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117VAC and LED's

Discussion in 'Electronic Basics' started by Dark Alchemist, Nov 19, 2003.

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  1. Back in the 80's I know I made a simple circuit with 1 led and a
    resistor connected directly to the 117vac pwr main and it worked without
    a hitch. Now all these years later I have forgotten what the formula
    was I used and the values needed. Anyone have any ideas?

    Thank you.
     
  2. Jim Large

    Jim Large Guest

    You probably used Ohm's law to work out what size
    resistor would give you around 20 milliamperes at
    that voltage. You might have ignored the voltage
    drop of the LED (it's nearly insignificant compared
    to the supply voltage), and you apparently ignored
    the fact that most LEDs have a low (less than ten
    volts) peak reverse voltage rating.

    If you do it again, you might want to use a couple
    of extra components: A rectifier diode in series
    with the LED and a high value resistor in parallel
    with it (the LED) will protect it (the LED) from
    seeing full line voltage when reverse biased. You
    might also want to work out how much power is
    dissipated in the ballast resistor. (HINT: a 1/8
    watt metal film part is going to get toasty warm!)

    -- Jim L.
     
  3. If I remember right I used a long resistor so it was either 3 or 5 watts
    to dissapate all that energy and your right I used something like a
    1N1004 (man its been so many years) but it worked for a really long
    time. Since I wish to replace a neon bulb with an led what is the
    easiest route?

    Thanks again.
     
  4. Jim Large

    Jim Large Guest

    Well, I worked out that the ballast resistor would have to be
    around 6800 ohms, and it would have to be rated for at least
    two and a half watts.

    -- Jim L.
     
  5. Or use a capacitor, about 0.44uF I think, suitable voltage, and any old
    diode inverse parallel to the LED. Saves a lot of power.
     
  6. so c1 to the anode of the led and a 1n1004 connected in parallel? D1
    anode connected to L1 cathode and D1 cathode connected to L1 annode?

    What is this type of circuit called as I don't think I have seen this
    before without at least one resistor.
     
  7. The 0.44uF cap will limit the RMS current to about 3.3mA. You still need the
    diodes, though.

    One thing to consider is that there can be a big inrush of current to the
    cap when the circuit is first powered. Without a resistor, this can destroy
    the LED. I'd say stick with the resistor, or use both a resistor and a
    capacitor.

    The following should work:

    1uF 250V
    1k 1W
    ___ ||
    +-|___|----||---+--->|----+
    line| || | |
    ---- | |
    - V LED
    ---- ^ -
    neutral| | |
    +---------------+---------+
    created by Andy´s ASCII-Circuit v1.24.140803 Beta www.tech-chat.de

    Regards,
    Bob Monsen
     



  8. ---------+ ----------+ ----------+ ---------+
    | | | |
    | | | |
    .-. .-. | |
    R 68k | | R 6k8 | | R 6k8 .-. R .-.
    | | | | | | 6k8 - 120k | |
    1/4W '-' 1.5W '-' 3W | | 1/4 - 3W | |
    | | '-' '-'
    | | | |
    | | | |
    | V LED +-----+ +----+-+ |
    NE .-. - | | | A A |
    ( o) | si - LED V LED V | +-+
    '-' | ^ - - +-)--+
    | | | | | A A |
    | | +-----+ +----+-+ |
    | V 1N4007 | |
    | - | |
    | | | |
    ---------+ ----------+ ----------+ ----------+

    fig. 1. Old fig.2 Normal LED fig.3 Normal LED fig.4 LED in
    fashioned, simple, at about 20mA. The at 20mA. Si diode a bridge. Both
    low(est) power 1N4007 blocks the bypasses reverse half cycles
    consumption. reverse voltage. voltage of LED. used.

    created by Andy´s ASCII-Circuit v1.24.140803 Beta www.tech-chat.de

    To give a direct answer to your question, One LED + one resistor does not work. There are no LEDs I am aware of that can resist over 170V reverse voltage. If your memory is correct, I guess you got a "bipolair" LED. That's two LEDs in one package connected similar to fig. 3. Neglecting the LED voltage with respect to the 127Vac mains, you can calculate the resistor using Ohms law. Knowing a general purpose LED likes to feel 20mA, the resistor becomes:

    R=127/20=6k35.

