multi white led circuit

"jay" schreef in bericht news: snipped-for-privacy@f6g2000cwb.googlegroups.com...

As you tell nothing about the specs of the LEDs, it will be hard to calculate the value of any current limiting components.

I nevertheless can assure that such an amount of white LEDs on so small a surface will become hot, very hot.

Powering LEDs with AC is always a bad thing as LEDs are DC devices. You will need a rectifier of some kind.

Using (only) a capacitor as a current limiting device is risky. Depending on the momentary voltage when switched on, the inrush current can become high enough to blow the LEDs. I'd prefer resistors or electronics.

petrus bitbyter

I won't be a whole lot of help, but if I might ask, where are you buying white LEDs from? The best price I found on them was $1.97ea in lots of 100. If all 640 of those LEDs are white....

I bought white leds from A dealer on ebay. 100 - 10000 mcp 5 mm cost me near

14.00 usd. They are a slightly blueist tint and draw 30ma max. I made a pcb with 24 of them running at 12 for the group that is almost as good as a spotlight. Lights up the trees across the road from my house. They do run hot and the way I designed the pcb with oversize pads and clinched leads, I add a silicone transistor insulator and heatsink compound on the pcb bottom - then attach the board to a small alluminum heat sink. If you dont heatsink the led's, they will go into thermal runaway. JTT.

you don't seem to have addressed the fact that LEDs only want to conduct in one direction.

inrush current may also be an issue. That resistor is going to dissipate 12W If it's connected across the 110V

Bye. Jasen

Jasen,

thanks for the reply.......any suggesti> > > I am wanting to make a light source for a video projector.

The problem with using lots of leds in series is that if one led fails you have to hunt through a string of two or three dozen led's to find the one that failed and took out the whole string. If you want to avoid that headache you'll have to use a transformer or switching power supply to give you a lower voltage to run the led's. I'd get hold of a power supply from a discarded computer. If your led's run 3.5 volts for example, put three in series to run off twelve volts. That's 12-10.5=1.5 volts across your current-limiting resistor for each series of three leds, and calculate resistor value to give you your target current. The power dissipated in each resistor will be tiny, so you could use surface-mount resistors. Resistors will dissipate about an eighth of the total power. Of course, you'll have a couple hundred of these series, each with its own resistor, but in the long run you may avoid headaches. Just about any power supply from a discarded desktop computer should have enough power for your light source. It will draw less than 100 watts. No matter what you finally choose to do, use current limiting. You're buying trouble if you don't. And the cap business for limiting current is very risky. You will get nasty surges on startup, with unpleasant results.

You sure cannot do without a rectifier. Which will give you a pulsating DC-voltage. As you still did not tell a thing about the LEDs you want to use, let's suppose they require Iled[A] at a voltage of Vled[V].

The maximum amount of LEDs you can chain: N=120/Vled

If you do so you will sure blow the LEDs so you chain N-2 LEDS and use a resistor to limit the current. R=2*Vled/Iled[Ohm] It has to dissipate Pr=2*Vled*Iled[W] Which is the minimum power the resistor must be able to handle. I advise to take one that can handle twice that amount. You will need a resistor for every string of LEDS. If you expect the mains voltage to vary, it will be wise to reduce the number of LEDS in the chain and increase the value of the resistor accordingly.

Suppose the maximum current for the rectifier is Irec[A], then the number of LED-chains per rectifier: M=Irec/Iled I'd use only half that amount and take an extra rectifier if necessary.

If I rate the number a LED-chains on 20, your lightpannel will dissipate Pp=20*120*Iled[W].

For example, suppose Iled=20mA, Pp=48W which can be doubled easily depending on the LEDs. Anyway, almost all of that power will produce a lot of heat, which will reduce the lifetime of your lightpannel or even may be able to destroy it. So in the construction you will need to add the necessary cooling.

petrus bitbyter

here's an idea.

put enough leds in series to exceed the peak AC voltage (160V) and feed them from a voltage doubler with capacitors picked to limit the current to what the LEDs want.

use 10A recticier diodes in the doubler and a 22 ohm resistor in series with the input of each chain to limit the inrush to what the LEDs want. put a 100K resistor across each string of LEDs to discharge the caps when it's powered down.

doubler leds ||C1 -[22R]-+---||----+--->|------+-->|-->|-...->|-->|--+---. | || | | | | [100K] | || | | | | .-->|-' .---||--+----[100K]-----------' | | | | ||C2 | | | | | -------+---+---------+---------------------------------' C1 controls the amount of current, C2 limits the ripple. (and may not be needed)

this setup will exhibit a 60Hz flicker (view with a fixed pitch font, eg paste into notepad)

Bye. Jasen

Your scheme will work with modification, but it's not safe, and you really don't get an advantage using a cap. If you are bound and determined to do it:

1N4004 Led1 LED20 Ac ---+---[R]---||---+ 1N4004 Led1 Led20 | Ac -------------------------------------+ 16 strings of the above. Resistors will be 120 - Vf*20/I ohms. I'd get 100 1.8K 2 watt resistors and put 2 in series for each string to spread the heat a bit. You'll save $5.04 using resistors vs capacitors, and have 20 left over by buying 100 resistors for $10.00 vs 32 capacitors for $15.04 (Mouser). And no ugliness of shorted capacitors or high current surges. The IN4004 diodes are needed in either case. Finally, 3600 ohms will accomodate a good range of Vf. At Vf = 4, 11 mA; at Vf = 2.8, 17 mA. Your Leds will have Vf > 2.8, < 4.0

Ed

Ed, should that be 120 or 170 (is it rms or amplitude)?

