ginos Posted September 23, 2010 Report Share Posted September 23, 2010 Hallo, I am using K thermocouples and I want to measure temperatures between 30℃ and 60℃ which corresponds to 1,203 – 2,436mV voltage range. I use the following Data Acquisition system: Connector block SCXI-1102B Controller PCI-6034E The specification of the above are: SCXI-1102B has 2 possible input signal ranges: ±10V (gain=1) and ±0,1V (gain=100) PCI-6034E has 4 possible input signal ranges: ±10V (equals gain=0,5), ±5V (equals gain=1), ±0,5V (equals gain=10) and ±0,05V (equals gain=100) I want to know how each device amplifies the signal and what will be the total amplification of the signal, in order to calculate the code width. Thank you very much, Giorgos P Quote Link to comment
ShaunR Posted September 23, 2010 Report Share Posted September 23, 2010 Hallo, I am using K thermocouples and I want to measure temperatures between 30℃ and 60℃ which corresponds to 1,203 – 2,436mV voltage range. I use the following Data Acquisition system: Connector block SCXI-1102B Controller PCI-6034E The specification of the above are: SCXI-1102B has 2 possible input signal ranges: ±10V (gain=1) and ±0,1V (gain=100) PCI-6034E has 4 possible input signal ranges: ±10V (equals gain=0,5), ±5V (equals gain=1), ±0,5V (equals gain=10) and ±0,05V (equals gain=100) I want to know how each device amplifies the signal and what will be the total amplification of the signal, in order to calculate the code width. Thank you very much, Giorgos P You already have most the information to calculate the code width. Look at the spec for your device again and find the resolution. code width = range/(gain x resolution) Quote Link to comment
ginos Posted September 23, 2010 Author Report Share Posted September 23, 2010 Thank you for your answer ShaunR, The resolution of PCI-6034E is 16 bit. I know the formula you mentioned. What I really want to know is how the gain is applied. For example i assume I take a temperature measurement at 30 oC. This temp equals with 1,203mV. This input signal enters the SCXI. The SCXI uses the range ±100 mV (gain 100). So the signal will be 1,203mV*100=120,3mV ??? Then the signal is amplified in the DAQ board. If the above is correct then the DAQ board will use the ±500mV (gain 10) Again the signal will be 10 times amplified 120,3*10=1203mV=1,203V? The code width with the above setting will be 1V/216=15,26uV Assuming that I want to calculate what will be the minimun temperature that the DAQ board will detect. For K thermocouples 1oC is 40uV. So if I have ΔΤ=1oC according to the above amplification the 40uV will be amplified to 40mV. So 15,26 (uV)/40 (mV/oC)=0,3815*10-3 oC Please can you verify if the steps of amplifications are correct because I feel that 0,3815*10-3 oC is a very small temperature to detect. Thank you very much ShaunR for your time. Giorgos P Quote Link to comment
ShaunR Posted September 23, 2010 Report Share Posted September 23, 2010 Thank you for your answer ShaunR,The resolution of PCI-6034E is 16 bit. I know the formula you mentioned.What I really want to know is how the gain is applied.For example i assume I take a temperature measurement at 30 oC. This temp equals with 1,203mV.This input signal enters the SCXI. The SCXI uses the range ±100 mV (gain 100).So the signal will be 1,203mV*100=120,3mV ??? Then the signal is amplified in the DAQ board. If the above is correct then the DAQ board will use the ±500mV (gain 10)Again the signal will be 10 times amplified 120,3*10=1203mV=1,203V?The code width with the above setting will be 1V/216=15,26uV 1/(65356 x 10) = 1.53E-6 uv so yes. the codeword is correct. Assuming that I want to calculate what will be the minimun temperature that the DAQ board will detect.For K thermocouples 1oC is 40uV. So if I have ΔΤ=1oC according to the above amplification the 40uV will be amplified to 40mV.So 15,26 (uV)/40 (mV/oC)=0,3815*10-3 oC Not quite. Thermocouples are non-linear. The K-type is especially wobbly around 0°C. You need to use thermocouple tables (or polynomial approximations) to calculate the temperature for a particular voltage. But for the K type your analysis is correct but only for that temperature. Don't assume that it will remain at that as you increase and decrease in temperature. Please can you verify if the steps of amplifications are correct because I feel that 0,3815*10-3 oC is a very small temperature to detect.Thank you very much ShaunR for your time. Giorgos P Thermocouples produce very small voltages. You can see this from your Thermocouple range (1.4-.1.2)/ 30 = 0.04 mv/°C. This is why they use characterisation tables rather than an approximation. Its very important to minimise errors and introduce compensation if possible if you are looking for accuracy. Take a long hard look at your hardware spec (noise, temperature stability etc) and make sure it is capable Quote Link to comment
