AD590 Chip......>>>>>> ADC raw data kelvin to degrees Celsius how do I the ma
sam_sam_sam
Posts: 2,286
·Here is the· conversion·· ·1 Kevin = -272.15 degrees Celsius
Here is the data sheet to the AD590 Chip
http://www.analog.com/en/other-products/militaryaerospace/ad590/products/product.html
Thanks to anyone that can help
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Sam
Post Edited (sam_sam_sam) : 4/26/2010 4:14:29 PM GMT
Here is the data sheet to the AD590 Chip
http://www.analog.com/en/other-products/militaryaerospace/ad590/products/product.html
Thanks to anyone that can help
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Sam
Post Edited (sam_sam_sam) : 4/26/2010 4:14:29 PM GMT
Comments
deg2C = deg2K - 27215 ' in units of 1/100 degree
Your puzzle may have more to do with how your ADC returns the data, and what language you are using.
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Tracy Allen
www.emesystems.com
I can use / 12 and get 25 wich is a little off it should closer to· 30
I am using a BS2 for this Demo Board
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Sam
Post Edited (sam_sam_sam) : 4/26/2010 4:55:49 PM GMT
Let's assume zero offset.
' value ADC returned here
kelvin = ADC ** 64674 ' calibration: 300 Kelvin @ 304 ADC
celsius = kelvin - 273
fahrenheit = kelvin * 9 / 5 - 460
Do you have a 1kΩ resistor to convert the current from the AD590 into Volts, being read by an ADC at 1mV per bit? If you use a 10k resistor instead, then the reading can be more accurate.
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Tracy Allen
www.emesystems.com
I have 1.2 k ohm on there now
Do you have a 1kΩ resistor to convert the current from the AD590 into Volts, being read by an ADC at 1mV per bit?
I try this and see what happen
If you use a 10k resistor instead, then the reading can be more accurate
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Sam
·········································································································@·about 80* F
I changed the resister to 10k ohms and now the·result value is 1530·· |
When the resister was 1.2k ohms it was 304···························· ········|··
Is·this what I should get as a value changing the resister to 10k ohms ·········
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Sam
I believe there must be something wrong electrically with your circuit. Do you have at least 5 volts across the AD590 sensor?
Math for the AD590 is very simple. The sensor produces (nominally) 1 microAmp per Kelvin. So at 300 Kelvin, the voltage across a 1000Ω resistor should be 300 mV, or across 1200Ω it should be 360mV, or across 10000Ω it should be 3000mV. Simple
mV = µA * R /1000 with µA = Kelvin, R in Ohms
300 Kelvin is near 27 Celsius or 80 Fahrenheit.
You report that the result value is 304mV with 1.2 kOhms, which would solve backwards to
Kelvin = µA = mV / 1200 = 304 * 1000 / 1200 = 253 Kelvin
That is -20 degrees Celsius, I don't think so. The result would make much more sense if you had a 1000Ω resistor, which is what I was assuming you had above. The result is even worse with the 10kΩ resistor:
Kelvin = µA = mV * 1000 / 10000 = 1530 * 1000 / 10000 = 153 Kelvin! -120 Celsius!
That bad result could be explained if there is not enough voltage across the AD590. Or maybe I am wrong about the numbers reported by your ADC. Does 304 mean 304 milliVolts?
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Tracy Allen
www.emesystems.com
With a 1.2k Ohm· = 1.2 volts @ about 80* F
One is
With a 10k Ohm = 2.345 Volt @ about 80*F
The voltage reading where taken with a volt meter
I hope this helps
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Sam
Please draw out or describe your circuit, including how you are powering the AD590. Are you measuring those voltages with the MAX186 program you posted above, or with a multimeter?
Those values don't make sense. For example, 1.2V across a 1.2kΩ resistor implies a current of 1 mA, which would be produced by an AD590 at 1000 degrees Fahrenheit (as if it could go that high). Do you see the problem?! At 80 °F, across 1.2kΩ, you should expect near 360mV.
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Tracy Allen
www.emesystems.com
I have it the way the data sheet shows it on the top of page 7
·I found the problem why my voltage was so off was that my input ground·was not hook up on the MAX186 Chip
Now I am getting 324.1 millivolts @ about 80*F which is still of by a little I just have to find the right value resistor to get 300 millvolts
Thank for your help
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·Now wanting to learn Spin· Thanks for any··that you may have and all of your time finding them
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Sam
Post Edited (sam_sam_sam) : 5/1/2010 4:20:35 PM GMT