# Instrumentation Amplifier

Problem:
http://img443.imageshack.us/img443/2926/capturelgw.jpg [Broken]

I need help getting this started. Does anyone understand that R1, R2, R5, R6-11 are 100k, but then whats 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.

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berkeman
Mentor
Problem:
http://img443.imageshack.us/img443/2926/capturelgw.jpg [Broken]

I need help getting this started. Does anyone understand that R1, R2, R5, R6-11 are 100k, but then whats 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.

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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?

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berkeman
Mentor
*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 cant 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 cant seem to calculate that given the constraints I mentioned above...

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

thanks.

berkeman
Mentor
i need help with part B identifying the wiper potentiometer...

i cant 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 cant 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?

berkeman
Mentor
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?

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

berkeman
Mentor
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
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?

well that way im getting (12/220)*20 = 1volt?

was that correct? +1volt -1volt offset?

berkeman
Mentor
well that way im getting (12/220)*20 = 1volt?
was that correct? +1volt -1volt offset?
I don't get exactly +/-1V -- close, but not exactly 1.

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?

berkeman
Mentor
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.