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Measuring current output from a floating current source

by yapatel
Tags: current, floating, measuring, output, source
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yapatel
#1
Mar25-14, 04:28 PM
P: 2
Hi -

I am by no means an EE, so please be aware I may have some fundamental flaws in my knowlege.

I have a battery powered isoluation unit that takes some voltage in and provides either a voltage or current source output. I am giving it 1V as an input and according to the conversion ratio, I should be getting 1mA out, however I am having trouble measuring it.

I took a 100kOhm resistor and connected the two output leads of the isoluation unit (red and black) to the resistor (so now the resistor is in series with the output of the isolation unit). Then I took another set of leads (red and black with a BNC connection) and plugged it up to my oscilloscope and set the input impedance of the scope to be 50Ohms. This gives me a measure of the voltage across the resistor.

I know that using V=IR I should be able to take the voltage shown on my oscilloscope, divide it by 100kOhm and get my current value. However when I do this I get a reading of 50mV on my oscilloscope, which means my current is 0.0005 mA. But this is not right since I should be getting 1mA out.

I checked the 1V input I am giving the isolation unit and it is showing up as a 1V signal on my oscilloscope. I am confident I am doing something wrong in measuring this current but am not sure what. Help/advice?
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berkeman
#2
Mar25-14, 06:02 PM
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P: 40,707
Quote Quote by yapatel View Post
Hi -

I am by no means an EE, so please be aware I may have some fundamental flaws in my knowlege.

I have a battery powered isoluation unit that takes some voltage in and provides either a voltage or current source output. I am giving it 1V as an input and according to the conversion ratio, I should be getting 1mA out, however I am having trouble measuring it.

I took a 100kOhm resistor and connected the two output leads of the isoluation unit (red and black) to the resistor (so now the resistor is in series with the output of the isolation unit). Then I took another set of leads (red and black with a BNC connection) and plugged it up to my oscilloscope and set the input impedance of the scope to be 50Ohms. This gives me a measure of the voltage across the resistor.

I know that using V=IR I should be able to take the voltage shown on my oscilloscope, divide it by 100kOhm and get my current value. However when I do this I get a reading of 50mV on my oscilloscope, which means my current is 0.0005 mA. But this is not right since I should be getting 1mA out.

I checked the 1V input I am giving the isolation unit and it is showing up as a 1V signal on my oscilloscope. I am confident I am doing something wrong in measuring this current but am not sure what. Help/advice?
You need the measuring device to have a much higher input impedance than the resistor that you have connected across the current source. Try using a 1kOhm resistor and use your 1MegOhm input DVM leads across that 1kOhm resistor. You will get 1V/mA from the resistor.
meBigGuy
#3
Mar25-14, 08:18 PM
P: 1,074
Saying in different words:
1ma across 50 ohms = 50mv. The scope 50 ohm input is effectively your measurement resistor. Set the scope to high impedance. The isolation unit would then need to supply 100 volts to get 1 ma across 100K. Drop the 100K to 500 ohms and you should see 0.5V.

yapatel
#4
Mar25-14, 08:56 PM
P: 2
Measuring current output from a floating current source

Hi Berkemen/meBigGuy -

I appreciate the concise response. I have a few theoretical questions (just to help understand this better):

1. Berkemen - When I try your approach, my oscilloscope can't measure the voltage of the signal because the signal is too high in amplitude for the screen - and since Agilent scopes use the digital screen to measure it, it just shows me some random "greater than" value. I can do your approach with a standard Agilent multimeter, but don't understand how to measure it using an oscilloscope due to the aforementioned issue.

I'm still fuzzy on the meaning of impedance. I have read a lot about it and understand the math, but not the physical implications of it. Any explanations you can offer are appreciated.

2. meBigGuy - I originally calculated the current using the formula you state: 50mV/50ohms = 1mA. However I would like to measure the current output using a 100KOhm load on the output. As I mentioned above, the oscope can't measure the voltage on that because the volts/div don't fit (i.e. can't get 100V to fit onto the screen). How do you overcome such an issue when measuring with the scope? is it even possible since there is a built in 50ohm resistor in the scope?

Apologies in advance if anything is terribly incorrect or novice.
meBigGuy
#5
Mar26-14, 11:00 PM
P: 1,074
If the scope input is 50 ohms only, then you can simply put the 100K resistor in series with it. (or find a 99.950K ohm resistor). When there is 100V on the resistor, you will see 50mV on the scope.

Does your source actually produce 100V for a 1V input?
berkeman
#6
Mar26-14, 11:58 PM
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P: 40,707
Quote Quote by yapatel View Post
1. Berkemen - When I try your approach, my oscilloscope can't measure the voltage of the signal because the signal is too high in amplitude for the screen - and since Agilent scopes use the digital screen to measure it, it just shows me some random "greater than" value. I can do your approach with a standard Agilent multimeter, but don't understand how to measure it using an oscilloscope due to the aforementioned issue.
1V is too big for your oscilloscope? 1mA through 1kOhm is 1V.


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