Mark

Hi Mark,

You raise an excellent point. He needs to recognize that the max current rating for his LEDs is 30 mA DC, which converts to 21.21 RMS. Using 120 is fine in the computation, as long as he recognizes that. Otherwise, using 170 as you suggest is the way to go: R = (170 - Vf*20)/I

Ed

Hmm... Did Jay buy the same LEDs James bought? Otherwise we still don't know a thing about the LEDs. Nevertheless, most LEDs have different values for continuous current, average current and peak current. Both last values have to be taken into account when using rectified, unsmouthed sine.

petrus bitbyter

Right - we know nothing about the LEDs, other than they are white. The formula using Vf and I rather than specific figures will work, but you do have to pay attention to the units you are using if you want to get things exact. It is simpler, however, to use the continuous max as the peak limit + or - a few percent. In fact, unless there is some good specific reason to the contrary, it is better to drive LEDs well below max, in my view.

But this is going away from the op's question and the answer, which is predicated on "if you are bound and determined to do it" where "it" was running a string directly off AC with no diodes and a cap and resistor to limit current. The alternative you posted is going to save his LEDs from excessive reverse voltage. If he gets sick enough of the flicker, he'll be willing to take additional steps. Note that despite your first reply - which was the first answer he got - he did not post specs on his LEDs. And he still hasn't, despite your second reply to him. Sometimes, you can't help people. If he can't or won't post the specs for his LEDs, peak, average, continuous are all irrelevant to him, as is a formula that uses any of those specs. The point I wanted to make to the op was the relative cost/benefit of using resistor vs capacitor and resistor.

Ed

Ok. Use either 880 or 920 LEDs.

*Make strings of 40 Leds per string*. You will use either 22 strings (880 LEDS) or 23 strings (920 LEDS) DO NOT change those numbers when using the circuit below. DO NOT change the parts specified.

View in fixed font:

----- AC---+ +---+---i|LM317|o-+-----+---} }---+ | | | --a-- | | | | ---- | | | [202R] [R] [R] +-|~ +|-+ |+ | | | | | BR | [C] +-----+ [40Leds] [40Leds] +-|~ -|-+ | | | | | ---- | | [24K] | | | | | | | | AC---+ +---+-------+-----------+---} }---+

R = 100 ohm, 1/4 watt CAT# 1.8K-1/4 (buy 30) $ 1.50

24K = 12K + 10K + 2K in series, 1/2 watt CAT# 12K-1/2 10 for $ 0.50 CAT# 10K-1/2 10 for $ 0.50 CAT# 2K-1/2 10 for $ 0.50 202R = 200 ohm + 2 ohm in series, 1/2 watt CAT# 200-1/2 10 for $ 0.50 CAT# 2-1/2 10 for $ 0.50 C = 1000 uF 200 VDC CAT# EC-1021 $ 1.25 BR Bridge rectifier CAT# FWB-46 2 for $ 1.00 LM317 Adjustable voltage regulator CAT# LM317T $ 0.50

The prices and catalog #'s for the above parts are from Allelectronics:

Pins on the LM317 are marked in the diagram as follows: i = Vin, o = Vout, a = Adj

There will be approximately 168 volts across the capacitor at 120 volts AC. You cannot treat this circuit casually - it can bite.

The circuit will work on AC from ~108 to ~132 volts The LM317 will regulate the voltage at its Vout pin to 149.76 over that range.

The 40 LEDS in series will drop 148 volts (40*3.7). The current through the LEDs will be limited to

17.6 mA by the 100 ohm resistors. You could replace the 100 ohm resistor with an 80 ohm and a 10 ohm in series, to get 19.5 mA. DO NOT go below 90 ohms total.

As was mentioned by another reply, all those LEDs are going to produce a lot of heat. It's your experiment, but I have the sinking feeling you will be disappointed. Assembling 900 lEDs to replace a projector bulb will entail a lot of work, and I don't think you will be able to get the same amout of light from the same footprint - in other words, the incandescent bulb will take up less area than the LEDs. I don't know how you will manage the heat produced by all those LEDs crammed into a small space.

Ed

Duh - I forgot a critical part. You are going to need a big heatsink for the LM317. Allelectronics does not have one big enough, nor do they have an LM317K (which is the TO-3 package and can take higher temperatures). Something like 567-401-A from Mouser will work. Mouser also sells the LM317K.

Ed