• posted

George,

LED's typically have a fairly low reverse voltage rating (5V or so) and the parallel diode limits the maximum reverse voltage the LED will see to .6V or so.

```--
James T. White```
• posted

I wanted to thank everyone for their input on my previous post. I ended up implementing the following circuit to light an LED from a

24vac source:

+----||---+ DIODE

What is the rationale for having the LED and the diode in "anti-parallel"? One reason given was to limit the reverse voltage. What does that mean? Can someone please explain the above circuit to me?

thanks, george economos

• posted

The 1/4 watt rating is just what is guaranteed, the resistor can probably dissipate several times that without failure. I once tried to use a 1/4 watt 1/2 ohm resistor as a 3 amp fuse, but it didn't work. At

5 amps (12 watts) , it just got red hot and smoked off all the paint, but still measured 1/2 ohm when it cooled off.

-Bill

• posted

The reversed diode _will_ limit the reverse voltage seen by the LED. It will limit it because the reversed diode will conduct during that half-cycle at under 1V across it. So that will be the most that the LED can see when it isn't conducting, itself. If that reversed diode weren't present, then the reverse voltage would be 24*sqrt(2) or about

34V worst case because the reverse leakage current through the LED is so small that the 2K2 resistor wouldn't drop any of it. It's a reasonable arrangement.

Jon

• posted

What power rating did you pick for the resistor? At the nearly 34V peaks of the AC sine wave, the resistor will be dissipating over 1/2W. Since the signal is AC, the average dissipation should be about half that. If you picked a 1/4W resistor, chances are it's hot as a firecracker and won't last long. Use at least a 1/2W resistor.

• posted

As he'd find out quickly. Just from RMS, 24^2/2200 is just as you say, about 1/4 watt. I'd go 1 watt for the resistor, though. 1/2 watt is still too close for comfort. Do I hear a bid for 2 watts?

Jon

• posted

immagine it isnt there. when the 24Vac is positive and the swiitched return is negative no noltage will flow through the LED (assuming it's an ideal diode) and so by ohm's law there's no voltage lost in the resistor. The peak of 24VAC is about 32V so the LED would experience 32V LEDs aren't built for more than about 5V reverse so it'd die. the antiparallel diode acts as a "pressure releif" by conducting and allowing to most of voltage to be lost in the resistor so the LED sees less than 1V reverse (which it can withstand easily) Bye. Jasen

• posted

You are using AC, so the LED (like any diode) conducts only during one half wave, not during the other.

When the LED conducts, LED and resistor form a voltage divider, limiting the voltage the LED sees (and the current that flows through it) to a safe value.

If the second diode where not present, no current would flow through the circuit during the second half wave, where the LED does not conduct. Thus no voltage would drop across the resistor (remember Ohms law) and the LED would be exposed to the full peak voltage of the power supply (sqrt(2) * 24 V = 34 V). That's what is meant by reverse voltage. However, LEDs don't like reverse voltages in excess of 5 V or so. Thus in the absence of the second diode the LED would be destroyed.

The second diode creates a path for the current to flow during the second half of the cycle and limits the reverse voltage to 0.7 V or so. This is safe for the LED.

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