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Connecting a LED directly to a 230v isolation transformer

carbonblack

Jul 16, 2022
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What should be the voltage across the resistor if measured with a multimeter and what shoud be the resistance of the resistor? (Is it the average of the peak voltage?)

And is it gonna be still AC voltage across the LED?
 

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Martaine2005

May 12, 2015
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In an ideal world, 230V - 2V = 228V across the resistor. But we don’t live in an ideal world and 230V will certainly vary (quite a bit).
In your schematic the current through the LED is 20mA so an absolute minimum of 5W resistor but larger is highly recommended as suggested above. Lowering the LED current allows for smaller wattage resistors. 1mA you might get away with 1/4W but 1/2W would be recommended.
What is it for? A power indicator?.
It will probably be flashing visually to AC frequency which isn’t nice and is very crude. You could add an inverse parallel diode or a series diode. But I don’t see the point.


Martin
 

carbonblack

Jul 16, 2022
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In an ideal world, 230V - 2V = 228V across the resistor. But we don’t live in an ideal world and 230V will certainly vary (quite a bit).
In your schematic the current through the LED is 20mA so an absolute minimum of 5W resistor but larger is highly recommended as suggested above. Lowering the LED current allows for smaller wattage resistors. 1mA you might get away with 1/4W but 1/2W would be recommended.
What is it for? A power indicator?.
It will probably be flashing visually to AC frequency which isn’t nice and is very crude. You could add an inverse parallel diode or a series diode. But I don’t see the point.


Martin
Won't the LED act as a half wave rectifier in this case and the average voltage will be less than 228V.
Vpeak = 230/0.707=325V
Vpeak after LED = 325V - 2V = 323V
Vavg = Vpeak/π = 323V/π = 103V

Isn't 103V the voltage across the load?
 

Bluejets

Oct 5, 2014
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Mains connect LED not recommended at all.

Led outer casing and therefore the AnodeCathode come out into the real world so to speak and if the LED blows, which is almost a given, mains live terminals will be exposed and extremely dangerous.
 

Harald Kapp

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And is it gonna be still AC voltage across the LED?
Yes of course. Why not?

This setup will not work at all. On the negative part of the sine wave (negative with respect to the forward direction of the LED) the LED will be reverse biased by 325 V (230 V × sqrt(2)). The LED will break down much earlier and will be destroyed.
 

carbonblack

Jul 16, 2022
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I
Yes of course. Why not?

This setup will not work at all. On the negative part of the sine wave (negative with respect to the forward direction of the LED) the LED will be reverse biased by 325 V (230 V × sqrt(2)). The LED will break down much earlier and will be destroyed.
I'm not going to practically do this. I just needed a theoretical explanation. That's all. Thank you for replying.
 

crutschow

May 7, 2021
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An efficient way to limit the LED current is to use a series capacitor of the proper size, instead of a resistor.
For example for 20mA of peak current at 50Hz, 230Vac, the value should be about 200nF.
The capacitor must be non-polarized with a 400V minimum voltage rating.
A inverse connected parallel diode (or another LED) must be added to provide the reverse current of the capacitor.

A small series resistor (e.g 1kΩ, 1/2 W) should be added to limit the peak transient current when power is first applied.

LTspice simulation below:

upload_2022-7-16_9-10-58.png
 

Bluejets

Oct 5, 2014
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I'm not going to practically do this. I just needed a theoretical explanation.

Mains LED drivers these days (those encased within lamps) use a dedicated driver chip along with rectifier and other associated parts all mounted on a heatsink base.
Quite an involved bit of technology available for a few dollars, latter due to mass production in some Eastern country or other.
Such as this......
https://www.daraz.pk/products/led-b...-15w-9w-7w-5w-3w-i119698518-s1273868049.html?
 
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