# Powering LED Array

1. Aug 10, 2010

### jgordo32

Hey, so I have a large array of LEDs. I think I have 168 of them.

I want to power them from my wall in the US (120V AC). The forward bias for each LED is 2.8V. The current rating is 25mA.

I plan to arrange them in chains of about 42 to get 117.6V drop over all the LEDs. That leaves me with 2.4V to drop over a resistor. So to get 25mA I would need a 96 ohm resistor in each chain (2.4V / 96Ohms = 25mA). The diodes should act as rectifiers by themself.

First question, does all of that make sense or am I missing anything?
Second question, will there be any points in my circuit that are super dangerous?

Thanks!

2. Aug 10, 2010

### Mike_In_Plano

Aside from the obvious safety implications, here are a couple of things to consider:

The AC voltage may be 120 volts RMS, but that translates into peaks that are over 160volts ( sqrt (2) x 120 ). Also, the AC voltage varies considerably. So, you'd need a lot more "ballast" resistor to regulate your current. Perhaps R = 60 volts / .025Amp, R=2.4k at a couple of watts.

LED's are diodes, but they're not meant to block 160V. If some diodes leak more current than others, than you'll be faced with LEDs being driven into avalanche. I don't know whether they can take that. The safe thing to do would be to add a couple of 1n4006 rectifiers. One across the diode chain to protect it from reverse leakage currents and another in series with the resistor to act as the rectifier.

3. Aug 10, 2010

### jgordo32

Thanks for the response mike!

First point: I see what you mean about 160V being too much. But I'm questioning your solution (no offense, haha). If I put in a bigger resistor, sure it will handle the peak better. But what about the valley? Won't it draw the current well below the 25mA rating and make the LEDs super dim, if they turn on at all. I wouldn't normally care since the polarity is switching so fast, but my circuit already has them turned on only half the time. Wouldn't putting a larger resistor on result in them only being on... about 1/4 of the time?

Second point: I was under the impression that since I have 42 LEDs in series, no single diode is responsible for stopping all 160V that need to be rectified at peak. Is that a misunderstanding. Furthermore, I don't understand what you mean by put one across the LED chain AND one in series. So I have a chain of leds with a resistor at the end, I put one in series with that to act as a rectifier, the other one goes across to the other chains or what? Please elaborate more on your second point, I seem to misunderstanding a lot here.

Thank you muchly!

4. Aug 10, 2010

### vk6kro

You could rectify the mains with a bridge rectifier and then filter the DC resulting. Bridge rectifiers are very cheap and you can get them rated at 1000 volts and 30 amps for a small cost.

Assuming a load of 1600 ohms, (160 volts and 100 mA) a simulation indicates that you need a capacitor of about 50 uF to get reasonable filtering without having too much current charging the capacitor. This results in about 15 volts of ripple on the 160 volt supply, so that would probably not be visible.

The extra brightness this should give the LEDs would make up for the extra complexity of the circuit.

You would need 4 resistors each in series with 42 LEDs. These resistors should be about 1.8 K and about 5 watts rating, to be safe.

When designing LED strings, it is a good rule to allow about 25% of the supply voltage across the series resistor. Otherwise, the LED current is too dependent on the exact value of the supply voltage.

However, using high voltages always carries some risk. The 160 volts DC is quite capable of giving a bad shock, so proper construction practice to keep the high voltage covered securely when power is applied is essential.
I remind you of this because other people view this and an extra warning never goes to waste.
All wiring, including the connections to the LEDs, needs to be enclosed in a rugged plastic or metal box. Metal boxes need to be grounded.

5. Aug 10, 2010

### jgordo32

@vk6kro
So, you're proposing that I hook up my AC power supply to a bridge rectifier as show on wikipedia's article http://en.wikipedia.org/wiki/Diode_bridge" [Broken]. Then I connect a 50 uF smoothing capacitor across my two inputs. That will result in a relatively smooth 160V DC input you claim? Can you show me how to do the capacitor calculations?

As for the resistors, I put a 1.8k in each chain of 42 LEDs. Unfortunately, I've noticed that resistors @5W are pretty expensive. So correct me if I'm wrong but if I put 2x900 ohm resistors, that will do the same job of limiting current, but allow me lower power ratings on the resistors yeah? I like your point about the 25% rule, I noticed that high dependence in my calculations earlier, but sort of just accepted it as a fact of life haha.

