Mixing different leds in series

[methias]

Hi all.
I am a newby when it comes to LED arrays and rather than destroy what I purchased I am asking for some advice on this project.
I am trying to make a light bar with different color led's on the same string.
I have a Mean well APC 12-350 Constant current driver that gives 9-36V DC @350mA 12W **broken link removed** .

I am using 1W High power Epistar chips (red & Blue). 6 red led's FV = min 1.8V tip 2.4V @ 300mA and 6 blue FV = min 3V tip 3.2V @ 300mA that I got from Ebay http://vi.vipr.ebaydesc.com/ws/eBay...Trk=1&lsite=0&ittenable=false&domain=ebay.com

I would like to run them in one series if possible.

I found a led array wizard at **broken link removed** and averaged the two different Typical forward voltages (2.4 + 3.2) /2 =2.8V .

Am I on the right track?

The diode forward current (if I am reading the ebay data correctly) is 300mA. http://vi.vipr.ebaydesc.com/ws/eBay...Trk=1&lsite=0&ittenable=false&domain=ebay.com .
I entered the following data:
35 Source voltage **broken link removed**
2.8 diode forward voltage **broken link removed**
300 diode forward current (mA) **broken link removed**
12 number of LEDs in your array



This is it's solution:
Solution 0: 12 x 1 array uses 12 LEDs exactly

+35V
R = 4.7 ohms
The wizard says: In solution 0:

• each 4.7 ohm resistor dissipates 423 mW
• the wizard thinks 1W resistors are needed for your application **broken link removed**
• together, all resistors dissipate 423 mW
• together, the diodes dissipate 10080 mW
• total power dissipated by the array is 10503 mW
• the array draws current of 300 mA from the source.

I assume that the LED driver will adjust the voltage to 35 V DC and the above solution is telling me that the circuit will draw only 300mA and just over 10.5 watts.



Is this correct or am I over simplifying things?

Thanks in advance for any and all assistance offered..

 

[ronsimpson]

Your LED driver will output 350mA if possible. It needs the stack of LEDs to add up to (9 to 36 volts). The voltage is not important in this case.
You don't need any resistors because you are using a constant current power supply. They are only used with a constant voltage supply.

Remember LED are a current device not a voltage device. You have the right power supply.

 

[4pyros]

Your LEDs are 300ma and your supply is 350ma so it will not work.

 

[methias]

Your LEDs are 300ma and your supply is 350ma so it will not work.

The listing FV says "if =300mA" . I took that to mean the typical voltage would be x.x if the amperage is 300mA.
Am I off base with that concept?
Will too much amperage fry the led's or shorten their lifespan?
If so any ideas on how to reduce the amperage to acceptable levils or do I need a different power supply?
Again thank you for your input.
I just don't grok all of this.

 

[ronsimpson]

Is there a control where you can turn down the supply? Set it to a lower current.
At 300mA the LEDs are going to get hot with out a heat sink.

New idea:
If the voltage is 30V and you places a 300 ohm resistor across the stack of LEDs then 100mA will go to the resistor and 250 will go to the LEDs.
That is 3W in the resistor. Just a idea! 600 ohm for 50mA and 1.5Watts

 

[KeepItSimpleStupid]

What the power supply spec says is that if you add up all of your Vf''s and their + % tolerances and it > 9V or < 36 V the current delivered will be 350 mA. No extra series resistors are needed. When you add a series resistor, you add another voltage drop: e.g. 0.350*Radded.

Your LED's have a range of operation. Lifetime is basically defined as a specific decrease in brightness.

To reduce the current delivered to your LED's you have to add a PARALLEL path. That parallel path would have to consume 50 mA.

Since it isn't critical, you can use the computed voltage drop, say it's 10 V and use a resistor of R=10/0.050 ann put it in parallel. The Power required for the resistor is (I^2)*R.

It's not 100% ideal, but it should work, otherwise you need a constant current sink of -50 mA.

 

[methias]

Is there a control where you can turn down the supply? Set it to a lower current.
At 300mA the LEDs are going to get hot with out a heat sink.

