NI 6009 analog Input problem

[hedehodo]

Hi,

I am developing an application with USB DAQ 6009. I am trying to get an AC signal whose amplitude is peak to peak around 2V and it is a triangle shaped AC signal. I noticed that there's a DC noise on the signal. After some research i measured a DC 1V3 at all analog inputs when there's nothing connected.

Is that normal or is my device malfunctioned? Because i have another brand new one and that one either has 1V3 DC at its analog inputs

Regards,

CK

[hedehodo]

Hi,

I am developing an application with USB DAQ 6009. I am trying to get an AC signal whose amplitude is peak to peak around 2V and it is a triangle shaped AC signal. I noticed that there's a DC noise on the signal. After some research i measured a DC 1V3 at all analog inputs when there's nothing connected.

Is that normal or is my device malfunctioned? Because i have another brand new one and that one either has 1V3 DC at its analog inputs

Regards,

CK

[The_Red_Barron]

Howdy CK!

If you don't have any signal connected to the DAQ device then the input voltage will likely float to a non-zero value.  To verify that your USB-6009 is reading correctly tie the input to ground and you should measure zero volts.

NOTE: If you are taking a differential measurement (the default configuration) then you will need to tie the differential pair together or tie both inputs to ground to measure zero volts.

Regards,

[hedehodo]

Hi Barron,

Thx for the reply but the relevant input is not empty.The measuremnt is over a 10K resistance. I totally lost my confidence in this product.  Because for example i don't have such  a problem when i measure a battery.

[John_P1]

Hi CK,

Can you attach a connection diagram?  What exactly are you seeing?  It's probably expected given the USB 6009's input circuitry:

When nothing is connected, you should expect to measure a voltage of around 1.4 Volts.

Best Regards,

[Prof Q]

Hi John,

Thanks for the post I'm having a similar issue in a class that I'm teaching where the students are using the USB 6009 to measure the voltage across the middle of a resistor bridge. The input circuitry is interfering with bridge voltage/resistance. How bad would it be for me to open up the module and remove the 30.9k and 39.2k resistors from a couple of channels just for this lab. I looked at the board and it would not be hard but I don't want to do it if it will cause some other issue. I realize it would probably void the warranty on the boards but I'm ok with that, not my first rodeo:)   Thanks - Dave  

[John_P1]

Hi Dave,

Those resistors are necessary to pull the voltage to an appropriate value before it is passed to the internal amplifier on the 6009.

The better solution would be to implement a voltage follower (a.k.a. unity gain buffer) so that the resistance of your bridge would no longer be a factor. 

Best Regards,

[Prof Q]

Thanks for the quick reponse. Just a little bit dissapointed that the input impedance is so low on these inputs in comparison to a voltmeter. Why is it designed this way? I'd like to pass the info along to my students.

[John_P1]

I'll start by saying that the 6008/6009 has quite a different design from most of our other DAQ boards. Many of the higher-end boards have input impedances in the 100 GOhm range (even the myDAQ, which is marketed and discounted for students, has >10 GOhm input impedance).

The 6008 and 6009 however use a resistor network to divide-down the voltage to a suitable level for the components used on the board.  This is the reason for the 144 kOhm input impedance spec.  The resistors have to be low enough so that the impedance of the ADC (which has quite a bit of variance) does not significantly affect the voltage divider circuit.  The ADC used on this particular board has a typical input impedance of 6 MOhm for RSE, and 7 MOhm for Differential.

Even if the 6008/6009 input impedance was higher, if your source impedance is too high and you are measuring across multiple channels you could start to see ghosting due to inadequate settling time (which is dependant on your source impedance) between channels.  Regardless of input impedance, the voltage follower is a good idea for any multiplexed measurement of high source impedances if you want to sample at relatively high rates.

I wasn't employed by NI at the time of the original design of the 6008/6009, but I do know that various compromises are sometimes made when trying to balance performance and cost.  Depending on the use-case, 144 kOhm should be perfectly adequate for many applications, so it was likely decided that this would be acceptable given whatever tradeoffs were present at the time of the original design.  To make a significant increase to the impedance of this device retroactively would require a complete re-design of the board. 

If you'd like, feel free to post to the idea exchange as we are always looking for feedback.  With your current hardware however you will need to implement the voltage follower to measure from a circuit with a significant source impedance.  On the bright-side, the act of implementing the voltage follower should demonstrate the practical use-case for such a circuit and should hopefully be a good learning experience for your students.

Best Regards,

[Teaclassification]

Hi John

I am using NI USB 6009 for measurement of resistance of sensor.

My problem is i am getting different value of resistance of a sensor by multimeter and USB 6009.

I have a number of sensors having different resistance values (ranging from 1 k ohm to 20 Mohm).

the problem is that what should be excitation current (I assume 0.2 A). In that case the range of measuring resistance -100 ohm to 100 ohm. How can I increse the range of resistance?  

Please give guidlines

Thank you

Regards,

Debashis Dutta

[Frank-L]

The best way to do a resistance measurement with the USB 6009 would be to do a Voltage Divider with the resistance that you are trying to measure, and a known resistance across the 5V on the USB-6009 and Ground. You would measure the voltage between the known resistor and the unknown resistor using an analog input. You would then use the Voltage Divider equation to calculate the resistance of your unknown resistor. This is by far the easiest way to do this calculation. On page 22 of the 6008/6009 User Manual and Specifications, you can see an example of this being done.