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LED driver without IC and Transformer

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What heatsink do you have on the LEDs? A 1W LED will get very hot without a heatsink!
 
How big?
 
Hi,

A quad of quick points...

First, R2 is there to prevent a surge when the unit is first plugged in. 18 ohms only limits the current surge to around 30 amps while 100 ohms comes in around 6 amps. Of course 100 ohms uses more power and requires a larger resistor, so the choice is yours. There are also PTC thermistors used for this purpose that have resistance that gets lower after the circuit operates for a while. You may want to think about this because this is not your typical 20ma off line LED circuit.

Second, a full wave bridge is used for better utilization of the capacitor. The capacitor is probably the most expensive part in the circuit, so you want to get all you can out of it. What this means is that for a full wave bridge that drives the LED at around 200ma and uses 3uf, it will require DOUBLE that value (6uf) for half wave operation to get the same 200ma average current. Using two more cheap diodes (you need a second diode anyway for half wave) is a small price to pay for two times the current.

Third, the 3uf cap should be rated for AC operation, and its rating should be no less than 1.5 times the voltage of the line, or better yet 2 times.

Fourth, the electrolytic has to have enough capacitance to maintain the voltage the LED needs, but with no resistor between that cap and the LED this probably wont happen anyway. The LED limits the voltage the cap can charge up to, and with the cap being only 220uf it can easily discharge quickly below the LED voltage, meaning it probably wont do anything much. We can look into this more. The better design uses a series resistor so that the cap can have some flexing room while the LED draws current between half cycles. True, the resistor uses some power too, so a tradeoff is required between a really good value and a value that works more or less ok.
 
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I would also strongly suggest placing a zener diode across the capacitor - to prevent the capacitor 'exploding' if the LED goes O/C (as the OP has already killed capacitors in this way). It's voltage simply needs to be higher than the LED, so it has no effect on the operation unless the LED fails.
 
Nigel, As to Zener protection; I have never seen a open LED. It must happen I just have never seen one and I have seen millions of LEDs. I have seen someone build the circuit and not put in the LED. Why try the circuit with out a LED?...........If a LED at 3V uses about 1Watt then a 6.3V Zener will use 2W and a 12V Zener will use 4W. A 250mW Zener will over heat and most likely short to 0 volts. I don't see a need for a 2W Zener, just let it overheat in the very rarer case of the LED going O/C.

MrAl,
4) I agree. Lets leave the electrolytic out. That solves the Zener problem.
3) "X" capacitor and "Y" capacitor Agree.
2)Agree
1)Yes Yes Some one here did a big study on that a couple of years ago. With a small LED I would not but with a 1W LED a PTC thermistors starts to look good. Again-leave out the electrolytic and the duration of the current spike gets very short. (30uS)
 
I have never seen a open LED. It must happen I just have never seen one and I have seen millions of LEDs.

I can't comment on 1W LED's, but certainly the normal failure mode for small LED's (visible or IR) is going open circuit. I've replaced many IR ones in remotes, and a smaller number of visible ones over the years (in particular the Throrn TX10 TV chassis used one as a reference on the CRT base, which used to go O/C, and of course normal indicator LED's which go O/C).
 
Hi,

I like the idea of leaving out the electrolytic cap if the 100Hz blinking wont be a problem. I dont think it will change the surge current too much though because the LED still conducts.

I would assume that the LED can fail either open or short...that covers all cases and it's not just about the LED anyway, it's partly about the wiring that supplies the LED with power. Copper wire or copper trace or solder joint, if it opens the LED looks open anyway.

I read Externets reply again and is see i misunderstood it on the first read. Two LEDs back to back are nice together, as long as the application budget can tolerate a rise in cost for the second power LED, and of course as Ron points out the ever present blinking because of the physical separation distance between LED dies, in addition to the usual 100Hz fluctuation.

Another issue that comes up with these things is the failure mode of the series capacitor. If it dies short, we've got big problems. This usually ushers in a fast blow fuse of some type.
 
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