ginos Posted September 23, 2010 Author Report Share Posted September 23, 2010 Thank you for your answer ShaunR. I was mostly interested in the gain part of the scxi and pci6034e that you mentioned first. I have also a couple of questions about absolute accuracy in the DAQ system. Shall i post them here or create another topic? Quote Link to comment
ShaunR Posted September 23, 2010 Report Share Posted September 23, 2010 Here's fine. Quote Link to comment
ginos Posted September 23, 2010 Author Report Share Posted September 23, 2010 OK! I am trying to calculate the overall uncertainty of DAQ System. I know about the accuracy calculator in the ni.com and I am trying to check my results using this tool. So the calculation formula is: Absolute Accuracy = +/-[(Input Voltage x % of Reading)+ Offset +System Noise +Temperature Drift] In my measurements there is a temp drift, from 29-31°C, but the above formula is for enviroments form15-35°C. So the temperature drift is zero. At first I set to the accuracy calculator: DAQ device to PCI-6034E and SCXI module to none. Then I set for example 0.001203V and Average of 100 readings.The result is the following: According to the E Series User Manual (page 38 gain+resolution PCI-6034E) Absolute Accuracy=0,001203*0,0588/100+ (28,9 +2,75)*10-6=0,03235mV which the same with the accuracy calculator. When I I set to the accuracy calculator: DAQ device to PCI-6034E and SCXI module SCXI-1102B, I do not get the same results about the absolute accuracy of the DAQ device. Accuracy Calculator: Me: DAQ Absolute Accuracy=0,1203*0,0588/100+ (100 +5,04)*10-6=0,175mV which I a very large number. I assume that the input value is 0,1203 because the gain of the scxi 1120b is 100. What I am doing wrong? Sorry my posts are a bit tiresome!! Quote Link to comment
ShaunR Posted September 23, 2010 Report Share Posted September 23, 2010 OK! I am trying to calculate the overall uncertainty of DAQ System. I know about the accuracy calculator in the ni.com and I am trying to check my results using this tool. So the calculation formula is: Absolute Accuracy = +/-[(Input Voltage x % of Reading)+ Offset +System Noise +Temperature Drift] In my measurements there is a temp drift, from 29-31°C, but the above formula is for enviroments form15-35°C. So the temperature drift is zero. At first I set to the accuracy calculator: DAQ device to PCI-6034E and SCXI module to none. Then I set for example 0.001203V and Average of 100 readings.The result is the following: According to the E Series User Manual (page 38 gain+resolution PCI-6034E) Absolute Accuracy=0,001203*0,0588/100+ (28,9 +2,75)*10-6=0,03235mV which the same with the accuracy calculator. When I I set to the accuracy calculator: DAQ device to PCI-6034E and SCXI module SCXI-1102B, I do not get the same results about the absolute accuracy of the DAQ device. Accuracy Calculator: Me: DAQ Absolute Accuracy=0,1203*0,0588/100+ (100 +5,04)*10-6=0,175mV which I a very large number. I assume that the input value is 0,1203 because the gain of the scxi 1120b is 100. What I am doing wrong? Sorry my posts are a bit tiresome!! Your input value is still 0.001203. Gain is not included in this calculation, only the reading which already has the gain applied by the internal processing of the device. This is a "black-box" calculation. Subsequently your calculated value is in error by a factor of 100. Quote Link to comment
ginos Posted September 23, 2010 Author Report Share Posted September 23, 2010 Someone told me that in order to relate this digitized signal accuracy to the original signal, they divided it by the gain given by the SCXI-1120B module. This is why there is a factor of a 100 difference between the two. Is this correct? Quote Link to comment
ShaunR Posted September 24, 2010 Report Share Posted September 24, 2010 (edited) Someone told me that in order to relate this digitized signal accuracy to the original signal, they divided it by the gain given by the SCXI-1120B module. This is why there is a factor of a 100 difference between the two. Is this correct? Well. I'm no DSP expert. But that seems a bit simplistic, possibly a rule-of-thumb?. What do they mean by digitised signal accuracy? If you mention accuracy to me I think in terms of a compound of additive errors (as you can see from your calculation example which is derived in terms of temperature, reading an offset). I'm aware of aperture, quantization and clock errors for ADCs. Possibly he/she is referring to those in a general way. But those are mainly measured in bits rather than voltage, so it depends on your range rather than gain. What exactly are you trying to get to? You have the measurement accuracy of your system. You have the codeword size. These I can understand would be important to you for measuring temperature. Are you trying to break down the accuracy into every single error contributor in the system ? If so. this could be a very, very long thread Edited September 24, 2010 by ShaunR Quote Link to comment
