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Line voltage to 5VDC supply, transformerless

Discussion in 'Electronic Basics' started by HC, Dec 18, 2008.

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  1. HC

    HC Guest

    Hey, all, I tried to build a circuit that I found online and had a
    problem with it. I e-mailed the creator but didn't get a reply so
    here I am.

    What I have is a 120VAC device that switches on and off through
    existing circuitry; it is a fully functioning device and circuit.
    Periodically the existing circuitry will power up the device for a
    period of time. I want a circuit that will detect when the current
    device cycles from off to on to off and, upon returning to the off
    state, will enable a second device for a configurable period of time.
    That is, I have a circuit (I did not create or build) that controls a
    120VAC load. I want to introduce another 120VAC load that will run
    for a set amount of time each time the original load turns on and then

    My goal is to use a 555 timer in mono-stable configuration controlling
    a 5 VDC relay to switch the 120VAC line voltage to my second load. I
    found this circuit:
    Look at the third circuit under AC Line powered LEDs.

    I bought all the components specified except the 270 ohm 2 watt
    resistor. I had to buy two 560 ohm 1 watt resistors and run them in
    parallel (measured 273 ohms). The thing works..but the resistors get
    hot as sin in about 30 seconds and the whole time the voltage is
    creaping up. When I say hot I mean they are painful to touch and give
    off an odor. I added two more resitors for a total of four, two in
    series, two series in parallel and still they get hot (painful). With
    the four in parallel/series I measured a little over the 560 ohms the
    capacitors were designed to be. Still, that's over twice the
    resistance the orginal circuit calls for and I thought that would
    reduce the current and therefore the heat. Apparently I am wrong. :)

    I did some searching in here before making this post and the best I
    found as an alternative is on Microchip's site:

    It's TB008.

    I can do that (see Figure 4 in that PDF) but I only have 1.5 uF
    capacitors in that voltage range (specially purchased for this
    project) and, mostly, I hate to abandon an idea that fails without
    understanding the failure. Plus, the Microchip shematic doesn't
    indicate the wattage of the resistors (that I noticed) whereas the
    first circuit I listed I'm wondering if I can get away with
    1/4 watt resistors or not. I bet I make popcorn. And, as I get
    nervous as hell handling 120VAC in these kinds of situations
    (electronics where wires are bare and alligator clips are in use), I
    thought I'd see what experience might be out there to learn from.

    I could get transformers but that will make my project much more bulky
    and I'd like some finesse here. :)

    Thank you for your time and help.

  2. Phil Allison

    Phil Allison Guest


    ** Yep - you are.

    Did you use a film cap for the 1.5 uF or an electro ??


    Scroos doen to the secion headed

    " 7 Cheap Death ( or How Not to Design a Power Supply )

    The utter moron who created and published that circuit ought to be taken out
    and shot.

    ...... Phil
  3. HC

    HC Guest

    Hey, Phil, thanks for the reply. I used a "xicon radial metallized
    polyester film capacitor" according to the label on the baggy from
    Mouser. I took care to get a non-polarized cap. I would think an
    electrolytic would have gone of like a grenade. :(

    I looked at the site you referenced and I see the point there, I
    believe: tying the ground to neutral at the device is silly
    (dangerous). So, if I use the schematic they mention and if the
    device was installed by a trained professional so that the polarity
    was properly connected (hot to the positive side, neutral to the
    negative side, earth left connected to the chassis for its intended
    purpose (to trip a circuit breaker if a short-circuit happens), then
    this could work. I think.

    My problem is that I need three low voltage DC sources from 120VAC:
    one to power the 555 timer circuit all the time, one to take the
    120VAC feed to the load and make it an input-level voltage for the 555
    timer circuit mentioned, and one to run a relay with a 5 VDC coil
    (since I already bought some).

    So, if I build the circuit in section 6 of your referenced web page,
    hardwired it myself to be sure the polarity was right, then couldn't
    that work?

    Thank you for your time and information. The site you linked to maded
    sense and I see the danger inherent in the Microchip design.

  4. Phil Allison

    Phil Allison Guest


    ** You got some pathological aversion to using a small transformer or wall
    wart ? ??

    Please explain.


