Hi!
I normally use the 200 uA source of my labjack U6 pro to read from 2 Pt100 sensors. This time I want to read 4 Pt100 sensors so I bought 2 LjTickResistance-1K. I connect the red cable to the ground of the LJTres and the other two black wires of the sensors to Vs and outA. Additionally, I connect Vs to a free analog input to measure the real voltage of the source.
I use the following python code:
d = u6.U6()
d.getCalibrationData()
feedbackArgument = u6.AIN24(PositiveChannel=0, ResolutionIndex=13, GainIndex=0)
ainBits = d.getFeedback(feedbackArgument)
v = d.binaryToCalibratedAnalogVoltage(gainIndex=0,bytesVoltage=ainBits[0])
feedbackArgument = u6.AIN24(PositiveChannel=2, ResolutionIndex=13, GainIndex=0)
ainBits2 = d.getFeedback(feedbackArgument)
vref = d.binaryToCalibratedAnalogVoltage(gainIndex=0,bytesVoltage=ainBits2[0])
ohm=(1000*(vref-v))/v
Temp = (2.586904*ohm - 258.821046)
The result of ohm gives me 10.53 ohms which equals to -231°C (My room is around 25°C). If I connect the Pt100 to the 200 uA source (Red = Ground, Blacks = 200 uA and AIN) It gives me the correct temperature using the same calibration constants.
Could anyone spot the problem?
Thank you
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Sounds like you have a 3-wire RTD. That means 2 wires connect to one side of the sensor and 1 wire connects to the other side of the sensor. Figure out which side has 2 wires and twist them together to use as a single wire. Then the wiring for an RTD is simply to connect one side to Vref and the other side to VINA. This gives you a buffered voltage divider as shown in Figure 2 on the following app note (R1=Runknown=RTD, R2=1000):
https://labjack.com/support/app-notes/signal-voltages-out-range
Assuming the resistance of the LJTR-1k is exactly 1000 the equations for Runknown are:
Vout = Vref*1000/(Ru+1000)
Ru = ((Vref-Vout)*1000)/Vout
For a PT100 you expect that it will be 100.0 ohms at 0 degC and the resistance will change about 0.385 ohms/degC (most common coefficient). Say room temperature is 22 degC and putting your fingers on the sensor warms it up to 25 degC. At 22 deg C you expect 108.47 ohms and at 25 degC you expect 109.63 ohms. The voltages you expect are:
22C => 2.255361V
25C => 2.253003V
For a PT1000 you expect that it will be 1000.0 ohms at 0 degC and the resistance will change about 3.85 ohms/degC (most common coefficient). Say room temperature is 22 degC and putting your fingers on the sensor warms it up to 25 degC. At 22 deg C you expect 1084.7 ohms and at 25 degC you expect 1096.3 ohms. The voltages you expect are:
22C => 1.199213V
25C => 1.192577V
Note that on the T7 there you can use the AIN-EF system to do the math, but the above is needed for the U6 and even on the T7 is still useful for understanding and troubleshooting:
https://labjack.com/support/datasheets/t7/ain/extended-features/rtd
Thank you for your fast response. What you mention I did it already and everithing works fine except that the resistance of the cable is adding 2°C approx. to my final measurement. My question here is: how can I connect the 3rd wire of the Pt100 to the LJTickResistance as to eliminate the resistance effect of the lead? Do you have an idea? I am sure it is possible to do it with a voltage divider, what do you think?
Cheers
You could use the 3rd wire on your RTD and do some differential measurements and change your math to separate out the wire resistances, but that is usually much more detail than is justified. 2 quick solutions:
1. If you know extra resistance is causing a 2degC error I would simply adjust by 2degC in software.
2. Use fatter wire to decrease the effect of the wires.
So your cable resistance is about 0.77 ohms? Sounds like a lot. Are you sure that is the source of your error?
24 AWG wire (which is quite small) is specified with a typical resistance of 0.026 ohms/ft, so 0.77 ohms would be about 30 ft of such cable (15' or 5m out-and-back).
18 AWG wire (which is quite small) is specified with a typical resistance of 0.006 ohms/ft, so 0.77 ohms would be about 128 ft of such cable (64' or 20m out-and-back).
https://en.wikipedia.org/wiki/American_wire_gauge
A few other possible sources of small errors:
- AIN accuracy of the U6.
- Error in the Vref value used in your math, whether assumed or measured.
- Error in the assumed value (1000) of the fixed resistor in the LJTR.
