Wiring LEDs with different V & mA?

[bwackv]

Hello,

I'm trying to wire
20 white LEDs 4v @ 50mA
7 UV LEDs 4v @ 100mA
to a 12v wall source

This is for an art project on which I'm working. The 20 white LEDs are on one panel, the 7 UV on the other, I want to wire them all to the 12V source and plug it in to the wall and have both panels light up...

How do I figure out the way to do this? I don't know how to calculate for the different LED types. The LED series parallel array wizard also says about the UV LEDs: the wizard thinks the power dissipated in your resistor is a concern

I am including screenshots of what it says for the separate white and UV wiring.

Should I order lower power LEDs?

Best,

Bwack

 

[Alec_t]

Have you considered the safety aspects of this? UV radiation is damaging to eyes and skin. Safer to just use two different visible wavelengths.

 

[Gryd3]

You can do two things...
You can find a higher voltage power supply and simply add more LEDs into series xD
You can also just buy a 1 watt resistor. It's a concern because it will run very hot, but there is not other option for you right now unless you decide to use 8 UV LEDs. (You can always hide one of them)

The alternative is using an LED driver... but this is probably the best way for you to start out

 

[bwackv]

You can do two things...
You can find a higher voltage power supply and simply add more LEDs into series xD
You can also just buy a 1 watt resistor. It's a concern because it will run very hot, but there is not other option for you right now unless you decide to use 8 UV LEDs. (You can always hide one of them)

The alternative is using an LED driver... but this is probably the best way for you to start out

But how do I calculate what I actually need? Won't the resistor requirements change because what I actually want is 20 + 7 (the wizard calculates separately)? (I can add an 8th by the way, thanks)

This project uses UV sensitive paints that is why I'm using UV.

 

[Gryd3]

But how do I calculate what I actually need? Won't the resistor requirements change because what I actually want is 20 + 7 (the wizard calculates separately)? (I can add an 8th by the way, thanks)

This project uses UV sensitive paints that is why I'm using UV.

Calculation is done as follows:

Take power supply Voltage.
Begin subtractive the Voltage of each led you wish to put in-line with each other to make a string. (All mA ratings of these LEDs should match. Voltage does not have to)
When you can no longer subtract an LED voltage from the supply, you use the left over supply voltage and the mA rating for your LED and calculate a resistor.
That's what the wizard does for you... it them simply connects all of these strings in parallel to light them at once.

There is no simple calculation because they can be hooked up a myriad of ways, and each way has it's benefits and draw-backs.

 

[bwackv]

So I have to use two different sources because the mA are different?

 

[Gryd3]

So I have to use two different sources because the mA are different?

Nossir,
When you put things in a string (series) the mA are the same through all parts.
When you put things in parallel, the mA get's added together...


So a couple strings with 50mA LEDs will pull 100mA total from the supply.
Add a couple more strings of 100mA LEDs, which is 200mA total, and now the supply needs to provide 300mA.

One power supply can do it, you just need to add the mA of each string to find the total and get a supply that can handle it.

 

[bwackv]

I see. So I don't care how it is done as long as the left panel lights the 20 white LEDs and the right panel lights the 7 (or8) UV LEDS using a wall plug 12v, what's the easiest way to do this?

 

[Gryd3]

I see. So I don't care how it is done as long as the left panel lights the 20 white LEDs and the right panel lights the 7 (or8) UV LEDS using a wall plug 12v, what's the easiest way to do this?

Easiest way is using that calculator you shared in the first post.
I would use Solution 1 for the white LEDs. (Not solution 0... simply because the 1Ω resistor does not really do much... and 3x 4V LEDs is equal to the 12V you are using... leave a little extra.
Just build a bunch of strings with 2 LEDs and one resistor each... then attach the + of all the strings together, and the - of all the strings together before attaching to the power supply.
It's not the most efficient, but is the easiest.
Each string can be tested by itself as you go to make sure it's done properly.

You will need a 12V supply to power 10 of the White strings (at 50mA each) and 4 of the UV Strings (at 100mA each)

So get something that puts out more than 1000mA.. although 1200mA to 1500mA is ideal.
(You can go bigger, the LEDs *With the Resistors* will only take what they need... so you could use a 5000mA and be fine. You do not need to adjust your circuit to use different mA rated power supplies. You would have to adjust if you change the voltage of your power supply from 12V to something else though.)

 

[bwackv]

I see. I'm such a noobie sorry for these questions, but does this mean I need more than a 12V power supply?

 

[Gryd3]

I see. I'm such a noobie sorry for these questions, but does this mean I need more than a 12V power supply?

You gotta start somewhere...
No, not at all.
A 12V power supply simply tells you it puts out 12V... but you also need to know how many mA it is capable of reaching...
It may be able to light an XMas light, but it won't be able to start a car!
All power supplies have two specs:
- Voltage Rating which is set, or adjustable.
- Amperage rating in A or mA which is the Maximum the power supply can provide.

So, you need a 12V power supply rated at 1000mA or higher

 

[bwackv]

Oh I see! Ok thank you so much for being so patient with me.

 

[cjdelphi]

An alternative option is to buy a couple 1w LEDs to get the same amount of light

 

[BobK]

Do not use the solutions with 3 LEDs in series. This is garbage, it will do nothing to regulate the current in the LEDs.

In order for a resistor to regulate the current it must drop a couple of volts. The solutions with 3 LEDs in series have 4 volts on each of 3 LEDs giving 12V leaving nothing for the resistor to drop. If the voltage is a little high, they could carry enough current to destroy themselves. Whoever wrote that wizard does not know what he is doing.

Bob

 

[Tha fios agaibh]

So, you need a 12V power supply rated at 1000mA or higher

I've got a dumb question:
An AC wall wart that's rated 12v 500ma will produce 12v when it's fully loaded and about 15% higher voltage at no load.

Would it then be advised to not use a much larger supply because it won't be loaded enough to render the correct voltage?
I know if circuit voltage is critical, it's wise to use a regulator, but would you ever size (load) a standard wall wart to find its voltage sweet spot?

 

[Gryd3]

I've got a dumb question:
An AC wall wart that's rated 12v 500ma will produce 12v when it's fully loaded and about 15% higher voltage at no load.

Would it then be advised to not use a much larger supply because it won't be loaded enough to render the correct voltage?
I know if circuit voltage is critical, it's wise to use a regulator, but would you ever size (load) a standard wall wart to find its voltage sweet spot?

Not a dumb question and certainly right on the nose!

Wall-wart AC-DC power supplies are often not 'regulated', and the voltage printed is what is provided under load. This would also be a much worse problem if using Solution 0 which was already advised against... Luckily though, switching power supply would not cause this kind of problem.
Total current draw is expected at 400mA for the UV, and 500mA for the whites. 900mA in total. I would still advise a 1200mA supply, although going way overboard with a 'transformer' style wall-wart with no regulator could cause problems.

If in doubt, measure it with a multi-meter!

Thank you 'The Fios Agaibh'