Cree Q5 in series

[FATTYTRIATHLETE]

Hi,

3 Questions please.

Can I run 4 Cree Q5s in series off a 13.2v 4500mah NIMH battery ?

Is there a way of calculating the run time I might get ?

How do I step down the 13.2v if I want to run 1 or 2 of the Cree MC-E ?

Many thanks.

 

[odessit]

1. Yes you can

2.
Here is the data sheet
http://www.cree.com/products/pdf/XLamp7090XR-E.pdf
If you are driving them direct each LED will get

V.led= 13.2 / 4 = 3.3V

From the chart on page 7, 3.3 V is roughly 400 mA

So, to calculate runtime:

T.runtime= 4500 mAh / 400 mA = 11.25 hrs

This is rough, best case scenario estimate. Actual runtime will be a bit lower and the brightness will be a lot dimmer at the end

3.
To stepdown voltage you would need one of two things
a. Cheapest is to get some resistors. See http://led.linear1.org/led.wiz
Unfortunately, the LED will not be regulated and depending on setup will waste a LOT of power as heat.
or
b. Best is to get a buckpuck current regulator (this circuit will regulate amperage and stepDOWN the voltage), but it is more expensive.
http://www.ledsupply.com/wired-buckpuck.php
Advantage is about 90% efficiency (again at best) and constant brightness.

I am building similar setup (3x Q5) with 14.4V 1200mAh battery using http://www.ledsupply.com/03023-d-e-1000p.php
you may not need external pot (to dim/brighten LED), but i chose to have this option.

 

[JWP_EE]

I would not connect the 4 LEDs directly across the 13.2V battery. You need some series resistance to prevent thermal runaway. And the fully charged battery may be 14.5V or higher.

The 3.3V across the LED is at the junction temperature of 25 deg C. As the junction temperature goes up the voltage across the LED goes down. About 3mv/deg C. As the voltage across the LED goes down the current goes up. More heat - voltage goes down - current goes up and thermal runaway.

This may not happen if things work out just right, but it is a real possibility.

 

[odessit]

you are right, 1 ohm resistor is needed according to LED calculator with 4 LEDs @ 400ma. This resistor will also dissipate only 160 mW of power.

 

[yellow]

I dont think this underpowered setup will drive the led by 400 mA,
but anyways: use a driver

Led in series, SHARK, 13.2 V battery


also simplifies lower output: no disconnection of individual led, but lower current of the whole setup,

 

[odessit]

Yellow, why do you think that 13.2V battery will not drive some LEDs direct? The Voltage is within reason, Amperage is within reason. The battery will not be too overloaded for a big voltage drop...
If anything, a driver will increase a load on the battery to keep the performance constant.

 

[gillestugan]

or get regulated output with a $3.5 driver using this setup:
http://www.candlepowerforums.com/vb/showpost.php?p=2560909&postcount=59
voltage drop to driver is only 0.2V

 

[FATTYTRIATHLETE]

Thanks very much for everyone's help.