Resistors for parallel LED circuit

Jade5051

That guy...
I originally posted this in the advanced ATS thread because it's for my ATS but it's actually a different subject matter so here we go...

Need some help with LEDs...I need a specific sized resistor to make sure each circuit path is getting the appropriate voltage/wattage etc. There will be 6 paths and each path will contain 3x3w LEDs. I want to run them in parallel with resistors on each path similar to this...(again 6 PATHS 3x3w LEDs on each path).

786f6db71d5673f00024a9096d370147.jpg


Here is the run down...

- 24V 2.5A 60W MeanWell driver.
- 18x3w LEDs (Reverse Voltage: 5.0 V, DC Forward Voltage: Typical: 2.2V Max: 2.4V, DC Forward Current: 750mA)

If I put this 3w Watt 330 Ohm Metal Film Resistor on each path will I be fine?

http://www.ebay.com/itm/3w-Watt-330-.../191793626972?

If not, can you please tell me what resistor I need or please explain the math so I can figure it out cause it's been to long since I had to make sense of ohm's law.
 
[moved]

You'll be better off here in the DIY forum. These guys should be able to take care of your question.
 
Sorry, but it's not as simple as you think. Balancing current is fairly involved and there's a reason it's not done very often. You'd be way ahead selecting a more appropriate driver setup. Look into the LDD series with a sufficiently large power supply
 
Avoid paralleling at all cost. Really, just use two drivers. If you must parallel, reduce the number of parallel segments.

At a stated Vf max of 2.4V (this seems very very low, are these red LEDs?), you can run about 9 LEDs in series. You have 18LEDs, so thats two channels. This is assuming your Meanwell driver is a constant current driver and can be limited to 1.5A.

(If you wanted 6 parallel channels at 750mA, you'd be SOL anyway, since 6*0.75 is 4.5A, more than your driver can accommodate)


If you must parallel, ideally you will need to know the I/V curves of your LED with the Vf variance (how much extra current will they conduct with a tiny voltage increase). The best application guide I've seen is this: http://www.ti.com/lit/an/slva325/slva325.pdf - If you however do not know the I/V curves, we can do a rough voltage drop calculation:


If we do an approximate calculation with the two-series approach, there is 21.6V max Vf, from a 24V supply, we have 2.4V potential over the resistor. Using the average Vf of 2.3V*9=20.7V, with a potential of 3.3V. A basic solving of ohms law for 750mA at 3.3V gives you 4.4ohms. Target something less than that, say 3 ohms, where each resistor needs to be clipped to a heatsink since its dissipating 2.5 watts entirely as heat.
 
I hope I understand the problem correctly =P

The DC forward voltage is the voltage drop across each LED. You wan't the excess voltage to be 'absorbed' by the resistor, so that the sum of the voltage drops through each loop of your parallel circuit is 0 (Kirchoffs law). You're starting with 24 VDC through each loop of the parallel sequence. You have a 2.2-2.4 VDC drop across each LED. The leftover voltage in each loop is ~22 VDC. If they are operating at 750 mA, you can figure out the resistance required in each resistor via ohms law (V=I*R). This comes out to roughly 30 Ohms.
 
I hope I understand the problem correctly =P

The DC forward voltage is the voltage drop across each LED. You wan't the excess voltage to be 'absorbed' by the resistor, so that the sum of the voltage drops through each loop of your parallel circuit is 0 (Kirchoffs law). You're starting with 24 VDC through each loop of the parallel sequence. You have a 2.2-2.4 VDC drop across each LED. The leftover voltage in each loop is ~22 VDC. If they are operating at 750 mA, you can figure out the resistance required in each resistor via ohms law (V=I*R). This comes out to roughly 30 Ohms.

Thats also 16.5Watts of space heater for each loop :)
 
As many have stated, normally constant voltage w/ current limiting resistors is not recommended.
not disagreeing here at all..
but for the sake of discussion this may help a bit:
http://led.linear1.org/led.wiz


Now somewhere, someone said this calc is "inaccurate".
wouldn't mind someone explaining to me why...

http://led.linear1.org/led.wiz

3=3W led in series is a poor choice for a 24V ps..
Coming close to matching the ps voltage eliminates most of the waste heat..

At 24v 9x2 w/ 2.2ohm 2W resistor..
together, all resistors dissipate 2475 mW
together, the diodes dissipate 33750 mW
total power dissipated by the array is 36225 mW

3x6 array at 12V ps:

each 6.8 ohm resistor dissipates 3825 mW
the wizard thinks the power dissipated in your resistors is a concern Help
together, all resistors dissipate 22950 mW
together, the diodes dissipate 33750 mW
total power dissipated by the array is 56700 mW
the array draws current of 4500 mA from the source.

23w heater...
bottom line is you are taking efficient emitters and making them inefficient, besides the problem of load sharing..
 
