Opamp offset question

The LT1013 is good. It was preferred when I did space stuff. But it's too expensive

Reply to
Klaus Vestergaard Kragelund
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That is not possible I am affraid

Reply to
Klaus Vestergaard Kragelund

We have a 12bit ADC, so raw performance is not good: We do use oversampling, so that can get the performance back on track

Reply to
Klaus Vestergaard Kragelund

Good comments

About the ADC reference it is taken from the SMPS with post LC filter. So there is a little difference between the reference voltage ripple and the voltage fed to the pullup resistor for the PT1000

One option to get better performance is to sync the PT1000 sampling to a time slice when the microcontroller has steady current consumption, but that makes the SW more complicated and susceptible if changes are done in other places in the SW

Yeah. problem is that we can have wide temperature range, so I am afraid the calibration is not valid

Reply to
Klaus Vestergaard Kragelund

could you toggle between two currents ?

Reply to
Lasse Langwadt Christensen

Probably some horrible temperature gradient times a huge thermocouple coefficient. Silicon has huge TC coeffs versus any metal, for instance.

It should be much smaller in an isothermal test.

Cheers

Phil Hobbs

Reply to
Phil Hobbs

If you can turn off the pullup, maybe you can autozero. Turn it off and measure offset.

We like to do RTDs like this:

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The measurement is purely ratiometric on the Susumu thinfilm resistor.

Reply to
John Larkin

tirsdag den 30. november 2021 kl. 02.22.04 UTC+1 skrev John Larkin:

an extra resistor on the ground side puts the differential voltage in the middle of the input range, and makes the noise "impedance" of the two signal wires similar

Reply to
Lasse Langwadt Christensen

Not just calibration of the electronics - those cheap sensors have significant hysteresis.

Reply to
Spehro Pefhany

Yes, could be. So you are thinking like syncronous detection? Shifting the signal to a higher frequency avoiding DC errors

Reply to
Klaus Vestergaard Kragelund

something along the time lines of

two currents I1 and I2

I1 * R + offset = V1 , I2 * R + offset = V2

(I2 * R + offset)-(I1 * R + offset) = I2*R-I1*R + offset- offset => V2-V1 = (I2-I1)*R

R = (V2-V1)/(I2-I1)

Reply to
Lasse Langwadt Christensen

That's nice, so in effect the dynamic resistance is found between two load lines :-)

Reply to
Klaus Vestergaard Kragelund

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