.

Vout = Aol( Voltage on V+[Non Inverting] - Voltage on V- [Inverting])

So if Aol =10.
Voltage on V+[Non Inverting] = 0V
Voltage on V- {Inverting]) = -1

So Vout = 10( 0 -(-1) = +10

E

#### Xenon02

Joined Feb 24, 2021
355
hi,
Ref your Vout = Aol(V+ - V-),
The equation is intended to be read this way.

Vout = Aol( Voltage on V+[Non Inverting] - Voltage on V- [Inverting])

So if Aol =10.
Voltage on V+[Non Inverting] = 0V
Voltage on V- {Inverting]) = -1

So Vout = 10( 0 -(-1) = +10

E
Okey

Using this equation Vout = Aol(V+ - V-), and knowing that V- = 0V because V+ is grounded and V+ = V-. the Vout should be Vout = 0V.
More over, I read that where I plug Vin it decides what will be the Vout. For example 1V connected to positive input OP Amp, the Vout will be also 1V (or more the thing is that it is positive). If I connect Vin = 1V to negative input of OP Amp then the Vout is negative.
For both feedbacks it is not a typical sinlge feedback, so why connecting Vin to negative input of OP Amp then the Vout must be inversed ? V+ is now not grounded. The value of V+ and V- can be anything.
I've had the same issue with positive feedback, because there I also could calculate V+ and V- that they could be equall V+ = V-. But the problem was that because when I connected Vin = 1V to negative input of OP Amp then the output was also positive. And that was a problem and I don't know why. Even though the V- and V+ can have any value but must be equal, and Vout is decided by this equation Vout = Aol(V+ - V-). Not only by Vin connected to positive or negative input of Op Amp.

#### Jony130

Joined Feb 17, 2009
5,419
For a finite value of Aol, the (V+ - V-) will not be equal to 0V. Only for the case when Aol =∞. We will have V+ = V-

#### Xenon02

Joined Feb 24, 2021
355
For a finite value of Aol, the (V+ - V-) will not be equal to 0V. Only for the case when Aol =∞. We will have V+ = V-
Okey.
So how it works with both feedbacks (for now let's just ignore the fact of "stronger" feedbacks).
V+ = V- can be calculated for positive feedback and for negative feedback. For negative it works fine but for positive feedback it can be calculated but it is somehow wrong. Why ? If V+ and V- can determine the Vout, then why we say that Vin determines what is the Vout ?
For both feedbacks the value of V+ and V- can be anything, but for one example it will be okey but for the another it won't. Because Vin = 1V connected to positive input of Op Amp cannot give negative V+ and V-. Because Vout must be positive. I don't get this part. Why Vout must be positive ? V+ and V- can have any value and they determine the Vout.

#### MrChips

Joined Oct 2, 2009
27,114
Vout = Aol(V+ - V-)

“Same current flows in R1 and R2.”
So write this in the form of an equation.
Following up on what I posted,
V- = V+ = 0
Hence Vout = 0.
We know this is incorrect. Hence something is wrong.

As for the current equation, use Kirchhoff’s Current Law, the sum of currents in all branches equals zero. I am going to spell it out for you.

Vin/R1 + Vout/R2 + V-/R- = 0

#### MrChips

Joined Oct 2, 2009
27,114
Let us examine:
Vout = Aol(V+ - V-)

A large number multiplied by a large number gives a large result.
A small number multiplied by a small number gives a small result.

What do you get when you multiply a large number by a small number?

#### Xenon02

Joined Feb 24, 2021
355
Following up on what I posted,
V- = V+ = 0
Hence Vout = 0.
We know this is incorrect. Hence something is wrong.

As for the current equation, use Kirchoff Current Law, the sum of currents in all branches equals zero. I am going to spell it out for you.

Vin/R1 + Vout/R2 + V-/R- = 0
Okej.
V- = 0 so Vin/R1 = -Vout/R2.

So why doesn't it work for positive feedback, and why this : Vout = Aol(V+ - V-) is incorrect ?

What do you get when you multiply a large number by a small number?
Something in the middle ?