Unregulated high current woes

LED Flashlights – General Info
1

This is an attempt to get some more insight into the current measurement error caused by the inclusion of the DMM itself, when dealing with high-draw unregulated lights (DD).

Meet my two Voltcraft DMMs:

- One with a 10A range (nicknamed "Ten", to avoid confusion between range and readout figures below),
- and an older one with a 20A range (nicknamed "Twenty").

Using an 18650 unprotected Flame and thick quality leads, I measured the current drawn from battery by my unregulated TrustFire F15 on High. Taking the reading about 5sec after connect for the current to stabilize somewhat, and allowing the setup a 5min rest between measurements.

Battery no-load voltage before first current measurement ... 4.00V.

Measuring with both Ten and Twenty in series ... 2.45A (Ten) and 2.43A (Twenty), i.e. 2.44A (average)
Measuring with Ten alone ... 2.75A
Measuring with Twenty alone ... 3.15A (falling)

Battery no-load voltage 45min after last current measurement ... 3.99V.

Thus, adding Twenty to the circuit already containing Ten, induces a current drop from 2.75A to 2.44A (-11%),
while adding Ten to the circuit already containing Twenty, induces a current drop from 3.15A to 2.44A (-23%).

Considering that the introduction of the first DMM into the circuit (the way current is normally measured) has a relatively bigger impact (exact amount unknown) on current than the introduction of the second DMM (since by the introduction of the second DMM the first DMM has already raised the total circuit resistance), I'd venture a rough estimate that the actual DD current draw is 20-40% higher than measured (depending on circuit and DMM internal resistances).

Also, given the DMM and battery, the more current a direct driven light is pulling, the bigger the relative current drop provoked by the ammeter and thus the larger the measurement negative error.

2

If you want accurate readings, or at least readings that don't substantially affect the current draw of the light to take them, use a current shunt and measure the voltage across it... If you don't have a precision current shunt handy, something like 7 inches of #12 copper makes a good 1mv/A shunt, or 16 inches of #18 copper for a 10mv/A shunt. Cut the wire a bit long, then put it in series with an accurate meter, and trim it until you find the right length to make the voltmeter you have across the shunt match the ammeter you're using to calibrate it.

--Bushytails

3

That's actually the way the current measurement in a DMM is arranged. Using a calibrated, temperature stable shunt of resistance as small as the voltage measuring module sensitivity allows.

They key is a lower-resistance stable shunt, which in turn requires a voltmeter of higher sensitivity then the one in the DMM. What I mean to say is, that with a given DMM you can't go lower with an improvised external shunt then they went in the factory with the internal calibrated one. If the voltmeter sensitivity allowed it, they would have built in a lower resistance shunt in the first place, since low resistance in current measurement is a valued parameter in a measuring instrument.

4

Much of the inaccuracy with cheap meters is the probe leads, several feet of #20 or thinner wire with cheap banana jacks. Using your own current shunt bypasses these. Also, you can easily construct a shunt of much lower resistance than the stock one - it just will read with lower precision. For example, using a .001 ohm shunt (1mv/A), I can measure up to 200A on the 200mv scale on my cheap HF meter - but with one less digit of precision than the 10A scale. 200.0A instead of 20.00A. And, of course, the voltage drop is much lower than the meter's internal shunt.

--Bushytails

5

Yes, to expand the current measuring capability, that's an acceptable workaround.

6

And to reduce the voltage drop. By using a .001 ohm shunt instead of a .01 ohm shunt, you get a tenth the voltage drop across the shunt, and much less effect on the emitter's draw. Getting rid of the probe leads makes a big difference too.

--Bushytails

7

Regardless, what my test was meant to show is the approximate error level that a normal measurement with a DMM (what everybody does) involves.