New idea:
If the voltage is 30V and you places a 300 ohm resistor across the stack of LEDs then 100mA will go to the resistor and 250 will go to the LEDs.
That is 3W in the resistor. Just a idea! 600 ohm for 50mA and 1.5Watts

The supply is sealed so no control.
I have two aluminum u channels/heat sinks that I am using to mount the led's to.
12" x 3/4" wide and 3/4" tall.
I plan to stagger 6 on each channel / heat sink.
There will be about 1/8" between each bar.
The whole array will be suspended with free airflow all around.

 

[KeepItSimpleStupid]

The voltage can be anywhere from 9 to 30 V. That's called the compliance voltage. Your string will have a voltage. You can measure it briefly and use it to size the resistor or just estimate it from Vf of all of the LEDs. Using 30 V will underestimate the size of the resistor. 30/0.05; actual/ 0.05 and 9/0.050. The resistor required will fall somewhere between that range. There will be temperature effects, but they should not be a problem, You could use 30/0.05; measure V and then re-compute.

Fixed some typos. 0.5 to 0.05.

 

[Misterbenn]

I agree with KeepItSimple

 

[methias]

The voltage can be anywhere from 9 to 30 V. That's called the compliance voltage. Your string will have a voltage. You can measure it briefly and use it to size the resistor or just estimate it from Vf of all of the LEDs. Using 30 V will underestimate the size of the resistor. 30/0.05; actual/ 0.5 and 9/0.050. The resistor required will fall somewhere between that range. There will be temperature effects, but they should not be a problem, You could use 30/0.5; measure V and then re-compute.

Many thanks to Keepitsimple, Misterben, 4Pyros, & Ronsimpson for your input. I think I now have a handle on this.
I will post an update once I get it all assembeled and (hopefully) working.

 

[methias]

Alright.. Just finished assembling the first one . I even got the reflectors on (used a hot glue gun). Turns out my soldering is much better than I remember from 35 years ago. For temp I made the connection w/ wire nuts (gonna fix that before final) and plugged it in. Those 1W led's are brighter than I expected. Yes I know when I add the parallel resister things will dimm a bit. Actual volts DC is 29.5 / 0.05 = 590 ohm resister that I will PU tomorrow. Ronsimpson you were off by only 10 ohm's. I'm impressed. Thanks again.
My strip is kinda ghetto but 12W LED strip complete for < $30.00 each (I got enough for 4 strips) not too bad.
I know that for many of you guys this is a simple project but I'm jazzed that is works.
Many thanks all.
A shrine has been built (from pocket protectors) in your names to remember this day.

 

[MrAl]

Hi,

When connecting LED's in series you add up all the voltages to get the total voltage. But the current will be the same in every LED.

If you have a constant current driver then the current will be the same with several LED's even if they all have a different individual voltage.

If that constant current is too high for the LED's then you just add a parallel string. That means making two strings instead of one, and then the current splits between the two strings so each gets less current. If the current is then too low though you may want to think about using one string and a parallel resistor to take some of the current away from the LED string. That will reduce current to the LED's and thus protect them.

 

[methias]

Hi,

When connecting LED's in series you add up all the voltages to get the total voltage. But the current will be the same in every LED.

If you have a constant current driver then the current will be the same with several LED's even if they all have a different individual voltage.

If that constant current is too high for the LED's then you just add a parallel string. That means making two strings instead of one, and then the current splits between the two strings so each gets less current. If the current is then too low though you may want to think about using one string and a parallel resistor to take some of the current away from the LED string. That will reduce current to the LED's and thus protect them.

Yes that was one of my initial conserns ^. after plenty of help from this forum I decided to go with the recommendation to put a resister in parallel to take out the extra current (50mA) and keep things in line.
People here have been very good to me and I do appreciate everything everyone did to help me get this right. As I said ^ all I need now is a 590 ohm resister to wire in parallel and I'm done.

 

[methias]

OK, I'm missing something.
I ran down to radio shack and picked up a 570 ohm resister 1/4W (where I believe my issue is) and soldered it in (parallel). The damn thing got quite hot and I unplugged it at 165 degrees F.
I did some searching and that seems to point to getting a resister with twice the watt capabilities (12W supply) or 25W +/- and I found lots of 25W and 3 to 8 ohms or one at 750 ohms.
Something tells me that the best solution is just get a different power supply @ 300mA for this assembly and then get some more 350mA LEDs to match this supply.
So much for me being all jazzed about my success….
any thoughts???

