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07-14-2011, 03:26 PM #1
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chipdouglas
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Math help...Running 2 LD's in series

Wazzzzz UP!!!!
so basically i am monkey see, monkey do. I can figure out now resistor value for the ddl circuit, only after years of headache lol. Now i am on to a new endeavor...

I don't necessarily just want the answer. But help getting the answer. Search is not my friend.

i've decided on running 2 LPC 826 diodes in my projector. i will be running them in series. from my understanding it is safer.

I need help figuring the resistor value and resistor placement.

What i have:
2 LPC 826 diodes
1 Flexmod P3 (already set to 550ma.)
Power supply is 12vdc.

How do i calculate which resistor to use. and do i need one for each LD? also does the resisitor go on the + pin or - pin?

thank you much!!!

Michael.

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Last edited by chipdouglas; 07-14-2011 at 03:28 PM.

07-14-2011, 03:34 PM #2
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123splat
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Re: Math help...Running 2 LD's in series

Why would you need a resistor on the LD's if you are wiring them in SERIES??

I can see using a balance resistor to even out the current in a parallel branch, but the current in a series branch, such as this, willbe limited by the supply (the deiver).

What purpose would the resistor(s) serve?

07-14-2011, 03:36 PM #3
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chipdouglas
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Re: Math help...Running 2 LD's in series

I don't know lol... i'm pretty sure i have read that if you run ld's in series, you need a resistor. i could be wrong.

isn't this in series? http://laserpointerforums.com/f65/8-...ild-64296.html

michael.
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Last edited by chipdouglas; 07-14-2011 at 03:46 PM.

07-14-2011, 03:48 PM #4
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Re: Math help...Running 2 LD's in series

I think it's in parallel that they are needed to balance the current load in the parallel branches. (from days of working with LEDs), if you put two LEDs in parallel, you can run into a situation where one LED will draw more current than the other (particularly if they are different colors). Then one will be brighter than the other. The other problem (potentially) is that if one LED flakes out on you, then the other is in danger of current overload (super nova), then you loose both LEDs. I THINK the same concerns apply for LDs. Is NOT an issue in series brances. If you loose one, in series, the other goes out from lack of power, not overload damage.

I don't think a resistor in series would hurt, other than dropping the current, but I cann't see where it would help either...

Just my 2 cents, good luck.

07-14-2011, 03:55 PM #5
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Re: Math help...Running 2 LD's in series

dV = IR
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07-14-2011, 03:57 PM #6
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Re: Math help...Running 2 LD's in series

^^^thats greek to me =(
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07-14-2011, 03:59 PM #7
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Re: Math help...Running 2 LD's in series

^^ wha??? what's that supposed to mean? dV would either be equal to dI*r or I*dR... and how does that apply to this discussion??

Sorry mike, you jumped in before me.

Last edited by 123splat; 07-14-2011 at 04:00 PM.

07-14-2011, 04:53 PM #8
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Re: Math help...Running 2 LD's in series

You don't need a resistor for series. Only parallel configurations require a resistor to balance the current. The problem with series is that reds are case negative, so the heat sinks need to be electrically isolated from each other. Otherwise you short past the second diode.
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07-14-2011, 05:31 PM #9
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Re: Math help...Running 2 LD's in series

Would i still keep my driver set at 550ma? also, once the heat sinks are isolated, do i run my driver output leads to LD 1. then connect LD 2 + to LD 1 + and LD2 - to LD1 - ?

thank you
michael
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07-14-2011, 06:06 PM #10
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Re: Math help...Running 2 LD's in series

Quote:
 Originally Posted by chipdouglas Would i still keep my driver set at 550ma? also, once the heat sinks are isolated, do i run my driver output leads to LD 1. then connect LD 2 + to LD 1 + and LD2 - to LD1 - ? thank you michael
Er, Uh,,, I think that would be "in-Parallel".

