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Current discrepancy

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ABCD:
I was attempting to measure current on a 9VDC circuit powering an LED (in series with a 680-ohm resistor).  I understand that the meter is placed in series with the circuit under test.  By Ohm's law, the current flowing out of the resistor should be 13.23 mA (9/680), right?  When measuring with two different meters, I get 10.61 and 10.81 mA.  So at least the meters are in agreement.  I have crocodile clip patch cable attached to the resistor, to the + probe, and another one connected to the - probe and the positive terminal of the LED.  I'm not sure if that's a factor (the cable), and I don't think burden voltage is either at 9V.

I measured the battery and resistor values with each meter and using those numbers with Ohm's law I get 13.72 and 13.77 mA.  The meters in question are the EX330 and AM220 from Dave's reviews.  I got burden voltage data from the PDF comparison sheet.

Since both meters are in close agreement, I don't think that they're damaged, so I'm thinking there's some theory I haven't grasped yet and/or technique.

Am I missing something here?

baljemmett:

--- Quote from: ABCD on May 21, 2011, 10:12:00 pm ---I was attempting to measure current on a 9VDC circuit powering an LED (in series with a 680-ohm resistor).  I understand that the meter is placed in series with the circuit under test.  By Ohm's law, the current flowing out of the resistor should be 13.23 mA (9/680), right?

--- End quote ---

Not quite -- the voltage across the resistor will be 9 volts minus the forward voltage drop of the LED.  In this case, your LED appears to be dropping about 1.7 volts (which is a pretty typical sort of figure), which would give a current of (9-1.7)/680 = 10.7mA -- corresponding with an average of your readings.  If you measure the voltage across the resistor you should see that it's around 9-1.7 = 7.3 volts, and Ohm is once again satisfied :)

ABCD:
Thanks, that explains it excellently.

Zero999:
What's the battery voltage with the LED connected?

The battery voltage will drop below 9V as it discharges and under load so the actual battery voltage when the load is connected needs to be used in the calculation.

vk6zgo:
Actually,baljemmet answered the question,the forward voltage drop of the LED has to be subtracted
from the 9 volts supply,so:-
 
                  I is not=  9/680,

                  I = (9-LED volt drop)/680,-----------hence the reduced current.

All the other comments are red herrings.

VK6ZGO

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