Multifunction DAQ

[Sworks]

Hi,

does any one know the possible samplingrate of the DMM-input ?(not the 'normal' analog inputs). I leared that it is possible to use also the DMM-input in a VI.

Reason: I want to do a DC-current measurement with logging. A samplerate of 1 kHz would be great, but I'm not sure if this is possible. I can't find anything about this in the manual.

thanks,

Georg

[Stephan_D]

Hi Georg,

the myDAQ is shipped with a DMM-function, that is true.

I've look in the manual by myself. Unfortunately, the sampling rate are not specified.
The chip, which is used to implement the DMM function doesn't specify a sampling rate for the DMM. The only information, which i can give you, is that the sampling rate of the DMM is approximately 1 Hz. It is definitly not fast enough for a 1 kHz measurement.

Regards,

Stephan

[Sworks]

Hi Stephan,

thanks for the reply. Not the answer I wanted ;-)

Ok, so I have to find another solution.

Thanks ones again,

Georg

[crossrulz]

You can use a small resistor as a current sense.  I typically use a 0.1 Ohm resistor to sense the current of a circuit.  So the current would be I = V/0.1 = V*10.  You can even setup a DAQmx scale in MAX so that the voltage is automatically converted into your current for you.

[Sworks]

Thanks for the input, crossrulz,

I will test this. I only have to take care about the max. power of the resistant, isn't it ? But with only 0.1 Ohms, this is quite low...

regards, Georg

[crossrulz]

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Sworks wrote:

Thanks for the input, crossrulz,

I will test this. I only have to take care about the max. power of the resistant, isn't it ? But with only 0.1 Ohms, this is quite low...

regards, Georg

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I should clearify a couple of things.  You will want the current shunt to be on the return side of the circuit (low common mode voltage so you don't blow up your card).  The wattage of the shunt resistor does need meet your power requirements.  If you are expecting, for example, 10A, then you will want P = I*I*R = 10*10*0.1 = 10W and your DAQ voltage will be V = I*R = 10*0.1 = 1V.  Some circuits care about this voltage drop, so be careful about that.