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M. Z. Aziz Khan

Analog Discovery 2 with Current Measurement Capability

Question

Would it make the Analog Discovery 2 more powerful device if it is added the feature to measure current too? Furthermore it can still draw the IV-Characteristic curve using XY draw function in Oscilloscope for which we use some low value resistor etc. But will it be not more easy to use and be precise if it has built-in current measurement capability and can draw the IV-Characteristics directly for device under test?

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Hello,

Thank you for your opinion.

Currently you can use the custom Math channel to calculate the current value on a given resistor and adjusts the additive test circuit offset error, like:
M1 = C1 / 0.0471 - 0.0012 (for ~47mOhm resistor and 1.2mA error)

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On 2/24/2016 at 0:23 PM, attila said:

 

 

Hello,

Thank you for your opinion.

Currently you can use the custom Math channel to calculate the current value on a given resistor and adjusts the additive test circuit offset error, like:
M1 = C1 / 0.0471 - 0.0012 (for ~47mOhm resistor and 1.2mA error)

Hi attila,

I too would like to know how to do this, and so I wanted to ask if you could provide screenshots on how to do this?

Sincerely,
JohnB

Edited by JohnBee

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On 3/20/2017 at 3:16 AM, attila said:

Just to be sure I'm understanding this correctly, is this approach based on using two channels?
That is to say, one channel measuring each side of the resistor?
And then using a math-channel to calculate the difference i.e. the voltage drop across the resistor?

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The scope inputs of AD are differential.
Connecting C2+/- over a resistor this will measure the voltage drop on it. Then using math channel divide this value with the resistor value you will get the current, M1 =C2/R
Having the C1- connected to ground and C1+ to a test point this will measure the voltage relative to ground, like single ended input.

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Okay thanks for that, I think I have a better understanding now. That said, what would be the proper way to position the resistor in my circuit? 
Would I place the resistor in series on the positive side of the circuit, or across both sides(+ and -)?

With that being said, and just to be sure I'm understanding, connecting C1 to ground would be the wrong approach to take in this particular case?

 

Edited by JohnBee

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Hello @JohnBee,

I apologize for not making sure that you received a response back much earlier in the year. To answer your question, you need to have C1+ and C1- on opposite sides of the resistor so that the voltage difference on the two sides of the resistor is observed. The resistor itself will need to be attached to the voltage source (Vcc) and ground (Gnd).

As attila mentioned, the analog inputs are differential, so the value of Channel 1 on the Scope view inside of the WaveForms 2015 software will be the calculated value of (C1+ minus C1-). So, if C1- is connected to ground, this turns the equation into (C1+ minus 0V) for a positive valued result. Thus, when you add the custom math channel (by clicking on the Add Channel button with the green plus sign on the right hand side of the WaveForms GUI and select "Math: custom" and choose to divide C1 by the value of the resistor (either the nomimal value or as determined by a DMM), you will end up getting a positive current result.

However, if instead C1- was connected to the positive side of the resistor and C1+ was connected to the grounded side of the resistor, WaveForms 2015 would measure the scope input as a negative voltage, so when you calculate the current via {I = V/R}, you will end up with a negative current value. You can check out a picture version of this in the Forum Gallery here.

Naturally, the current itself will always flow in the same direction with positive current flow from the more higher voltage to lower voltage (and the electrons 'flowing' from a lower voltage to a more positive voltage), but we can observe a negative current by (in a sense) changing our measurement perspective.

Let me know if you have any questions.

Thanks,
JColvin

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