Measuring High Current using cFP-AI-110 ?

[duhtrev]

I am trying to measure current coming out of a wind generator and am having difficulties knowing/understanding exactly what I need to measure that current.  My understanding is that I need to place a "precision resistor" in series with the wind generator and then using my cFP-AI-110 I can measure the voltage across the precision resistor and calculate the current.  My problem is that I am expecting currents of up to 40 amps (@ 12V).  And with that type of current, I'd burn up any typically resistor.  So two questions:1. Is my theory correct in what I need to do to measure the current?2. Can anyone suggest a "precision resistor" that is capable of taking 480W and doesn't cost a fortune?Thank you for your time,Trevor

[Pie56694]

Hi Trevor,Cool application!&nbsp; Your concept of operation is correct.&nbsp; All current measurements on a cFP-AI-110 module are voltage measurements of the voltage-drop across an internal 100? resistor.&nbsp; Take a look at page 6 of the <a href="http://digital.ni.com/manuals.nsf/websearch/DE9C12E83CD51846862570830078E19A" target="_blank"> FP-AI-110 and cFP-AI-110 Operating Instructions</a> for more information. &nbsp; A current measurement of 40A would destroy the unit instantly. The cheap way to take a current measurement of 40A with power resistors is to put a few identical power resistors in parallel.&nbsp; The current should divide evenly between them, and you may then take a voltage measurement with the cFP-AI-110.&nbsp; The caveat is that your generator must not regulate voltage, i.e. the output voltage must be linearly proportional to the output current with a purely resistive load.&nbsp; I can't recommend a particular power resistor, but I found a few online for about $10 / 100W.&nbsp; You'd need five of these at least.There are also other options.&nbsp; If your signal is AC, you could use a transformer to step-down the current.&nbsp; You could also use a converter to bring your signal into a measurable range.&nbsp; You could use an analog current probe.&nbsp; I hope these ideas help.&nbsp; Let me know if you have any additional questions.

[AnalogKid2DigitalMan]

Such DC applications use what is also known as a current shunt, here's some basic info on them, be aware of high side and low side configurations. Floating differential inputs on the data acquisition setup are probably desired. Also, wattage is&nbsp;different than what you calculated- say you use a 0.1 ohm shunt at 40 A, that would be W=&nbsp;I^2R = 40^2*0.1 = 160W. The voltage drop across the shunt would be E=IR = 40*0.1 = 4 volts, a good range for a 5V full scale setting on the DAQ. Hope this helps.
&nbsp;
<a href="http://en.wikipedia.org/wiki/Shunt_(electrical" target="_blank">http://en.wikipedia.org/wiki/Shunt_(electrical</a>)
<a href="http://www.rc-electronics-usa.com/current-shunt.html" target="_blank">http://www.rc-electronics-usa.com/current-shunt.html</a>
&nbsp;
-AK2DM

[Chris R.]

AnalogKid is right that your 12V does not come in to the shunt power dissipation equation. I see no need to burn 100W in a shunt though. I'd use something around 1 milliohm, which would dissipate 1.6W and develop a 40mV drop, perfectly measurable with a decent DAQ board or DMM. One example of a suitable part is the CSL series from IRC (http://www.irctt.com/file.aspx?product_id=15&file_type=datasheet) but I'm sure there are others if you look around. Chris