Discussion in 'Electronic Basics' started by George Economos, Jan 16, 2006.

1. ### George EconomosGuest

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:

+----|<---+
24VAC 2K2 | LED | 24VAC Switched
o-------/\/\/\------+ +---------o Return
Switched | |
+---->|---+
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

2. ### Jonathan KirwanGuest

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

3. ### James T. WhiteGuest

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.

4. ### Anthony FremontGuest

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.

5. ### Jonathan KirwanGuest

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

6. ### Bill BowdenGuest

As he'd find out quickly. Just from RMS, 24^2/2200 is just as you
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

7. ### Jasen BettsGuest

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

8. ### Dr Engelbert BuxbaumGuest

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.