    Under NO circumstances would I ever suggest that any *novice* like you
    build an electronic circuit that operates from the AC supply without one.

    ...... Phil
  5. We had a discussion about this a while ago, and I have used similar
    circuits for some applications where the entire circuit is contained in an
    enclosure and the interface to the outside world is an optoisolator or
    relay. The Microchip circuits are particularly bad because of the incorrect
    use of a fuse in the neutral leg.

    As Phil pointed out, you probably used a polarized electrolytic capacitor,
    which will break down during reverse voltage excursions and cause excessive
    current flow. You were probably lucky that it did not explode. The 270 ohms
    limited the current to about 1/2 amp which would be about 70 watts in the
    resistors. They would get seriously hot almost instantly.

    I have had problems with the series resistor getting much hotter than
    expected, even with properly sized metal film capacitors, and it may have
    been due to high frequency harmonics on the AC line (possibly if used on an
    inverter). The best solution is really a small wall-wart supply that is
    relatively fail-safe and isolated, and can be had very cheap (or free).

    Otherwise, you can use a power resistor to drop the line voltage. You can
    use a series diode to eliminate the reverse voltage of the applied AC, or a
    full wave bridge for better efficiency. You can get 30 mA out if you use a
    resistor of R=120/.03 = 4000 ohms, which will dissipate about 3.6 watts.
    That is a lot of heat, and even a 10 watt resistor will get too hot to

    There is a linear regulator that will accept rectified line-voltage up to
    450 VDC, and will put out 3 mA continuous or 30 mA pulses.

    It should be possible to make a very small line operated switching supply
    for 1 watt or less, but I have not found any commercially available. The
    difficult part would be making such a small transformer with the number of
    turns necessary. About the smallest line operated switchers I have come
    across are about 5 watts, and they are about 2" x 3", like these:

  6. HC

    HC Guest

    No, nothing pathological. It's just that since I need one to drive
    the logic circuit and maybe also the relay, and another to transform
    the original load supply voltage, that would mean I would need TWO
    transformers and, if there is a way to elegantly do it without that
    overhead, I'd like to do it. It's a terrible waste to put two
    southern-engineered wall-warts into play (there won't be any
    electrical outlets, this will all be hard-wired) when there might be a
    better, more economical solution.

    It's interesting that you use the word pathological because it
    describes a mentally disturbed condition. I say it's interesting
    because you have a mentally disturbed condition, Phil. Listen, you
    chastised me for a fair question once before and I didn't get all
    nutty on you when you replied to me today because you seemed to have
    some useful information. I'm willing to let bygones be
    bygones...chalk it up to you bein' on the rag at the time or
    whatever. But this is twice, Nut Job. Listen to me carefully, I'm
    sorry you have a small dick, or you're fat, or you have beady eyes, or
    AIDS or whatever is wrong with you. For whatever you have that you
    think is wrong with you, I'm sorry, but I didn't make that happen, so
    don't take it out on me. I come here seeking information, prostrating
    myself before those who know more. I don't need you telling me with
    callouts that I'm a *novice*. I know I don't know it all. Hence the
    post here. It's called LOGIC, I highly suggest you look into it.
    Listen. I read your posts from years back, prior to when you first
    roasted my ass used to be a nice guy and handle questions
    with some respect and knowledge and care. I don't know what happened
    to you...your wife left you, died of cancer, maybe. Something bad
    happened, or maybe your mind just rotted...some kind of disease.
    Maybe you woke up one morning and decided you like other men. I don't
    know...but your latter posts are full of hate and self-loathing thinly
    disguised as sarcasm to those of us online who seek knowledge from
    those we believe know more than us. To attack us for not knowing is
    pathetic. Get a cup of coffee. Find a comfy spot in the house (tell
    your mother to get her fat ass out of the way if she's already in that
    spot) and take some time to think about yourself and your goals and
    your life and get your head right with yourself. Really, man,
    straighten your life out. Don't waste our time and don't take it out
    on us.

    And Merry Christmas!

