How do I calculate the accuracy of the NI 9239 AI module

[gammalight]

Hello all,

I am in the market for an AI module that can help me resolve angles to an accuracy of 1/10 of a degree.  I am going to have a resolver and I want to do the work of a resolver to linear converter by reading in the signals and doing the math with an NI 9239.  I am just trying to figure out if I can get the accuracy I need out of it.  How do I go about figuring this out.  I don't want someone to do it for me just some help to step through it.

The 9239 has 24 bit resolution and I am using it from +/- 10V, so a 20 volt range.  This should give me an accuracy of

20/((2^24)-1) = 1.19x10^-6 V/bit

Is this correct, so I should be able to get an accuracy of a 1/10 of a degree?

[yenknip]

You could also use an encoder wheel and the counter input on the DAQ device. To gain additional accuracy you could then gear the encoder by 10:1 or 100:1

Measuring very small voltages often results in erroneous values being read from induced currents or nearby interference. If you still want to use the analog input, then you could still gear up your measuring device, which would decrease your potential error ratio

*EDIT - Spacebar went all strange and posted the unfinished message O_o

Message Edited by yenknip on 10-14-2008 04:09 PM

[gammalight]

Hey Ed,

Thanks for the reply, your method for resolving sounds pretty slick, but I am going to use a resolver and just resolve the angles in SW my self.  I figured out that the 9239 will provide me the accuracy I need.  This is how I figured it out, with the help of an NI engineer.

The modules accuracy:

voltage range = 20 V

bits of res = 24 bits

20/((2^24)-1) = 1.19 muV

I need to be able to detect a 1/10 of a degree of change in angle so the output from the resolver is a sin and cos component

therefore

resolver excitation voltage = 7 V

coupling factor = .5

.5*7*sin(.1) = .006 V

so a 1/10 degree change will induce a .006 V change in voltage, when I can detect a 1.19muV change in voltage with the module.  So I will be able to detect that change.  Q.E.D

[johnsold · Accepted solution]

You need to check at the peak, not at the zero crossing, because the voltage changes most slowly at the peak.

0.5*7*(sin(90) -  sin(89.9)) = 5.33E-06 V

This is the change you must be able to detect to get your 0.1 degree resolution.  Of course when one phase is at peak the other phase is near zero.

Lynn