Author Topic: Resistor selection for instrumentation amplifier  (Read 2178 times)

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Offline AQUAMANTopic starter

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Resistor selection for instrumentation amplifier
« on: October 05, 2016, 11:56:26 am »
I have an LT1167 instrumentation amplifier, which requires an external resistor to set the gain
http://cds.linear.com/docs/en/datasheet/1167fc.pdf

This is for a precision application and I need very low temperature dependence. I need a gain of less than 10 so will be buying a 10k resistor. The gain is not important but the repeatability of the measurements is
Does anyone have any recommendations on what resistor to buy?

I would buy an AD624 which has set gains and  tells your the temp coefficient of each gain setting, but its €35 and a huge package

 

Offline trophosphere

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Re: Resistor selection for instrumentation amplifier
« Reply #1 on: October 05, 2016, 02:49:49 pm »
Can you quantify "low temperature dependence" or rather how much are you able to tolerate? There is a wide range from 100ppm/C for thick film to that of 2ppm/C for thin film (Ex. Susumu or Caddock). If that is not enough then try Vishay as they have some speced for 0.05ppm/C.
 

Offline AQUAMANTopic starter

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Re: Resistor selection for instrumentation amplifier
« Reply #2 on: October 05, 2016, 03:45:15 pm »
Can you quantify "low temperature dependence" or rather how much are you able to tolerate? There is a wide range from 100ppm/C for thick film to that of 2ppm/C for thin film (Ex. Susumu or Caddock). If that is not enough then try Vishay as they have some speced for 0.05ppm/C.
Have just seen the vishay ones, thanks
 

Offline Brutte

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Re: Resistor selection for instrumentation amplifier
« Reply #3 on: October 06, 2016, 06:43:24 am »
This is for a precision application and I need very low temperature dependence.
It says:
G(Rg) = 1 + (49.4k/Rg)

So that means:
"dG/dRg"=-49.4k/Rg^2
(where the thingy in quotes is the differentiation symbol)

which at your 10k attains:
"dG/dRG"(10k)=4.94e-4 [1/ \$\Omega\$]

so you can approximate your G with affine function of dRg.
G(Rg+dRg)~G(Rg)+"dG/dRg"*dRg

Going further we know that Rg is a function of temperature:
Rg(T+dT)~ Rg(T)+"dRg/dT"*dT
where T is temperature and dT is the deviation from T.

The "dRg/dT" is also known as TCR (linear temperature coefficient of resistance)

Concluding:
G(T+dT) ~ G(T)+"dG/dT"*dT
 

Online David Hess

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Re: Resistor selection for instrumentation amplifier
« Reply #4 on: October 06, 2016, 04:36:41 pm »
The datasheet says that the gain drift with a perfect resistor is 20ppm/C typical and 50ppm/C maximum so using a perfect resistor may only halve the gain drift versus a 20ppm/C or 50ppm/C resistor.  In the error budget analysis, they use a 10ppm/C resistor.
 

Offline rob77

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Re: Resistor selection for instrumentation amplifier
« Reply #5 on: October 06, 2016, 06:28:16 pm »
a vishay foil resistor would be a waste of money... simply just go ahead with a 15ppm/C smd resistor there.
 

Offline bson

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Re: Resistor selection for instrumentation amplifier
« Reply #6 on: October 07, 2016, 12:04:06 am »
If the thermal response is repeatable and not particular fast changing you could also calibrate for temperature, then co-measure the temperature and factor it out from the measurement for a thermally linearized data point.  This of course assumes minimal thermal hysteresis and an error manifesting itself as a repeatable value deflection rather than noise.  (Although you can always characterize the hysteresis - the real key is predictable repeatability.)
 


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