 

[methias]

If I got say 3 25W 200 ohm resisters is that about the same as a 25W 600 ohm resister?
Or is it 12 1W 600 ohm that is the same as a 12W 600 ohm?

 

[4pyros]

In a word no;
Three 200 ohm 25 watt resisters in series equals 600 ohms with a 25 watt capacity. Both the ohms and the watts add.
12, 600 ohm 1 watt resisters can not add up to 600 ohms now matter how you hook them up.

 

[KeepItSimpleStupid]

You need at least a 2 Watt resistor. from I^2*R. 0.050*0.050*600

Now you can take two 1200 ohm 1W resistors and place them in parallel and get a 600 ohm 2 W resistor.

The formula for parallel resistors is 1/Rt=1/R1+1/R2+....1/Rn; The wattages would add.

Two resistors of the same value is equivalent to the resistance divided by 2. eg. 8 ohms in parallel with 8 ohms is 8/2 or 4 ohms. if the resistances are UNEQUAL, you have to use the more general formula above.

Yep, a 1/4 Watt resistor would get quite hot. It's like 1/6 to 1/8 the power dissipation required.

A 300 ohm 1/4 Watt in SERIES with a 300 ohm 1/2 Watt would be equivalent to a 600 ohm 1/4 Watt resistor. Resistors in series add. The one with the lowest wattage in the string would act like a fuse.

 

[MrAl]

Hi,

As others have mentioned you need a resistor that can handle the power. I meant to say something about this but i had to do something yesterday and could not get back soon enough. Story about this nutty adventure at a later time

Anyway, a resistor has a Ohmic value like 100 Ohms, but it also has another 'rating' or rather specification, and that is the power handling capability. The resistor must be able to handle the power developed in it due to the voltage across it or the current through it.

Knowing the voltage across it we get the power from:
P=V^2/R

where V is the voltage in volts and R is the resistance in Ohms.

Knowing the current through the resistor we get the power from:
P=I^2*R

where I is the current through it in Amperes and R is the resistance in Ohms.

So you can use either of these, and it is quite easy to do.

For example, say you have 10 LEDs in series each 3.2 volts. That's ten times 3.2 equals 32 volts. With a resistor of 100 Ohms, the power would be:
P=32^2/100=1024/100=10.24 watts.

To get a more safe value for the power we would double this, and that would mean we'd need a 20 watt resistor at 100 Ohms.

Doubling the power rating means we get a resistor that runs cooler, but going even higher like 4 times the rating means an even cooler running resistor. Of course this assumes that the resistor gets fresh air through vent holes in the case of the unit it is installed inside of. If there is no fresh air supply then other means have to be employed such as heatsinking to the side panel or something like that.

 

[KeepItSimpleStupid]

I think you post is a "bit confusing" because you picked numbers out of the air. The LED SERIES resistor is chosen from what's left over from 32 V, so if the supply was 42 V, you would get (42-32) or 10 V for V. For V^2/R; you would get 100/100 or 1 Watt. Yea, I used your resistor.

The resistor limits current, not power of the string.

The OP's problem was completely different. A Shunt path (PARALLEL RESISTOR) which, sort of, looks like what you were trying to do.

 

[MrAl]

Hi,

The last post he posted talked about the parallel resistor, so i did too. And yes, i gave a simple example where the voltage across the LED's was 32 volts but that's just an example. If a resistor is placed across a 32 volt supply for example then the power is P=32^2/R, and that's exactly what i gave.

If we want to talk about a series resistor that is entirely different of course, but i assumed that was worked out already or non existent actually because i thought he had a constant current regulator not a voltage regulator. It sounded like that so i went with it. I think he does have a constant current source because "350ma" was mentioned and he actually only needed "300ma" so we had to shunt 50ma. That's where the parallel resistor comes in. Since he hooked one up and it got too hot, i talked about the parallel resistor and it's required power rating