How Can I Connect a 230V AC to a 4V LED?

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To connect a 230V AC to a 4V LED, using a transformer to step down the voltage is recommended for safety and efficiency. A transformer with a 24V output can be used, requiring a 1500-ohm resistor to limit current to 20 mA for the LED. A diode, such as the 1N4007, is suitable for rectification, while a capacitor rated for at least 40 volts is necessary to smooth the output. The choice of capacitor size, like 330uF, balances charging current and ripple, but alternatives like 1000uF are also acceptable. Proper calculations ensure the circuit operates effectively without excessive heat or inefficiency.
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hi,
im looking to connect a 230V AC to an 4V LED. i was suggested a thyristor could be used then a rectifier to convert AC to DC then using a resistor to drop the voltage to 4V. i dan't seem to understand how a thyristor can aid this process. Any suggestions, please help?
 
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I suppose you could wait until very late in a mains cycle before turning the SCR on.
This would give you brief pulses at some low maximum voltage.

[PLAIN]http://dl.dropbox.com/u/4222062/late%20SCR%20triggering.PNG

For example, if you turned on the SCR at 177 degrees and the mains changed polarity at 180 degrees, you would get about 12 volts peak, but this would only be available for a very short time... 3 degrees in 360 or 1% of the time. So, you would need to use a large capacitor with a very large peak charging current to make use of this very small pulse.

You would also need some pretty elaborate switching circuitry to control the SCR to such a small angle, accurately.

It would be much better and safer to use a transformer to get 12 volts or so.
 
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hi,
thanks for your replies, you also wrote to me about the pulsating LED.
I though about the transformer and found quite a small on to reduce the 230 to 6v, i just need guidence on how to choose a rectifier and a capacitor, as you said it makes the LED brighter, and the brighter the better :) please help.
thank you
 


If you have a 6 V secondary on a transformer, you can run a single LED like this:

[PLAIN]http://dl.dropbox.com/u/4222062/half%20wave%20LED%20driver.PNG

This uses minimum components and should give about 20 mA into a 3.5 volt LED

You can vary the size of the resistor to alter the LED current.

Almost any small power diode would be OK for the diode. 1N4007 diodes are usually cheap.
There will be some ripple on the output, but not enough to matter.
 
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thanks for the help. i got a transformer the only one available has an output of 24V, 250mA. Am i correct in sayin that a 220 ohm resistor will work for a 4v LED in this case? and what capacitor can be used? could the same one be used?
 


That will be very inefficient.

The resistor would have to be 1500 ohms and the capacitor should be rated for 40 volts, at least.

Here is how you work it out:

The 24 volt transformer output has a peak value of (1.414 times 24) or about 34 volts.

You lose 0.6 volts in the diode and there is a 3.5 volt drop across the LED.

So, the resistor has about 30 volts across it and it is carrying 0.02 amps (20 mA).
R = 30 / 0.02 = 1500 ohms
The power the resistor uses will be (30 volts times 0.02 amps) or 0.6 watts. So, you need a 1 watt resistor.

Incidentally, you could run 7 LEDs in series with a 470 ohm resistor from that transformer.
 


hi again,
i was jus wondering why you used a 330uf capacitor in the circuit, and how did you come to choosing it?
 


That was just a compromise between charging current and final ripple.

I ran a simulation of it and juggled the capacitor size to get the optimum value.

Cost usually goes up with capacitance, too, so it pays to get the capacitor that is just good enough for the job it has to do.

It is not all that important, though, and if you could only get 1000 uF capacitors, one of those would be fine.
 
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