# Sign with 175 LEDs

Discussion in 'LEDs and Optoelectronics' started by kevbo423, Jan 22, 2013.

1. ### kevbo423

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Jan 22, 2013
Thank you so much guys. Kris and Steve you were both such an excellent help and I cannot thank you enough. Just a few simple questions. Will I have to worry about heat with the sign? Also, do laptop power supplies operate with DC or AC current?

2. ### KrisBlueNZSadly passed away in 2015

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Nov 28, 2011
DC.

Heat probably won't be a problem. But you can calculate the amount of power dissipated by each component in the sign from the formula P = V I where P is power in watts, V is the voltage across the component, and I is the current through the component, in amps.

So for example if your LED string is running at 20 mA and the current limiting resistor has 3V across it, the power dissipated in that resistor will be 3 * 0.02 which is 0.06 watts or 60 mW (milliwatts).

A typical small component without a heatsink has a thermal resistance of around 100 degrees Celsius per watt, which means that for every watt that it dissipates, its temperature will rise by 100 degrees Celsius above ambient. In this example, 60 mW will produce about 6 degrees Celsius heating, so if ambient temperature is 25 deg. C, the component will run at about 31 deg. C which is not a problem.

Heating due to power dissipation can start to become a problem if the LED string current is a lot more than 20 mA and/or when there is a lot of voltage dropped across the current limiting resistor (because there are few LEDs in the string). In any case you can use that formula to calculate the dissipation and estimate the temperature rise.

Adding heatsinking to a component will reduce the thermal resistance to ambient, so that for each watt dissipated, the temperature will rise by less than 100 deg. C, for example, 30 deg. C or less, depending on the amount of heatsinking.

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Jan 22, 2013
4. ### KrisBlueNZSadly passed away in 2015

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Break one of the wires to the power supply (by convention, the positive one) and connect the switch across the break.

5. ### kevbo423

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Jan 22, 2013
So the schematic would look like this? Where the wire coming off goes to the middle prong on the switch.

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6. ### kevbo423

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Jan 22, 2013
Ok. I just went through and completely planned out this circuit.

Supplies:
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59 - Blue 5mm LEDs (20-30mA) (3-3.2V)
5 - Green 5mm LEDs (20-30mA) (3-3.2V)
63 - White 5mm LEDs (20-30mA) (3-3.2V)
48 - Red 5mm LEDs (20-30mA) (2-2.2V)
9 - 1/2W 330 Ohm Resistors
25 - 1/4W 120 Ohm Resistors
1 - (Unknown) 240 Ohm Resistor
1 - Switch
1 - 19V Laptop Power Supply
1 - Laptop Power Supply Jack

Going through everything I found that I will have:
--------------------------------------------------------------------
9 - Series of Red LEDs
11 - Series of Blue LEDs
1 - Series of Green LEDs
11 - Series of White LEDs
1 - Series of four White and one Blue LEDs
1 - Series of three Blue and two White LEDs
1 - Series of three Red and two White LEDs
*All series contain five LEDs

For all the calculations that I did, I assumed that I will be using a 19V power supply, and running all the LEDs at 25mA (the middle that they are all rated for) and using the highest rated forward voltage.

I used the online LED calculator to find out the resistors that I needed for each of the solid series of LEDs.

------------------------------------
Red LEDs need 1/2W 330 ohm Resistors
Blue, Green, and White LEDs need 1/4W 120 ohm Resistors

For all the series of LEDs that are not made up of one single color, I did the calculations by hand, using the information that you provided me.

------------------------------------
(W)(W)(W)(W)(B) - needs a 1/4W 120 ohm Resistor
(B)(B)(B)(W)(W) - needs a 1/4W 120 ohm Resistor
(R)(R)(R)(W)(W) - needs a (Unknown) 240 ohm Resistor

A few questions that I have:
------------------------------------
1) For the series made up of different colored LEDs; does it matter what order I put them in?
2) For the resistor that I calculated for the three red and two white LED series; how do I figure out what Watt value I should use?
3) How would you recommend wiring each series to the power supply?
4) What will be the best way to prevent a short circuit? Should I use something such as shrink tubing, or is there a better way?
5) Do you know a website that I can buy these components for cheap? I am currently planning to buy my LEDs from Joe Knows Electronics here
6) With a three prong switch, how exactly do I wire that up in my circuit?
7) When calculating what resistors I would need, I used the highest rated forward voltage for the calculations. Should I have done that? Or should I have picked something in the middle? For example, with a forward voltage rated between 3-3.2V, what value of forward voltage should I use?