  7. Phil Allison

    Phil Allison Guest

    "Halfwit Cunthead "

    No, nothing pathological.

    ** Oh I am absolutely sure it is !!!


    Electrocute yourself - ASAP


    ..... Phil


    You cannot legally hard wire home made devices to the AC supply.

  8. HC

    HC Guest

    LOL! You are so weak! Really, laughing out loud! Oh, and they say
    entertainment bargains don't exist anymore! Ha! This is good stuff!

    My sides hurt or I'd write more....

  9. HC

    HC Guest

    Hey, Paul, thank you for your reply. I did not use a polarized
    capacitor unless I'm a complete doofus (I tried to be careful about
    choosing a non-polarized capacitor and got, according to the baggie
    before me, a "xicon radial metalized polyester film capacitor").
    That's from Mouser's baggie description. FWIW. I believe that is not
    a polarized capacitor.

    I would gladly go to a small wall-wart but here's the problem I see:
    I have 120VAC supply that needs to supply my watch-dog 555 timer
    circuit, my 5VDC relay, and also a separate 120VAC circuit that is the
    trigger. It goes like this: a 120VAC circuit comes to life
    periodically. It is energized for a while and then shuts off. When
    it shuts off I want my device to come to life and run for a while,
    powering another 120VAC device. So, I thought a 555 mono would do for
    the "come to life and run for a while" thing, and it would power the
    coil on a 5 VDC relay which would switch the 120VAC to my newly added
    device. But that means I need a way to feed the 555 a trigger that
    follows the 120VAC feed to the first device...and that means I would
    need two transformers: one for the power to the logic circuit/relay
    which needs constant power and one to feed the 555 a state signal (off-
    on-off). I could do it, and I suppose the space is available
    but...the environment is such that less space is better and the
    environment is slightly corrosive so it'd be great if the whole thing
    was small and could easily be encapsulated in epoxy or some other
    protection. :-/

    Thank you for your time.

  10. How much power are you interested in pulling from this circuit?

    You can analyze the circuit by using the 'phasor' method, where the
    impedance of the cap is Z = 1/(2*PI*f*C)

    Using Zout = sqrt(Zc^2+R^2), the current will be

    I = 120/sqrt(Z^2 + R^2) ~= 67mA

    The power dissipated in the resistor is therefore 1.23W. So, the
    resistors will get hot, but should not fail.

    The other thing to note is that the zener is passing any current your
    circuit isn't using. If you have a 1k load, then the 5mA through that
    will lower the current through the zener to about 60mA. Thus, the
    power dissipated by the thing will be

    60mA * 5.6V / 2 = 168mW

    So, you need a 1/4W zener.

    Since the power only flows into your load 1/2 of the time, however,
    you can only use 1/2 the maximum current above, a whopping 33mA.

    You should also use a fuse with this circuit to prevent a fire if the
    cap fails closed. Use a 'slo blo' fuse, because the initial blast of
    current to charge up the cap can be something like 120/270 = 444mA. A
    slow blow 250mA fuse should be fine.

    The maximum voltage across the cap is about 170V in both directions,
    so you need either a single non-polar cap that is designed to
    withstand that voltage (and more), or two polar caps that are arranged
    in series, facing opposite ways, and that can withstand twice that
    voltage (initally; it'll eventually calm down to 170V due to leakage).

    Note that a little transformer will be safer, more efficient, and much
    cooler than the circuit above. You can get fairly small transformers.
    To get equivalent current to the circuit above, you would need about a
    1/2VA transformer. The page below has a 1/2VA transformer that is less
    than an inch tall. See the 161C10 here:

    It is less than 1 in on a side.

    You can get 100mA at 5V out of it if you wire it in parallel, rather
    than the 33mA you'll get from the circuit you picked.

    Bob Monsen
  11. Jasen Betts

    Jasen Betts Guest

    they will get hot there is no way to cure that that doesn't
    compromise the circuit, if it produces too much heat you could try
    reducing the capacitor.

    the circuit shown is unsafe in that it can present a high voltage on
    the plug after it's unplugged.

    use a transformer instead
    you could use a switched mode-powersupply but they are complex to design
    and build so you'd be buying one pre-built, possibly you could
    scrounge one from an old phone charger etc...

