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Explain Circuit Please

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

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  1. 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. 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. 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. 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. 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 Bowden

    Bill Bowden Guest

    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 Betts

    Jasen Betts Guest

    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. 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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