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Reduce voltage of DC adapter without reducing current

Discussion in 'Electronic Basics' started by suputnic, Jun 5, 2006.

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  1. suputnic

    suputnic Guest

    The OP's problem is that he doesn't understand that in order to bump
    The current I measure is after ignition....I'm guessing that the
    somewhat crypic comment
    means that a very large current is drawn during ignition, and that the
    reversed zeners and resistors are stopping this current from being
    drawn. If so then back to the drawing board.
     
  2. suputnic

    suputnic Guest

    means that a very large current is drawn during ignition, and that the
    (replies to self)

    No actually AFAIK the zeners don't resist current flow, so stiil not
    sure why the attempt mentioned above didn't work.
     
  3. Guest

    Lucky! It's been like a furnace out here, and it's gonna get hotter.
    Are you near Auckland? Christchurch? Someday I'd love to go skiing in
    South Island...
     
  4. I would hav ethought that was obvious from the first post.... Why is
    the OP an idiot? He has a legit question... (And if it took you this
    long to figure it out, who is the idiot? :) )

    As for the OP's Q, why not use a regulator?

    http://www.onsemi.com/pub/Collateral/MC33269-D.PDF

    Look up part MC33269T-3.3G - 3.3V regulator, only needs a couple of
    caps, total cost < $1 US. WIll take nearly any reasonable DC in.

    --Yan
     
  5. redbelly

    redbelly Guest

    Wasn't obvious to me. At first I thought the OP was talking about
    heating a single cup of water or something small like that. Didn't
    think about it deeply enough at first to realize a couple of 1.5V
    batteries wouldn't heat up even that small amount.

    Mark
     
  6. It just struck me as funny.... And I needed a laugh... :)
     
  7. suputnic

    suputnic Guest

    I'm in Auckland.
     
  8. suputnic

    suputnic Guest

    Thanks but I don't think any circuit with resistors is going to work,
    see the post by John Fields.


     
  9. ehsjr

    ehsjr Guest

    Oh - I thought your DC adapter was 5.9 volts.

    Here are some of the options I have tried:
    (good info snipped along with the misleading
    current measurements)

    Ok - but your current readings are meaningless.
    Not your fault - it's the nature of the circuit.
    I believe the igniter is drawing *much* higher
    current, but for brief moments. The net effect
    is that your current measurements mislead you.
    What is meaningful is the works/does not work
    observation.

    Many/most of these DC adapters have current
    capability well under 1 amp - and my bet is
    you need far more current than your adapters
    can provide. Try this: CAT# DCTX-5200 from
    http://www.allelectronics.com/

    It will give you 5 volts at up to 2 amps.
    You can put a couple of 1N540x 3 amp diodes
    in series between the adapter and the igniter
    to put the voltage in the right neighborhood.

    Ed
     
  10. Jasen Betts

    Jasen Betts Guest

    John fields knows his stuff, but look at the circuit.
    you see the resistors aren't in the path that the current must
    take to power the igniter.

    the only device blocking the current is the LM317.

    I'm fairly sure John was referring to circuits where the
    resistors are in the supply path.
     
  11. John Fields

    John Fields Guest

    ---
    22mA

    Please bottom post.

    It's not that resistors keep it from working, I'm guessing that the
    current required from the source, during ignition, is so high that
    almost anything you put in series with the igniter is going to drop
    the voltage to the point where the arc won't strike.

    Just considering that the spark will dissipate 12 watts at 12Kv, if
    it has 1 milliampere in it, means that the input current to the
    igniter must be greater than 4 amps at 3 volts during ignition.
    Easy for the batteries to supply, with their low impedance, but
    considerably harder for anything else.

    So, it may also be that the source itself (the wall-wart) is
    incapable of supplying the short-term ignition current requirements
    of the igniter. At least the higher voltage ones. I recall you
    said the 5V one worked. no?

    So what is it you're trying to do? Outfit a bunch of heaters with
    some junk wall-warts you've got hanging around?
     
  12. redbelly

    redbelly Guest

    As do we all. Can't fault ya fer that.

    Mark
     
  13. suputnic

    suputnic Guest

    So what is it you're trying to do? Outfit a bunch of heaters with
    That is correct.
    I suspect this may be incorrect, because every wall wart I have tried
    lights the burners no problem, and most are rated at under 1 Amp.

    The higher the adapter voltage, the faster it sparks and starts.
    However once I get up to 12V nominal adapters, it keeps sparking after
    ignition. I am going to use up all my 9V and less adapters to start
    with.

    I am a bit concerned that overvolting the input by such an amount (9V
    nominal = 11V actual, compared to 3V specified) may shorten the units'
    life. Or will it just be dissipated as heat?

    I may look into the voltage regulators and other circuits mentioned by
    others for the high voltage adapters, thanks all for your assistance.
     
  14. suputnic

    suputnic Guest

    Thanks for your input, but as I say below every wall wart works no
    problem, and most are rated under 1 Amp. I don't think the diodes will
    work because they will have a resistance of a few ohms, and the
    resistance needs to be kept much lower than this as John Fields has
    noted in a couple of places.
     
  15. ehsjr

    ehsjr Guest

    Your reasoning is wrong. Diodes do not have a resistance
    of a few ohms.

    You have two alternatives to try - the diodes and the voltage
    regulator circuit. Report back after you have tried them.

    You can even get a kit from Dick Smith Electronics - K3594
    that will give you 3V out with 12V in if you are reluctant
    to try to build one from scratch. With the confidence you
    gain from the $6.49 kit, you'll be able to build a regulator
    from the schematic I posted from scratch.

    Try the diodes first - they are only 9 cents apiece from
    Dick Smith - Z3204. Figure 1 volt drop per diode. The
    exact drop will depend on the current drawn - these 1N4007
    diodes drop ~1 volt at 1 amp.

    Ed
     
  16. suputnic

    suputnic Guest

    I tried a 12V nominal (16V measured) adapter with a 1N4148 small signal
    diode attached forward biased to the positive lead. It fired up the gas
    a few times then stopped working. When it did work it kept sparking
    after the gas was lit. I managed to get a voltage masurement, it was
    about 9V.
     
  17. suputnic

    suputnic Guest

    Oops that diode is only good for 100mA I've found out. I'll have to get
    some rectifier diodes and try them.
     
  18. almo

    almo Guest

    Thank God!

    I can't believe I had to read this far down before somebody mentioned
    using a 3v regulator :)))
     
  19. suputnic

    suputnic Guest

    Chaining forward biased rectifier diodes in series to the positive lead
    does the trick, but it's a bit of a messy solution. I still would like
    to know why adding a reverse biased zener diode to the positive lead
    does not work.
     
  20. suputnic

    suputnic Guest

    Oh I think I know why. I tried a 15.5V measured adapter and a 10V Zener
    diode, but under load this combination would drop to less than 3V. I
    will try a smaller Zener diode.
     
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