ginos Posted September 24, 2010 Author Report Share Posted September 24, 2010 Are you trying to break down the accuracy into every single error contributor in the system ? If so. this could be a very, very long thread That wouldn't be such a bad idea!!!! Ok before I post here I posted in the NI forum (I hope you were not offended because I appreciate your valuable answers) but nobody gave me an answer until I told you before. So he said: I went over your calculations and you have calculated the accuracy correctly. The DAQ absolute accuracy at .5 to -.5 is .175mV but they have converted this value so that it has meaning to the orginal signal. The original signal acquired by the SCXI-1102B was multiplied by a gain of 100. After this gain, the signal is sent to the DAQ card to be digitized into your computer.In order to relate this digitized signal accuracy to the original signal, they divided it by the gain given by the SCXI-1120B module. This is why there is a factor of a 100 difference between the two. Jim St National Instruments Applications Engineer What do they mean by digitised signal accuracy? Does this answers your question? I hope that you follow this thread because I believe that I will hit back with more questions:D Thank you ShaunR Quote Link to comment
ShaunR Posted September 24, 2010 Report Share Posted September 24, 2010 So. He is saying the internal processing already accounted for the gain in the reading which you negated by including it in your calculation. Sounds familiar Quote Link to comment
ginos Posted September 25, 2010 Author Report Share Posted September 25, 2010 So. He is saying the internal processing already accounted for the gain in the reading which you negated by including it in your calculation. Sounds familiar Sorry my friend but I have to disagree with you. I believe what you are saying is different from what he suggested. From what I understand you suggest not to take into account the gain in this calculation, but take into account the reading of this gain. I believe this is wrong for two reasons: If you take a closer look at the two images that I have posted, you will see that in the first one the range in the DAQ board is ±0.05 (gain=100) while in the second one the range is ±0.5 (gain=10), which means that the accuracy calculator takes into account the gain in the calculation. The calculation 0.001203*0,0588/100+ (100 +5,04)*10-6=0.1057mV (which you told me) is different from the result of the accuracy calculator (0.0018mV) if the input value is as you suggested below Your input value is still 0.001203. Gain is not included in this calculation, only the reading which already has the gain applied by the internal processing of the device. This is a "black-box" calculation. Subsequently your calculated value is in error by a factor of 100. So the calculation you suggested me would not be in error by a factor of 100, but it would be decreased by 70mV from what I had calculated. Anyway, I have to agree with you on the thing that this is a "black box" thank you ShauR for you thorough answers, you helped me a lot Quote Link to comment
ShaunR Posted September 25, 2010 Report Share Posted September 25, 2010 Sorry my friend but I have to disagree with you. I believe what you are saying is different from what he suggested. From what I understand you suggest not to take into account the gain in this calculation, but take into account the reading of this gain. I believe this is wrong for two reasons: If you take a closer look at the two images that I have posted, you will see that in the first one the range in the DAQ board is ±0.05 (gain=100) while in the second one the range is ±0.5 (gain=10), which means that the accuracy calculator takes into account the gain in the calculation. The calculation 0.001203*0,0588/100+ (100 +5,04)*10-6=0.1057mV (which you told me) is different from the result of the accuracy calculator (0.0018mV) if the input value is as you suggested below So the calculation you suggested me would not be in error by a factor of 100, but it would be decreased by 70mV from what I had calculated. Anyway, I have to agree with you on the thing that this is a "black box" thank you ShauR for you thorough answers, you helped me a lot I fail to see where in Absolute Accuracy = +/-[(Input Voltage x % of Reading)+ Offset +System Noise +Temperature Drift] gain is used since it is a sub-component of "Reading". I took your word on the 100+5.14 since I didn't have that info (neither could I find the 28,9 +2,75 in the spec pages you pointed me to (which is where the 70mv lies) if that is the "system noise" and offset) . But it was glaring obvious that 0.1203 was incorrect. Perhaps I should have said "about 100" But you have an answer you are happy with so that's good. Quote Link to comment
ginos Posted September 25, 2010 Author Report Share Posted September 25, 2010 Specs are here NI 6034E (page 4) Quote Link to comment
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