Very low power, power supply

You don't need a power supply. Put all the LEDs in series with each other, and with a 5.6k resistor. Get a resistor with a power rating of about 5 watts.

Pearce Consulting

had

Here is the usual way of doing this, with a capacitor to drop the voltage. You will have to connect all 6 LEDs in series. Make sure the capacitor is X2 rated for connection across the AC line, and all parts are insulated to prevent a shock hazard. The typical .47 uF X2 rated cap is about a half inch thick, but you could put two smaller, thinner capacitors in parallel.

BTW, if you don't use a filter capacitor, the flicker becomes noticeable.

don't

had

So

to

And because you forgot to include a 1N4004 diode, the 6 LEDs will be subjected to 160V peak in the reverse direction, causing them to breakdown since they're only rated for 5V reverse x6 = 30V.

I don't think I'll be asking for your consulting services in the near future.. ;-)

Or a 250Vdc/160Vac 0.47uF cap. (assuming 60Hz) (about 50p from Farnell), then it doesn't get warm.

But either way I wouldn't be happy unless the LEDs were covered. Blow one up and there's two little prongs sticking out with mains on them.

Paul Burke

Hi Don,

D>

What a bad advice. Lots of wasted power, lots of wasted LEDs. LEDs dont like exessive reverse voltage, typically only about 5 volts.

So better use a 470nF capacitor capable of 150V AC, a small bridge rectifier (needs only to survive the summed LEDs forward voltage) and a little 100 ohms resistor.

(switch to fixed pitch font to view)

o------. --- 470 nF 150V --- | 120V AC | 60Hz .-. | | 100R 0,25W | | '-' | .-----. the LEDs -----| ~/ +|- ->|->|->|--| o-----------|~/ -|-------------| '-----'

You can vary the LED current with the capacitor.

HTH Wolfgang

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The overcurrent isn't the problem, it's the flicker. Most white LEDs are rated for 30mA max, 100mA peak. But the flicker is really noticeable especially 60 Hz half wave.

pernews.com>) about 'Very low power, power supply', on Thu, 17 Mar

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The usual white LED is rated at 5V max in the reverse direction, this arrangement far exceeds that voltage. As for power, the max dissipation is about 3.3V * 30mA or 100mW. 5V at 20mA would be 100mW, but the breakdown voltage is usually much higher than that, some say the LEDs will break down in the tens of volts range. In that case, the dissipation would far exceed the 100mW max. That's plainly not acceptable.

You need some resistance in series to prevent the inrush current zapping the LEDs and/or the cap. For 120 V mains, 470 ohms 0.5 W may be enough.

A 5K, 2W resistor, and an antiparallel 1N4004. Put the LEDs in series. (and the 1N4004 in parallel with the series string, but opposite polarity.) 5.1K might be easier to find - it's not terribly critical. But it _will_ dissipate two watts, so you might want to use a 3W R.

Cheers! Rich

Or, just get a low-profile night light:

Cheers! Rich

Also: Make 2 banks of series out of the LEDs. Hook the two banks in reverse parallel.

----/\/\/---!!-----+------------ ! ! --- V ^ --- ! ! --- V ^ --- ! ! --- V ^ --- ! ! --------------------+-----------

This way, the current has a path for both directions.

or use 2 leds anti-parallel, or build de rectifying bridge out of leds, feeding a resistor or....

(or I would buy a ready made one for a few $ in the local DIY shop:-)

Chris W wrote: > I am looking for a power supply to power a small LED night light. I > want it to run of the 120V AC and I need it to be very small. I > don't want the completed night light to protrude more than 1/2" from > the outlet when it is plugged in. I could probably live with 3/4" if > I had to though. Since this is to power LED's it would be nice if it > had a current limiting feature. It only needs to run 6 20mA white > LEDs. So far the smallest I have found is about a 1" cube. It takes > 120V AC to 5V DC. >

You have to run the LEDs off DC or you will never be able to get 20mA average current through them at 60Hz sinusoidal without significant over-current.

I read in sci.electronics.design that Watson A.Name - "Watt Sun, the Dark Remover" wrote (in ) about 'Very low power, power supply', on Thu, 17 Mar 2005:

The reverse current is limited by the series resistor or capacitor. For the resistor, it's 90 V/5600 ohms = 16 mA. Whether that's acceptable or not can be found on the data sheet.

Red/amber low-current LEDs can glow at 1mA, but I'm not sure about the "white" ones. What's wrong with using a 120V neon indicator and integral dropper? It will last for up to ten years. Or go for a self-indicating, internally soft-lit translucent plug (used as nursery/corridor courtesy light)?

Hi Ken,

Ken Smith schrieb:

Well, this will work. But imagine what happens if one the LED fails open... *BOOM*

Wolfgang

If one LED fails, the circuit is broken. After the "BOOM" the circuit is still broken.

Actually there is no "BOOM" the back biased LEDs will fail shorted and the series RC will still limit the current.

In article , Watson A.Name - \"Watt Sun, the Dark Remover\" wrote: [...]

If you use two strings anti-paralleled, the flicker is very hard to detect. The frequency is 120Hz and most of the time there is light produced. The eye detects spikes in light better than dips.

fairly

them

cause

The typical LED will handle 100mW with noticeable heating. That's because most of the heat has to be conducted thru the two leads, which are not very heavy. I'd say that if you made it dissipate "only" 0.4W, it would get mighty hot and melt.