Lastly, what do you mean by bad shock? I plan to use a standard plug from the wall which has plastic sheathing and solder the leads onto a PCB. Then do the rectifier on the PCB too. The LEDs I plained to just wire the leads to each other and daisy chain them. I will use heatshrink to cover almost all of the open wire so it shouldn't be a problem, but I am an amateur. So I''m worried my heatshrink will not be 100% safe. Is a bad shock like it will hurt a lot, or like it will send me to the hospital? If it's hurt a lot, I think I'll take the risk with my decent heatshrink option. If it's hospital, I might need to look into another option.

Thanks a lot for your reply, wish I had found these forums earlier in my design process =)

Edit: Also, can you link me to some parts? I found a 600V, 25A bridge rectifier on mouser for $2, and 910Ohm resistors @3W (assuming I use 2 of those per chain) at mouser for$0.45 each. Those parts sound good?
http://www.mouser.com/ProductDetail/Fairchild-Semiconductor/DFB2560/?qs=sGAEpiMZZMuL5UzuAmrlvNNa//Gjhnx3" [Broken]
http://www.mouser.com/ProductDetail/Panasonic-Electronic-Components/ERG-3SJ911V/?qs=sGAEpiMZZMvhlCB8CTbT5F%252bwr9A4KnIaWy6ZdnRYoso%3d" [Broken]

Last edited by a moderator: May 4, 2017
6. Aug 11, 2010

### vk6kro

So, you're proposing that I hook up my AC power supply to a bridge rectifier as show on wikipedia's article here. Then I connect a 50 uF smoothing capacitor across my two inputs. That will result in a relatively smooth 160V DC input you claim? Can you show me how to do the capacitor calculations?

Yes, that is the circuit.
I used a simulator, but there are simple formulae to do this calculation.
This is one of them:
Capacitance in µF = (Load current * 0.0083 * 10^6) / ripple voltage
C= 0.1 amps * 0.0083 * 10 ^6 / 15 volts = 55 µF

The 0.0083 figure is for 120 Hz, full wave rectified. It is just the period of a 120 Hz sinewave. You could rewite the formula like this:
Capacitance= (Load current * 8300) / ripple voltage

I found a 600V, 25A bridge rectifier on mouser for $2, and 910Ohm resistors @3W (assuming I use 2 of those per chain) at mouser for$0.45 each. Those parts sound good?
Yes those parts sound OK. I was thinking of the ceramic 5 watt types which are very reliable although a bit more expensive.

Lastly, what do you mean by bad shock? I plan to use a standard plug from the wall which has plastic sheathing and solder the leads onto a PCB. Then do the rectifier on the PCB too. The LEDs I plained to just wire the leads to each other and daisy chain them. I will use heatshrink to cover almost all of the open wire so it shouldn't be a problem, but I am an amateur. So I''m worried my heatshrink will not be 100% safe. Is a bad shock like it will hurt a lot, or like it will send me to the hospital? If it's hurt a lot, I think I'll take the risk with my decent heatshrink option. If it's hospital, I might need to look into another option.

If I could achieve one thing today, it would be to convince you that ANY risk of electric shock is totally unacceptable.
160 volts DC from a 50 uF capacitor could kill you. If it killed somone else, you could get a lawsuit that would ruin you for years to come.
You might end up in hospital, but electric shocks can stop your heart working and even if it starts again, parts of it could be dead and you could be crippled for life because of this.

So, I would urge you to think about safety. Heatshrink is not good enough for this voltage.

7. Aug 11, 2010

### jgordo32

If I could achieve one thing today, it would be to convince you that ANY risk of electric shock is totally unacceptable.
160 volts DC from a 50 uF capacitor could kill you. If it killed somone else, you could get a lawsuit that would ruin you for years to come.
You might end up in hospital, but electric shocks can stop your heart working and even if it starts again, parts of it could be dead and you could be crippled for life because of this.

So, I would urge you to think about safety. Heatshrink is not good enough for this voltage.

