Custom LED light and driver

Discussion in 'The Projects Forum' started by goody1928, Apr 19, 2014.

1. goody1928 Thread Starter New Member

Apr 19, 2014
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I am building a off road light. My input voltage is between 12.7 and 14.4 depending onI'm going to use 6 t6 led in 2 parallel 3 series fashion for an Needed output voltage at 9.9v and I need an current of 2.8 amps. The circuit I'm researching is on instructables here http://m.instructables.com/id/Circuits-for-using-High-Power-LED-s/?ALLSTEPS on page 5 shows the diagram oI was going to use a 3 watt .18 ohm resistor to give me a 2.8 amp output. On the following page the circuit included a zener diode and I would like that one because its even more efficient. Does everyone think this circuit will work or any ideas to change it for the better. I will have a lot of heatsink

2. Alec_t AAC Fanatic!

Sep 17, 2013
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Can't identify 'page 5' or any other page. Which particular circuit do you have in mind?

3. AnalogKid Distinguished Member

Aug 1, 2013
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The circuit is a classic, and will work as drawn.

The zener does not improve efficiency. Technically it decreases it, but not enough to matter, or barely measure. It is there solely to protect the MOSFET gate in case your V+ exceeds 20 V for some reason.

You're running out of headroom, but I think you'll be ok. Each of your 3-LED strings will need a ballast resistor to even out small differences in the forward voltage of each LED, so that one string doesn't "starve" the other. You can use the same 0.18 ohm resistor as in the regulator.

Your design current is 3.33 A, not 2.8 A. 0.6 / 0.18 = ?

Change R to 0.22 ohm.
Calc voltage drop at 2.8 A.
Calc voltage drop at 1.4 A (current in one LED string with a 0,22 ohm ballast resistor.

The sum of these two voltages plus 9.9 V for the LEDs is your absolute minimum source voltage requirement. Anything above this is the headroom your regulator has to work with, and is the voltage used to calculate the heat dissipated in the MOSFET.

ak

4. goody1928 Thread Starter New Member

Apr 19, 2014
27
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Sorry for my slow reply I had plans easter and was unable to check. Thank you for the reply and I know how to solder and understand most circuitry its just been a while. My LEDs each require 2.8a for maximum efficiency per lumen so I was going to run 2 sets of 3 in series. Do I need 2 circuits or can this circuit produce 5.6 amps at 9.9v with enough efficiency to cool with a small heatsink. I can use a massive heatsink as well just trying to keep this under \$100 bucks is all

Also in step 6 of my link source 1 there is no diode at all is that just as efficient or do you recommend the diode?

Last edited: Apr 21, 2014
5. goody1928 Thread Starter New Member

Apr 19, 2014
27
0
What size heatsink Would you recommend for heat dissipation also should I upgrade the lm317 to a 338 for the beefier internals to withstand more heat?

6. goody1928 Thread Starter New Member

Apr 19, 2014
27
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Sorry o linked the whole all steps on one page thing. Step #6 is what I was talking about and source 2 vs source 3... I can build this circuit no problem I just need the hand designing it is all and I can find all the formula from searching myself. I don't expect a lesson from anyone either but if you give me one I will happily read it. I'd you only feel like saying use part a and part b and solder I'll gladly give it a shot

7. ronv AAC Fanatic!

Nov 12, 2008
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You can do it either way. I would be tempted to build 2 of them. Usually people use .6 volts not .5 volts as the voltage to begin turning on the transistor in your circuit so if you built 2 you could use a .25 ohm resistor (3 watt) which may be easier to find. With one each the FET will dissipate about 12 watts when the battery is charging so you should look for a heat sink with a temperature rise of 4 to 6 C per watt. 7 to 10 Sq. inches of surface area.

8. goody1928 Thread Starter New Member

Apr 19, 2014
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So how do you change the ref voltage to 6 instead of 5. I'll be buying everything at digikey

9. ronv AAC Fanatic!

Nov 12, 2008
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No need to change anything. The base emitter voltage is more like ,6 volts than the .5 volts in the instructables.

10. goody1928 Thread Starter New Member

Apr 19, 2014
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http://www.digikey.com/schemeit/#rtn

here is my scheme it set up im going to order parts pending you agree with my set up

R1 - 100k ohm 1/4 watt CFR-25JB-52-100K
R3 - .22 ohm 3 watt PNP300JR-73-0R22
Q2 - LM350T
Q1 - 2N5088BU
3x LED - 3.3v @ 2.8A

if i switch the R3 to a .12ohm 3W like this PNP300JR-73-0R12 then can i run 6 leds 3 in series with 3 more connected parallel with the same circuit? or even 9 with a .75 ohm 5 watt resistor in R3 like this MPR5JB75L0

Last edited: Apr 23, 2014
11. goody1928 Thread Starter New Member

Apr 19, 2014
27
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also just to clarify this voltage regulator will push a constant 3.3 V per LED correct basically only use what the led requires by automatlically burning up the unused voltage inside of Q2?

Apr 19, 2014
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I meant .075

Apr 19, 2014
27
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14. Bernard AAC Fanatic!

Aug 7, 2008
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Subject to review, heat sink seems adequate, but do not put it inside of a box; maybe box to a side of heat sink.

15. goody1928 Thread Starter New Member

Apr 19, 2014
27
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Using my circuit what will the constant voltage be with an input of 14.4v what determines this voltage?

16. goody1928 Thread Starter New Member

Apr 19, 2014
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bump to see if i can find an answer

17. Bernard AAC Fanatic!

Aug 7, 2008
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It might help if you posted the schematic that you plan to use, & name the LED's. They look like 9W LED's from your specs. Order list only shows 3 LEDs & you call for 6. If you are planning for 6 LED's, you do not list ballast resistors, or are you planning on 2 seperate ckts?
For your question: Voltage across 3 LED's will be about 9.9 V, depending on differences in LEDs & temperature, any leftover V is dropped by Q2 & R3.

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18. goody1928 Thread Starter New Member

Apr 19, 2014
27
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i can remake the circuit no problem my question is can i add aditionall LEDS in parallel to the original 3 that are wred in series so here is a circuit

http://www.digikey.com/schemeit/#sf2

19. wayneh Expert

Sep 9, 2010
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You need a resistor in series with each parallel leg. Otherwise one leg can hog all the current. Depending on your voltage supply, you may want these resistors anyway to reduce the amount of heat dissipated in the MOSFET.

Oh, and you'll need the right value and power rating of R3, to adjust for double current.

20. Bernard AAC Fanatic!

Aug 7, 2008
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393
Your schematic confuses me, LM350t is a voltage or current regulator, & not a FET. Vdd must be negative or LED's are backwards. Q2 good for 3A, just bairly sufficient for one string, can be paralleled but why not make a reg for each string.