need help selecting the proper sized resistor for my L.E.D. light array

[alf51]

What size resister (OHM & WATTS) and what type of configuration (Number of total diodes, type of connections, number in series) do I use with 4.2amp 24v DC output with 2.4v to 2.8v LED if 700 ma. I need the value for my resisters in ohms and watts.

I believe you divide the output 4.2 by the current limit .700 to get a 6 across for ballast and a series of 7 or 8 strings but I'm not sure and I will be purchasing my resister from All electrics which list their resisters in terms of ohms and watts
Could you give me the answer to these questions and explain how you arrived to the answers in terms that a none tech person like myself can understand Thank You for your help

 

[Boncuk]

Hi alf51,

hope that answers your questions.

Boncuk

 

[alf51]

This really helps

Thank You so much for your help. So if I'm reading it right I will be using 48 diodes in my array 6 parallel and 8 strings with a 1.8 ohm 2 watt resister at the head. I have another problem to solve if you don't mind helping me with. Its the same situation but using a combined multiple diode voltage of the 2.8v and 3.8v at a 3 to 1 ratio powered by a much larger 300 watt 24 amp 12v. I know I will need many diodes for this project but I would feel more confident in tackling it if I could get the same help with the resister and configuration. Thanks again for your time I'm glad there is someone out there willing to help.
Alf51

 

[MikeMl]

Whenever you parallel series strings of LEDs, each string needs to have its own current limiting resistor.

 

[alf51]

Boncuk
I think I get most of it but I'm still confused on how you came up with the .666A result I'm a little slow on understanding.

 

[Boncuk]

Boncuk
I think I get most of it but I'm still confused on how you came up with the .666A result I'm a little slow on understanding.

The voltage the resistor has to drop is 1.2V. With a resistor of 1.8Ω the current is calculated according to Ohm's law: I(A)=U(V)/R(Ω).

Putting in the numeric values: I=1.1988V/1.8Ω --> I=0.6666~A

Boncuk

 

[Boncuk]

Hi alf51,

I do not suggest to use LEDs with different forward voltages in a string. They will not be of equal brightness - even if they have identical forward currents.

You can mix LEDs using 2.8V LEDs and 3.8V LEDs in separate strings calculating the current limiting resistor as shown.

Boncuk

 

[alf51]

Thanks Boncuk

 

[alf51]

If I have a 1.5 amp 12v power supply what type of resistor would I need to allow only 1.4 amps and 10.8v to pass, dissipating .1 amp and 1.2v

 

[cowana]

12-10.8 = 1.2v

1.2v / 1.4A = 0.86 ohms

1.2*1.4 = 1.7W of heat generated

 

[alf51]

Boncuk, I have one last question on the 6X8 array you configured for me. Why did you choose a 2w resistor instead of a 1w which as far as I can figure at a P=.7984w the nearest would be 1.8 ohms 1w which are available at digikey

 

[Diver300]

Resistors get extremely hot when running at their rating. It never hurts to get a larger size as it will run much cooler.

A 2 W resistor running at 0.8 W will only just burn your finger......

 

[alf51]

You show on your calculations a 1.2v drop. I come up with a 1.6 drop. Could you please double check your figures and tell me where I'm going wrong. I also have a question on selecting 2.8V at your indicator. The rating was given as a veritable between 2.4 and 2.8 I have noticed others using the median of 2.6 as there multiple factor which is correct?

 

[Diver300]

I make it 1.6 V as well.

You should really allow for anywhere in the range 2.4 - 2.8 V. If you do that, it would be better to have fewer LEDs in each string, so that there is much more voltage drop, and the current is better defined.

However you should really have a proper current control circuit for LEDs that large.