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AD7792/AD7793 with 2-wire PT100

Question asked by Bernardo.Ramos on Oct 15, 2017
Latest reply on Oct 23, 2017 by JellenieR

Hi!

 

I plan to use an AD7792 or AD7793 with a 2-wire PT100 sensor.

 

The connections will be like this:


   +--- IOUT1
   +--- AIN1(+)
   |
[PT100]
   |
   +--- AIN1(-)
   +--- REFIN(+)
   |
[Rref]
   |
   +--- REFIN(-)
   +--- GND

I need to calculate the reference resistor value.

 

Example: PT100, 2-Wire, 0°C to +100°C
Resistance = ~ 100 ohms (0°C) to ~ 140 ohms (100°C)
Iexc = 1mA

 

Voltage across RTD:
At 0°C: 1mA x 100 ohms = 100 mV
At 100°C: 1mA x 140 ohms = 140 mV

 

Voltage in the comparator (PGA=16):
At 0°C: 16 x 100mV = 1.600V
At 100°C: 16 x 140mV = 2.240V

 

Reference Resistor:
Rref = 2.240 V / 1 mA = 2.24 k ohms
Vref = 2.2K x 1mA = 2.2 V

Vtotal (IOUT to GND) = 140mV + 2.2V = 2.34V (OK, under 3.3V)


But at the voltage comparator we will have both positive and negative values:

At 0°C: 1.600V - 2.200V = -0.600V
At 100°C: 2.240V - 2.200V = 0.040V

 

Is this OK?

If yes, would it be better to use an 1.8 kohm resistor? (it would be next the middle of the range)

If no, should I use a reference resistor of 2.7 kohms?


Here is the calculation for a 2.7 kohms resistor:

Vref = 2.7K x 1mA = 2.7 V

Vtotal (IOUT to GND) = 140mV + 2.7V = 2.84V (OK, under 3.3V)

Voltage comparator difference:
At 0°C: 1.600V - 2.700V = -1.100V
At 100°C: 2.240V - 2.700V = -0.460V


Which would be the best reference resistor value for this case?

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