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I'm hoping to get some guidance with a circuit I designed. I have a current source which operates between 200uA and 400uA. In order to measure it, I used the following circuit before feeding it into an ADC.

Vadc = 5*(5*(i*500)-0.5V)

enter image description here

I've built a number of these circuits and all of them appear to be within 1 degree at room temperature. Now after designing the circuit, I was asked to calculate the drift over 100C and to evaluate the precision of the circuit to see if I could accurately report temperatures at 0.1C with calibration. For the drift, the data sheet calls out the following values:

  • 500, 2K resistors, 5ppm/C *100C * 100% = 0.05% = RL% drift
  • Dif Amp: INA2331A: 2ppm/C *100C * 100% = 0.02% = G1/G2% drift
  • 2.5V Ref: REF1925A: 25ppm/C *100C * 100% = 0.225% = Vref% drift

Now for the 2K/500 ohm resistor divider, since they have the same 5ppm, I think I can ignore that drift. Therefore I'd get the following equation:

((RL% *Gain1) +G1% + Vref%/5)*Gain2 +G2% = total drift

((0.05% *5)+0.02% + 0.225%/5)*5 + 0.02% = 1.595%

Is that accurate in how I would do that calculation?

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1 Answer 1

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Your system equation is OUT= G2 * (G1*RL - 2.5*R1/(R1+R2))

you need

delta(OUT)/delta(t) ~~ d(OUT)/d(t) = d(G2)/dt *(G1*RL - 2.5*R1/(R1+R2)) + G2 * d(G1*RL - 2.5*R1/(R1+R2))/dt

where t is tempereture.( d(G2)/dt equal to drift G2 amplifier)

the first element is the effect of drift of G2 in that bias point, the second is the effect of other drift with gain G2. ...

you can assign d(OUT)/d(t) ~ delta(OUT)/delta(t)

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  • \$\begingroup\$ We do have mathjax here, I highly recommend using it. \$\endgroup\$
    – Hearth
    Commented Sep 15, 2021 at 5:42

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