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converting USB 5v to 3.6v

Discussion in 'Electronic Basics' started by Mark Lin, Jul 4, 2003.

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  1. Mark Lin

    Mark Lin Guest

    Hi group:
    I have a little circuit that takes 3.6 battery. I dont know its
    current draw, but the original battery was a Saft 3.6 volt with 50 mA
    output. The battery is running out way too fast, so I have the idea
    of getting power from my USB port. I know it's possible because they
    also sell the USB powered version of the circuit but it wasn't
    available at the time when I bought it. I know USB port can put out
    maximum of 5V and 500 mA(with data port used) or 100mA(without using
    data port). Can some enlight me on building the voltage converter and
    maybe I need a current regulaor(?).

  2. Kasper

    Kasper Guest


    What about using a 3,6V Zenerdiode and a resistor ??

    +5V----RESISTOR------ZENERDIODE (-|<-)---- GND
    3,6V out to GND
  3. Larry Hatch

    Larry Hatch Guest

    DIODEs drop .7 volts, two will drop 1.4. 5v-1.4v=3.6v Will that work?


    Will that work?
  4. Mike Ring

    Mike Ring Guest

    It will work, and save power, but you are vulnerable to a short circuit
    which could take out the supply.

    With a conventional if the load shorts it does not short the supply, so if
    power consumtion is not the main consideration it's prolly better

    Mike R
  5. Mike Ring

    Mike Ring Guest

    But if you use a series current limiting resistor the voltage across that
    resistor will vary with losd current and you won't have a constant voltage

    If your load is costant, just use the appropriate series resistor and
    forget the semiconductors.

    If it's not, the zener diode _in parallel_ with the load will soak up
    current variations within design limits, ie how much current is available
    and how much you are prepared to "waste".

    Incidentally I've just looked at your numbers again and you quote .7 volts
    as diode drop. IIRC (I'm long retired), that a geranium diode drop, a
    silicon diode, much more bulletproof and common, drops 1.4 ish volts, so
    one of them will do in series - but the vulnerability to resultant damage
    from a short still applies.

    You can also use a series zener of any voltage to get a required volts
    drop, but the s/c condition still applies (and when I last looked zeners
    didn't go down to 1.4 volts, I don't know if thats because it's impossible,
    or given the standard diode drop, it would be silly)

    Mike R
  6. Mark Lin

    Mark Lin Guest

    So using a silicon diode with 1.4 volts drop would be a good
    choice? Can I purchase the part in Radio Shack? So I guess the
    scheme would like this:
    power ---> diode ---> circuit?
  7. I disagree with Mike.
    As a rule of thumb, I'd point
    - 0.7 V drop on a silicon diode;
    - 0.5 V drop on a germanium diode;

    Go for two 1N4007 (silicon diodes) in series like this



  8. Mike Ring

    Mike Ring Guest

    Many aplogies for duff gen: Horowitz and Hill (what I stole on early
    retirement) says that it's .6 - .7 for a geranium and .9 - 1.0v for a
    silicon (.9 for the 1N4007)

    I used to use the 1N4006 for everythig, steering, quenching, rectifying
    being small and bulletproof, etc.

    I think my 1.4 V may have been the foreard drop of a LED (good for giving a
    light as well as a volts drop - IIRC the backwards drop of a LED was a
    handy 2V, but I stand to be corrected on all that, it's more music theroy
    and piano for me these days.

    Any how, experiment!

    Mike R
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