Quick question...

[Tabish]

If I power a PHR / rkcstr driver (set at 100ma) using 2x 10440 batts (from dx).

How long before the voltage drops to where its too low for the rkcstr.

My guess:

Fully charged = 4.2 x 2 = 8.4v

phr @ 100ma uses 4.5v ??? add 2.25v the driver burns up

4.5 + 2.25 = 6.75v

8.4 - 6.75 = 1.65 (max voltage drop allowed before voltage is too low)

8.4 - 1.65 = 6.75v (lowest i should let the voltage for both bats hit)

That means 3.375 V per cell

SO BASICALLY:

The draw is 100ma. The battery is just over 300ma. (trustfire rates at 600)

How long before a 4.2v cell drops to 3.375v at that current draw and battery capacity?

Just curious because I want to put it inside jayrobs pen host but I was told 2 10440's cannot power a phr / rkcstr.

I don't mind if it runs 5min. So long as it RUNS AT ALL.

 

[ndrew2505]

cant tell you an exact time but it does take a while... if you have an lpm you can test it. the power will start to sag when the batteries get drained enough.

 

[jayrob]

Yes, it will work... (at full charge)

But with a load on the batteries, the voltage immediately drops. As far as just how much, and for how long the driver will stay in regulation, will simply depend on the quality of the batteries. And the charger for that matter...

See Igor's comments here:
http://laserpointerforums.com/f38/8x-br-diodes-42806-3.html#post582195

 

[Tabish]

Ok. Thanks for the replies.

Jay, aypal:

Eventually I'll upgrade it it with a flexdrive and a 6x or 8x.