# led sign

Discussion in 'The Projects Forum' started by popto, Dec 29, 2009.

1. ### popto Thread Starter Active Member

May 1, 2009
44
0
Hi,
I am making an LED sign and I think that I have everything worked out but since I have never done anything like this I want to make sure. I don't want to solder in 100+ leds then blow them all as soon as I turn them on. I am using an old laptop charger for power whose output is 19.5 volts, 3.34 Amps. I am going to try to make strings of 5 leds in series with a 100 ohm resistor and put lots of those strings in parallel. I got the resistance with this calculation:

R = (V1 - V2) / I

V1 = supply voltage
V2 = LED voltage
I = LED current

R = (19.5 - [5 x 3.5]) / 0.02
R = (19.5 - 17.5) / 0.02
R = 2 / 0.02
R = 100

The amount of current that all those leds will need should be 2.12 amps

A = 0.02 x 106
A = 2.12

That is what I got so in theory this should work but I would like to make sure it will work before I spend hours making my sign.

2. ### mik3 Senior Member

Feb 4, 2008
4,846
69

Apr 5, 2008
19,169
3,836
Hello,

Do you intend to have 106 X 5 = 530 leds on this powersource?
Then the calculations are correct.

Greetings,
Bertus

4. ### popto Thread Starter Active Member

May 1, 2009
44
0
No, there are just 106 leds in the whole project. wouldn't 530 leds be 10.6 amps?

Oh, one other thing. What would happen if later I decided to add 100 more leds to the sign. That should make the total amperage 4.12. this far exceeds the amperage on my power supply. what would happen then? would the leds just get dimmer of would the power supply overheat or what?

Thanks for the quick replies

Apr 5, 2008
19,169
3,836
Hello,

If you have 106 leds, you can divide it by 5.
You will have 106 / 5 = 21 strings of 5 leds and 1 single led.
You need to make 21 strings of 5 leds with the 100 Ohms resistor
And 1 string with 1 led and a resistor of (19.5 - 3.5) / 0.02 = 800 Ohms (820 is the nearest E-series value).

The total current will be 22 X 0.02 = 0.44 A

Greetings,
Bertus

6. ### popto Thread Starter Active Member

May 1, 2009
44
0
oh, I didn't realize that in series the leds would just use up 0.02A per string but it does make sense. I guess that also answers my other question as well, I don't really have to worry about the amperage at this point.

Thank you very much

7. ### SgtWookie Expert

Jul 17, 2007
22,199
1,801
You should measure your power supply under no-load conditions, and also with a load applied. The laptop charger may not be internally regulated.

Also, such power supplies/chargers are not rated for use outdoors. If your sign is going to be used outdoors, you must use a supply in an enclosure that is NEMA rated for the environment of the installation.

8. ### popto Thread Starter Active Member

May 1, 2009
44
0
I will try that. Does it make any difference if the power supply is internally regulated? What does "internally regulated" mean and what will if effect.

I will definitively be using it inside but thanks for the tip.

Thanks

9. ### Wendy Moderator

Mar 24, 2008
21,495
2,961
Regulated vs. unregulated means the if the power supply is unregulated it will change it's voltage (usually a lot) when it has a load. A regulated power supply won't. Generally a unregulated power supply will have the rated voltage written on the case when the current is the same as on the case.

Both exist, the unregulated are much more common and cheaper.

LEDs, 555s, Flashers, and Light Chasers

10. ### popto Thread Starter Active Member

May 1, 2009
44
0
Hi,
I tried the power supply with a load and it didn't change at all, so I guess it is regulated.

Thanks

11. ### Wendy Moderator

Mar 24, 2008
21,495
2,961
Cool, this means you can use simple resistors instead of regulators. The regulator is built into the power supply.

12. ### popto Thread Starter Active Member

May 1, 2009
44
0
I started assembling my sign and after some mistakes I managed to get 7 strings of 5 LEDs soldered in and hooked up. After every string of LEDs I test the whole circuit and when I tested after the 6th string I noticed that after about 4 minutes the circuit started getting fairly warm. The leds themselves didn't seem to get hot, just the underside of the circuit board. So I thought that it may be the solder that I am using. Some of the solder that I have is Radio Shack 0.022 diameter rosin-core solder and I don't have much of it. I live in italy so I cant find any more of it. Since this project takes a lot of solder I have been using some thicker older solder that came with an old soldering iron that I found in the garage. I can tell that it isn't as good as the radio shack stuff because it isn't shiny when it cools.