Nonetheless, if you want to use the 3rd wire, consider that including the resistance of each length of wire the total is:
2*Rw + Ru + R2 or 2*Rw + Ru + 1000
You said your sensor has 1 red wire and 2 black wires (BlackA and BlackB). Presumably the 2 black wires connect to one side of the RTD and the 1 red wire connects to the other side of the RTD. Connect as follows:
1. RTD-BlackA connected to LJTR-Vref, and also a small jumper from LJTR-Vref to U6-AIN0.
2. RTD-BlackB connected to U6-AIN1.
3. RTD-Red connected to LJTR-VINA which goes to AIN2.
A single-ended measurement of AIN2 gives you the voltage across the fixed resistor in the LJTR. If you assume this is exactly 1000, the you can determine current from this:
I = Vain2 / 1000
A differential measurement of AIN0-AIN1 gives you the voltage drop across 1 length of wire, which is assumed to be the same across the other length of wire. A single-ended measurement of AIN0 gives you entire excitation voltage, and if you subtract the AIN2 measurement you have the voltage across both lengths of wire and Ru:
AIN0-AIN2 is the voltage across 2*Rw + Ru
The differential measurement AIN0-AIN1 is the voltage across one Rw, so you can determine Rw:
Rw = V/I = AIN0-AIN1 / I
Excellent! Thank you for all this information I will try it and I will post back the results. Thank you very much for the fast reply.
I looked closer at the 3 errors I mentioned plus a 4th:
Rfixed assumed value error: The LJTick-Resistance uses 0.05% resistors (with excellent 10 ppm tempco BTW). I believe that translates to a roughly 0.05% error in determining Ru, which would be +/-0.05 ohms, so about +/-0.1 degC error.
Vout measurement error: If using the +/-10V range, the calculation is pretty much the same as above so about +/-0.2 degC error.
Vref feedback measurement error: Error band is +/-2 mV, which is +/-0.08% of 2.5V. I believe that translates into a roughly +/-0.08% error in determining Ru. If Ru=100 that is +/-0.08 ohms or about +/-0.2 degC error.
Vref assumed value of 2.50V error: The LJTR has a great reference, better than the LJTD, with an initial max tolerance of only +/-0.04%. I believe that translates into a roughly +/-0.04% error in determining Ru. If Ru=100 that is +/-0.04 ohms or about +/-0.1 degC error.
I take 3 things of interest from this. First, the error due to the assumed value of the fixed resistor in the LJTR is only about +/-0.1 degC. Second, all these errors are about an order of magnitude less than your 2 degC error, so these errors likely do not explain your error. Third, the reference on the LJTR is so good you are better off assuming it is 2.50 than using an AIN to measure the actual value.
I also did a little more math to come up with full equations.
I still like the connections I suggested. I like doing the differential measurement of Vwire at the top, since the Vout at the bottom comes through the LJTR.
You said your 3-wire PT100 RTD has 1 red wire and 2 black wires (BlackA and BlackB). Presumably the 2 black wires connect to one side of the RTD and the 1 red wire connects to the other side of the RTD. Connect as follows:
1. RTD-BlackA connected to LJTR-Vref, and also a small jumper from LJTR-Vref to U6-AIN0.
2. RTD-BlackB connected to U6-AIN1.
3. RTD-Red connected to LJTR-VINA which goes to AIN2.
The equation for the resistance of each length of wire is:
Rw = R2 * (AIN0-AIN1) / Vout
Rw = 1000 * (AIN0-AIN1) / AIN2
Use a differential measurement to get the small voltage AIN0-AIN1, whereas AIN2 is a single-ended reading of the output voltage from the LJTR.
The equation for Runknown (RTD) when we include 2 Rw resistances in the equation winds up being:
Ru = ( VrefR2 - 2VoutRw - VoutR2 ) / Vout
Vout is AIN2 single-ended, R2 = 1000, and Vref = 2.50, so we get:
Ru = ( 2500 - 2VoutRw - 1000Vout ) / Vout
For completeness, here is what I come up with for a 4-wire RTD measurement (or any resistance) with the LJTick-Resistance.
On the top side of the RTD one wire goes to Vref and the other wire goes to AIN0. On the bottom side one wire goes to LJTR-VINA and the other wire goes to AIN1. Now a differential measurement of AIN0 (versus AIN1) gives us the voltage across the RTD (Ru). A single-ended measurement of AIN2 (Vout from the LJTR) is the voltage across Rfixed and tells us the current through the RTD.
Ru = Vu / I
Vu = AIN0 differential
I = AIN2 / R2
Ru = R2 * AIN0diff / AIN2 = 1000 * AIN0diff / AIN2 (for the LJTR-1k)
Here is a little table showing how much wire has resistance equivalent to 0.25C for a PT100 or PT1000 RTD:
PT100 0.25C PT1000 0.25C
AWG ohms/ft ft for 0.096 ohms ft for 0.96 ohms
16 0.0040 24.0 240.0
18 0.0064 15.0 150.0
20 0.0102 9.4 94.1
22 0.0161 6.0 59.6
24 0.0257 3.7 37.4
So for example, if using a PT1000 with 20 gauge lead wires, it would take 94 feet of wire to have enough resistance to look like 0.25C of error, thus counting the wire out and back 94 feet of 20 gauge wire would have enough resistance to cause 0.5C of error.
Thus it takes a substantial length of wire before you need to worry about wire resistance errors, and a PT1000 is much more immune to wire resistance errors than a PT100.