Greeeaaat. So let's talk remedies. The way I see it... here are my options:
1. Drop the voltage down to a safer level
2. Better protection solution

I'd prefer the second option but I'm lost as to how I should pursue this. Let me give a little more detail on my setup. So I'm living in a collegetown apartment where things could get messy. We're not throwing huge parties or anything, but I'm sure there will be a drunken idiot near this contraption at least a few times.

As for details on the contraption, it will be a fold up pong table. So its a 2' x 8' piece of wood with folding legs on the bottom. I drilled 3mm holes in it and planned to just wedge the LEDs in so they're flush with the top of the table. Then I would do all the wiring underneath the table and cover all bare wire with heatshrink. The PCB would be taped to the bottom of the table with the plug running to the wall.

Any ideas on better protection or should I start looking into getting a 24V AC/DC converter?

8. Aug 11, 2010

Well Id think of better ways than just heatshrink. Possibly bolt up a wooden panel underneath the table to physically block any access to the wiring. Or it might well be easier just to have the converter, you can probably just pick one up for $50 or so. 9. Aug 11, 2010 ### jambaugh 25mA x 168 = 4.2 Amps. You could hook them all up in parallel with a 2.8 volt power supply rated at 5 Amps. 168 = 8x3x7= 2x84 = 3x56 = 4x42 =6x28=7x24= 8x21 = 12x14 I would suggest you work in blocks each block of LED's in parallel and the blocks themselves in series. 2 Blocks at 5.6V x 84 LED's at 2.1A , 3 Blocks at 8.4V x 56 LED's at 1.4A, 4 Blocks at 11.2V x 42 LED's at 1.05A, 6 Blocks at 16.8V x 28 LED's at 0.7A, etc. That way, you can use safe voltage and off the shelf power supplies and if one or two LED's fail you won't loose the whole business. 10. Aug 11, 2010 ### jgordo32 Thanks jambaugh, that much I could handle on my own =P Is it reasonable for me to build a converter from 120V AC --> ~9V DC? 11. Aug 11, 2010 ### vk6kro Very low voltages require a lot of extra resistors and more complex wiring. It is accepted, I think, that voltages below 40 volts are safe. So, if you had a transformer delivering 28 volts and this was rectified and filtered, you would get a voltage of about 38 volts. (28 times 1.414 minus two diode drops of 1.2 volts) Take 75% of this to be the voltage across the LEDs so this gives you 10 LEDs per string. So, you could have 17 strings of 10 LEDs So, you would need 17 * 390 ohm resistors each dissipating 220 mW. Half watt ones would be OK. The filter capacitor would then have to be about 1000 µF (or more) but it would only need to be rated at 50 volts or so. This is to get 10% ripple. Last edited: Aug 12, 2010 12. Aug 11, 2010 ### jgordo32 @vk6kro Is there any advantage to using a transformer and leaving it in AC just to rectify as opposed to just using a converter to something like 40V DC? Also, are the 2 diode drops from the bridge rectifier you want me to put in? Also also, since I'll have my last string with only 8 LEDs, would the 390 ohm resistor be big enough? Lastly, what do you guys use to do circuit simulations? Could someone point me to a good beginners tutorial on that? Thanks so much, this has been a great learning resource already 13. Aug 12, 2010 ### vk6kro I fixed a typo in the previous reply. Yes, a converter that gave 40 volts out would be OK as long as the output was totally isolated from the primary. Two diode drops would be 1.2 volts and this is lost in the bridge rectifier because there are two diodes in series with the output at any one time. You would have to recalculate the final resistor. I get about 639 ohms. You could put a 560 and an 82 in series, or just try a 680 ohm resistor. One simulator is the free one from Analog.com. It is called LT Spice 4. It is very powerful, so it looks scary when you first see it. However the actual controls you use all the time are very simple. It is excellent software so I suggest you get a copy and have a play with it. 14. Aug 12, 2010 ### jgordo32 as the output was totally isolated from the primary. I don't know what that means, could you explain? Also, could you point me to a good converter? I've never purchased one of those before. Thanks. 15. Aug 12, 2010 ### vk6kro "Fully isolated" means that there is no input wire that goes directly to the output. Even if this is the neutral wire, this would not be fully isolated. It is best that you get this near where you live. You should be able to ask if the device has an output that is fully isolated from the mains input. If they don't know what this means or just say "sure it is", go somewhere else. They should check it for you. Get an assurance that you can bring it back if you find it isn't fully isolated. 16. Aug 13, 2010 ### jgordo32 @vk6kro Could you explain why not having an input wire connected to the output matters? For example, if I want to take it back for not being fully isolated, how will I know? Unfortunately, a google search for "fully isolated" did not return any helpful results, so I still don't really know what you're talking about. Do you think a radio shack would carry them? Maybe microcenter? I'm only just getting into electronics so I don't really know where to buy things besides the online places like mouser, digikey, and jameco. Thanks! 17. Aug 13, 2010 ### jambaugh Using AC directly will cause a couple of problems. One is you're only getting half the cycle as light and so effectively you're halving the value of your LED's. What's the price of a brig rec compared to the price of half your LED's? The other is you'll get a 60Hz flicker which can cause eye strain, headaches and in some very rare cases epileptic seizures. http://en.wikipedia.org/wiki/Flicker_(screen)" [Broken] http://books.google.com/books?id=_5mdSJfur8cC&pg=PA290&lpg=PA290&dq=60hz+flicker+induced+seizures&source=bl&ots=gY3askFcns&sig=8yIG1vnhHprpU2dO44M56LgQYms&hl=en&ei=ZFJlTKqMAcWclgf_z-3VDg&sa=X&oi=book_result&ct=result&resnum=5&ved=0CCkQ6AEwBA#v=onepage&q&f=false" [Broken] Last edited by a moderator: May 4, 2017 18. Aug 13, 2010 ### vk6kro Could you explain why not having an input wire connected to the output matters? For example, if I want to take it back for not being fully isolated, how will I know? Unfortunately, a google search for "fully isolated" did not return any helpful results, so I still don't really know what you're talking about. You would measure this with a multimeter on "ohms". If you don't already have one, they are available cheaply at almost any electronics store. http://www.harborfreight.com/7-function-digital-multimeter-92020.html Just measure for any connection between the output wires and the input plug. Often the packaging for these power supplies would give information about whether the secondary is isolated from the primary, or not. This may be in the form of a warning not to touch the output wires. It matters because if you are not very careful, you can end up with the active side of your mains supply connecting to your LEDs. Radio Shack sell an adapter for running grounded 3 pin plug devices on a 2 pin outlet. Radio Shack and other stores have their own websites and you can easily check these for availability or prices. I couldn't see anything suitable from Radio Shack, but you might find something. Frys have a 24 volt 40 watt transformer for$11 which would just need a bridge rectifier and a capacitor. FRYS.com #: 4411656 This plugs into the mains so you don't need to do any mains wiring.
It doesn't say if the secondary is isolated from the primary. You could ask that.