After I noticed that the circuit was getting hot I decided that it might be the solder and soldered string 7 with the radio shack solder and when I tested the circuit that part didn't seem to get very warm.

I didn't think led were supposed to warm up so I assume that this is a problem. Is it possible that the solder is warming up or do I need to add a larger resistor instead of 100 ohms.

13. ### Wendy Moderator

Mar 24, 2008
21,495
2,961
You probably need to add larger resistors, but leave the ohms the same. Resistors are rated in watts too. What is the voltage across these 100Ω resistors? This will also tell you LED current. The voltage across the resistor can be calculated into current via Ohm's Law (I = V / R), with 100Ω you just move the decimal place.

The wattage of the LED is calculated with this formula:

P = V X I

Generally a resistor should be twice the wattage that is needed.

As a guess you used ¼W resistors, you may need ½W.

You can also use multiple resistors to spread the wattage around. If you have 2 resistors in series or parallel the resistance will change, but they will split the wattage between them.

Using 4 identical resistors like this the resistance will stay the same, but they will handle 4X the wattage.

Did you look at that article I suggested, it goes through all this.

LEDs, 555s, Flashers, and Light Chasers

I wrote it to help beginners. You are interested in chapters 1 and 2.

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14. ### popto Thread Starter Active Member

May 1, 2009
44
0
Hi,

The average voltage across the resistors is 3.3 volts. One of the resistors read 2.4 volts but I noticed that one of the leds was dimmer then the others and the voltage across it was 4.5 or something like that. When I replaced that led the voltage across the resistor for that string read 3.4 volts.

I = V / R
I = 3.3 / 100
I = 0.033

P = V X I
P = 3.3 X 0.033
P = 0.1089

So the resistor wattage should be about 0.2178 watts? The resistors I am using are the little ones and the last color ring is gold. Should I get bigger ones.

15. ### Wendy Moderator

Mar 24, 2008
21,495
2,961
Nope, .25W = ¼W. You're good.

That LED with the high voltage is bad.

Your current is 33ma, this sounds a bit high to me. What is the spec on the LEDs for current? If it is 20ma you need to increase the value of the resistors to adjust.

Does this help?

16. ### popto Thread Starter Active Member

May 1, 2009
44
0
The leds should run at 20 mA, 3.0-3.6 volts. How much will the extra current effect the leds' lifetime? Before starting to assemble the sign I tested one string of 5 leds in series with the 100 ohm resistor and after 3 hours they were just a little warm. Now when they are on the circuit board hooked up in the same way they get warm right away. I soldered one whole section with RS solder and it stays relatively cool. Based on those facts I assume that the old solder must be the problem for heating.

Back to the current. assuming that I lower the Current down to 20 mA:

I = V / R
0.02 = 3.3 / R
R / 0.02 = 3.3
R = 3.3 / 0.02
R = 165

So the resistance that I should use to get 20 mA is 165 ohms? I have never tried to do algebra on the computer before so I may have messed up.

Thanks

17. ### popto Thread Starter Active Member

May 1, 2009
44
0
I almost forgot, Yes it does help! I had done the calculation to get the current for the leds but I hadn't thought twice about it being 33 mA.

Thanks again!

18. ### popto Thread Starter Active Member

May 1, 2009
44
0
I just went over the calculation that I did to get the resistor for my leds and inserted 3.3 (which is what I actually measured across the leds) instead of 3.5 (which is what I assumed the leds whould use) and when I redid the problem the resistance came out 150 ohms. using that resistance:

I = V / R
I = 3.3 / 150
I = 0.022

That is pretty close to the rated current. I guess changing that will help solve the heating problem as well.

Thanks

19. ### Wendy Moderator

Mar 24, 2008
21,495
2,961
Remember, the drop across the resistors is an accurate measurement of current.

If the LEDs are rated for 20ma, then that is what you should aim for. Increasing the current over spec dramatically shortens the lifespan of the LED, and the brightness increase isn't linear, which is to say you're not getting that much more light.