Mains LED with blocking cap

Use a PTC thermistor (or just a resistor and put up with the energy loss) in series with the cap.

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Paul Hovnanian     mailto:Paul@Hovnanian.com
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definition: recursion; see recursion.

Any problems driving an LED from the mains using a blocking cap as a constant current source?

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Dirk

http://www.neopax.com/technomage/ - My new book - Magick and Technology

Many :-) Lack of isolation. Lack of safety. No (inrush) current limit. (You want your LED to emit light, I guess, not to explode?) Fire in the house after just one failing component :-)

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Kind regards,
Gerard Bok

The inrush can be dealt with using an inductor. As for safety, a completely sealed and grounded unit although a 1:1 xformer might do both jobs

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Dirk

http://www.neopax.com/technomage/ - My new book - Magick and Technology

Reverse voltage limit

Well, with a diode or two of course. I was mainly thinking of using it as a cheap constant current source. Bearing in mind peak currents

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Dirk

http://www.neopax.com/technomage/ - My new book - Magick and Technology

Neither a capacitor nor an inductor are suitable to create a constant current source, imho. In a transformerless mains design a capacitor can only reduce the amount of power in the real current limiting device, usually a resistor.

Hard to tell. With 230 Volt AC mains overhere, I am glad there is some kind of transformer between mains and a led :-)

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Kind regards,
Gerard Bok

An ordinary diode in reverse-parallel with the LED can take care of that, that is if the OP really meant a single LED. Otherwise, a FW bridge rectifier in series with the cap and the LED. A resistor can limit the inrush current.

Assuming that the OP meant a current setting device rather than a true constant-current source, a capacitor can act as one because it presents a constant reactance at the mains frequency. I've used caps that way myself and have seen commercial products using them.

The thing to watch out for is the initial current surge when the circuit is switched on in mid-cycle. A series resistor can reduce that inrush current. I've seen resistors as low as 47 ohms in series with a cap and a string of 20mA LEDs in lamps meant for

230V 50Hz AC. That represents a theoretical worst-case inrush of more than 6A if the lamp is switched on when the AC is at one peak. Apparently it doesn't often cause a failure.

Wouldn't a spike blast straight through the capacitor... (I = C.dV/dT)

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John Devereux

Only if there is no noise or harmonics on the line.

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You can't fix stupid. You can't even put a Band-Aid? on it, because it's
Teflon coated.

Works fine until your air conditioner clicks off and the fast-rise line transient takes out your device. Or lightning, or your neighbor's air conditioner or welder or the power company flips a switch or or or.

Caps make great AC current limiters if you have FULL control of both ends of the system.

C'mon, too complicated by far. All you need is a cap of the right reactance and a single diode to prevent reverse breakdown on the led / string of led's.

You can also use two leds reverse parallel instead of the added diode. Have built several in the past and the idea works fine. The cap must be one of those X or Y (?) or polypropylene caps, whatever, rated for the full ac voltage. You can add a 1/8w low ohm resistor to act as a fuse if required.

Not a new idea. In the old days of tube series heater strings, a cap was sometimes used in place of a dropper resistor as it provided a constant current and eliminated the need for a heater switch on surge thermistor and dropper. Much more efficient as well...

Regards,

Chris

I certainly wouldn't recommend such a system for something that needs to be highly reliable. But it seems to be good enough for a quick lash-up that wouldn't cause a disaster if it does break down.

In fact, during an event that lasted for a few days last year, I ran some three dozen lamps of the type I mentioned from a square-wave UPS. The application required switching the lamps on and off many times in an hour, at times several times a minute. The switching was done with non-zero crossing solid-state relays. None failed.

AC line transients will kill the LEDs. One fix is a series R-C.

John

So, the 'ballast' is the capacitor? Some issues: the LED diode does no current limiting, so it has to be a safety-rated AC capacitor (has to fail safe). Power transients cause overcurrent. Bad-power-factor items on the same circuit create harmonics and add to the LED current.

Since you need a reverse shunt diode anyhow, make it a 12V Zener and put a bit of resistor in series with the LED. And, maybe a fusible resistor in series with the capacitor (if the resistor keeps it safe, the capacitor needn't have the fails-open safety rating).

Yes.

Hope This Helps! Rich

It'd be cheaper and safer to go to Radio Schlock and get a switching wall wart:

Good Luck! Rich

--- Looking at a single white LED with a 3.5V drop across it and 20 mA through it makes it look like a 175 ohm resistor:

E 3.5V R = --- = ------- = 175 ohms I 0.02A

If you're running 230V mains and using a capacitor as a lossless voltage dropper, then the equivalent circuit is:

175R 230AC>--[C1]--[LED1]--+ | 230AC>----------------+

If the RC is to allow 20mA through it with 230V across it, then its impedance must be:

E 230V Z = --- = ------- = 11500 ohms I 0.02A

So, since:

Z² = R² + Xc²

we can rearrange and solve for Xc:

Xc = sqrt(Z² - R²)

= sqrt(11500R² - 200R²)

~ 11499 ohms.

At 50Hz that would be a capacitance of:

1 1 C = ---------- = ---------------------- = 2.77 e-7F 2pi f Cx 6.28 * 50Hz * 11498R

or about 0.277µF, so our circuit now looks like this:

0.277µF 175R 230AC>--[C1]---[LED1]--+ | 230AC>-----------------+

If we put in another LED in parallel opposition to LED1:

0.27µF 175R 230AC>--[C1]-+--[LED1>]--+ | | 230AC>-------+--[