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Current measurement using NI 9219

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Hello, I am trying to do low-current measurement using NI-9219.

Basically, the current will be carried by the jet produced by a process called electrospinning (Please, look at figure).

The current will be from micro to nanoampere, so I think I measure the voltage drop across the resistor and then calculate the current using Ohm's law.

My concern is the setup design (Figure). If I follow this diagram, I am worried that the conductive collector causes serious measurement flaws as itself is a huge resistor since the current is very low (micro-nano). 

I am new to electric stuff, so please be generous.

Thanks.

 Capture.JPG

 

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Message 1 of 9
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Hi Danny,

 

have you considered to use an amplifier?

Best regards,
GerdW


using LV2016/2019/2021 on Win10/11+cRIO, TestStand2016/2019
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You can lose your external resistor.  The NI 9219 module is capable of measuring currents directly. The NI 9219 computes current from the voltage that the ADC measures across an internal shunt resistor.  According to the datasheet, the input current range of the module is +/- 25 mA, which gives you resolution down to 3 nanoAmps at best.

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Message 3 of 9
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Hi,

 

since the current is very low (micro-nano).

When measuring the current directly please consider keep those accuracy, stability and noise numbers in mind…

(Resolution is something different than accuracy.)

Best regards,
GerdW


using LV2016/2019/2021 on Win10/11+cRIO, TestStand2016/2019
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How much voltage is applied and expected at the measurement point. You may be exceeding the channel to ground ratings of the 9219.

Perhaps an instrumentation amplifier with isolation may be the preferred route to take.

 

-AK2DM

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"It’s the questions that drive us.”
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Message 5 of 9
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Thank you for the reply.

1. Maybe it sounds stupid question, but how input range of 25 mA give me 3 nanoAmps? 

2. Then new diagram will be like this?

Capture.JPG

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Thank you for the reply.

From other literature, they used this setup in diagram with the resistor of 20 Mega ohms. The measured current was about 0.04 micro Amp. So it gives me about 0.8 V. 

Capture1.JPG

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Solution
Accepted by topic author Danny6802

Resolution = input range / (bit depth - 1) = (0.025 + -0.025) / (2^24 - 1) = 2.98 x 10^-9

 

No.  Current is never measured on a parallel leg.  Current is measured in a serial circuit.  Your current flows from the supply V+ to the tool, to the collector plate, and then to ground, which is also your supply V- (COM) reference.  Just insert the NI 9219 current + and - terminals in line somewhere, so that all current also flows through the module.

 

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Thanks for your help!

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