Use Analog Discovery 2 to measure current

[Paul Chang]

Dear Sir,

Below picture is our job. (I would like to know the maximum current that AD2 can handle. Does it damage the AD2?)

 

Best Regards,

Paul

[JRys]

Hello Paul,

The Analog Discovery 2 (AD2) measures voltage. Its input impedance is approximately 1M ohm.

It's not recommended to work directly with high voltage because if something were to go wrong, the voltage could make it way back to the operator. Its best to use an isolation signal conditioning module between your test rig and the AD2 inputs. The 8B series from www.dataforth.com has 1500 volts of isolation and will provide some protection. However, these may be a problem for you because their bandwidth is less than that of the AD2. 

Going back to your question, what I see is a 100:1 voltage divider. If the discharge voltage hits 10000 volts, the voltage at the AD2 terminals could reach 100 volts, which exceeds the 25 volt maximum input range and damage the unit.

John

[Paul Chang]

@JRys Thanks for your anwser.

[Paul Chang]

Dear Sir,

The schematic shown below : 

In my experiment, series 300M and other resistance with AD2, connect AD2 to usb gnd of PC.

But the voltage brtween 1+1-  pin of AD2 is much bigger than expected.

Is there any possible that resistance of AD2 was combined with PC to form a bigger R then cause bigger voltage than expect?

BTW,I didn't connect any node to GND of AD2.Would it be the problem?

Which node should I connect the AD2's GND?

 

Best Regards

Paul

[JRys]

Hello,

Based on your wire diagram, the AD2 is measuring Vhv.

I revisited your first post showing the corona tube. I believe if you were to reverse the resistor position of the 10M and 1G ohm resistors and lower the value of the 10M ohm resistor to 1M ohm, then the AD2 should measure 9-10 volts with a 10,000 volt discharge. Put the 1M ohm resistor lead to ground so that the (-) input of the AD2 is also at ground. Measure the voltage across the 1M ohm resistor.

V = ( 1M / (1G + 1M) ) * 10000

V = 10

and

V = ( 1M / (1G + 1M +10M) ) * 10000; includes tube resistance of 10M ohm

V = 9

To calculate current, divide the measured voltage by 100k ohms

Best regards,

John

 

 

 

[Paul Chang]

 

Hi @JRys:

The question of my measurement is that voltage of AD2 is about 2 to 3 times than expected.

For example: expect value Vad2 = 9v , measured Vad2 = 24.7 v much bigger and I don't want this happen.

 

Best Regards,

Paul

[JRys]

Hi Paul,

I'm sorry but there's something about your setup that I'm not understanding. Perhaps you could seek out an Electrical Engineer to debug it. For more information about the AD2 input circuitry refer to the following page: https://digilent.com/reference/test-and-measurement/analog-discovery-2/hardware-design-guide

Take care,

John