# Driver/ resistor in 4 led series?



## Whiplash87 (Dec 20, 2014)

So I am a total noob in the sense of high power leds. Ive wired 5mm 6mm and smd leds without issues using resisitors but the math and logic of high power leds is a bit confusing. To my understanding a driver maintains a smooth constant voltage to the leds to maintain performance and a resistor limits voltage to prevent overpowering and burning up the led? Please correct me if im wrong. 

My project is wiring four white cree xpg leds in series. Number on the led star says 3211 on it. I have a variable volt power supply at 1200ma with 3/4.5/6/7.5/9 and 12v output. I have The leds soldered with 20 gauge wire and set at 12v. They are lighting but the power supply is getting hot so im assuming something is wrong and I have stopped using it till I ensure it is safe. I cut 1.5/1.5" little aluminum heat sinks and secured them with arctic silver thermal paste. Im using these as terrarium lighting. Any tips/pointers or stickies on info on how to safely wire these up would be greatly appreciated.


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## Anders Hoveland (Dec 20, 2014)

Whiplash87 said:


> To my understanding a driver maintains a smooth constant voltage to the leds to maintain performance and a resistor limits voltage to prevent overpowering and burning up the led?


Just the reverse, actually. A driver maintains a smooth constant current, adjusting voltage to try to keep a constant current (mA) through the circuit. A resistor limits current, but the amount of current able to flow through a resistor is dependent on the voltage (or actually the _voltage drop_ taken up by the resistor, after subtracting the voltage taken up by the LEDs).

Normally a driver is used to power LEDs. A resistor _can _be used, but this is much more problematic, for numerous reasons.



Whiplash87 said:


> Any tips/pointers or stickies on info on how to safely wire these up would be greatly appreciated.


The driver typically has a voltage range on it, for example 4-12v. The total voltage taken up by the LEDs in series should be within the voltage range of the driver. Sometimes the driver will just say 24v, without explicitly indicating a lower limit, but if you just power 1 or 2 LEDs, they will likely burn out. If the voltage from the driver cannot drop below the rated voltage for each LED, the current (mA) flowing through the circuit will be too great. A typical LED takes up 3 volts, more or less, so 4 LEDs in series will take up 12 volts. But as long as you stay within the voltage range, you can use any number of LEDs with the constant current driver. That's why it is called "constant current"– the voltage may be different depending on the number of LEDs you put in the circuit, but the current will always remain the same. 




Whiplash87 said:


> My project is wiring four white cree xpg leds in series. Number on the led star says 3211 on it. I have a variable volt power supply at 1200ma with 3/4.5/6/7.5/9 and 12v output. I have The leds soldered with 20 gauge wire and set at 12v.


1200mA seems like a high level of current. What is the current rating of your LEDs ?
Unless each LED is designed for at least 3.8 watts, you are using too much current.


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## Whiplash87 (Dec 20, 2014)

The Leds are Cree XP-G R5 7.5K acording to the order sheet I was emailed. These were bought like 4/5 years ago This is the data sheet:

http://www.cree.com/~/media/Files/C...d Modules/XLamp/Data and Binning/XLampXPG.pdf
Sorry this was the easiest way to post the data


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## Epsilon (Dec 21, 2014)

Anders Hoveland said:


> 1200mA seems like a high level of current. What is the current rating of your LEDs ?
> Unless each LED is designed for at least 3.8 watts, you are using too much current.



1200mA is fine, they are rated to 1500mA and when mounted directly to copper, can be run at 2500mA+.
But, the leds will probably need more than 12v when connected in series at 1200mA.

I do not work for Taskled, but he makes top notch drivers.

You can go a few ways:
_Linear driver_
Such as http://www.taskled.com/lflex.shtml

You can set the current at certain setpoint and the output will max at that current (and have certain modes) as long as the voltage is high enough. Voltage IN needs to be close to Voltage OUT (12v and 11.5v for example).

_Buck driver_
Voltage IN needs to above Voltage out (around 2v). The power IN needs to sufficient to not overload your power supply.

What you can do is use 12v IN and 2 series 2 parallel connected leds. Add a small (0.3Ohm, 1W) resistor in series of each led string to autoballance the two strings.

_Boost driver_
Voltage IN needs to under Voltage out (around 2v). The power IN needs to sufficient to not overload your power supply.
This option will probably not be feasable, since your supplied current is 1.2A max and you will need more at 9v to supply sufficient power to driver.


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## Whiplash87 (Dec 21, 2014)

Thank you both for the assistance. With a linear driver would I need a higher voltage? How do I ensure that voltage in is close to voltage out? I clearly need to educate myself on the math involved with these projects


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## Epsilon (Dec 21, 2014)

Measure the voltage of the voltage IN. If this is 12.0v steady then maybe you do not even need a driver but a small 1W resistor of around 0.3Ohm. But if it isn't like power adapters which has 4.5-6-9-12v selectors on it, they are around 16v unloaded @ 12v setting, you will need a driver.

4 leds in series will need around 3.3v each at 1.5A.


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## Whiplash87 (Dec 21, 2014)

Thank you so much!


