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Taking 9volts from a wall wort to 1.5volts

Discussion in 'Electronic Basics' started by Bullwinkle Jones, Sep 23, 2004.

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  1. Hey guys. I've got the following components:

    2n3906 xistors
    2n3904 xistors
    100ohm resistors
    18komm resistors

    I'm wanting to take 9volts from a wall wart and make it into 1.5volts (or

    I'm thinking I could just use the xistors as diodes somehow.. I'm just not
    sure how to hook it all up!

    A schem on ABSE would be nice, too..

    OR if you could point me to a website that would actually explain the basics
    so I could figure it out myself!

    Thanks a lot!
  2. Thats DC 9vDC to 1.5vDC.
  3. Clarence

    Clarence Guest

    How much current at 1.5 Volts?
    How much current it the Wall Wart rated at?

    There is also the obvious question, Why? What are you going to use it for?
  4. Not much current needed.. The wart can provide 300ma at 9v. It's going to
    be used to power a walkman instead of using batterys. The walkman uses 1 AA
    battery normally.. I'm thinkin the current draw won't be much at all.

  5. View with courier font.

    +9V in--+----+---+
    | | |
    100 | |
    | | |
    100 | |
    | |/ |/
    | |> |>
    | | |
    | 100 100
    | | |
    | +---+-- 1.2 vout
    | |
    | 100
    \| |
    <| |
    | 100
    | |
    0v in --+--------+-- 0v out

    Both transistors 2N3904.
    Parallel the lower resistor with 18k resistors to trim the voltage
    upward (about 60 mv per added resistor).
    If you short the output, some parts will get hot.

    Theory: The upper two resistors supply base current to turn on the
    two paralleled pass transistors that supply the load current through
    their emitters (and balancing and soft current limiting resistors).
    When the output voltage rises to two diode drops, the voltage at the
    mid point of the lower divider reaches the turn on threshold voltage
    of the lower transistor, whiuch shunts all the extra base drive for
    the upper transistors ot ground, holding the output at about two diode
    drops (or whatever the division ration of the lower divider is.
  6. Clarence

    Clarence Guest

    Without more information this is the best and simplest for now.

    Like how much current does the walkman draw?

    For about25 ma This will work.

    Posted on ABSE
  7. When I simulated this circuit, it wasn't very stiff with changes to the
    load. Thus, I played with it in spice, and this is the result.
    (view with courier font)

    | |
    | ,----. ,----. +----+----+
    .-. .-. .-. .-. .-. .-. .-. .-.
    | | | | | | | | | | | | | | | |
    | | | | | | | | | | | | | | | |
    '-' '-' '-' '-' '-' '-' '-' '-'
    '----' '----' | +----+----+
    | |
    | |/
    | |> output
    | o---------------
    | | ,---.
    | .-. .-. |
    All Resistors | | | | | |
    are 100 ohms | | | | | |
    | '-' '-' |
    Transistors | '---' |
    are 2N3904 \| |
    <| |
    | ,---'
    | ,---. |
    | | .-. .-.
    | | | | | |
    | | | | | |
    +--------------------o--------------' '-' '-'
    GND '---'

    created by Andy´s ASCII-Circuit v1.26 beta

    This one was stiffer around 1.5V. Its really the same idea, using
    negative feedback to control the output, which I like.

    The only real change is the three resistors on the collector of the
    rightmost transistor. They are there to dissipate some of the load, so
    you don't need two pass transistors and their emitter resistors. Between
    1mA and 150mA load on the output, the circuit works with 1/4W resistors,
    and the output simulation was between 1.47 and 1.44 Volts, resp. The
    pass transistor stays below 430mW; max for the part is 650, but it
    should be heat sinked anyway to keep it cool.

    Bob Monsen
  8. Clarence

    Clarence Guest

    Looks to me that this one will saturate the series pass before you reach
    regulation. at 150 ma load

    Okay at 75 ma!
  9. The pass transistor doesn't saturate... those three resistors are
    parallel, so 33 ohms at 200mA is 6.6V, which makes the collector 2.4V.
    It can go down to 1.8 V, I guess, so it should be able to go up to about
    210mA before saturation. Using a 2N4401 would enable it to go even lower.

    I simulated it at loads between 1mA and 200mA. It stays within 10mV of
    1.5V in that range. The limits are those three resistors, which get to
    1/4W at 50mA. Thus, the limit for 1/4W resistors is 150mA. Adding
    another one causes the pass resistor power dissipation to go up a bit,
    but its good up to 650mW, so its still in the safe zone (if you can keep
    it cool.) That means, it can go up to 200mA before the resistors top
    out. At that current, though, the regulation feedback circuit can't keep
    up. The voltage at the output drops to 1.42V

    I'd say its easily good up to 150mA in the configuration shown above.
    Above that, I'd go with a regulator IC, cause it'll have current
    limiting and overtemp protection, which this one obviously doesn't have.

    Bob Monsen
  10. Clarence

    Clarence Guest

    Okay. The design was constrained to use only the transistors on hand, and the
    two values of resistors. So he got what he needed.
  11. That really sucks. The usual 2N3904 will go into thermal runaway if you
    get anywhere near the 310 mW dissipation that they were rated for. Been
    there, done that! Use at least two in parallel. Prefereably three.
  12. [snip]

    Why can't you people simply READ the datasheet? The _Absolute_Maximum_
    current for the 2N3904 is 200 mA, and coming anywhere near that will
    cause the gain to drop _drastically_. Just don't push it!!!

    Yeah, no doubt it'll go into thermal runaway!
  13. According to the data sheet, they go up to max 650mW, derate 5mW/C above
    ambient, max 200mA. Perhaps you are reading a different datasheet. I
    believe these parts are manufactured by different suppliers.

    Look at the "Power dissipation vs ambient temp" graph, and recall that I
    specified a heatsink for higher powers. At 430mW, the device is within
    spec below 150F (65C)

    Anyway, we don't know how much power the device requires, so we are
    shooting in the dark about limits. He may only need 10mA. He may need
    300mA. We don't know.

    Bob Monsen
  14. No, thermal runaway happens when current increases with temperature.
    That doesn't happen here, because of the feedback.
  15. Clarence

    Clarence Guest

    He was quite clear, it's a Walkman, I got one and measured it. The ONE I
    measured took an average of 19ma. I allowed 25ma, and assumed a motor start
    current of three times the average.

    I usually limit rise in a small device to 25 degrees C, so from 22 the max I
    allow is 47 degrees C. It may be built into a small plastic pill bottle.
  16. Clarence

    Clarence Guest

    How about posting your simulation?

    Which Spice do you use?

    I know I can do a simulation that can not be built, and always have to do some
    manual checks to prove to myself it's a sane solution.

    I'd Like to compare results, and his circuit is simple enough to make it
    meaningful and quick.
  17. (snip)> > When I simulated this circuit, it wasn't very stiff with
    changes to
    I agree about overheating the transistors. My design gets stiffer
    when you parallel more transistors and emitter resistors, while
    cooling them off. I forgot to mention the modularity of the pass
    section in my first post.

    Without the emitter resistor, it is harder to get parallel transistors
    to share equally.

    I also doubt that the player needs a very stiff supply. A single AA
    cell is not so stiff, either.
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