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Help needed for simple circuit troubleshoot

Discussion in 'Electronic Basics' started by Lambing Flat, Apr 14, 2005.

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  1. Lambing Flat

    Lambing Flat Guest

    G'day all :)

    I am very much a beginner at this electronics lurk so I am hoping that
    someone can help me figure out what is wrong with my version of the
    circuit shown in the following diag.
    http://www.trainweb.org/toenailridge/circuit.jpg

    The circuit's function is to smoothly accelerate a G gauge battery
    powered locomotive to running speed once the circuit is turned on, and
    then to smoothly decelerate it back to stop once the circuit is turned off.

    My problem is that it accelerates very nicely to speed in about 3 secs
    once the switch is turned on, but when the switch is turned off it takes
    about 2 minutes to stop! I have obviously done something wrong, as the
    deceleration is supposed to be at the same rate as the acceleration, but
    I can't figure out what my mistake is :-(

    Here are photos of both sides of my version of the circuit, (apologies
    for my poor standards of soldering ;-)
    http://www.cia.com.au/bullack/Circuit1.jpg
    http://www.cia.com.au/bullack/circuit2.jpg

    The only difference my version has to the circuit diagram is that I have
    temporarily moved the connection to the base tag of the MOSFET from the
    "power always" side of the on/off switch to the other side, as part of
    ongoing experiments to try to find out why it won't slow down! Made no
    difference at all, I'm afraid......

    I have carefully checked all other connections and components, and as
    far as I can tell, (other than as mentioned above) everything is
    connected and in the correct place and there are no shorts.

    As part of my careful checking of everything I have established that the
    batteries produce 7.5v and that is the voltage present at the collector
    leg of the MOSFET once the switch is turned on and the circuit powers
    up. The voltage drops to 5v on the emitter leg of the MOSFET.

    Hopefully the fine brains trust here will be able to point me in the
    right direction!


    --
    James McInerney

    My G gauge garden homage to the now long gone railways of Tasmania's
    west coast, the "Rurr Valley Railway"
    http://www.cia.com.au/bullack/rvrtitle.html

    also http://www.cia.com.au/bullack/ , where the steam era NSWGR
    secondary lines live on in HO at bucolic "Lambing Flat"

    and http://members.optusnet.com.au/lambingflat/ for the family stuff!
     

  2. You haven't done anything wrong. The problem is the circuit was designed by
    a fellow newbie, so by design it takes approximately 100 times longer to
    turn off than to turn on. If you want to make the delays approximately the
    same...

    Then replace the 470k ohm resistor with a 1k ohm resistor and hook it up
    such that the top side of the resistor goes to the right side of the on/off
    switch (which also hooks to the left of the 4.7k resistor). The bottom side
    of the 1k resistor should go to the negative terminal of the battery pack
    just as the current 470k resistor does. If the timing is still not to your
    satisfaction after this modification you can play with adjusting the value.
    Larger resistance will take longer to stop, lower resistances than 1k will
    not make much difference. You can also play with the value of the 4.7k
    resistor or the 470uF capacitor if you like.
     
  3. Lambing Flat

    Lambing Flat Guest


    Thanks Fritz, I'll give that a try and post my success or failure later :)

    --
    James McInerney

    My G gauge garden homage to the now long gone railways of Tasmania's
    west coast, the "Rurr Valley Railway"
    http://www.cia.com.au/bullack/rvrtitle.html

    also http://www.cia.com.au/bullack/ , where the steam era NSWGR
    secondary lines live on in HO at bucolic "Lambing Flat"

    and http://members.optusnet.com.au/lambingflat/ for the family stuff!
     
  4. Bob Monsen

    Bob Monsen Guest

    If you can use a double-throw switch instead of a single-throw switch, you
    can simply connect the other throw to ground. When you switch it from the
    ground throw to the Vcc throw, it'll charge up in 3 seconds. Then, when
    you switch it back, it'll come down in the same amount of time.

    The N-MOSFET is hooked up as a follower. This means two things. First,
    it'll hold the load at between 2 and 4 volts below the battery voltage.
    That means your load is going to get less voltage, and thus less power
    with a given current flow. Second, if the load is drawing an amp, that
    means you are wasting between 2 and 4 W in the N-MOSFET, so it'll get hot
    and waste your battery time.

    If you are willing to build a somewhat more complex circuit, you can
    avoid this power loss, and thus improve the life of your battery. Here is
    an example circuit:

    Vcc
    .------.
    | |
    | |
    | ---
    | / \ 1N4001
    Vcc-----. | ---
    | [LOAD] |
    o | |
    \ |\ o------'
    SPDT o---[4K7]--o--[330K]---o------| \ ||--'
    | | | \ ||<-.
    o | | | ------||--|
    | | | | / |
    | 470uF --- [R] .-o| / |
    | --- | | |/ |
    | | | | |
    | | | '----------------o
    | | | |
    | | | [0.1R]
    | | | |
    | | | |
    GND------o----------o-----------o--------------------'


    When the switch is thrown to Vcc, the current through the load will
    gradually increase over a few seconds to whatever maximum you set with R.
    I don't know how much current your load will draw, so I can't set it up
    exactly. You'll have to adjust the value of the resistor marked R in the
    picture to ensure that the MOSFET turns on all the way, but doesn't get
    too much above the voltage required. Measure the current when your load is
    connected directly across the battery, using a multimeter. Then, assuming
    the current is I amps, you want to set the voltage at the non-inverting
    input to be I/10 when the SPDT switch is connected to Vcc. That will be

    R = (334.7k*I)/(10*Vcc - I)

    So, for example, if your max current is 2A, and Vcc is 9.6 then

    R = (334.7k*2)/(96-2) = 7121 ohms

    You can use any cheap single supply opamp. I added in the diode to prevent
    the mosfet from getting fried by back-emf from the motor. With this, you
    don't have to use a super high voltage mosfet like the guy specifies; you
    can use one that is about 30V ds without a problem.

    Regards,
    Bob Monsen
     
  5. Lambing Flat

    Lambing Flat Guest

    Thanks Bob :)

    I really appreciate your help. I'm learning electronics in my old age
    by doing ;-), since I somehow managed to avoid learning any electronic
    theory when I was younger! Your circuit looks very interesting, I think
    I understand most of it ;-) so I'll build it and see if I can add a bit
    more to my knowledge base.....

    --
    James McInerney

    My G gauge garden homage to the now long gone railways of Tasmania's
    west coast, the "Rurr Valley Railway"
    http://www.cia.com.au/bullack/rvrtitle.html

    also http://www.cia.com.au/bullack/ , where the steam era NSWGR
    secondary lines live on in HO at bucolic "Lambing Flat"

    and http://members.optusnet.com.au/lambingflat/ for the family stuff!
     
  6. Jim Douglas

    Jim Douglas Guest

    Your web site is incredible! I have spend the last 45 minutes checking out
    all that work! This stuff is beautiful and I admire the work and research
    you have done! I have this site in my favorites and will check back often
    to see what's new.

     
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