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Old 12th November 2004, 08:21 AM   #1
AndrewT is offline AndrewT  Scotland
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Default LED on mains voltage?

I want to monitor the presence of voltage on each of the three phases in my house (230/415Vac). We suffer very frequent power cuts.
I can put in a diode followed by a resistor then a LED.
The resistor dissipates quite a lot of power. or
I could use a constant current source but it would need to be Hi Voltage. or
I could wire in three transformers & then LED at low volts.
I don't like any of my solutions.
Can anyone suggest a cheap, safe & simple circuit?
regards Andrew T.
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Old 12th November 2004, 08:36 AM   #2
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I dare saw you could put a diode in anti-parallel across the LED, and then use a class-X2 capacitor as the current limiter.

A 33nF, 250v ac rated capacitor should do the trick. Don't forget the reverse-connected diode across the LED to limit the reverse voltage!
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Old 12th November 2004, 09:24 AM   #3
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That was going to be my exact suggestion! However, if he is experiencing frequent power cuts, might there be lots of spikes on the lines? These would pretty much pass straight through the cap blowing the LED. Or do X2 caps have 100 ohms inside them, and this would prevent the damage?
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Old 12th November 2004, 09:31 AM   #4
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I don't think mains transients would blow the LED. LEDs can take very considerable pulse currents through them provided that they are of short duration. (Think of the high pulsed forward currents used in cycle LED lights).

It would be easy enough for the guy to try out anyway.
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Old 12th November 2004, 10:52 AM   #5
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You can check my softstart and how I drive both a LED and relays.

As noted, beware of transient protection becasue they go just right through the cap.

http://home.swipnet.se/~w-50719/hifi/sst01
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Old 14th November 2004, 04:09 PM   #6
AndrewT is offline AndrewT  Scotland
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Hi All,
Do I understand this correctly?
build a string of diode, cap, and LED. The diode only lets half the AC waveform through, the cap acts as a voltage reducer without power loss (no heating) and then the LED lights up.
Q1: Do I need a resistor in parallel with the cap as shown in Peranders schematic?
Q2: Do I need a resistor in series with the LED?
Thanks so far
Andrew T.
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Old 14th November 2004, 04:17 PM   #7
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Almost

Connect the cap is series with a LED and Diode That are parallel ,but opposite Anode to cathode and cathode to anode.

As the cap charges on the positive swing, it'll flow current through the led, as it discharges and charges neg, it'll flow current through the diode (preventing too much reverse voltage, damaging the led)
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Old 14th November 2004, 04:26 PM   #8
AndrewT is offline AndrewT  Scotland
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Hi Brian,
the diode//led pair see the full AC waveform but each only flow current in one direction. Obvious really!!
thanks
Andrew T.
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Old 14th November 2004, 06:52 PM   #9
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Quote:
Originally posted by AndrewT
Hi All,
Do I understand this correctly?
build a string of diode, cap, and LED. The diode only lets half the AC waveform through, the cap acts as a voltage reducer without power loss (no heating) and then the LED lights up.
Q1: Do I need a resistor in parallel with the cap as shown in Peranders schematic?
Q2: Do I need a resistor in series with the LED?
Thanks so far
Andrew T.
Q1: It's common pratice to have a bleeder because it's quite unpleasant if you touch 100 nF or more charged with 300 V DC worst case.

Q2: You should have a transient protection of 3.4-6.8 volts, Zener 1.3 watts or a tranzil.

(Cap//bleeder) + (Res 1-2 Watts) + (Zener//(Res+ LED))......(//= in parallel with)

The resistor may be 100 or something whoch will limit the LED current at zener voltage.

Cap = 100-470 nF(250 VAC/630 VDC) at 230 VAC

Bleeder = 3-5 X 100-470 kohms, 0.6W metall or high voltage types, like 1250 volts total resistance, 220k-1.5M ohms

Res 1-2 W = 470 ohms- 1k

Zener = 1.3 W, 3.4-6.8V, or tranzil 600W

Res(LED) = 100-220 ohms 0.25 W

LED = any kind
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Old 14th November 2004, 11:33 PM   #10
cpemma is offline cpemma  United Kingdom
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Let's get back to basics. The capacitor has the nice property of being a resistance (actually reactance) without generating heat, so is the main voltage dropper.

Reactance = 1/2.Pi.f.C, so a 0.33uF will be 9645R on our 50Hz. Add a 1k series resistor to limit surge as the capacitor first charges, led current will be 22mA RMS on 230V AC but only half the time, the resistor will produce 0.48W so no big problem, a 1W will do.

You still need a 1N4004 in inverse parallel with the led to protect it from reverse current.

Another thought would be to just run SPCO relays on the mains, with a battery-powered buzzer to act if the coil loses power.
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