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LED on 230V

frimer

New Member
Hi

I'm planning to run a LED of mains power in a project. My idea is to use a series capasitor insted of a resistor. The circuit needs to fit in a very small space and i'm having prolems finding a capacitor rated for 230V AC that is small enugh. Does anyone have an idea of how a more regular capasitor would stand up to the mains voltage. I plan on putting in a small resistor in series in case the the capasitor should decide to die.

Anyone have any input?
 

marcbarker

New Member
I'm planning to run a LED of mains power in a project.
I'd use a neon + resistor myself.

Maybe this is a mass-manufactured product where every cent counts.

I say, if you must use an LED, I suggest you ditch the capactor and use only a resistor like a neon does. Get the most efficient mA / millicandela LED available. So as to reduce the current down to 'neon current'.
 
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frimer

New Member
Even with a very bright led I would have to disipate several watts, that is a lot of heat. How hot does neons get. Mybe I can cast it into something to make it look like a LED
 

Hero999

Banned
How small is small?

Please provide dimensions.

The capacitor can be made half the size if you use a bipolar LED which lights when the power is applied in either direction, it also means you don't need another diode to protect the LED.
 

frimer

New Member
I need to fit it through a 13mm diameter hole. my plan was to use a 12mm diameter pcb. My plan is for a 0,33uF capasitor, a small signal diode and aprox. 470 ohm in series for safety. The capasitors I have been looking at are much larger than this.

The only capasitors with 250V AC rating (I would like 500V) I have been able to find are class X1, X2 and Y2 and they are enormous.

I think I can reduce the value of the capasitor to 0,22uF without much trouble.

I should probably also put about 1M ohm across the capasitor. I just did a test setup with a capasitor i pulled from an old computer monitor and got quite a jolt from the capasitor after it was unpluged.
 

RODALCO

Well-Known Member
I have used 47nF 1 kV capacitors with success. They are bigger than a 5 Watt resistor.
You need a 820 or 1 k.ohm 1 Watt resistor in series to limit the inrush current.
It yields in about 5 mA LED current which is fine for indicating purposes.

Most >5000 MCAD LED's are already quite bright with 1 or 2 mA current through it, you don't need 20 mA hence a oversized cap.

Go with neons whcih you can run at 680 µA and a 270 k ¼ Watt seies resistor for 240 Volts
The neon itself drops about 50 - 70 Volts.
And you can get blue and green neons as well now these days.
 

RODALCO

Well-Known Member
1n4007

or,

go with resistors and a 1N4007 diode in series about of which I have posted various topics in this forum.

The 1N4007 reduces your power by 0.707 and doesn't get hot, then 2 x 22k ohm 1 Watt in series.

And insulate/sleeve all bare wire ends. you dealing with lethal voltages here.
 

marcbarker

New Member
what about a 'phase controller' + resistor? SCR turns on after the line voltage (pos half cycle) is falling, when voltage drops below say 50 V, the SCR turns of for the remainder of the half cycle (about 10 % duty cycle or less). Adding a bit of silicon and a couple BJT's might be cheaper than a 1 kV capacitor. Don't forget an ordinary 250 VAC rated capacitor is going to be a bit too fragile, the fail mode is going to be a s/c. A 'X' rated cap (not a 'Y' rated) cap that's self-healing according to the data sheet, probably can't be relied on either.

I'd say go for a neon!
 

frimer

New Member
what about a 'phase controller' + resistor? SCR turns on after the line voltage (pos half cycle) is falling, when voltage drops below say 50 V, the SCR turns of for the remainder of the half cycle (about 10 % duty cycle or less). Adding a bit of silicon and a couple BJT's might be cheaper than a 1 kV capacitor.
This sounds interesting, I did a quick google on SCRs, I'm guesing it would be something like the schematic i atached to this post. Maybe someone can help me a bit with this circuit.
 

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ericgibbs

Well-Known Member
Most Helpful Member
I should probably also put about 1M ohm across the capasitor. I just did a test setup with a capasitor i pulled from an old computer monitor and got quite a jolt from the capasitor after it was unpluged.
The UL standard requires that a 1 megohm is fitted across capacitors when used in this way also a 47R in series is specified.
 

taz128

New Member
Hello, in the image bellow you can find a circuit that I have runing for more than 2 years with no problem. This circuit is for 120 volt, but I have change the capacitor for 0.22uF @400v for 240 volt and the resitor is 1K 1w.
Hope this help you,
taz128
 

frimer

New Member
The UL standard requires that a 1 megohm is fitted across capacitors when used in this way also a 47R in series is specified.
UL standards bring brings up a whole lot of questions, but that will have to wait for later. Right now I'll just have to find a way to do it.

or,

go with resistors and a 1N4007 diode in series about of which I have posted various topics in this forum.

The 1N4007 reduces your power by 0.707 and doesn't get hot, then 2 x 22k ohm 1 Watt in series.
Will this get warm after constant use or maybe even hot?
 

RODALCO

Well-Known Member
Led

Nope, have used this set up for 20 + years with no problems.
If you have a good LED, you can use 2 x 33 k ohm resistors and reduce the current eve nore.

just test it out for yourself.

I use white LED pilot lights on time switches with a 1 Watt 100 k Ohm resistor and 1N4007 and they are still perfectly bright.
 

RODALCO

Well-Known Member

RODALCO

Well-Known Member
frimer, as you probably have worked out by now is that I like LED's a lot for indicator purposes, hence various threads on this forum.

Interesting I find that most people are obsessed to run LED's at 20 mA's.
No need for that with the new high efficiency LED's.
Most run perfect at a couple of mA's and for reliability use a resistive circuit, capacitors will work but a spike can still pop the LED.

Attached photo's I just took today from my white (13000 MCAD) led with a 100k Ohm 1 Watt resistor on 230 Volts and 1N914 in anti parrallel ( no 1N4007 used here ).
LED current 2.3 mA.
power dissipation in R 0.53 Watts.
If space allowes use 2 x 47 k ohm 1 Watt in series to spread any heat developed.

The red led (5000 MCAD) is controlled via a 39k ohm resistor and a 1N4007 diode

For comparison the ON / OFF status is visualised.
 

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