Connect with us

HELP to make LEDs work with AC current

Discussion in 'Electronic Design' started by Mark Main, Oct 8, 2008.

Scroll to continue with content
  1. Mark Main

    Mark Main Guest

    I have a transformer that is reducing the AC supply voltage down to
    25.5VAC, which connects to a thermostat that connects to a Honeywell
    water valve for my home boiler system, which then completes the
    circuit back to the transformer. I don't see an amperage limit marked
    on the transformer--it's about baseball size, so it's not small.

    Only 1 of the poles of the relay is used and I would like to use the
    other poles of the relay switch to power a red LED (either a standard
    LED or a Bi-Polar LED).

    This is my home, safety is top of the list (e.g. in my research I've
    seen some people offer designs that don't use resistors with the LEDs
    or they connect diodes straight across the power supply, both of which
    don't seem like they maximize safety).

    I want the design to still be safe when one of the parts does
    eventually fail.

    I need help with the design, and I'd especially like to know the
    formulas that were used in the design because I eventually want to
    create some LED lighting under my kitchen counter that uses 110V AC...
    So if you can help me calculate how I alter the design for 110V vs.
    this current 25.5V design that would be great.

    I'd like to know how to do this using standard LEDs and also how to
    also do it using the bi-polar LEDs.

    Thanks for any help... I'm new and learning and have only done a few
    basic projects so far.
  2. Guest

    From Google:

    The usual disclaimers apply; use at your own risk; I am NOT an
    electrical engineer qualified to make assessments re: electrical

    Folks, any comments on the safety of those circuits? Electronic
    current-limiting countermeasures recommended?

  3. Guest

    Even simpler (no bridge rectifier):

    I recall seeing a schematic similar to the above, but I just can't
    find the exact link anymore...

  4. The reactance of the capacitor at 50 Hz = 1 / 2.pi.f.C =
    1 / (314 . 22.10^-8) = 100000000 / (22 x 314) = 14475 Ohm, say 15 kOhm.
    The current at 230 V = 230 / 15000 = 15 mA.
    You need no resistor.

    Hopefully I did the math right.
  5. Rich Grise

    Rich Grise Guest

    Hopefully, yes, but you _do_ need a resistor to limit the inrush of
    charging current through the cap when you switch it on.

    And at 25V, you shoudn't need to do any tricks - say your LED is
    1.2V Vf, subtract that from 25V and calculate the resistance at
    that voltage that gives 10 mA. Also calculate the power, and size
    the resistor accordingly.

    To run it on AC, put a 1N400x (or any GP power diode) in antiparallel
    with the LED to limit its reverse voltage.

  6. That is very true.
    How inductive do you think the mains feed is?
    Could be all sorts of caps from filters in parallel.
    So for a peak of 230 x sqrt(2) = 325 V and a max charge current of say 100mA
    (most LEDs can handle that as pulse) Z would have to be 3250 Ohm.
    Then at 15mA eff the power would be .7 W, so a 1W resistor of 3k3.
    So indeed 4k7 / 1W would be fine too.

    And, in case no diode in the box, use 2 LEDs in anti parallel.
  7. Phil Allison

    Phil Allison Guest

    ** Both are VERY dangerous and NOT what the OP needs at all.

    He has a 25 volt transformer !!!!

    Piss the hell off - you ridiculous shithead.

    ....... Phil
  8. Phil Allison

    Phil Allison Guest

    "Rich Grise"

    ** LEDs respond to the average value of DC current flow - which is about
    10% less than the rms value when sine waves are involved.

    PLUS: a LED is a *diode* and only conducts half of the time with AC feed -
    so the average current is about 45% of the value found by the simple calc
    you suggested.

    The series resistor's power dissipation WILL have to be calculated using
    rms values.

    ** Correct.

    ...... Phil
  9. Mark Main

    Mark Main Guest

    I've read the thread and I still can't figure out how to do the math
    to calculate what the relationship should be between the resistors and
    the capacitors when trying to drop down from my 25.5V to an LED of say

    I've also done some other reading, and in this example (see links
    below) "Roff" advises connecting the diode in parallel to the LED and
    have both connected in series to the resistor.

    Right now I'm seeing lots of information, but I can't organize all
    this info into actual math formulas that help me decide what
    resistors, capacitors, and diodes I should be selecting; and how to
    safely arrange them.

