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Driving a led with mains.

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Sigurthr

Wed Sept 18 2013, 12:29AM

Registered Member #4463 Joined: Wed Apr 18 2012, 08:08AM
Location: MI's Upper Peninsula
Posts: 597

A series capacitor sized to the needed impedance will act as a capacitive ballast limiting current without wasting energy as heat. Most LEDs (all I've ever seen) don't tolerate more than a few volts when reverse biased though, so what you need to do is rectify the mains with a bridge that has a reverse breakdown voltage higher than Vpk of the mains (170 or 340V). You need full wave and not half wave though as a series capacitor looks like an open circuit to a half wave rectifier since only half of the sine will have current flow and thus no polarity change on load side of the cap as no current is allowed to flow.

A series ballast cap on the HOT line feeding into a full wave rectifier bridge which then drives the LED with the pulsing DC should work. If you don't want flicker then add a filter capacitor between the outputs of the rectifier, but you'll have to add a current limiting resistor between the filter cap and the LED as the filter cap will convert it from a current source to a voltage source.

kimbomba

Wed Sept 18 2013, 03:57AM

Registered Member #3854 Joined: Fri Apr 29 2011, 03:45AM
Location: Mexico
Posts: 95

Dr. Slack wrote ...

johnf wrote ...

Just use two leds back to back and a capacitor of the right value to get the current you need 1/2x pi x f x c will get the capacitance you need

... and it will send the led domes flying across the room the second time you turn it on. You also need a series resistor to limit the inrush current that the capacitor can take if it's switched on at peak mains voltage. Consult the LED data sheet for the largest surge current it can stand, and choose the resistor accordingly. You might be surprised at how large the resistor has to be to limit the current to 100s of mA, or less.

You are right, I just connected the led with the capacitor and a rectifier bridge and sometimes when you turn on, it lights up normal. Some other times it burns up (when switching is at peak mains voltage). So, is there a solution to this without adding a resistor that consumes power?

Bjørn

Wed Sept 18 2013, 06:42AM

Registered Member #27 Joined: Fri Feb 03 2006, 02:20AM
Location: Hyperborea
Posts: 2058

Transformerless Power Supplies: Resistive and Capacitive
Microchip application note - AN954

Dr. Slack

Wed Sept 18 2013, 07:02AM

Registered Member #72 Joined: Thu Feb 09 2006, 08:29AM
Location: UK St. Albans
Posts: 1659

but you'll have to add a current limiting resistor between the filter cap and the LED as the filter cap will convert it from a current source to a voltage source.

No. Look at my schematic. There is no series resistor between the filter cap and the LEDs, yet the LEDs are driven constant current by the high mains voltage through the high impedance of the series capacitor. Now if you were to detatch the LEDs from the cap, let the cap charge, then reconnect the LEDs, then you'd be in trouble without series R where you suggest. But you don't use it like that, do you?

You are right, I just connected the led with the capacitor and a rectifier bridge and sometimes when you turn on, it lights up normal. Some other times it burns up (when switching is at peak mains voltage). So, is there a solution to this without adding a resistor that consumes power?

No

But if you use a capacitor to be able to tolerate a big switch-on transient, then the resistor can be small, so dissipate very little power.

Choice between a full bridge rectifier and a single anti-parallel diode is efficiency. I was making an 'always on' background LED light for a window-less hall, so wanted max efficiency.

See Bjorn's link. It has stuff on safety, as well as doing the sums for component size and ratings (safety, we don't need no stinkin' safety!) I think I'll go back to my light PSU and add a fuse, if not a varistor.

Sigurthr

Wed Sept 18 2013, 01:45PM

Registered Member #4463 Joined: Wed Apr 18 2012, 08:08AM
Location: MI's Upper Peninsula
Posts: 597

wrote ...

but you'll have to add a current limiting resistor between the filter cap and the LED as the filter cap will convert it from a current source to a voltage source.

Oh, you're right, my mistake. For some reason I was thinking the initial surge current from the filter cap would be an issue, but that's only true if the load disconnects from the source, as you pointed out.

Ash Small

Wed Sept 18 2013, 03:28PM

Registered Member #3414 Joined: Sun Nov 14 2010, 05:05PM
Location: UK
Posts: 4245

Dr. Slack wrote ...

What you've got there Ash is a rectifier for the mains, so 170 or 340 volts on the cap depending on which side of the pond you live. Then all the power gets dissipated in the resistor.

OK, here's a revised design. C2 has a value 1000 times C1, voltage across both capacitors is 340V, voltage across C2 is 340x1/1001=~0.34V. The resistor in series with the LED dissipates ~0.001% of the power dissipated in my previous design?

1379518134 3414 FT157237 Led Mains Indicator

EDIT: Adding a bleed resistor to C1 would be advisable, in the interests of safety, maybe?

And maybe a fuse between the mains and the first resistor?

Steve Conner

Wed Sept 18 2013, 03:42PM

Registered Member #30 Joined: Fri Feb 03 2006, 10:52AM
Location: Glasgow, Scotland
Posts: 6706

This design is completely wrong. Insert Grumpy Cat meme here.

The most obvious mistake is that you have a half-wave rectifier with a capacitor in series. No current will flow at all.

Beyond that, you have 2 diodes in series doing exactly the same thing. One of them is redundant.

Ash Small

Wed Sept 18 2013, 04:37PM

Registered Member #3414 Joined: Sun Nov 14 2010, 05:05PM
Location: UK
Posts: 4245

OK, Now I'm back to this:

1379522223 3414 FT157237 Led Mains Indicator

C2 is 1000 times the value of C1, so the value of the resistor in series with the LED's is small, so doesn't dissipate much power.

Experimentonomen

Wed Sept 18 2013, 05:58PM

Registered Member #941 Joined: Sun Aug 05 2007, 10:09AM
Location: in a swedish junk pile
Posts: 497

Commercial "light bulb" led bulbs are NOT transformerless as you may think, no they have a small smpsu inside them.

Driving leds from the mains usinf just resistas and capacitors is outright wrong and stupid!!! It is NEVER done commercially, there always a smpsu or iron transformer based drive circuitry based used.

Dr. Slack

Wed Sept 18 2013, 06:15PM

Registered Member #72 Joined: Thu Feb 09 2006, 08:29AM
Location: UK St. Albans
Posts: 1659

Ash, have you actually looked at Bjorn's link? Very thorough, informative, the thread should have stopped with that.

I'm going to choose not to accuse you of trolling, yet. If C2 is 1000x C1, then the voltage across C2 will be 0.24v rms (non US) or half that (US), not enough to light a LED. Use the provided information, and think.

I have 50 LEDs in series, being driven at 20mA 160v from a series C from 240v AC into a bridge rectifier with a 120uF 330v photoflash cap across it, as I've drawn earlier in the thread, lighting my windowless hallway. Quite efficient, quite Christmassy. Runs cool, no transformer, so not at all stupid.

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