Hello All.

As we look upon the difference amplifier, it resembles very closely the (non)inverting amplifier referenced to a non-zero voltage, with the exception of an additional potential divider on the non-inverting input. I tried to find out any explanations as to why this potential divider on the non-inverting side has been employed, but could not find any satisfactory answers.

Also, I was unable to deduce anything from the gain/transfer function of the difference amplifier by removing the non-inverting-side potential divider (R2=0 and Rg=∞), as it makes the equation inconclusive.

Any hints/guidance regarding the same is really appreciable.

 

hi abbas,
What are the Gain equations for the Inverting and Non Inverting inputs of a OPA.?
E

 

hi abbas,
What are the Gain equations for the Inverting and Non Inverting inputs of a OPA.?
E

Hello, ericgibbs.

I am not sure if this is what you are asking, but what I mean by (non)inverting amplifier circuits is these:

 

Hello, ericgibbs.

I am not sure if this is what you are asking, but what I mean by (non)inverting amplifier circuits is these:

Hi..not sure

The difference amplifier does not resemble both non inverting and inverting?? Look again

Now, after looking then apply superposition

1. short V2 source... Vout = (-Rf/R1) * V1

2. short V1 source..Let X = Rg * (Rg+R2) * V2 Then Vout = (1+Rf/R1) * X

"I tried to find out any explanations as to why this potential divider on the non-inverting side has been employed, but could not find any satisfactory answers."

If no divider then only number 2 changes above ... X = V2

Superposition!!

 

Hi..not sure

The difference amplifier does not resemble both non inverting and inverting?? Look again

Now, after looking then apply superposition

1. short V2 source... Vout = (-Rf/R1) * V1

2. short V1 source..Let X = Rg * (Rg+R2) * V2 Then Vout = (1+Rf/R1) * X

"I tried to find out any explanations as to why this potential divider on the non-inverting side has been employed, but could not find any satisfactory answers."

If no divider then only number 2 changes above ... X = V2

Superposition!!

Yeah! That's pretty simple!

And, through Superposition only, Difference amplifier resembles closely the (non)inverting amplifiers.
1. Short V1 and it resembles an inverting amplifier.
2. Short V2 and it resembles a non-inverting amplifier (except the factor X= Rg/(Rg + R2)).

That's what I am trying to ask! Why is that factor 'X' needed? Does it yield any advantage over the simple (non)inverting amplifier configurations referenced to a non-zero voltage?

The only advantage I can come up is the freedom (up to a certain degree) to reference the difference-amplification to a non-zero voltage (by putting a voltage source between Rg and ground), as is implemented in most In-Amps.

Are there any other advantages to this topology?

In other words, if one needs to perform differential amplification between two voltages but does not need to change the reference voltage for the differential amplification, i.e., the reference is invariably set to ground, can one employ the simpler (non)inverting amplifier topologies, which require lesser discrete components, and so, lesser error?

 

Yeah! That's pretty simple!

And, through Superposition only, Difference amplifier resembles closely the (non)inverting amplifiers.
1. Short V1 and it resembles an inverting amplifier.
2. Short V2 and it resembles a non-inverting amplifier (except the factor X= Rg/(Rg + R2)).

That's what I am trying to ask! Why is that factor 'X' needed? Does it yield any advantage over the simple (non)inverting amplifier configurations referenced to a non-zero voltage?

The only advantage I can come up is the freedom (up to a certain degree) to reference the difference-amplification to a non-zero voltage (by putting a voltage source between Rg and ground), as is implemented in most In-Amps.

Are there any other advantages to this topology?

In other words, if one needs to perform differential amplification between two voltages but does not need to change the reference voltage for the differential amplification, i.e., the reference is invariably set to ground, can one employ the simpler (non)inverting amplifier topologies, which require lesser discrete components, and so, lesser error?

Hi sorry late response : thought you understood

"I tried to find out any explanations as to why this potential divider on the non-inverting side has been employed, but could not find any satisfactory answers."

I think that divider is there so we can attenuate the voltage source and amplify that attenuated signal...think it's attenuated so it's possible to get answer like R2//R1 (V2 - V1)...If V2 is not attenuated the can get answer like this??

Anyways dunno

You're studying op amps?

Maybe you can answer you question yourself..With superposition in mind, please how to make circuit to give (Va - Vb) ?

We know that Vb will be placed on the "-" terminal of op amp and if " the resistors are the same = Rx) then Vout = -(Rx/Rx) * Vb = -Vb

Now what about for Va? We know it will be connected on the "+" terminal...Connect in directly to op amp with 1 single resistor...
and we already chose the resistors (Rx) on the "-" terminal to be equal...
Vo = (1+Rx/Rx) * Va = 2Va

so Vo = 2Va - Vb but we want Va - Vb (can we attenuate the Va signal by 2 before applying it to the "+" terminal of the op amp?)

