Question about current draw

[Zatoichi]

I've just tried a 18650 in my Jetbeam III M, and decided to measure the draw to compare between CR123 primaries and the 18650. I'm a newbie in this area, and am not sure how to interpret the results.

With CR123's it measures 0.64mA (battery voltage 2.97v)
With the 18650 it measures 0.53mA (battery voltage 4.06v)
(voltage measurements without load)

Does this look normal? And could someone point me to a good source of information explaining how this works, or summarise for me please. I tried search terms like 'current draw' and 'understanding current draw' but didn't find what I'm after. Cheers.

 

[bigchelis]

I've just tried a 18650 in my Jetbeam III M, and decided to measure the draw to compare between CR123 primaries and the 18650. I'm a newbie in this area, and am not sure how to interpret the results.

With CR123's it measures 0.64mA (battery voltage 2.97v)
With the 18650 it measures 0.53mA (battery voltage 4.06v)
(voltage measurements without load)

Does this look normal? And could someone point me to a good source of information explaining how this works, or summarise for me please. I tried search terms like 'current draw' and 'understanding current draw' but didn't find what I'm after. Cheers.

I use current draw to estimate my runtime. I use 80% of the cell capacity. Say the 18650 you will use is AW 2200mAh cell at 80% of it you can use 1760mAh.

Then the put into the formula for runtime.
1760mA divided by current at the tail 530mA = 3.320
3.320 multiply by 60 minutes = 199 minutes is the runtime!!! approximately!!!

 

[Russel]

You may want to try looking up Ohms law.

http://en.wikipedia.org/wiki/Ohm's_law

Russ

(I'm sure others, much more educated than I, will chime in soon.)

 

[Mr Happy]

I've just tried a 18650 in my Jetbeam III M, and decided to measure the draw to compare between CR123 primaries and the 18650. I'm a newbie in this area, and am not sure how to interpret the results.

With CR123's it measures 0.64mA (battery voltage 2.97v)
With the 18650 it measures 0.53mA (battery voltage 4.06v)
(voltage measurements without load)

Does this look normal? And could someone point me to a good source of information explaining how this works, or summarise for me please. I tried search terms like 'current draw' and 'understanding current draw' but didn't find what I'm after. Cheers.

Probably you mean 0.64 A rather than 0.64 mA? And the 2.97 V is per each CR123A, rather than the two together (which would make 5.94 V)?

What you are likely seeing is that the light draws slightly less current on the lower voltage 18650 than on the higher voltage of 2 x CR123A. Typically this suggests the light is not using full regulation but only partial regulation.

To draw more conclusions would require more measurements under more rigorously documented and controlled conditions.

 

[bigchelis]

You also can use the amp draw/current at the tail to measure how effective the regulation is of the LED.

If a LED is regulated say from 4v to 24v at all voltage levels the watts of power will be the same. Apply more voltage and current goes down. Less voltage and current goes up.

say 1 cell 4.2v x amp draw at the tail (.53A) = 2.2 watts of power.
say 1 cell 3v x amp draw at the tail (.640A) = 1.92 watts of pwer
as you can see regulation works, but its not perfect. If it was perfect it would be that exact same watts at both voltage levels. In my application there was a range from 5.5w~6w of power from 4v to 12v of regulated supply.

 

[Zatoichi]

Thanks everyone, that's all very helpful in making sense of it.

Mr Happy, yes I meant amps.

bigchelis, for your calculation would it need to be the voltage of the cell (or cells) under load x amp draw = watts, to be fairly accurate?

I'm going to read ohm's law now... I may be some time.

 

[bigchelis]

Thanks everyone, that's all very helpful in making sense of it.

Mr Happy, yes I meant amps.

bigchelis, for your calculation would it need to be the voltage of the cell (or cells) under load x amp draw = watts, to be fairly accurate?

I'm going to read ohm's law now... I may be some time.

For my low current applications under 1amp I use the volts of the cell under no load. I do not think a IMR 18650 will sag under load with 1amp draw. (not much at least)
MrGman and myself tested the lumens readings of my P60 drop-in P7 direct drive and when we applied one 18650 the 350 lumens out the front were low to say the least.,
but the 4.2v saged to 3.5~3.7v.
With 2 cr123 primaries 6V. They saged to 3.7~3.9v and 450 lumens out the front.
Then with a regulated 4.5volts with no drop or sag we got 680 out the front lumens, but you will need 4 to 6 18650's in parrallel in order to avoid the sag and I think this is part of the reason the Leggion II does so well under load and is a top performer. It also has way better heatsinking to provide the 742 lumens for 2 hours.

 

[Zatoichi]

Thanks again bigchelis. These examples give me a better handle on it than just reading theory.

 

[bigchelis]

Thanks again bigchelis. These examples give me a better handle on it than just reading theory.

No problem.

Hopefully this weekend we can test the Maglight P7 1D running a single IMR 18650 (4.2c 1600mAh) vs I KD D cell (4.2v 5000mAh). It would be good to see how the cells sag under these new bigger hosts. I sure hope the 1D from KD doesn't sag

To further measure sag with high current applications I want to also test my 2D Mag P7 running 3 NiMH Tenergy 5000mAh each cell. That is 4.2v and 5000mAh which is the same as 1D from KD, but we will see the results.