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# Indirect Measurement of the Current Draw of a Device 2012-10-15

Sometimes we want to know the current draw of a device for various reasons. One reason would be to estimate the run time of the device when it is running on batteries. It is often very inconvenient to try to break the battery connections and insert a current meter or sense resistor, so here is an alternate method that does not require breaking the current flow.

To accomplish this, we perform a series of very simple tests and from the results we can calculate the current draw. The tests do not involve inserting anything into the circuit in series with the batteries as most current tests require, so it makes it much easier to make the measurement overall.
If you have any questions feel free to PM me here to my screen name.

First i'll outline the procedure and then follow with an example where the device is a remote control transmitter that runs on four AA cells.

1. Measure battery no load voltage. This is with device turned off. Call it Vb.
2. Measure battery loaded voltage, with device turned on. Call it Vo.
3. Turn device off, load the battery with a large value resistor (R3) like 10k. Note the battery voltage and call it Va. Now the idea is to get Va to equal Vo by finding a resistor that pulls the battery voltage down as much as the real original device did, but sometimes this is not too convenient because we might need a power resistor. To make up for this, we decrease the value of the resistor only so far down and then go to the next step.
4. Calculate the value of R1 from:
[LATEX]R1=((Vb-Va)*R3)/Va[/LATEX]
where Va and Vb found above, and R3 is the load resistor from step 3.
5. Finally, calculate R2 from:
[LATEX]R2=(Vo*R1)/(Vb-Vo)[/LATEX]
and this is the equivalent device load resistance.
6. Double check to make sure we got it right from:
[LATEX]Vx=Vb*R2/(R1+R2)[/LATEX]
and compare Vx to Vo, they must be the same or nearly so. If they are the same, we did it right.
7. Calculate the device load current from:
[LATEX]I=Vo/R2[/LATEX]
and that is the equivalent device current draw.

Now here is the example i promised...

A certain device (a radio control transmitter) uses four AA cells. We want to estimate the current draw from the four cells so that we can estimate the run time when out in the field.

With the device turned off, we measure across all 4 batteries:
[LATEX]Vb=5.756 volts[/LATEX]

With the device turned on we measure again:
[LATEX]Vo=5.460 volts[/LATEX]

Now we turn off the device and connect a 100 ohm resistor (R3) across the batteries. We measure:
[LATEX]Va=5.695 volts[/LATEX]

and we calculate the value of R1 from:
[LATEX]R1=((Vb-Va)*R3)/Va[/LATEX]

and filling in the values for the variables we get:
[LATEX]R1=((5.756-5.695)*100)/5.695[/LATEX]

which comes out approximately equal to:
[LATEX]R1=1.071115[/LATEX]

Now we calculate R2 from:
[LATEX]R2=(Vo*R1)/(Vb-Vo)[/LATEX]

and after filling in the values and working it out we get:
[LATEX]R2=19.76 ohms[/LATEX] approximately.

Now we double check by using:
[LATEX]Vx=Vb*R2/(R1+R2)[/LATEX]

and with the values we have found so far we get:
[LATEX]Vx=5.46[/LATEX]

which is exactly what Vo equals, so we did it right.

Lastly, we calculate the current draw from:
[LATEX]I=Vo/R2[/LATEX]

and we get:
[LATEX]I=0.276 amps[/LATEX]

approximately, so the estimate for the current draw of the transmitter is 276 milliamps.

Note we never had to insert a resistor or current meter in series with anything.
Willen
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MrAl
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