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Low cost mains power supply

Discussion in 'Embedded' started by Nomen Nescio, Apr 8, 2013.

  1. Nomen Nescio

    Nomen Nescio Guest

    In most DIY stores you can find these cheap Chinese digital timers to
    turn you lamps or appliances on at specified time. The thing I'm
    wondering is, how do they power the digital circuitry from
    110VAC in a device costing only $5-$10? Are they using a transformer or
    are they directly powering the digital circuitry from the AC voltage?
    Nomen Nescio, Apr 8, 2013
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  2. For a small amount of power (and efficiency in the single digit
    percentage range), a rectifying bridge, a zener for regulation, a
    current limiting resistor and a capacitor or two is all you need. A
    switch and fuse are probably a good idea too.

    Isolation is key for safety on something like that.
    Robert Wessel, Apr 8, 2013
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  3. Nomen Nescio

    Pete Guest

    They don't use a transformer. They usually use an X2 rated capacitor to
    drop the voltage, then some rectification and regulation. This might be
    as simple as a single series diode, a resistor, a zener, and a smoothing

    Or, there are small IC's around that basically only need the mains in
    via an X2 capacitor on one side, and deliver regulated 5v out at the
    other side.

    Pete, Apr 9, 2013
  4. Nomen Nescio

    Stef Guest

    In comp.arch.embedded,
    Google for [capacitor dropper] and you will find lot's of examples.
    Efficiency is not that bad, power factor is terrible.

    And keep in mind that this type of circuit is not isolated, so your 'low
    voltage' side is live AC!
    Stef, Apr 9, 2013
  5. Nomen Nescio

    Jasen Betts Guest

    I've got one here it has a capacitive dropper and a tiny 4mAh NiCd cell (I
    assume NiCd because the charge rate printed on it is C/10) the AC switch
    is a 48V relay and there's a TO92 device which probably switches it.
    image in ABSE
    Jasen Betts, Apr 9, 2013
  6. Nomen Nescio

    Robert Macy Guest

    I one time needed a super efficient 5Vdc [allowed to be directly
    connected to the mains] power supply but had NO room for the size of
    caps needed for the normal drop stage. So, I designed up a small
    circuit that connectd the mains to the rectifier diode starting at
    zero crossover and DISCONNECTING when the mains input went above
    around 7V and started to take the 5 Vdc output voltage up to 5.1, or
    so. thus I had efficiency AND regulation.
    Robert Macy, Apr 9, 2013
  7. Nomen Nescio

    Joe Chisolm Guest

    Look at some of the smart meter app notes from Atmel, Microchip, etc.
    They have examples of small offline supplies.
    Joe Chisolm, Apr 9, 2013
  8. Nomen Nescio

    John S Guest

    The mains voltage rises to about 7V input (5V output) in about 100us and
    to about 7.1V input (5.1V output)in 111us. So you have about 11us to
    charge your storage capacitor. Then, the storage cap must hold up the 5v
    for about 16ms until the next rise (this assumes 170V peak, 60Hz,

    I must be missing something very important and I would appreciate your
    help in understanding what it might be.

    Many thanks,
    John S, Apr 9, 2013
  9. Nomen Nescio

    Oliver Betz Guest

    I consider it useful to mention mains voltage (and frequency) when
    presenting such calculations.
    I also think that "7.1V" is somewhat tight.

    The Harris HV-2405E used this principle, of course with wider

    Oliver Betz, Apr 9, 2013
  10. Nomen Nescio

    Arlet Ottens Guest

    They were mentioned (170V peak, 60Hz, half-wave) in the part that you
    And of course, it all depends on the current consumption. With very low
    currents used, the cap would only discharge slowly.
    Arlet Ottens, Apr 9, 2013
  11. Nomen Nescio

    Oliver Betz Guest

    correct, sorry!

    Oliver Betz, Apr 9, 2013
  12. Nomen Nescio

    rickman Guest

    It may not be isolated, but it can be current limited and so not lethal.
    rickman, Apr 9, 2013
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