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AD590 and 6008: advice needed

I have to measure an outdoor temperature with a AD590 and a 6008.

The range that I expected to measure is [263K...313K] in other words
[-10C..+40C].

I was thinking to send +5V provided from USB6008 to power AD590 (located
at about 5 meters from DAQ board), then use an AI (configured as RSE, in
order to minimize ground loop since AD590 is a NRSE source) with a 10k
resistor (0,5%), connected from AI+ and GND input, so to have a
conversion of 10mV/K.

Is it a good solution?

Then the voltage range should be 2.63V...3.13V.

What do you suggest at this point?

I was thinking a conditioning with an differential amplifier (Av=4.42)
so to scale 2.63...3.23V to 0...5V and get a better resolution.

Or you suggest to connect simply the only resistor?

What do you suggest? Any hint/suggestions is really welcomed.
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Rossi,

That sounds like a pretty decent approach to me. I would probably recommend scaling the measurement to 0-5V in order to take full advantage of your DAQs resolution, just keep in mind that resolution isn’t going to be the only limiting factor in your measurements accuracy, but also the precision of resistor that you use. So I would determine if the precision of the resistor is going to be good enough that scaling up to the 5V scale makes sense. If it is not, then you may not want to go through the hassle of the extra circuitry to scale the signal.

-GDE

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Hello,

depending on the accuracy required, consider to use +2,5V ref (accuracy 0,25%) instead of +5V power supply (typ +5V, min +4,85V, max?).

Of course you need to adjust the input scaling.

 

Bye

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GDE [DE] ha scritto:

> the only limiting factor in your measurements accuracy, but also the precision
> of resistor that you use.

And if I create a NI-DAQmx current channel (instead of a voltage
channel) could improve the precision?
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Hi rossi,

Even when creating a current channel, you still have to use an external shunt resistor since the USB-6008 does not have current inputs.  As such, this will not improve the accuracy of your measurement since the driver simply takes a voltage reading and then divides by the specified resistor value to return a current.  Hope this helps.

Best regards,
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