    Nearest common value 6k8. You also have to take the dissipated power into account:

    P=20*127=2540mW

    Nearest common value 3W. It's not only the resistor that has to be able to withstand this heat, it also has to get rid of the heat. So air flow or heatsink may be nessecary. Which brings me back to the most oldfashioned, simple and cheap solution: The neon lightbulb.

    Common neon lightbulbs (fig. 1) ignite at 90V, glow at 60V and cease when the voltage sinks below that 60V. There are some different current ratings most of them in the range 1-2mA. Using Ohms law the series resistor for 1mA should be:

    R=(127-60)/1=67k

    Nearest common value 68k. Dissipated heat:

    P=1*60=60mW

    so a 1/4W or even an 1/8W will do. Overall power consumption is 127mW which hardly can be beaten by a LED. As a neon bulb has no polarity, both half cycles will be used.

    A general purpose LED (fig. 2) does best at about 20mA. But, being a diode it will conduct current in only one direction. As its reverse voltage usually has a maximum of 10V, you need ean extra diode - the 1N4007 - the block this reverese voltage. Neglecting the LED voltage and calculating for 20mA the series resistor becomes:

    R=127/20=6k35

    Nearest common value 6k8. Power consumption looks like:

    P=20*127=2540mW

    So 3W. But, the diodes only conduct one half cycle. The other half cycle current through the resistor will be blocked. Power is dissipated only half the time so a 1.5W resistor will do. Of course, light is only emitted when the LED conducts, so it only lights half the time. At 60Hz you may not see the flicker. Even at 50Hz flicker can hardly be seen or is acceptable.

    A less economical solution to prevent the LED from being blown by reverse voltage is connecting a cheap silicium diode antiparallel to the LED (fig. 2). You'll get not more light, but power consumption doubles and you have to rate the series resistor for 3W. If you can afford it (not for the money but for the heat) it's better to use two LEDs antiparallel or find a so called bipolair LED.

    Another way to reduce heat is lowering the current. A lot of LEDs may give light enough for your purpose at only 10mA. Some high efficiency types seems to do well at 2mA. Of course you can rise the resistor value accordingly (and lower the power rating) but at lower currents flicker may become disturbing. A simple diode bridge with cheap (1N4148 type) diodes will reduce flicker and give almost maximum light. Almost. If you want to squeeze the last foton out of your current, use LEDs for the bridge as well.

    I do not think I wrote the last word about this subject. But I'am sure I have some ideas.

    petrus
     
  9. Bill Bowden

    Bill Bowden Guest

    I'm not sure if the resistor is needed or not. When you turn the
    thing on at the peak of the line (170 volts) the current is limited
    mostly by the wire resistance and inductance, maybe an ohm or so.
    This means the inrush current could be 170 amps and decays by the
    time constant of RC or 1uS, and falls to almost zero in 5uS.
    So, we have an average current of say 170/2 amps for 5uS which
    amounts to 850 microjoules of energy if the LED voltage is 2.
    I don't think a millijoule of heat is much to worry about, it
    probably won't burn out an LED.

    I might run a test using a 2uF cap to see if I can destroy an LED.
    It should survive.

    -Bill
     
  10. I've placed a 0.22uF and a bi-color LED (red and green dies connected antiparallel in the same package) in series before (on my 120V 60Hz mains). After plugging it in and unplugging it three or four times, the LED was effectively destroyed. Light output of both colors of the LED diminished every time it was plugged in. The resistor definitely is needed (although in general I agree with your prediction and would speculate significantly less than 1 kiloohms are needed).

    One possible aggravating circumstance which led to the LEDs destruction was that the circuit was connected to the mains using a normal two-prong plug. Contact bounce could have occurred multiple times in quick succession while being connected or disconnected. This would have opened up the ability for the LED to handle multiple 170A surges in a short period. Further, since the capacitor would be floating at certain points, it would even be possible for the capacitor to charge up to over a hundred volts, be disconnected, and reconnected when the mains had swapped polarity, resulting in up to double the estimated current.

    Howard Henry Schlunder


    in message news:...
     
  11. If you use a single 1N4004 as a half wave rectifier, you will see a lot
    of flicker unless you filter the pulsating DC with a capacitor of a few
    dozen uF. You could use a full wave bridge and get 120 pulses per
    second, fast enough to reduce the flicker.

    You could also replace the resistor with a .47 uF 250VAC non-polarized
    capacitor, especially rated for connection to the AC line. I think it's
    called X2. The cap doesn't dissipate a lot of heat like the resistor.



