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I'm using the following setup to measure a voltage across a resistor. I measure it both by a DAQ system and an oscilloscope. Below is the illustration for the setup:

enter image description here

DAQ is the data acquisition hardware which is: http://www.mccdaq.com/PDFs/manuals/USB-1616HS-BNC.pdf

I1 is a current regulator which is set to 4mA constant current. OS is an oscilloscope. Mains is the mains multiple outlet for power supplies. A1 is the DC power supply for the DAQ. A2 is the DC power supply for the I1. R is the resistor through which current passes.

Voltage across the R is fed both to the DAQ and the OS.

Here is the problem:

When i observe the voltage across the resistor with the oscilloscope, it is around constant 2 Volt with a very very low ripples. Voltmeter also measures 2 Volt. This is expected, since R = 500 ohm.

But when I make data acquisition I saw a very low mean value around 0.7 Volt. Then I plotted the data from the data acquisition. Below plot is what I obtained:

enter image description here

It looks like a 100 Hz pulse with 2V max 0V min. DAQ is obviously wrong. What might be the problem here?

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  • \$\begingroup\$ Could the DAQ be resetting at 100 Hz? \$\endgroup\$ – vicatcu Mar 5 '15 at 13:59
  • \$\begingroup\$ what do you mean? \$\endgroup\$ – user16307 Mar 5 '15 at 14:05
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It looks to me like you are using a differential input on your DAQ to measure the voltage across a resistor. It also looks to me like you are exceeding the common mode input signal on your DAQ because of an earth problem. You have one power supply feeding the DAQ and one feeding the current generator. If one power supply has an unearthed output it will likely show a large AC common mode voltage on both wires with respect to earth.

If the unearthed output is the current generator then the voltage across the resistor will be 2V but both sides will be flying up and down together at several volts with respect to earth.

Try connecting the negative output of A1 to the negative output from A2.

You can easily prove it's a common mode problem by trying to measure the voltage at the same end of the resistor. Clearly the voltage should be zero but, due to common mode voltage problems you should see something like the "crazy" signal you currently see.

Here is an idea about common mode noise: -

enter image description here

The common mode noise is the AC on both signals. Between the two signals is a constant offset - this constant offset is what you are trying to measure with your DAQ. I suspect the AC signals on both wires is probably higher than the power supply on your DAQ.

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  • \$\begingroup\$ both power supplies are not earthed. the voltage across the resistor is fed to a channel by a BNC cable. you can see in the manual there is a pin called Analog common. can i use this instead? \$\endgroup\$ – user16307 Mar 5 '15 at 14:35
  • \$\begingroup\$ im also kind of confused what you mean by "If the unearthed output is the current generator then the voltage across the resistor will be 2V but both sides will be flying up and down together at several volts with respect to earth." could you draw a plot to explain that if possible? i would be very glad. \$\endgroup\$ – user16307 Mar 5 '15 at 14:38
  • \$\begingroup\$ (1) Maybe analogue common will work. Fix it to 0V on A2. (2) picture added. \$\endgroup\$ – Andy aka Mar 5 '15 at 14:57
  • \$\begingroup\$ what is V1 and V2 in my setup? thnx \$\endgroup\$ – user16307 Mar 5 '15 at 15:06
  • \$\begingroup\$ V1 is one side of your resistor and V2 is the other side. V1 - V2 looks constant at 2V but if the AC signal on each side exceeds the power rails on the DAQ then crazy results will happen. \$\endgroup\$ – Andy aka Mar 5 '15 at 15:29

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