Thank you for all your help everyone. Without it, I would not have been able to do this project.

Last edited: Jan 25, 2013
7. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

25,301
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Jan 21, 2010
No
I^2xR -- you know I = 0.025 and that R is the resistance. 1/2W should be fine, in fact you can probably use 1/2W for all of them.

Light gauge hookup wire. Some red and some black would be ideal.

You can use anything that keeps bare wires from touching. Heatshrink is one. If your layout is such that they are not close enough to touch (even if subject to a bump or a knock) then you're probably OK.

Unless they're particularly expensive, freight will probably swamp any difference.

You could look on eBay for LEDs. However I'd suggest you reduce the current to a more normal 20mA.

Generally you use the middle connector and one other one.

I probably would have used something in the middle.

Actually, I would have looked at the datasheet, or measured the voltage at the current I wanted to use, but a middle value is probably close enough for government work.

8. ### KrisBlueNZSadly passed away in 2015

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Nov 28, 2011
All excellent answers from Steve, as usual.

I'm a bit confused about the switch. In your schematic in the earlier post, you described it as an "LED rocker switch". Do you mean that it has an LED built in? If so, it probably has two connections for the illumination LED and two or three connections for the switch.

Whether it has an LED or not, the switch part will have two or three connections and they will probably be sturdier than the connections for the LED, if present. As Steve said, you use the middle one and one of the end ones. If the switch isn't marked with ON and OFF or 0 and 1 positions, it doesn't matter which end connection you use. If it does, set the switch to the ON position and use the switch that has continuity to the middle pin when measured with a continuity tester or ohmmeter.

9. ### kevbo423

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Jan 22, 2013
The switch I want to use has an LED built it. It can be found at this link: http://www.amazon.com/SPST-Automoti...=1359134555&sr=8-8&keywords=led+rocker+switch

It has the description: "SPST Automotive Round Rocker Switch with Blue LED 12 V" It has three prongs on the bottom.

Also, the size of my circuit is going to 9in x 8in. If I use Steve's suggestion, how do I keep the LEDs mounted in place? Could I use hot glue or something along those lines?

And what watt value resistors should I get. The online calculator that I used said that 1/4W would be fine for the 120 ohm resistors. And I should use 1/2W for the 330 ohm resistors. What does the watt value have to do with resistors and should I just get 1/2W resistors for all of them? Thanks guys.

Last edited: Jan 25, 2013
10. ### KrisBlueNZSadly passed away in 2015

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Nov 28, 2011
OK that switch is different from what I thought it would be. It has three terminals; two of them are the switch itself, and the third one is for the LED.

The page you linked doesn't give enough detail to be sure about this, but I assume the silver-coloured pins are the switch, and the gold-coloured pin comes from the LED cathode, and should be connected to 0V (the negative side of the laptop power supply).

The switch probably has a built-in current limiting resistor for its own LED, but the value of this resistor would be chosen for a 12V supply rail, and at 19V supply, the LED will run much brighter than it should. You should be able to reduce the LED brightness by adding an external resistor between the gold-coloured pin and 0V. Try 330 ohms; that's about right assuming the LED is designed to run at 20 mA.

The silver-coloured switch contacts need to connect between the +19V from the power supply, and the positive rail of the LEDs in the sign, as you showed on your circuit diagram. You need to connect those pins the right way round, but the description doesn't tell me which way this should be. Try both ways. One way, the LED will be ON all the time; that's the wrong way, so reverse the connections to the silver-coloured pins if that happens.

I don't know the best way to mount the LEDs. You could try drilling holes in a piece of thin wood, such as hardboard or thin three-ply, or something similar, and sticking the LEDs through the holes. Make them a tight fit, and/or use hot glue, and wire them into strings on the back side of the board, using the rigidity of the wires to keep the connections apart. You can also insulate the LED leads using heatshrink tubing, "spaghetti" (thin hollow plastic, like insulation that has been stripped off a piece of wire), or actual insulation that has been stripped off a piece of wire. Steve probably has some other ideas too.

Yes, the wattage calculated by the series resistor calculator is the minimum wattage rating you need to buy. So if the calculator says all the resistors need to be rated at 1/4W or 1/2W, you can use 1/2W resistors for all of them.