19. Aug 13, 2010

### mdjensen22

If you get a DC adapter/power supply that has UL markings along with other safety agencies, it is a pretty safe bet that the output is isolated. This might be one of your safest bets - especially considering this is going on a table (liquid spills, etc.).

If you wanted to cut down on your overall power consumption, you could use a processor or other circuit to activate alternating strings very rapidly - this would make only half active at the same time (depending on how you do it).

You had mentioned you planned on running the LEDs at 25mA which is their rated current - is this their maximum current rating? If so, you might want to consider around 10mA or so...

20. Aug 15, 2010

### jgordo32

@mdjensen22
No 25mA is not their maximum. Here's their http://www.hebeiltd.com.cn/led.datasheet/330LB7C.pdf" [Broken]

I'm thinking about doing microcontroller animations later in my project. Although not with the intention of saving power =P However, I just wanna get the thing safe and working before I start messing with that.

@vk6kro and others
Ok, I'm convinced... I'm going to buy a DC converter. I think I'll do 28 chains of 6 LEDs. I'll need a 24V DC output. I'll be using 16.8W (.025A X 24V * 28chains = 16.8W).

So am I correct in thinking that all my circuit needs now is the DC adapter (can anybody help me find a good one online so I know what to look for?), my LEDs, and 28 x ~288 ohm 1/2W resistors?

Oh also, do the DC adapters have builtin surge protectors? How good are the surge protectors in the power strips from like target that cost \$15? Should I buy a fuse or something to protect against surges?

Thank you!

Last edited by a moderator: May 4, 2017