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gpigeon

basic electronics question re: LED's.

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I have in the past driven a normal LED form 240VAC by using a resistor. Now I cannot find exactly what resistor value I used.

However, one of my ref books gives the formula

R=(Vsource - Vled)/Iled

Does this formula work for 240VAC?

Will a flashing LED work on 240VAC as well?

If someone has an online reference to this subject I would appreciate a link.

Thanks & "Dankie".

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You will want to choose a resistor to limit the current through the LED to an acceptable value so the LED doesn't break.

10mA is a good rule-of-thumb for most LEDs unless you can get a datasheet for it.

I would probably limit the peak current to about 10mA, so for 230Vac you need to design for about 325V.

The LEDs forward voltage is negligible when the supply voltage is so high, so to get the resistor value, use

325V / 10mA = 32.5K (So I guess a 33k resistor would be the smallest I would use)

The above videos show bigger resistors, so I guess they use even less current, which just means a dimmer LED.

You also need to be careful that you get a resistor with a high enough power rating that it doesn't burn.

You can calculate the power lost in the resistor as 230^2 / 33K (Or whatever resistor you choose) = 1.6W

Since an LED is a diode, it will only conduct during 1 half of each AC cycle, so the average power through the resistor will be halved.

Giving about 0.8W dissipated in the resistor. I would usually use a resistor with a rated power of at least 2x what I expect it to be dissipating,

so in this case I would probably use a 33K, 2W resistor.

 

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1 hour ago, Stanley said:

You also need to be careful that you get a resistor with a high enough power rating that it doesn't burn.

That's what I recall about this, that for line voltage this is pretty much a bad idea. The LED runs at around 2V and 10mA, assuming the garden variety 0.25W kind. So on 230V you essentially have to drop 228V over that resistor, but that's just about the same as just dropping the entire line voltage, so that resistor will dissipate around 2.5 watts. That's a fairly hefty resistor, and it will get quite hot too!

Much better to use a capacitive dropper circuit. This is usually as simple as a 200nF capacitor (or thereabouts) in series with the LED, though please do some research about this. It does the same thing as the resistor, except using the reactance of the capacitor. And it is much more efficient (but it has a horrible power factor... but at 10mA that's not going to matter).

(Of course this response is for @gpigeon, I just quoted the applicable part from Stanley's response).

Edit: I also have to add, that with the resistor you'll be using around 2.5 watts to light up a 0.25W LED. That's about a 10% efficiency 🙂

Edited by plonkster

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2 minutes ago, plonkster said:

Much better to use a capacitive dropper circuit.

I agree, although I would probably use a combination of a capacitor and a resistor to be safe.

You can calculate the capacitance required in a similar way to the resistor calculation I posted above.

for a reactance of 33k at 50Hz, you would get 33k = 1/(2*pi*50*C)

Rearrange for C = 1/(33k * 314.15)

So C is about 96nF (100nF is a common value)

The current is very low, but you need to make sure the capacitor is rated for 250Vac.

You can add a resistor of a few k in series as well for a bit of extra safety.

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You will also need to add a parallel diode (e.g. 1N4007 or similar), connected in reverse to the LED, so that the LED is not exposed to a large reverse voltage when the mains voltage polarity is reversed.  Most LEDs will have a maximum reverse voltage rating less than 10V so will degrade quickly without this reverse diode protection.

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3 minutes ago, NigelL said:

You will also need to add a parallel diode (e.g. 1N4007 or similar), connected in reverse to the LED, so that the LED is not exposed to a large reverse voltage when the mains voltage polarity is reversed.  Most LEDs will have a maximum reverse voltage rating less than 10V so will degrade quickly without this reverse diode protection.

Yes, I was actually wondering about the reverse voltage rating of LEDs after my first post earlier, it isn't something you normally need to think about.

 The reverse diode is a good idea. Just remember that the resistor power rating will need to double then because it will be conducting during both half-cycles.

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Thanks for your replies gents.

I tried both the 68k & 135k resistors....both work ok and I cannot see any difference in brightness of the LED. Did not use a cap.or diode! Why do I need these? I underrstand the reverse polarity of the AC supply but it seems to work ok without them.

Comments????

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13 hours ago, gpigeon said:

Did not use a cap.or diode! Why do I need these?

The capacitor method is much more efficient. The capacitor lets through a small bit of charge on each cycle, but it simply holds the charge and does not waste it as heat (well, except for a tiny bit). So it acts like a resistor, but without the heat loss.

The reason for also adding an extra diode to this circuit is because most LEDs can't handle a potential 350V reverse voltage. The maximum reverse voltage is as low as 5V, look at the data sheet. When the LED is reverse biased, it acts like a zener diode, it clamps the voltage and conducts current. This may or may not damage the LED, but it will shorten its life. A 1N4007 diode in series costs a few cents (and you can pilfer them from blown household LED/CFL lamps, then they cost nothing).

Additionally, it's adviseable to still use a current limiting resistor even when using a capacitor.

See these for info:

https://en.wikipedia.org/wiki/Capacitive_power_supply

https://www.instructables.com/id/Single-LED-on-230V/

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