Understanding an absolute-difference circuit

Thread Starter

Menyman

Joined Jan 31, 2021
6
Hi PRASS,
Don't think there's any relevance here to comparators..

Here's what i'm thinking:
For V1>V2: diode is off, the second opamp acts as a standard differential amplifier.
For V2>V1: Not sure my analysis is correct (would be happy to hear your inputs)-
The first opamp acts as a diff amplifier, the second as a buffer –
that way the top resistor R2 is in a close loop all the way from the first opamp’s output, to its negative input.

Any chance I'm right?
 

Boatman47

Joined Jan 22, 2021
11
As a novice, I am still trying to figure out the intended function of the circuit. Are the input voltages DC or AC for comparison? Using a value of 1k for R1 and 2k for R2 I ran this in Multisim and the measured values bear no resemblance to the ones presumed.
 

PRASS

Joined Feb 22, 2018
30
Hi Menyman,

Yes very much so.
It's like a self oscillating circuit that uses it's own reverb as its gain to produce a linear digital signal from a analogue signal.

Awesome for making frequencies out of Hz readings. . . .
 

Thread Starter

Menyman

Joined Jan 31, 2021
6
The explanation for the discussed case is:
" For the case where V2 >V1 , diode D1 conducts, producing the composite amplifier system made up of both IC1A and IC1B , where VOUT =(R2 /R1 )(V2 −V1 ) "

So no, did not understand much from it :)
 

Thread Starter

Menyman

Joined Jan 31, 2021
6
Sort of.
Did you not understand the explanation given in the article about the circuit operation?
The explanation for the discussed case is:
"For the case where V2 >V1, diode D1 conducts, producing the composite amplifier system made up of both IC1A and IC1B, where VOUT =(R2/R1)(V2−V1 )"

So no, did not understand much from it :)
 

ericgibbs

Joined Jan 29, 2010
12,901
hi Menyman,
This LTSpice sim of the circuit should help you understand.
I have arbitrarily
chosen all resistor values as 5k.

If you want to see the other Test point waveforms, ask.
E
AAA 1120 09.58.gif
 

PRASS

Joined Feb 22, 2018
30
Hi ericgibbs,

VO should be coming out as a Linear Digital signal.
signal.png
It's probably becuase your using 2 LT1078's which is only a duel op amp & not a quad. The LM2902 is a quad op amp.
Perhaps try a LT1079 which is an quad op amp.

This means you have applied a 9V+ at pin 8 & ground at pin 4 on both yours whereas a LM2902 has 5V+ at pin 4 & ground at pin 11. You have agian powered the 2nd LT1078 in your diagram which would completely defeat the purpose of using it as a comparitor (absolute-difference circuit) & just turns it into a second op amp not a coposite amp.

This is the LM2902 from the artical. _____________________________________________ This is a LT1078A

LM2902....png LT1078.png
 

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PRASS

Joined Feb 22, 2018
30
@PRASS
Could you explain how you decided the circuit works in that way in post #10.?
What is the definition of: Absolute Difference Amp?
Hi Eric,

I didn't decide anything it's what a absolute-difference circuits is designed to do, a device that compares two voltages or currents and outputs a digital signal indicating which is larger !!!
Since you were using a Sine wave it should also be linear.
Artical from the same publisher years before hand.

?????????????
 

crutschow

Joined Mar 14, 2008
27,188
That's strictly an analog circuit, nothing digital about it.
Do you not see the diode ?
Yes.
A diode can be used in an analog circuit (and is used in many of them) without making it digital.
Do you really think that any circuit with a diode is digital?
 
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Thread Starter

Menyman

Joined Jan 31, 2021
6
hi Menyman,
This LTSpice sim of the circuit should help you understand.
I have arbitrarily
chosen all resistor values as 5k.

If you want to see the other Test point waveforms, ask.
E
Thanks for the simulation, we can see clearly that the output of the circuit indeed matches its purpose.
However I'm still a little struggling to understand how does the circuit operate when V2>V1.
If the left amplifier acts in that case a differential amplifier, how does the loop close?

1612207729130.png

Is the right amplifier acts as a buffer in that case (with R2 in its feedback)?
If so, won't R1 (which is connected to V2) changes the operation of this amplifier?

Thanks for your answers :)
 
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