Noob LED power question...

[dustjunkie]

If I wanted to power two LEDs at 350mA which of the two options would be better, or does it really make a difference?

1) LEDs wired in parellel, and a 700mA driver

2) LEDs wired in series, and a 350mA driver

My understanding is that in the first method I could use a lower voltage, but the amperage goes up, so battery life is reduced.

That sound right? Is there a down side to doing this?

Thanks!

dj

 

[Marduke]

Battery life is identical in both cases.

3.7v x 700mA = 2.59W
7.4 x 350mA = 2.59W

 

[LukeA]

If I wanted to power two LEDs at 350mA which of the two options would be better, or does it really make a difference?

1) LEDs wired in parellel, and a 700mA driver

2) LEDs wired in series, and a 350mA driver

My understanding is that in the first method I could use a lower voltage, but the amperage goes up, so battery life is reduced.

That sound right? Is there a down side to doing this?

Thanks!

dj

In parallel, the LEDs will invariably end up taking different currents, negatively affecting LED life and output. Series avoids those problems.

 

[gillestugan]

you may use a boost driver if you want to use battery voltage lower than output voltage. Some AA batteries have problems delivering high currents like 3A, but at 350mA you will be fine. t

 

[dustjunkie]

so, lets assume the following:

I am running the LEDs in parallel
Two LEDs at 350mA
Vf - 3.5V per LED
Battery - up to three 18650 3.7V 2400mAh

Assumptions (correct as needed...):

Boost pucks/drivers are to big/expensive
I will need to run at least two batteries in series (7.4V) so that my input voltage is sufficiently higher than my Vf.

Questions:

Is there any value to adding the third battery in series (11.1 input voltage)?
Will the potential difference in Vf really make a difference in brightness/LED life? 50,000 hrs is a LOT... 1/5 of that is a lot...
Could I run a second set of batteries in parallel for added battery life?
If the two batteries were at different charge states would the battery with the higher charge "charge" the other battery?
If so, is either of those last two a problem?

Thanks!

dj

 

[yellow]

see it that way:
1 led equals 1 Li-Ion cell (or 3 Ni-Mhs),
so to run two led in direct drive, You need 2 cells in series ...
also You need an additional driver --> add one cell to compensate

--> 2 led in series + driver --> 3 cells in series

(led always in series, to have the same current at each.
series or parallel makes no difference for runtime)

PS: right, Pucks are HUGE.
better get drivers from Sandwich Shoppe or Taskled (or the very cheaper ones from Dealextreme, Kaidomain, ...)


[edit] ... didnt pay too much attention to Your questions ...
2 led in parallel (why?, thats not good)
2 or 3 Li-Ion cells (again in parallel?)
if the cells are used in series, 7.2 V (2 cells in series) is a "standard" value, as were 4 cells in series. There are chargers and batteries available.
A higher input voltage than 7.2 V does not have any impact at all for brightness.
right: if the cells are put together in parallel and the charging state is different, the "full" cell does charge the "empty" one. imho that would be a short, the protection would kick in, adding problems ...

advise:
the led in parallel with 2 (or 3, or 4) cells in series + buck driver, or better
the 2 led in series with 3 (better 4) cells in series + buck driver
... depending on Your charging equipment.
[/edit]

 

[dustjunkie]

[edit] ... didnt pay too much attention to Your questions ...
2 led in parallel (why?, thats not good)
[/edit]

So, I probably should have said this first....

The whole point of this is that I want to run 3 LEDs (Vf 3.3 to 3.7) and use a power supply of less than 12 volts. Also, I don't want to use a boost puck/driver.

dj

 

[gillestugan]

yes, you should have..

If you want to make it as cheap as possible you can use a AMC7135 based driver for about $4 from dealextreme or kaidomain.
Use 3 li-ions in series. Connect one led to the driver output and the other 2 leds i series with the driver. Works great.

Else you can use a resistor, but this is less efficient.

 

[dustjunkie]

yes, you should have..

If you want to make it as cheap as possible you can use a AMC7135 based driver for about $4 from dealextreme or kaidomain.
Use 3 li-ions in series. Connect one led to the driver output and the other 2 leds i series with the driver. Works great.

Else you can use a resistor, but this is less efficient.

I think I know what you mean, but I don't understand how this works. Would you mind helping a dummy out?

dj

 

[gillestugan]

You can find a scheme in this thread. Just use a driver with one 7135 chip instead of eight and you'll have your 350mA. You can buy a 700mA och 1000mA driver and simply desolder the "extra chips" so you only have one left. They can be bought with 1,2,5 or 20 modes for $3-$5 at kaidomain or dealextreme.

 

[dustjunkie]

You can find a scheme in this thread. Just use a driver with one 7135 chip instead of eight and you'll have your 350mA. You can buy a 700mA och 1000mA driver and simply desolder the "extra chips" so you only have one left. They can be bought with 1,2,5 or 20 modes for $3-$5 at kaidomain or dealextreme.

Makes sense, thank you! Can I assume that I could do the same thing with a buck driver?

dj

 

[gillestugan]

No, it only works with AMC7135 based drivers as they are behaving like variable resistors.