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## SemiMan (Dec 22, 2014)

That 0.3R resistor will do almost nothing. Forward resistance of the LEDs in series is way more than that.


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## AnAppleSnail (Dec 22, 2014)

Heck, I have a 4-XM-L lamp run directly with a 1.8A 12V wall-wart plug over my basil right now. I mounted the LEDs to a CPU Heatsink, wired in series. The fan is connected across 3 LEDs in series, giving it about 9V to reduce noise & flow. I put a metal shroud around it, and it'll run nicely this way near ambient temperature... Until my fan fails.

No driver, no resistor, just a 12v power supply unable to deliver over 30W.


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## Whiplash87 (Dec 22, 2014)

Hmmmm wonder why my power supply is getting hot then.


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## SemiMan (Dec 22, 2014)

Because you are running it flat out continuously and it is inefficient and gets quite hot.

LED drivers can hit 80-90c continuously if needed.


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## Whiplash87 (Dec 22, 2014)

So a driver is the way to go, resister wont be as effective?


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## Epsilon (Dec 22, 2014)

SemiMan said:


> That 0.3R resistor will do almost nothing. Forward resistance of the LEDs in series is way more than that.


Leds do not have a fixed resistance, that's the problem. When one string has thermal runaway, it will take the other string with it.

The 0.3Ohm resistor is only to balance the strings, not to drop the voltage (which is futile with buck drivers). The idea is that when one string is getting more current, more voltage will drop over the resistor, stopping the current.


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## SemiMan (Dec 22, 2014)

There is one string .... Nothing to balance.

If there was more than one, than some arbitrary 0.3R pulled out of thin air is not going to help balancing much. Vf difference between strings could be much larger and the effective forward resistance of the led string discussed at the discussed drive current is already about 1.2R.

Semiman


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## Whiplash87 (Dec 22, 2014)

http://support.luxeonstar.com/custo...ng-driver-to-regulate-current-to-luxeon-leds-

This site had some great little articles on why a resistor is not practical. Helped spell it out a bit more for me. Thank you all again. I will be purchasing an appropriate driver. Will 1000ma get close to max brightness for these leds rated at 1500ma max? Or should i aim closer to 1200-1400ma?


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## Whiplash87 (Dec 22, 2014)

In this application is a linear or buck driver preferred with 12v 1500ma power supply leds are 4.9w max, 1500ma max drive, forward voltage 2.9


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## SemiMan (Dec 22, 2014)

A very good buck is always "better" with rare exception.

Sent from my SM-T320 using Tapatalk 4


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## SemiMan (Dec 22, 2014)

Whiplash87 said:


> http://support.luxeonstar.com/custo...ng-driver-to-regulate-current-to-luxeon-leds-
> 
> This site had some great little articles on why a resistor is not practical. Helped spell it out a bit more for me. Thank you all again. I will be purchasing an appropriate driver. Will 1000ma get close to max brightness for these leds rated at 1500ma max? Or should i aim closer to 1200-1400ma?



Stick to 1000. They will last longer.

Sent from my SM-T320 using Tapatalk 4


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## Epsilon (Dec 22, 2014)

SemiMan said:


> There is one string .... Nothing to balance.
> 
> If there was more than one, than some arbitrary 0.3R pulled out of thin air is not going to help balancing much. Vf difference between strings could be much larger and the effective forward resistance of the led string discussed at the discussed drive current is already about 1.2R.
> 
> Semiman


I'm not going to argue about it with you, I had two posts, one with two strings and one with only 1.

0.3Ohm is just an example but is in the ballpark for ballacing, which is only to protect the strings for large differences in current, I have done the math before but will do it again: 1A difference (one string 0.7A and other 1.7A for 2.4A total) gets you 0.3v + the extra Vf of the higher current. That is plenty to ballance the two strings. You can not put a led string at "1.2R", that is just not right, it varies and drops with higher current.

At the 1 string it is a safety, which is mainly to stop the current from rising when total Vf is very close to 12v. But that said, in this case the 0.3Ohm is just an example and very dependant on the exact conditions. In his case, he can probably run direct drive.

But that is with a ballance power supply, which he probably has not. In that case the voltage is around 16v unloaded, which for a brief instant on turn-on could send a large current through the string and kill one or more LEDS. If he powers it down at the mains side of the power supply and keeps the 12v side connected, he might be alright.


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## SemiMan (Dec 22, 2014)

Epsilon said:


> I'm not going to argue about it with you, I had two posts, one with two strings and one with only 1.
> 
> 0.3Ohm is just an example but is in the ballpark for ballacing, which is only to protect the strings for large differences in current, I have done the math before but will do it again: 1A difference (one string 0.7A and other 1.7A for 2.4A total) gets you 0.3v + the extra Vf of the higher current. That is plenty to ballance the two strings. You can not put a led string at "1.2R", that is just not right, it varies and drops with higher current.
> 
> At the 1 string it is a safety, which is mainly to stop the current from rising when total Vf is very close to 12v. But that said, in this case the 0.3Ohm is just an example and very dependant on the exact conditions. In his case, he can probably run direct drive.



