The gain is 7.24 (1+ 20/3.205), not 6.24. The output voltages assumed by the original poster are exactly corresponding to nominal Pt100 resistance values. Resistance at 850 °C is specified as 390.481 ohm.It sounds about right.. not very familiar with this cct, but looks like an instrumentation amp set to a gain of 6.24??, and a constant current source of 1mA.
According to the circuit function, that has been basically analyzed correctly by sky_123, Rpt100 = Vo/7.24*1000. Output voltage to resistance is of course a linear function. Resistance to temperature is not exactly linear according to the Pt100 characteristic, but a assuming a linear relation may be sufficient for many cases, particularly in a smaller temperature range. What you most likely mean is that the linear function has a constant term b, Rpt100 = a*t + b. You can easily draw a straight line connecting the 0 an 850 °C points, it will involve about 30 degree deviation in the middle of the range. For an exact measurement, you should refer to a polynomial or a table.Can somebody tell me how to find the RTD PT100 resistance and temperature from Vo?
It sounds about right.. not very familiar with this cct, but looks like an instrumentation amp set to a gain of 6.24??, and a constant current source of 1mA.
So, at 0 degrees, you'd get about (100ohms*0.001A)*6.24 = 0.624V and you're getting 0.72V
At 850 degrees you'd get ((0.4ohms*850degrees)+100ohms)*0.001A*6.24 = 2.7456V and you're getting 2.82V
Your values are a bit higher, maybe component tolerances mean the gain isn't quite 6.24 or something.. maybe it's an offset. What op-amps are you using? Are your results real or a simulation?
There should be A and B numbers given together with this equation. What's your problem in calculation then? Did you try a pocket calculator?I got this equation Rt = Ro [1 + At + Bt^2] for 0-850 degree C range. But I don't know t to calculate Rt.
Did you look into the AD application note?I wanted to get the value of rtd pt100's resistance and rtd pt100's temperature from Vo of the Instrumentation amplifier.
[syntax = c]
val1 = (100.00 - resistance)/100.00;
val2 = 4.0 * (-5.775e-7) * val1;
val3 = sqrt(15.27480889e-6 - val2);
val4 = -1.155e-6;
temperature = (-3.8093e-3 + val3) / val4;
[/syntax]
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