Mixing different leds in series

methias · Apr 12, 2014

MrAl said:
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.

OK
Lets see if I have this.
My Constant current driver provides 350mA (50mA of which I want to shunt off) and 12 Watts
The LED series string measures a voltage of 29.5 VDC when powered.
The LED's have a forward amperage of 300mA
That means I need a resister ohm value of (29.5 / .05 = 590 ohms) 590 (or 600 for round numbers ?)
To get the wattage it must be able to take I calculate;
(V^2 = 29.5 x 29.5) = 870
and so P = 870 / 590 = 1.47….. or 1.5 watts (rounded)
To be safe means a 590 (or 600 rounded?) ohm 3 watt resister (with air circulation).
If I can locate a 600 ohm resister 3 (or higher?) Watt and plug that in parallel I should be able to end up with 300 mA's running through the LED string and 50 mA through the resister. right?
As my constant current driver is a 12 watt driver and I am trying to run 12 x 1 Watt LED's will the resister reduce the wattage to the LED's?
If so then (and I think I've got it) I should reduce the number of LED's to compensate for the watt loss (-3 LED's for the 3 Watts eaten by the resister)? or did I take this one step too far?

Damn, I wish I had stayed and finished at Devry 35 years ago…

methias · Apr 12, 2014

KeepItSimpleStupid said:
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.

So 3, 1800 OHMS 1 WATT resisters in parallel is equal to having 1 600 ohm 3 Watt resister?
Damn, this is like being back in school but even old farts can learn.
This project seemed so simple when I started but I will not give up.
Either I am going to fry the whole lot of parts or get it right.
Any bets which happens first. My lady is betting on fry.

KeepItSimpleStupid · Apr 12, 2014

So 3, 1800 OHMS 1 WATT resisters in parallel is equal to having 1 600 ohm 3 Watt resister?

Yep for the most part.

To be safe means a 590 (or 600 rounded?) ohm 3 watt resister (with air circulation).

Depends on what's available. There are different sets of resistor values depending on tolerance. 1%, 5%, 10%, 20%
No set rule. You can use the closest value and sometimes, the next lowest is better.

If I can locate a 600 ohm resister 3 (or higher?) Watt and plug that in parallel I should be able to end up with 300 mA's running through the LED string and 50 mA through the resister. right?

In theory.

As my constant current driver is a 12 watt driver and I am trying to run 12 x 1 Watt LED's will the resister reduce the wattage to the LED's?

But that's OK. You also have to watch the definition of Wattage. You would be operating them at 300 mA which is correct.

If so then (and I think I've got it) I should reduce the number of LED's to compensate for the watt loss (-3 LED's for the 3 Watts eaten by the resister)? or did I take this one step too far?

Your surrounded by "limits" and "ranges". You bought a 350 mA driver and you need 300 mA. Oops!
The limit is the "compliance voltage" which means how many LED's can be placed in series.

What we did not cover is tolerances. In a good design, one would have enough slop, so that if all of the LEDs were shifted to the higher Vf of their range, the circuit would still work.

MrAl · Apr 13, 2014

methias said:
OK
Lets see if I have this.
My Constant current driver provides 350mA (50mA of which I want to shunt off) and 12 Watts
The LED series string measures a voltage of 29.5 VDC when powered.
The LED's have a forward amperage of 300mA
That means I need a resister ohm value of (29.5 / .05 = 590 ohms) 590 (or 600 for round numbers ?)
To get the wattage it must be able to take I calculate;
(V^2 = 29.5 x 29.5) = 870
and so P = 870 / 590 = 1.47….. or 1.5 watts (rounded)
To be safe means a 590 (or 600 rounded?) ohm 3 watt resister (with air circulation).
If I can locate a 600 ohm resister 3 (or higher?) Watt and plug that in parallel I should be able to end up with 300 mA's running through the LED string and 50 mA through the resister. right?
As my constant current driver is a 12 watt driver and I am trying to run 12 x 1 Watt LED's will the resister reduce the wattage to the LED's?
If so then (and I think I've got it) I should reduce the number of LED's to compensate for the watt loss (-3 LED's for the 3 Watts eaten by the resister)? or did I take this one step too far?

Damn, I wish I had stayed and finished at Devry 35 years ago…

Hi again,

Yes that is pretty much correct there, very good

Now we might want to get just a little more careful in that we dont mind if we drive the LED's just a little under 300ma. The lower we go on the current for the LED's the longer their life is, and also to mention a finer point the voltage across the set will drop slightly when the resistor is connected, although it should only be a little bit.

So say it drops down to 28 volts. 28/0.050=560 Ohms. Then 28^2/560=1.4 watts. Double that is 3 watts again.
They do make 5 watt resistors, and also 10 watt resistors are very common and not super expensive.
If you want to use three 1 watt resistors then yes use three times the resistance and 3 in parallel, or 1/3 the resistance and three in series.
If you want to use 1/2 watt resistors then 6 in parallel, each resistance would be 6 times 560 Ohms.
If you want to use 1/2 watt resistors in series then 6 in series each resistance one-sixth of 560 which is about 92 Ohms.

As i said though 10 watt resistors are very common and not too expensive and that would not get too hot. The higher the power rating the better here.

methias · Apr 13, 2014

Allright
Since my CC driver can provide up to 36 VDC the LED's will take what they need and the resister will have what it wants and everything should be happy, including me.

In the future I intend to pay much better attention to all components when buying. I turned a simple project into something not so simple.

I know I'm being redundant but Thank you all !

KeepItSimpleStupid · Apr 13, 2014

And Hey, you learned a few things in the process.

Making mistakes is one way to learn.

Sometimes, you can find the time to do it over, but not find the time to do it right. This time, you got lucky.

methias · Apr 13, 2014

KeepItSimpleStupid said:
And Hey, you learned a few things in the process.

Making mistakes is one way to learn.

Sometimes, you can find the time to do it over, but not find the time to do it right. This time, you got lucky.

I got lucky finding this forum, and you guys.

MrAl · Apr 14, 2014

methias said:
Allright
Since my CC driver can provide up to 36 VDC the LED's will take what they need and the resister will have what it wants and everything should be happy, including me.

In the future I intend to pay much better attention to all components when buying. I turned a simple project into something not so simple.

I know I'm being redundant but Thank you all !

Hi,

You are welcome

I also thought that it would be nice if you could find some LED's that could take 50 or 60ma and then you could wire them in parallel (with series resistor) and then you'd have those LED's lit up too. Three strings of 20ma LED's would do it too, as that would steal 60ma from the main string.
Just a thought.

KeepItSimpleStupid · Apr 14, 2014

Nice idea and you would only need 4 or so. 4 LED's, each with a resistor in series drawing 20 ish mA. As a bonus, you get useful light and not heat.

ronv · Apr 16, 2014

Or, If they are more than bright enough for you, you could put 2 strings (3 of each color) in parallel with a small resistor - say 3.3 ohms in series with each string. Down side is they would be about 1/2 as bright. Upside is the resistor could be low wattage.

Mixing different leds in series

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