    I'm especially safety concious when I hook this up to 110V, but even
    at 25.5V I want to be safe.

    Thanks to all who are helping.
  10. ehsjr

    ehsjr Guest

    For the 25V design, you need a 2200 ohm 1/2 watt resistor
    in series with the LED and a 1N400x diode in antiparallel
    with it. That is explained below.

    A simple approach may be best for you. An LED needs to have
    the current limited to a (relatively wide) range, and needs
    to be protected against reverse voltage. Placing a diode in
    antiparallel with the LED accomplishes the latter. To limit
    the current you can use a resistor, and since the range is
    relatively wide, precision is not necessary.

    To determine the resistance needed to limit current to 10 mA
    in a circuit with a 25 volt source, figure 25V/.01A
    That equals 2500 ohms. A close standard value resistor is
    2200 ohms. If you use a 2200 ohm resistor that limits
    current to no more than 25V/2200ohms which equals 11.36 mA,
    which would be fine for an LED. Such a resistor would need
    to dissipate about .284 watts, so you would use a 1/2 watt
    resistor. That is enough figuring in this case for your
    needs, but it is not the full story.

    Your current will actually be a little lower than the 11.36 mA,
    because the LED drops the voltage roughly 1.5 volts. So for
    completeness, you can re-compute using (Vs-Vf)/R where Vs is
    the source voltage and Vf is the voltage drop of the LED.
    Then you get (25-1.5)/2200 or ~10.68 mA

    The answer to your second question (about how to design the
    circuit for 110 VAC for under cabinet lights): DON'T.
    It won't be safe.

    Instead, get a wall wart with a safe low level DC output,
    and compute your resistor using the formula used above:
    Vs-Vf/I = R where I is current, R is resistance, Vs is
    the source voltage and Vf is the voltage drop of the LED.

  11. neon


    Oct 21, 2006
    LEDs are diode in essence depends on what diodes you buy depends on the forward drop and forward current. usualy 2v to 4v @10-20ma. you must know these parameters then it becomes simple no matter what the voltage source 1000v or 10v a certain amount of current must flow thru the diode to be visible . there are safefty precautions it is called power dissipation you may not exceed this parameter or puff goes the device. for ac just add a diode of the 1n4000 series to pevent reverse breakdown at 120v 240v ac. So now to limit the current add a resistor in series to limit the current. suppose 120ac add 48 2.5v LEDS in series then add a resistor of 500 ohms 1/2w too many LEDS? then 120/20leds 2.5v leds=50v then the resistor becomes 120-50=70v same current 0.010ma then R= 70/.010=7k or 6.8k now we need a bigger resistor and more power like 2w dissipation. any of the above LEDS voltage current can be used. good luck
  12. Mark Main

    Mark Main Guest

    Thank you. That gave me the info that I was seeking. I'll be rigging
    it up this weekend.
  13. What I like to do is use a bridge rectifier to have both halves of the
    AC cycle bcoming DC for the LED. A bridge rectifier will protect the LED
    from reverse voltage while also never giving the LED reverse voltage.

    Put the resistor upstream from the bridge rectifier, so that the
    resistor limits current if the bridge rectifier shorts. And 400 volt
    bridge rectifiers are cheap.

    Have a resistor of sufficiently high power rating that no more than
    50-60% of the power rating is actually dissipated into the resistor - and
    consider worst case input voltage. Measure what the transformer's
    or wallwart's output voltage is when loaded by nothing but either a
    voltmeter, or a voltmeter in parallel with a resistor drawing a few mA.
    That can be well above the nominal output voltage.

    I have seen what the failure rate of resistors is when power dissipation
    gets to about 60-70% of the rating. It is low, but I consider it
    significant. A few times I have already fixed things where the problem
    was a resistor failing while dissipating about 60-70% of rated power. In
    one case, the resistance decreased greatly.

    In critical applications, there are "flameproof" resistors. Some are
    even UL recognized components.

    One more thing - you can reduce heat production here: There are now
    plenty of LEDs that get plenty bright at just a few milliamps.

    - Don Klipstein ()
Ask a Question
Want to reply to this thread or ask your own question?
You'll need to choose a username for the site, which only take a couple of moments (here). After that, you can post your question and our members will help you out.
Electronics Point Logo
Continue to site
Quote of the day