LED signs/ 120 Volts

[joshua17ss2]

I am making a few LED signs and to save on space and power supplies, i wanted to run them off of 120volts ac.

I know that the set up will be 120 v outlet, diode bridge, capacitor, and roughtly 60 leds depending on color and voltage of that color.

I will need more then 60 leds to do the entire sign. so my questions is where do i split the power to set up additional stings of leds.
Do i need to split it before it hits the diode bridge and converts to dc and just include a second diode bridge
or can i split it after it hits the diode bridge, and just add the 2 or 3 sets i will need with just one diode bridge?
thanks for any help

 

[Hero999]

Why not simply use a mains adaptor?

It will hardly use any extra space, a 600mA 12V mains adaptor should be able power all those LEDs with no problem.

 

[joshua17ss2]

i had considered using a 48 volt supply and 12 volt supply, for the size of the sign i wanted to try and keep it very neat as well. 120 can do very large chains of LEDs with only a small resistor. if i went with 12, i could only do groups of 4 and each would need a resistor, it would get messy im figuring upwards of 180 to 200 leds for each of the 2 signs i need to make.

 

[Hero999]

Resistors are cheap, I can buy a box of 100 from my local supplier for 50p, that's US 77c cents.

I don't see how connecting them in many stings in parallel will be that messy.

It's certainly much safer to use a DC supply, even at 120VAC, you'll still need more than one string in parallel to power that many LEDs.

 

[tcmtech]

I dont see any problem with using direct line power to run that many LED's I have done it many times when making LED based lights for around the home and for friends.

Here is the basic version of the circuit.

You will need to calculate C1 to supply the correct current for the LED sets based on their total voltage drop and running current and your common line frequency.

R1 can be anywhere from 220 - 1K ohms and C2 should be around 100 uf with a 2x line voltage rating.

 

[Hero999]

I had a feeling you'd say that.

Well there's nothing wrong with that method providing it's build in a proper enclosure, sealed away from moisture and fingers and if it's metallic the appropriate safety earth/ground connections are made.

 

[joshua17ss2]

The enclosure is all plastic and will be hanging high in a window, where it hasent been touched in years and will go on not being touched long after its reinstalled up there. basically moddifying it from a single light bulb to all led for better visability.

For R1 that will get calculated once i have choosen the final colors for the sign mostly red is what im seeing.
For C2 I was planning on using some old disposably camera flash caps. they are rated for 330 volts, and about 80uf

could you supply alittle more info about C1, i wasnt going to include any caps on the mains side of the bridge, but i also dont wana fry leds 60 at a time

thanks

 

[tcmtech]

C1 works as the current limiter or ballast capacitor. your LED's wont last very long without it!

To calculate that you need to know what your line voltage is and what your total voltage drop from all the LED's in series is as well.
After that its (line voltage - LED voltage)/ (amps running current.)

The camera flash capacitors will work well and R1 is just to further smooth out the line ripple that C2 does not take out hence the wide variation in its value.

 

[joshua17ss2]

Well my line voltage will be a standard 120, so i would figure around 116 vac, 60 Hz, im hoping to get the led voltage to drop to almost 0, 47 leds, at 2.5 volts = 117.5, the leds should be running around 25ma.

Digi-Key - P10734-ND (Manufacturer - ECQ-U2A474ML)
Something like that ?

 

[tcmtech]

If you are running that many LED's in series you will only need about 100 Oms of impedance provided by C1 at 60 Hz so that works out to around roughly 26.5 uf.

I would recommend breaking that down into sets of 25 - 30 and using multiple strings of LED's.
With two sets of 30 LED's at 2.5 volts you would get 75 volts.

120 - 75 = 45
45 / .025 = 1800

1800 ohms of impedance at 60 Hz would require a 1.5 uf capacitor. Multiple sets could be powered off of the same rectifier and C2 circuit with proportionally larger values of C1.

3 uf would power two strings of 30 LED's at 25 ma each without problems. Just make sure that R1 on all strings are equal.

 

[joshua17ss2]

how long would the leds last if i were to omit c1 and just wire up 2 or 3 sets inside the display with each on their own rectifier and c2, r1.

one thing i could never wrap my brain around was electronics math. Im gonna check my 48 volt supplys when i get home, i think each is 350 ma, which if i did 10 sets of leds would be a big enought supply to run them all. would i need to add a capacitor across the 48 volt supply + and - to inprove function or smooth it out any ?

 

[tcmtech]

C1 is key to how the circuit works. Without it I doubt your LED's will last more than a few seconds.

They need limited current and will burn themselves out without a current limiting system in place.