  12. HC

    HC Guest

    Hey, Bob, thank you for the reply. Okay, I got a reply from the
    creator of that schematic I referenced and between what he said and
    what you say I understand this a lot better. He suggested using
    capacitive reactance so I looked up a calculator for that and tried
    some things (I did a lot of reading between yesterday and today) and
    here's what I came up with: 120 VAC / 1,768 ohms capacitive reactance
    (1.5 uF @ 60 Hz) = 68 mA. You took the time to post the formula so I
    did it that way, too: 1/ (2 * pi * 60 * 0.0000015) = 1,767.67 Okay,
    now divide the voltage by the capacitive reactance squared multiplied
    by the resistance (is that the restance of that resistor in series
    with the capacitor, the 270 ohm one in the circuit I referenced?):
    120 / sqrt(1768 ^ 2 + 270 ^ 2) = 120 / 1788.5 = 67 mA. And that gets
    us to the dissipation of 1.23 watts (0.067 ^2 * 270) 1.21 (I rounded
    my numbers off as I went along).

    Okay, so now I know a lot more than I did when I started. The relays
    I bought consume 66.5 mA (measured) which means this circuit won't be
    able to supply the power I need for my logic circuit and the relay.
    So, a transformer is a guarantee now. Thank you for the reference to
    the PDF, that thing should work fine.

    However, I still need a way to trigger my 555 from the 120 VAC circuit
    I'm trying to monitor. So, using the information above, could I do
    this: put two 1.5 uF caps in series for (1.5 * 1.5 / (1.5 + 1.5)) =
    0.75 uF capacitance, with a capacitive reactance of 1/ (2 * pi * 60 *
    0.00000075) = 3,535 ohms. 120 / sqrt (3535 ^2 + 270 ^2) = 120 / 3545
    = 34 mA which gives me 0.034 ^2 * 270 = 312 mW, about one fourth the
    power I was dissipating before, well within the energy dissipating
    rating of two parallel 560 ohm, 1 Watt resistors.

    I could then feed that into my 555 trigger and power the 555 and relay
    from the transformer. The Zeners I have are 1 Watt, so I should be
    plenty safe there (34 mA * 5.6 / 2 = 95.2 mW).

    I can put a fuse on it as you suggest, no sweat.

    Am I doing this right? I could even use some kind of an optocoupler
    between the 120 VAC bastardized "transformer/trigger" circuit and the
    555 logic circuit so they wouldn't have to share ground and there'd be
    no way the 120 VAC could come to full line-level through the 555 logic
    circuit. Sound right?

    Thank you for your time.

  13. HC

    HC Guest

    Hey, Jasen, thank you for your reply. I have gotten a lot more
    information now and have found that I will have to go with a
    transformer for my logic circuit and the relay. However, I still need
    a way to trigger the 555 when the 120 VAC monitored circuit cycles
    from off to on to off. I put up another post detailing my intent.
    Basically, if I put two capacitors in series (I already bought the 1.5
    uF caps) I can drop the capacitance and reduce the current, put a fuse
    on this thing, and trigger my 555 circuit with it. This thing will be
    hard-wired but your point is well taken about the cap retaining
    voltage. A few schematics I looked at had a 1 meg resistor in
    parallel with the cap to discharge it and I will definitely do that.

    Bob had replied with a link to a tiny standard transformer that will
    do what I need. The relay I have chosen draws 66.5 mA (measured), and
    the 555 circuit draws something like 2 mA (measured) so the
    transformer that Bob suggested, with a capacity of 100 mA, should work

    Yeah, the switch-mode would be a pill to design and my circuit is so
    silly-simple that the power supply would be about 95% of the project
    effort. :)

    Thanks again.

  14. IanM

    IanM Guest

    Lot of work. Programmable time delay relays with 120V AC supply,
    control and load circuits are easily available sub $100.