If parallel is a must a current mirror is better than trying to balance things with resistors, fusing, etc. still wastes a bit of heat but works very well, all that's really needed is three transistors of appropriate specs for your LED strings. (one for each string)

I've done quite a few for plant lights and some shop lighting, but for most things just using a single driver per string of LEDs is way more efficient and preferable. There are a couple threads on the designs of the circuit, and many variations that will get the job done.
 
Current mirror is a much safer idea (even though just as inefficient, still linear regulation)

sort of, it is quite a bit more efficient, just adds voltage drop to strings as needed to keep them roughly equal (self adjusting), the closer they are the less voltage drop, I measured only a watt or so on each transistor in my last setup running three strings of 15 at ~400mA per string. I should add that it is best to use a constant current supply but you could get the same effect with a voltage supply just way less efficent (I guess in that case it would be not much more efficient as fixed resistors, just more stable in the long run)

fixed resistors are really not an option when running more than 100mA or so. too much heat, compounded by the higher the voltage on the strings is.
 
Personally, I think you guys are all overthinking this.

When I built my LED fixture, I did a whole bunch of parallel strings and I never had anything complicated like a current mirror involved. While I did run an in-line fuse on each string, all I really did was check that the current going to each of the two strings was within a few tens of milliamps of each other. I did this by measuring the voltage drop across a 1ohm 5W resistor. Because the resistor is 1ohm, the voltage drop equals the current passing through the entire string. No calculations involved and because the circuit stays intact you see the TRUE current, not the current with the meter involved.

OP, I would run two strings of 9 or 10 LEDs. In each string, I would incorporate a quick acting fuse rated below the rating of your LEDs and a 1ohm 5W resistor in series. Use the resistor to measure the current in each string and if its an issue, measure the forwarding voltage on each LED at low current and change the higher one for the lower one in the opposite string. Keep swapping LEDs until the difference in current is less than ~50mA at the current you plan on driving them at. I used 50% of max current when I did mine.

Good luck,
 
Personally, I think you guys are all overthinking this.

When I built my LED fixture, I did a whole bunch of parallel strings and I never had anything complicated like a current mirror involved. While I did run an in-line fuse on each string, all I really did was check that the current going to each of the two strings was within a few tens of milliamps of each other. I did this by measuring the voltage drop across a 1ohm 5W resistor. Because the resistor is 1ohm, the voltage drop equals the current passing through the entire string. No calculations involved and because the circuit stays intact you see the TRUE current, not the current with the meter involved.

OP, I would run two strings of 9 or 10 LEDs. In each string, I would incorporate a quick acting fuse rated below the rating of your LEDs and a 1ohm 5W resistor in series. Use the resistor to measure the current in each string and if its an issue, measure the forwarding voltage on each LED at low current and change the higher one for the lower one in the opposite string. Keep swapping LEDs until the difference in current is less than ~50mA at the current you plan on driving them at. I used 50% of max current when I did mine.

Good luck,

Swapping LEDs to balance things is the PITA avoided with a simple current mirror.......or course its a game of do you want to spend the time measuring and swapping or just use a mirror to keep things in check......I'm lazy in that regard and happy to sacrifice a few watts for less messing around during setup, but yes, can be done more than one way! the beauty of DIY! and for what it's worth the majority, if not all the old LED builds I've followed using the fused parrelel strings are still running fine as far as I know, with high power LEDs the little differences are not as big a deal as they used to be 5 or 10 years ago when we all started this game........
 
A cheap 12 volt power supply & and a couple of constant current drivers would allow you to create two series circuits. For an ATS, the drivers don't even have to be dimmable. Look @ rapidled.com. I think you can buy a driver that will run 8-9 3w LEDs for $26. Two of those would solve your problem. A very small 12 VDC power supply will run these drivers. Rapidled, Radio Shack, and many places on the web have them for cheap.
 
Personally, I think you guys are all overthinking this.

When I built my LED fixture, I did a whole bunch of parallel strings and I never had anything complicated like a current mirror involved. While I did run an in-line fuse on each string, all I really did was check that the current going to each of the two strings was within a few tens of milliamps of each other. I did this by measuring the voltage drop across a 1ohm 5W resistor. Because the resistor is 1ohm, the voltage drop equals the current passing through the entire string. No calculations involved and because the circuit stays intact you see the TRUE current, not the current with the meter involved.

OP, I would run two strings of 9 or 10 LEDs. In each string, I would incorporate a quick acting fuse rated below the rating of your LEDs and a 1ohm 5W resistor in series. Use the resistor to measure the current in each string and if its an issue, measure the forwarding voltage on each LED at low current and change the higher one for the lower one in the opposite string. Keep swapping LEDs until the difference in current is less than ~50mA at the current you plan on driving them at. I used 50% of max current when I did mine.

Good luck,

I balanced my strings similarly. I omitted the fuse as I'm under-driving my diodes. This doesn't have to be complicated.
 
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