Actually you are arguing with me and this thread is about a 1amp power supply and one string. Either way the 0.3R is from thin air again since the LEDs are already 1.2R. That will go 1 string or 2. On one string it may drop the current 15-20%.

I can most certainly say it is about 1.2R as that is the approximate dV/dI at 1A and it does not vary much from say 700mA to 1.3A.

Your math example is too simplistic to be of much value in the real world .....which is where I work day in and day out.

Sent from my SM-T320 using Tapatalk 4


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## Epsilon (Dec 22, 2014)

SemiMan said:


> Actually you are arguing with me


Correct, but didn't mean to


> and this thread is about a 1amp power supply and one string.


No this thread was about finding a suitable setup for his 4 leds and 12v power supply. Which has more solution than one string in series, of which I also proposed a 2P2S setup.


> Either way the 0.3R is from thin air again since the LEDs are already 1.2R.
> That will go 1 string or 2. On one string it may drop the current 15-20%.
> I can most certainly say it is about 1.2R as that is the approximate dV/dI at 1A and it does not vary much from say 700mA to 1.3A.


You got me confused there by setting the 1.2R for the LEDS. Because it is 1.2R + Voffset then, which has a completly different behavior.


> Your math example is too simplistic to be of much value in the real world


No it is not, but getting a full plotted graph here was a bit much don't you think? The calculation I put there is exactly what the resistor does: Compensating slight differences between Vf's due to different samples and temperatures. It is there to catch the big current differences which will kill the LED.


> .....which is where I work day in and day out.


Please do not post this negativity, post factual things like the calculation of 1.2R you posted, this view I like. Not the lowering the esteem of someones post. If someone's wrong, back it up with facts or explanations why and leave it there.


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## Whiplash87 (Dec 22, 2014)

So math: 
Input voltage calc:
Vt + (Vf X LEDn) = Vin
2 + (2.9V X 4) = Vin
~13.6= Vin

# of LEDs
(Vin - Vt) / Vf = #LEDs
(13.6V - 2V) / 2.9 = #LEDs
11.6 / 2.9 = #LEDs
4 = #LEDs
Is my math sound? 
I need at min 13.6v Dc @ 1000MA for a buck driver?


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## SemiMan (Dec 22, 2014)

Vf(1.25)-Vf(0.75) = 0.15V approx. R=v/I = 0.15/0.5 = 0.3. 4 in series , R = 1.2 ohms.


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## SemiMan (Dec 22, 2014)

Not sure what your Vt is in your equations.


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## SemiMan (Dec 22, 2014)

And no it does not compensate nearly as well as you indicate as the forward voltage acts much more like Vdiode + Rdiode linearized over a region versus your calculation which is purely Vdiode.

A small resistor has a small effect as there is already a large one.


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## SemiMan (Dec 22, 2014)

2p2s with a linear regulator?


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## Whiplash87 (Dec 22, 2014)

Sorry total voltage for dc buck driver: 
http://support.luxeonstar.com/custo...n-series-with-a-buckpuck-or-powerpuck-driver-

How would I us you formula and the 1.2ohms towards determining a driver?


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## SemiMan (Dec 22, 2014)

Oh and if you are going to start with "I am not going to argue" and then argue ... Which is a dismissive and immature way to start a post, then don't be surprised if I lose patience with your posts which are misleading to others.

If someone calls me out on inaccuracy I usually research the topic to either confirm I know what I am talking about or that I don't. I don't blindly argue when I don't even understand what the technical information the other poster is posting (which was pretty basic).

Sorry admins had to be said.


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## SemiMan (Dec 22, 2014)

Got it whiplash did not know you were assuming a buckpuck. 

The Vf of the LEDs could easily be 3.5v at 1A so 16v min would be best and a little more would not hurt.

If this is AC you would be better off with a constant current LED driver, likely one rated for 1.05A with about a 9-18V output range. You could even get a dimmable one.


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## Epsilon (Dec 22, 2014)

SemiMan said:


> 2p2s with a linear regulator?


Possible in different circumstances, but wasn't suggested in this case:

edit: Ah you read the suggestion 


Epsilon said:


> _Buck driver_
> Voltage IN needs to above Voltage out (around 2v). The power IN needs to sufficient to not overload your power supply.
> 
> What you can do is use 12v IN and 2 series 2 parallel connected leds. Add a small (0.3Ohm, 1W) resistor in series of each led string to autoballance the two strings.





SemiMan said:


> And no it does not compensate nearly as well as you indicate as the forward voltage acts much more like Vdiode + Rdiode linearized over a region versus your calculation which is purely Vdiode.
> 
> A small resistor has a small effect as there is already a large one.


The effect of 20% (which still means 1.2A vs 0.8A) which you depicted is more than enough to make the setup stable. The goal is not to compensate the Vf differences in the LEDS which are there initially. It is to stop the uncontrollable drop of Vf due to heat and thermal runaway. The equilibrium will lie much more to a 50/50 current split than without the resistor. In which case it will end more in the area of 80-20 instead of 60-40.


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## SemiMan (Dec 22, 2014)

No it will not as the current change is much less. Your math is still wrong as you refuse to include the LED resistance.


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