    However if you *must* build it, consider rolling your own optoisolators
    with neon bulbs and phototransistors. You *could* also get there with
    simple resistive droppers into optoisolators (two in parallel with the
    source side LEDs in antiparallel so they conduct on alternate half
    cycles for a fairly steady oytput or a shunt diode for reverse
    protection accross the LED if you dont mind a pulsed output. Direct line
    derived low voltage supplies are over-complicated for simple on-off

    A commercial product would probably be non isolated with direct line
    sensing via a resistive dropper but the hobbyist without a dedicated
    isolation transformer fed workbench is well advised to stick to projects
    with integral isolation and with the non-isolated parts of the circuit
    kept as simple as possible. The experianced skip-rat would scavenge an
    old mobile phone charger for the isolated low voltage supply and look in
    larger switched mode power supplies for suitable optoisolators or roll
    their own as above.
  15. neon


    Oct 21, 2006
    I don't see the advantage of using caps reactance to drop the line as opposed to just a resistor and diode . it is cheaper this way . if anything a neon in series will greatly reduce the resistor power.
  16. Jamie

    Jamie Guest

    You need a one-shot timer which can be done via a 555.

    The remote relay or signal will simply hold the Reset line of
    the timer low while the first device is on!..
    When the first device turns off, it'll release the hold on the
    reset line..

    Look up one-shot timer.."
  17. Herman

    Herman Guest

    I have been using direct line operated supplies for years. In most cases I
    use a .047uF 600 volt cap as a current limiter. In order to keep the
    current requirements lower I use a low current 24 volt relay. Tell me again
    what you want to achieve via e-mail and I can give you a circuit and relay
    that works. I must apologize for the response you got from the news group.
  18. ehsjr

    ehsjr Guest

    A single wall wart can do it.

    Connect a 5.5 V regulated wall wart like Allelectronics CAT#
    DCTX 532 - to the AC input to the existing device at points
    A and B below.

    ---------- ---------
    |Existing |---A---|Existing |
    |Switching | |Device |
    |Circuit |---B---| |
    ---------- ---------

    The circuit below will give you an adjustable delay to the second
    device you want to add - maybe long enough. You didn't say how long
    the second device should run, so that's why there is a maybe.

    The 5.5V in the diagram below comes from the 5.5V wall wart plugged
    in to points A and B above.

    +5.5 ---+----+-[D2]-[16R]-+----+----+
    | | | | |
    [Rly1] [D1] [Rly2] [C1] [Rx]
    | |a | | |
    Gnd ----+----+------------+----+----+

    AC----o o--------o o-----AC to second device
    | | | |
    |< | <
    Rly1 N/C Rly2 N/O

    AC------------------------AC to second device

    Use a 1 Farad super cap for C1 (CAT# CBC17), a 16 ohm 5 watt
    resistor, 1N400x diodes and 5 volt relays with contacts rated
    to switch power on/off to the second device. The maximum time
    you can get the second device to run depends upon the coil
    resistance of Rly2 - the higher the resistance, the greater
    the time. If you don't get enough time you can add another super
    cap in parallel with C1. If it runs too long, you can reduce the
    time by adding a resistor at Rx.

    When the existing device turns on, the 5.5 volt supply turns
    on and energizes Rly1. The normally closed points on Rly1
    open, preventing AC from reaching the second device you are
    adding. Shortly after Rly1 energizes, Rly2 energizes. When
    power drops to the existing device, it also drops to the
    5.5V supply, and Rly1 drops out, and its points close completing
    the path for AC to reach the added device. Rly2 stays energized
    while the super cap discharges through it, until the cap voltage
    drops low enough for the relay to drop out. As long as Rly2
    is energized, the second device gets AC through the closed
    point on Rly1 and the normally open point on Rly2. D2 prevents
    the cap from discharging through Rly1. It depends on the
    individual relay, of course, but in general terms Rly2 will
    remain transfered for a bit over 2 time constants, and with
    a 1 Farad super cap, the time constant will be about equal to
    the relay coil resistance. Figuring a 130 ohm relay coil, the
    second device would be on for about 4-5 minutes. If you add a
    resistor (Rx), that will reduce the time constant. You need
    to compute the power dissipation for Rx if you add it.
    P = I^2R, I = 5/R

    Parts are available from Allelectronics:

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