Series would be Driver + to LD1 +, LD1 - to LD2 +, LD2 - to Driver -.
*** NO NO NO ***If you set the driver to 550mA, both LD's will have to share (split) that current, so, each diode might get about 260mA, give or take...****NO NO NO**** DISREGARD PREVIOUS, MY BAD, CYPARAGON IS RIGHT

Is this a test, or are you serious?
If you are serious, maybe you are really asking how to calculate the value of the balancing resistor(s) in a parallel circuit (one driver, two LD's with LD1+ tied to LD2+ and to driver + out, LD1- tied to LD2 - and driver - out).

Last edited by 123splat; 07-14-2011 at 06:37 PM.

07-14-2011, 06:09 PM #11
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Re: Math help...Running 2 LD's in series

In series they share the voltage but the current stays the same. 550mA is probably too high in a projector unless you want to be replacing diodes every few weeks. Keep in mind you'll also need a supply voltage of 8V or more.
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Last edited by Cyparagon; 07-14-2011 at 06:11 PM.

07-14-2011, 06:27 PM #12
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Re: Math help...Running 2 LD's in series

123splat, you got it right the opposite, DON'T follow that, you will risk killing the diodes!

Basically:

When run in series, (battery --> driver --> diode1 --> diode2 --> battery)
Current (I) is the same: e.g. set the driver as if you were running only one diode
Forward voltage on diodes gets summed up, e.g. you will need to supply more: Vf * 2 + Vd = what voltage you need, Vd means driver dropout. For two red diodes in series and a LM317, that will mean: 4V * 2 + 3V = 11V

When run in parallel, (battery --> driver ---> diode1 ---> battery)
.................................................. ......\-> diode2 -/
Current gets divided: e.g. set the driver to double the current, 1100mA if one diode takes 550mA
Voltage is the same as if you were running one diode (Vf + Vd) For two red diodes in parallel and a LM317 driver, that would mean 4V + 3V = 7V

This is done considering both LD's are the same!

Think of your electric circuit as a water pipe. Water flow represents electron flow, voltage represents the tension/pressure, current represents, well, current. When in series, the current remains the same and tension drops with load. When in parallel, all loads get the same tension and current gets divided between them.

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Last edited by Johnyz; 07-14-2011 at 08:26 PM.

07-14-2011, 07:33 PM #13
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Re: Math help...Running 2 LD's in series

Quote:
 Originally Posted by 123splat Er, Uh,,, I think that would be "in-Parallel". Series would be Driver + to LD1 +, LD1 - to LD2 +, LD2 - to Driver -. *** NO NO NO ***If you set the driver to 550mA, both LD's will have to share (split) that current, so, each diode might get about 260mA, give or take...****NO NO NO**** DISREGARD PREVIOUS, MY BAD, CYPARAGON IS RIGHT Is this a test, or are you serious? If you are serious, maybe you are really asking how to calculate the value of the balancing resistor(s) in a parallel circuit (one driver, two LD's with LD1+ tied to LD2+ and to driver + out, LD1- tied to LD2 - and driver - out).

i assure you it is not a test lol... i really have a hard time with this. but once i get it, i got it. but i'm able to get it by doing. so i have to ask a million little questions to put it all together. sorry.

michael.
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07-14-2011, 07:36 PM #14
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Re: Math help...Running 2 LD's in series

Maybe you missed part... but this is a projector build. There is no battery, and no lm317. There will be a SMPS and a flexmod.

Quote:
 Originally Posted by Johnyz This is done considering both LD's are the same!
And they seldom are. That's why balancing resistors are used - to make the IV curve more steep.
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07-14-2011, 07:37 PM #15
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Re: Math help...Running 2 LD's in series

tonight i am hopefully combining my 2 reds with my ar coated pbs. so if all goes smoothely i will post some pix in the multi media section.
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07-14-2011, 08:21 PM #16
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Re: Math help...Running 2 LD's in series

I used LM317 as a sample dropout value, with a Flexmod, the drop will be 1.5V according to user manual. Of course, even diodes from the same batch are slightly different, I assumed he knew about balancing resistors, I just wrote that in in case somebody in future will find this and would try to ram a LOC and a 405 together, for example.
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Last edited by Johnyz; 07-14-2011 at 08:24 PM.

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