 

hi abbas,
This image clip from the PDF I posted should help you understand the purpose of R2 and Rg. [ left hand circuit]
E
This tutorial link is useful
https://www.allaboutcircuits.com/textbook/semiconductors/chpt-8/building-a-differential-amplifier/

https://www.electronics-tutorials.ws/opamp/opamp_5.html

 

I tried to find out any explanations as to why this potential divider on the non-inverting side has been employed, but could not find any satisfactory answers.

If you understand the purpose of a differential amplifier, you can answer that question.

 

As we look upon the difference amplifier, it resembles very closely the (non)inverting amplifier referenced to a non-zero voltage, with the exception of an additional potential divider on the non-inverting input. I tried to find out any explanations as to why this potential divider on the non-inverting side has been employed, but could not find any satisfactory answers.

Because without it, the circuit is not a true difference amplifier. It's that simple. If you write out the gain equations for the (+) and (-) inputs, it becomes obvious.

 

This tutorial link is useful
https://www.allaboutcircuits.com/textbook/semiconductors/chpt-8/building-a-differential-amplifier/

https://www.electronics-tutorials.ws/opamp/opamp_5.html

Hello ericgibbs.
Thanks for sharing those links, have read those already.

This image clip from the PDF I posted should help you understand the purpose of R2 and Rg. [ left hand circuit]
E
View attachment 180578

I am sorry, but I don't think I clearly get what you are trying to hint.

 

Hi sorry late response : thought you understood

"I tried to find out any explanations as to why this potential divider on the non-inverting side has been employed, but could not find any satisfactory answers."

I think that divider is there so we can attenuate the voltage source and amplify that attenuated signal...think it's attenuated so it's possible to get answer like R2//R1 (V2 - V1)...If V2 is not attenuated the can get answer like this??

Anyways dunno

You're studying op amps?

Maybe you can answer you question yourself..With superposition in mind, please how to make circuit to give (Va - Vb) ?

We know that Vb will be placed on the "-" terminal of op amp and if " the resistors are the same = Rx) then Vout = -(Rx/Rx) * Vb = -Vb

Now what about for Va? We know it will be connected on the "+" terminal...Connect in directly to op amp with 1 single resistor...
and we already chose the resistors (Rx) on the "-" terminal to be equal...
Vo = (1+Rx/Rx) * Va = 2Va

so Vo = 2Va - Vb but we want Va - Vb (can we attenuate the Va signal by 2 before applying it to the "+" terminal of the op amp?)

Thanks,
Zeeus, for taking all that trouble.
That really healped.

 

If we try removing the non-inverting-side potential divider (R2=0 and Rg=∞), we get
Vout = [V2*{1+(Rf/R1)}] + [-V1(Rf/R1)]
(i.e., it effectively sums the gain equations of both the inverting and non-inverting amplifiers).

This definitely isn't the exact differential amplification. For this arrangement to work as a perfect difference amplifier, the gain terms of both the input voltages must be equal. That is,
1+(Rf/R1) = (Rf/R1)

which is only possible when
1<< (Rf/R1), or
R1 << Rf.

Now, this puts a practical limitation on this arrangement, as we all know that increasing the feedback impedance too high brings a lot of other problems with it.

This is what I deduce from it!

If someone has some more insights, it is really appreciable to share those.

 

hi,
Remember it is a Difference Amplifier, without Rg is it no longer a Difference Amplifier.
Check these links.
ps://www.analog.com/en/analog-dialogue/articles/deeper-look-into-difference-amplifiers.html

http://www.learningaboutelectronics.com/Articles/Difference-amplifier-circuit.php

Also I have done a LTS sim of the Amps with and without Rg, do you now see the importance of Rg.??

E

 

Hello ericgibbs.
Thank you for all your efforts.

ps://www.analog.com/en/analog-dialogue/articles/deeper-look-into-difference-amplifiers.html

Read that already.

http://www.learningaboutelectronics.com/Articles/Difference-amplifier-circuit.php

I am sorry, but it seems it doen't covers any aspect of my question.

Also I have done a LTS sim of the Amps with and without Rg, do you now see the importance of Rg.??

E

Thank you for the effort.
Yes, I do.

Vout = [V2*{1+(Rf/R1)}] + [-V1(Rf/R1)]

That's the exact difference!

 

hi,
Look at your first posted image.
Assume all the resistors are 10K.
So the Gain of the INV input is Rf/R1 = 1
The Gain of the NI input is 1+(Rf/R1) = 2

So we have to make the input to the NI, half the value of V2 [ its a voltage divider R2 and Rg]
So now the V2 signal divided by 2 and then amplified by 2 = = an overall gain of 1, which is the same as the INV for V1,, so we now have a Difference amplifier.

So Rg value is chosen to match the value of Rf, its that simple.

Do you follow that OK.

E
Footnote: I am assuming for this basic circuit that the Voltage Source impedances are both zero Ohms