p7 2x 18650, run time?

[spankone]

I've finished my build p7 led powered by 2 x 18650 li-ons any idea on runtimes? and how do you estimate runtimes?

 

[jasonck08]

How are you driving it? Whats the current? What is the capacity of the 18650's your using?

If its driven at around 2.8A and the 18650's are rated at 2500mah then you should get about an hour and a half of runtime.

 

[spankone]

yeh its using a DW driver that puts out 2.8a and batts are around 2500mah i think I'll be happy with 1 1/2 hurs runtime, I guess if I run 2s2p the run time will double?

 

[jasonck08]

Yes if you double the batteries, runtime will effectivly double as well.

 

[spankone]

thank you

 

[parnell]

Here is an online runtime calculator to help estimate: http://www.candlepowerforums.com/vb/showthread.php?t=190439&highlight=runtime+calculator

 

[Justin Case]

You need to measure the actual tail current draw or estimate it. Then knowing the capacity of your 18650 cells, you can calculate the run time.

To estimate the tail current draw, you need to have an idea of the actual LED forward current (let's say it's 2.8A), the forward voltage (let's say 3.5V), battery voltage (7.4V), and driver efficiency (85%). Then power delivered by the driver equals power drawn by the LED:

driver efficiency * Vbatt * Ibatt = Vf * If
0.85 * 7.4 * Ibatt = 3.5 * 2.8

Solving for Ibatt, we get Ibatt = 1.558A (you'll get the same # using the online runtime calculator). Dividing Ibatt into the cell capacity (let's say 2500mAh) gives estimated run time:

2500mAh/1558mA = 1.6h (or 1h36m, which is what the online run time calculator gives)

 

[spankone]

thanks I'm using this torch as my bike light and most rides last up to an hour so this should see me home.

 

[gcbryan]

yeh its using a DW driver that puts out 2.8a and batts are around 2500mah i think I'll be happy with 1 1/2 hurs runtime, I guess if I run 2s2p the run time will double?

I assume your runing 4p right now so 3.7v and 2800mA = 10.360 Watts
If you run 2s2p you will be a 7.4V and 1400mA = 10.360 Watts

How you your time double? Or am I missing something?

 

[ma_sha1]

I assume your runing 4p right now so 3.7v and 2800mA = 10.360 Watts
If you run 2s2p you will be a 7.4V and 1400mA = 10.360 Watts

How you your time double? Or am I missing something?

He is running P7, so the 2S2P is not referring to emitter like MCE.
I am guessing he is referring to 2S2P of 18650, that mean 4 x 18650 vs. OP of 2x18650, therefore double the run time?

 

[spankone]

He is running P7, so the 2S2P is not referring to emitter like MCE.
I am guessing he is referring to 2S2P of 18650, that mean 4 x 18650 vs. OP of 2x18650, therefore double the run time?

yeh thats it