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  12. Lord Garth

    Lord Garth Guest


    A reactance limiter does make much more sense.
     
  13. 1. You can get quite a current surge through the capacitor and the LED
    when you apply power. Add some resistance to limit the peak current to
    100 mA. You can probably get away with 200 mA (1K ohm resistor).
    Better - use a bridge rectifier. Then put a capacitor of perhaps
    100 uF or a few hundred uF across the LED. The current-limiting
    capacitor, of course, goes in series with either AC lead of the bridge
    rectifier. You should still have a resistor to limit surge current
    through the capacitor (and whatever switch is switching this), but the
    resistance requirement is down to just several ohms now. Downside - some
    LEDs (in general yellow-green, yellow, orange, most but not all red) do
    better when pulsed if the average current is only a few mA, but
    usually not an issue once the average current is around 20 mA.

    2. Capacitors that have AC line voltage across them need to be rated for
    such duty. Put 110 volts AC across any old .47 uF 200V capacitor and
    there is too much of a chance it will fail, with heating being part of the
    problem.

    - Don Klipstein ()
     
  14. At 1 amp the voltage across an LED is typically something like 15 volts,
    sometimes more. Looks like at high currents, an LED is a 10-15 ohm
    resistor. 10 amps will blow one in microseconds or less.
    Let me know... If you claim survival then I will risk one of my LEDs
    and some of my time to verify! 2 uF and no resistance, I would think
    there is even a slight chance of cracking the LED... maybe only mildly
    overkill to wear safety goggles! Acid test... Charge a 2 uF (or just .47
    uF) to 150 volts and discharge into an LED, to be sure that you aren't
    making contact with AC at moments when the instantaneous voltage is low.

    - Don Klipstein ()
     
  15. [snip]
    I've looked for a reference on what all the X, X2, and X2S
    designations mean, but I haven't found an explanation yet. I'd like
    to know which cap to use for those AC line voltage circuits.

    --
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    ###Got a Question about ELECTRONICS? Check HERE First:###
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    Subject: line with other stuff. alondra101 <at> hotmail.com
    Don't be ripped off by the big book dealers. Go to the URL
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  16. Dark Alchemist wrote:


    The circuit to replace a neon bulb with an LED looks about like this

    o ~
    |
    |
    ___
    ___ Capacitor, rated 1000 V DC /630 V AC
    | (blind resistance to yield 10-20 mA current)
    |
    \
    / Protection resistor
    \ 10 R, 125 mW
    /
    |
    _____|_____
    | |
    | |
    ___ ___
    /\ \ / DUS
    LED / \ \/ (1N4148 or similar)
    ---- ----
    | |
    | |
    ------------
    |
    |
    o ~

    Note that the capacitor needs to be a type that is certified for use
    across mains voltages. The value of the capacitor needs to be calculated
    such that the current through that circuit is 10-20 mA (depending on
    mains voltage and -frequency), 100 nF works well with 240 V, 50 Hz.
    Because of the voltage drop across the capacitor the remaining parts can
    be low power, as shown. Note however that each part of the circuit has
    galvanic contact to mains, adequate precautions need to be taken.
     
  17. X2 means the cap should not be used where failure can expose people to line
    voltages. Y2 means they can be used in this capacity. There is a picture at
    the bottom of this link:

    http://www.illcap.com/pdf2/Series/MKT.pdf

    There was a great tutorial I found a few months ago when I was looking at
    capacitors for some reason, but I can't seem to find it now. If I relocate
    it, I'll post it.

    Regards,
    Bob Monsen
     
  18. http://www.evox-rifa.com/technote_pdf/rfi_fact.pdf
     
  19. Problem with this circuit is the flicker is really noticeable.

    It is less noticeable if you replace the 1N4148 with another LED.



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    ###Got a Question about ELECTRONICS? Check HERE First:###
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    My email address is whitelisted. *All* email sent to it
    goes directly to the trash unless you add NOSPAM in the
    Subject: line with other stuff. alondra101 <at> hotmail.com
    Don't be ripped off by the big book dealers. Go to the URL
    that will give you a choice and save you money(up to half).
    http://www.everybookstore.com You'll be glad you did!
    Just when you thought you had all this figured out, the gov't
    changed it: http://physics.nist.gov/cuu/Units/binary.html
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