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Jan 22, 2013
12. ### KrisBlueNZSadly passed away in 2015

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That power supply looks suitable. At that price though, it will be the lowest quality you can find. Since this is a very non-critical application you may find it worthwhile.

Based on my experience of extremely low-budget Chinese laptop power supplies, you may find that the mains cable will break and/or short out and cause a fire at the slightest bending or twisting because there aren't enough conductors in it; they do this to save a few cents because copper is expensive.

The socket looks right too.

13. ### kevbo423

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Jan 22, 2013
Ok great. When I wire up the socket, how do I do that since it has three prongs on it. Thanks.

14. ### KrisBlueNZSadly passed away in 2015

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Nov 28, 2011
You should be able to figure that out by looking closely at the socket.

From the photo, I can see that one of the connection tabs is connected to the centre contact. This will be +19V to the switch.

The 0V connection tab is the one that connects directly to the part of the socket that makes contact with the outside of the plug barrel. This will probably also connect to all of the surrounding metal.

The third terminal, which you don't use, is some kind of switched connection. You should be able to see some kind of moving part that is pushed when the plug is pushed into the socket, or maybe a small contact that touches part of the plug. It's hard to say from the picture. Have a close look at it, and you should be able to figure it out.

15. ### kevbo423

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Jan 22, 2013
Ok great. Thanks Kris

16. ### kevbo423

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Jan 22, 2013
Hey guys. I've been working on the LED sign and it is coming along nicely. I have a question about the switch I'm using though. It is classified as a 16A 12V switch. Since I'm using a laptop power supply, which is 19.4V, does that mean I need a higher rated switch? Or is the 12V rating just for the LED that is built into the switch. Thanks for any help.

Switch: http://www.amazon.com/gp/product/B007B87RIC/ref=oh_details_o00_s00_i00

17. ### KrisBlueNZSadly passed away in 2015

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Nov 28, 2011
It means the switch is designed for automotive use with a 12V battery. Technically you're running it outside its ratings at 19.4V but your current is much lower than the 16A specified, and your load is non-inductive, so they will be fine. No need to worry.

18. ### kevbo423

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Jan 22, 2013
Hey guys. I'm almost done with the sign. I have all the LEDs wired up and they are extremely bright. I was wondering if I could use something like a variable resistor to be able to change the DC voltage coming from the power supply. Thanks for any help.

19. ### KrisBlueNZSadly passed away in 2015

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Nov 28, 2011
Yes you can connect a resistance of some kind in series with the current path, but it will dissipate significant heat, since it will be carrying the total current for all the LEDs.

Variable resistors (potentiometers) are available in high-power versions ("wirewound" - made from resistance wire, instead of a thin carbon track) but they're rather expensive. It might be easier to buy some fixed resistors and use trial and error to find the value you want.

Calculate a typical resistor value based on the total current your sign draws, and a voltage drop of, say, 1~3V, using Ohm's Law: R=V/I. If your sign draws 1.5A then this works out to values of 0.68 ohms, 1.3 ohms and 2 ohms respectively, with power ratings (from P=VI) of 5, 5 and 10 watts respectively.

Or you could get a lot of lower-value resistors and connect them in series as required - assuming 1.5A current, a 0.33 ohm resistor will drop half a volt and dissipate nearly a watt, so you could buy a pack of 0.33 ohm, 2W resistors and connect them in series until you get the brightness you want.

If you drop the voltage like this, different LED chains will dim by different amounts, depending on the proportion of the total voltage dropped by the series resistance in the chain (as calculated with the LED resistor calculator). Chains with a higher total LED voltage (and less voltage dropped across the series resistor) will dim more noticeably than chains with a lower total LED voltage (and more voltage dropped across the series resistor).

Another common method of dimming LEDs is pulse width modulation (PWM). The LEDs are turned ON and OFF at a frequency high enough that the flicker is not visible to the eye (100 Hz is fast enough) and the ratio of the ON-time to the OFF time is varied, to control the average brightness.

This method is very efficient, because the series component (usually a MOSFET) is either ON or OFF, so it dissipates very little power, and doesn't get hot. It also solves the problem of some chains dimming more noticeably than others.

A PWM circuit adds a significant number of components - in your case, at least one IC, a MOSFET, and probably about 10 small components.

20. ### kevbo423

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Jan 22, 2013
Ok perfect. Thanks Kris.