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Dropping 1V from a Regulated 6V Wall Wart

Discussion in 'Electronic Repair' started by Watson A.Name - Watt Sun, Dark Remover, Dec 29, 2003.

  1. I needed to drop 1V from a 6VDC 200 mA regulated wall wart, so I tried
    a 3 amp rectifier, but it varied by more than .2V over a range of
    loads. So I tried this: (view with courier font)

    + From
    wall
    wart >--+
    |
    +---+------+
    | |
    | |
    400 \ / 2SC2334 or TIP31
    ohm / |/ NPN power TO-220
    WW \<-----| Heatsink optional
    pot / |\
    | E\
    | |
    | |
    +----+-----+
    |
    |
    +------> +
    output
    to load
    - >--------------> -
    From wall wart

    This has some advantages and disadvantages. It's simple and cheap,
    and keeps the output at 5V within a tenth of a volt over a current
    range. But it has a minimum current below which it loses regulation
    and the output starts to go up to 6V, because the transistor is not
    conducting and the current is being supplied thru the ww pot. This
    circuit is sometimes used in the bias circuit for the output
    transistors in high powered amplifiers. Also Win Hill showed us here
    how to use a similar circuit to maintain the voltage steady for a
    current regulator circuit used on four AA cell rechargeable batteries.

    I'm thinking about putting a 5.1V zener on the output so that if the
    voltage climbs above that, it just shunts the excess current. Oh,
    yeah, I set the pot to various values to see what the output voltage
    was with various loads. The two resistances were 120 ohms for the
    upper and 280 ohms for the lower. I suppose the 400 ohms total could
    be raised to a higher value, but the transistor needs enough base
    current to do its job. There's only 1V available minus the .6V E-B
    voltage, so even at 400 ohms, that's not a lot of current.


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  2. Or just hang a dummy load on the output.. 1K0 or a bit higher (<2K0)
    ought to do it.

    Best regards,
    Spehro Pefhany
     
  3. Ross Mac

    Ross Mac Guest

    Why not a 5volt zener and a resistor?....Just a thought...
     
  4. Jim Thompson

    Jim Thompson Guest

    Try this...

    + From
    wall
    wart >--+
    |
    +---+------+
    | |
    \ |
    / |
    20 \ / 2SC2334 or TIP31
    ohm | |/ NPN power TO-220
    o------| Heatsink optional
    | |\
    | E\
    | |
    | |
    | |
    | |
    | |
    | o-----> +
    | | output
    | / to load
    | \
    | / 2.2K
    __|___/ |
    / /\ |
    / \--------o
    / \ |
    -------- |
    | TL430 /
    | \ 2.7K
    | /
    | |
    | |
    | |
    - >--o----------o------> -

    You may need a compensation cap between cathode and control pin.

    ...Jim Thompson
     
  5. John Fields

    John Fields Guest

    ....


    6VDC>------+-----------------+
    | |
    [R1] |
    | |
    +-----|+\ C
    | | >------B
    | +--| / E
    | | |
    | +----+---------+
    | |+ |
    [R2] [C1] [RL]
    | | |
    GND>-------+-------+---------+

    R1/R2 = 5

    C1 = Whatever you need to keep the thing from oscillating; 100µF is
    probably a good guess.
     
  6. John Fields

    John Fields Guest

     
  7. I read in sci.electronics.design that John Fields <[email protected]
    Not always. (;-)
     
  8. John Fields

    John Fields Guest

     
  9. I read in sci.electronics.design that John Fields <[email protected]
    OK, let's go on in the same vein. R2 is not always 5R1; it might be 5R6
    or even 47K. (;-)
     
  10. Active8

    Active8 Guest

    I'm thinking 4.3V Zener and a series pass transistor.
     
  11. Bob Parker

    Bob Parker Guest

    What about using a 5V low-dropout voltage regulator?

    Bob
     
  12. Wrong way.

    Best regards,
    Spehro Pefhany
     
  13. Active8

    Active8 Guest

    Why?
     
  14. Active8

    Active8 Guest

    NEver mind. I tried to cancel this but Gravity thinks I didn't post
    this.

    Still at TARFU level, I guess.

    OK 5.6V zener, solly.
     
  15. Active8

    Active8 Guest

    But Mr. Wizard, this *package* says 502 and this one's on
    psychedelics.

    Why won't this magnet pick up this floppy disk?
     
  16. John Fields

    John Fields Guest

    <GROANNNN...>

    OK, it arteried: "In order to assure a voltage of 5VDC across Rl the
    resistance of R2 should be five times the resistance of R1." ?^)
     
  17. Fred Bloggs

    Fred Bloggs Guest

    You almost had it- put the Vbe multiplier inside the feedback loop and
    buffer like so:

    Please view in a fixed-width font such as Courier.


    6V >---+------+---------+
    | | |
    | / |
    | 22 |
    | / |
    | \ c
    | | |/
    | +-------| TIP31
    | | |\
    | | e----+--> 5V
    | | |
    === | +----+
    | c | |
    | \| / |
    | |-----> \ |
    | /| / ===
    | e \ 1000U
    | | | |
    | | | |
    | | | |
    GND>--+------+---------+----+--> GND
     
  18. Jim Thompson

    Jim Thompson Guest

    [snip]

    See also "4V-Regulator.pdf" on the S.E.D/Schematics Page of my
    website, for a similar application.

    ...Jim Thompson
     
  19. John Fields

    John Fields Guest

     
  20. Fred Bloggs

    Fred Bloggs Guest

    Then you can work a common 5.1V zener into the equation like so:
    Use a 2N3906 for the pnp for Vbe,max=6V rating.

    Please view in a fixed-width font such as Courier.





    6V >---+--------+----------+-------------+
    | | | |
    | / / |
    | 22 51 |
    | / / |
    | \ \ c
    | | | |/
    | +----------|-----------| TIP31
    | | | |\
    | | | e----+--> 5V
    | c | |
    | \| | pnp |
    | npn |--+-----|-------- c e---+
    | /| | | \ / |
    | e | | ---- |
    === | | | 180 | |
    | | | +----/\/\---+ |
    | | | | | |
    | | | | | |
    | | / _/ / ===
    | | 1K /^ 5.1v 1.2K 1000U
    | | / - / |
    | | \ | \ |
    | | | | | |
    GND>--+--------+----+-----+-----------+------+--> GND
     
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