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Variable DC supply circuit - any critiques?

Discussion in 'General Electronics Discussion' started by ohnoezitasploded, Mar 7, 2014.

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  1. ohnoezitasploded

    ohnoezitasploded

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    Mar 7, 2014
    Hey All,
    I've got a gnarly old stage lighting lamp that I'd like to repurpose as a desk lamp - it would only need to go to 10% of full output or so.

    My first thought was just to wire it up to a wallbox dimmer. But I think the fact that I really only need 5-10% of the output is going to make it difficult to trim in exactly where I want it, and I'd like to be able to set it to having the filaments barely glow as a nightlight.

    Also, it has some metal contacts on the socket that I could cover up, but I was thinking it would be cool to leave them exposed.

    So what I need is a 300W 0-~16V variable DC supply.

    I was thinking about the schematic for a typical triac dimmer, which for reference looks like this (the real version would have an inductor so it doesn't also function as an AM radio :) )
    [​IMG]

    If I added a diode going into the diac D1, the triac would only turn on when the voltage was from +0-60V on the AC wave. Then I could put a big filter cap on it to smooth out the ripple current, and I would have DC current from 0-~45V. Like this:
    [​IMG]

    Then I could increase the value of R1 to control the top trim...

    Thoughts? Thanks for the help, I appreciate it!
     
  2. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    Jan 21, 2010
    No. you can't get DC by doing what you suggest.

    I'm not sure where you intend to place an inductor.

    I'm also not sure I like the combination of exposed metal contacts and mains... especially in a night light.

    edit: Actually, let me add a warning that if you connect a capacitor the way you suggest it is likely to fail *very* quickly, and possibly explosively.
     
  3. ohnoezitasploded

    ohnoezitasploded

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    Mar 7, 2014
    Yeah, I took a look at the schematic and realized I botched it, here's a second try:
    [​IMG]

    The word nightlight may have gotten you concerned-- I don't have kids, it doesn't have to be child-proof (since a fragile glass bulb would be a poor choice for this anyway), I'm just trying to preserve some of the industrial character of the socket. I figure that if I only supply it a maximum of 14V DC, it's going to be safe enough that I'm not going to shock myself, which is why I'm pursuing a variable DC output.
     
    Last edited: Mar 8, 2014
  4. ohnoezitasploded

    ohnoezitasploded

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    Mar 7, 2014
    Here's kind of a fuller explanation of how I'm thinking this circuit would work, as I understand it anyway:

    So in the typical triac dimmer schematic shown in the very first image, R1 + VR1 control the charging time of capacitor C1. When Diac D1 sees more than +30V or less than -30V, it opens, which switches triac TR1 on. Then the voltage reversing causes the capacitor to drain, which closes D1 again,

    So the resulting waveform out of the dimmer looks like this:
    [​IMG]

    What I'm proposing to do is put a diode before the Diac D1, so that only the positive waveform comes out of it. But now that I look at it again, the correct solution would be to just put a diode on the *output* of the triac, and filter that.
     
  5. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    Jan 21, 2010
    It would probably be a lot safer to get a small transformer and run it from a low AC voltage or even to power it from a DC power brick. Or if you want to go mains operated, just use a standard light dimmer.

    I'm not certain what benefit you get from your circuit here.
     
  6. ohnoezitasploded

    ohnoezitasploded

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    Mar 7, 2014
    Right, but I want it to be adjustable-- from just a bare glow on the filament up to say enough light to read by.

    Circuit like this:
    [​IMG]
     
  7. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

    8,393
    1,268
    Nov 28, 2011
    I don't think that latest circuit will work either. The triac circuit needs AC across it to work, and the diode will block current flow during the negative half-cycles. Actually I'm not sure that the phase control method in conjunction with a smoothing capacitor is workable, no matter how you rearrange the circuit.

    In your case, how about using a variable DC power supply? Either one that runs from the AC mains and has an adjustable or switch-selectable output voltage, or convert the AC mains to, say, 24V fixed, and use a variable buck converter to drop it down.

    Either of those methods gives you isolation from the AC mains, which is really important if there's exposed metal on there, as Steve has already pointed out several times!

    Your lamp won't draw much power at such a low voltage. You can calculate its resistance at full power using R = V^2 / P then calculate its current drain at a lower voltage from that. Its filament resistance will be somewhat lower at lower brightness though.

    What about the colour temperature at low power? Normally, incandescent bulbs look pretty yellow at low power. White LEDs are a good way to get low light levels with a good colour temperature.
     
  8. ohnoezitasploded

    ohnoezitasploded

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    Mar 7, 2014
    Here's a picture of the bulb, just to give some context of what I'm trying to do here:
    [​IMG]
    And the socket:
    [​IMG]

    The main problem is that it's a 300W lamp. I don't need to supply anything like the rated 120V, but at say 12v that's 25A, and I don't know of a way to supply that amount of current short of using a triac to provide a switching supply.

    The idea of using DC was appealing to me just to add a measure of safety, but I just realized that's going to be a whole lotta ohmic heating in the wires. So scratch that.

    Maybe I use the standard triac dimmer circuit, and just increase the value of R1 so that it tops out at 15% of the waveform? Thoughts?
     
  9. ohnoezitasploded

    ohnoezitasploded

    6
    0
    Mar 7, 2014
    Wait, my math is wrong!

    So if the lamp is 300W at the rated voltage of 120V, that's about 48 ohms. At a maximum of 15V, that's 0.3A. And 300mA is completely do-able from a buck-boost converter.

    Okay then! Sorry for the confusion on my part, I've been out of school for awhile haha :)
     
    Last edited: Mar 9, 2014
  10. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    2,772
    Jan 21, 2010
    My advice would be to connect the lamp to a bench power supply (do you know anyone with one?) and measure the current as you wind the voltage up.

    You'll find that the current at 12V may be in excess of 1A as the resistance of the filament increases as it warms up. Measuring the cold resistance would also be a great idea.

    The main thing you can get out of a variable power supply is the knowledge of what the lamp looks like at different voltages.
     
  11. duke37

    duke37

    5,361
    767
    Jan 9, 2011
    Try a car battery charger on 12V to get an idea of the glow.
    If you have a tame friend with another battery charger, you could put them in series to see what it looks like on 24V.

    You seem to think that DC is safe. I am not so deluded.
     
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