Understanding the Potentiometer Configuration in an Instrumentation Amplifier

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Discussion Overview

The discussion revolves around understanding the configuration of a potentiometer in an instrumentation amplifier circuit. Participants seek clarification on resistor values, gain calculations, and offset voltage related to the potentiometer's position. The scope includes theoretical analysis and practical application of circuit design.

Discussion Character

  • Exploratory
  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • Some participants question the significance of a -1.0 M resistor noted in the circuit diagram, suggesting it may be a typo.
  • There is a discussion about the role of resistors R3 and R4 in determining gain when the switch is open or closed, with some proposing specific gain formulas.
  • Participants express uncertainty about calculating the Vref offset voltage based on the potentiometer's configuration and the surrounding resistors.
  • One participant proposes using a voltage divider approach to estimate the voltage seen at the potentiometer, leading to further calculations regarding Vref offset.
  • There is a back-and-forth regarding the correct interpretation of voltage drops across resistors and the resulting offset voltages, with some participants refining their calculations based on earlier posts.

Areas of Agreement / Disagreement

Participants do not reach a consensus on several points, including the interpretation of resistor values, the calculation of gain, and the determination of Vref offset voltage. Multiple competing views and calculations are presented throughout the discussion.

Contextual Notes

There are limitations in the clarity of the circuit diagram and the assumptions made about resistor values and configurations. Some calculations depend on specific interpretations of the circuit layout, which may not be universally agreed upon.

Who May Find This Useful

This discussion may be useful for students and practitioners in electrical engineering or related fields who are interested in instrumentation amplifier design and the role of potentiometers in circuit configurations.

foobag
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Problem:
http://img443.imageshack.us/img443/2926/capturelgw.jpg

I need help getting this started. Does anyone understand that R1, R2, R5, R6-11 are 100k, but then what's that -1.0 M doing underneath? Also I think that R11 - 20k is a typo to mean R12 as the potentiometer.

Id appreciate some help, thanks.
 
Last edited by a moderator:
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foobag said:
Problem:
http://img443.imageshack.us/img443/2926/capturelgw.jpg

I need help getting this started. Does anyone understand that R1, R2, R5, R6-11 are 100k, but then what's that -1.0 M doing underneath? Also I think that R11 - 20k is a typo to mean R12 as the potentiometer.

Id appreciate some help, thanks.

Yes, the - 1.0M does look like a dangling typo. All the resistors are accounted for if you fix the other typo and make the potentiometer R12.
 
Last edited by a moderator:
ok so let me get this logic straight...

when its open, then what happens? there is no offset control, or the fact that the second OA2 amplifier is not used? thus the gain would be dependent on OA1 and final Vout at OA3?

and I assume by range they mean input voltage range right?*EDIT -

Wait as I see it, R3 is currently the gain resistor when the circuit is open, when it is closed than R4 and R3 come into play as they are in parallel?
 
Last edited:
foobag said:
*EDIT -

Wait as I see it, R3 is currently the gain resistor when the circuit is open, when it is closed than R4 and R3 come into play as they are in parallel?

Yep.
 
alrite so my attempt is as such:

the gain overall would be equal to (1+ 2R1/R3)*(R7/R5) when the switch is open

and when the switch is closed than RG = R4*R3/(R3+R4) , and the gain would be equal to:
(1+2R1/RG)*(R7/R5)

does this seem logical?
 
i need help with part B identifying the wiper potentiometer...

i can't seem to understand how to find the Vref offset voltage given the circuit, my textbook is pretty crappy showing me a +15v supply added to a 24k resistor followed by 100k and another 24k connected to -15v supply, the wiper is pointed towards the 100k resistor...the book goes on to say that Vref is between +10 and -10V but I can't seem to calculate that given the constraints I mentioned above...

how can I apply this to what I have in my circuit?

thanks.
 
foobag said:
i need help with part B identifying the wiper potentiometer...

i can't seem to understand how to find the Vref offset voltage given the circuit, my textbook is pretty crappy showing me a +15v supply added to a 24k resistor followed by 100k and another 24k connected to -15v supply, the wiper is pointed towards the 100k resistor...the book goes on to say that Vref is between +10 and -10V but I can't seem to calculate that given the constraints I mentioned above...

how can I apply this to what I have in my circuit?

thanks.

The position of the wiper just changes the input voltage to that opamp. What values does that input value take on based on the position of the wiper?
 
so i searched around..

the only thing that made some remote sense was that if i take the two outer resistors of 24k as a parallel voltage divider ((24*24/48)/((24*24/48)+24) i would get 0.333 * Vin = 5Volts..so that means out of the 15v supplied, 10 volts must be seen on the 100k variable wiper resistor?

with this logic could i use it on the circuit? so that the R10 and R11 resistors are in parallel voltage divider meaning that 0.33*Vin(12) = 4 volts dissipated over those resistors thus +8 and -8V will be seen as Vref offset?

is this correct?
 
*bump* any help/suggestions?
 
  • #10
foobag said:
so i searched around..

the only thing that made some remote sense was that if i take the two outer resistors of 24k as a parallel voltage divider ((24*24/48)/((24*24/48)+24) i would get 0.333 * Vin = 5Volts..so that means out of the 15v supplied, 10 volts must be seen on the 100k variable wiper resistor?

with this logic could i use it on the circuit? so that the R10 and R11 resistors are in parallel voltage divider meaning that 0.33*Vin(12) = 4 volts dissipated over those resistors thus +8 and -8V will be seen as Vref offset?

is this correct?

I may be misunderstanding what you are saying, but R10=R11=100k, and the pot is 20k. So what fraction of the overall +/-Vcc voltage is available at the pot?
 
  • #11
berkeman said:
I may be misunderstanding what you are saying, but R10=R11=100k, and the pot is 20k. So what fraction of the overall +/-Vcc voltage is available at the pot?

rite so i got a voltage divider between the two 100k and got 50k (50/150) 1/3 voltage dissipated, so 2/3 left on 20k resistor giving +8V, -8V? because of the 12V supplied 4v lost?

this is the only logic i could understand giving my textbook knowledge, wasn't entirely sure
 
  • #12
berkeman said:
I may be misunderstanding what you are saying, but R10=R11=100k, and the pot is 20k. So what fraction of the overall +/-Vcc voltage is available at the pot?

foobag said:
rite so i got a voltage divider between the two 100k and got 50k (50/150) 1/3 voltage dissipated, so 2/3 left on 20k resistor giving +8V, -8V? because of the 12V supplied 4v lost?

this is the only logic i could understand giving my textbook knowledge, wasn't entirely sure

The pot is 20k. Each of R10 and R11 are 100k, so the total series resistance is 220k. The same current is going through all 3 resistors, so what fraction of the total voltage drop do you get across the pot?
 
  • #13
well that way I am getting (12/220)*20 = 1volt?
 
  • #14
was that correct? +1volt -1volt offset?
 
  • #15
foobag said:
well that way I am getting (12/220)*20 = 1volt?

foobag said:
was that correct? +1volt -1volt offset?

I don't get exactly +/-1V -- close, but not exactly 1.
 
  • #16
well i got I = V/R = 12V/220(total resistance series 2-100k, 1-20k)

then V = R*I = 20k * I found from above, roughly 1.09 volts.

that should be right?
 
  • #17
foobag said:
well i got I = V/R = 12V/220(total resistance series 2-100k, 1-20k)

then V = R*I = 20k * I found from above, roughly 1.09 volts.

that should be right?

Yes, I think so. That's what I get: +/-1.09V for the pot wiper.
 

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