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discrete voltage regulator circuit

Discussion in 'Electronic Design' started by gearhead, Apr 13, 2008.

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  1. gearhead

    gearhead Guest

    check out this voltage regulator:

    http://homepage.sunrise.ch/mysunrise/joerg.hau/mot/voltreg.htm

    I used this circuit on a motorcycle I owned. I made a couple of
    changes: flipped all the polarities around in order to switch the low
    side of the field, and substituted a mosfet for the darlington. It
    worked pretty well.
    Now I have a couple of questions.

    1) About cap C1 that rolls off the higher frequencies so T1 doesn't
    oscillate: How does one calculate the rolloff frequency? -- it's
    worth knowing for practical and theoretical reasons.

    2) I think it would make sense to put a cap in series with R4, making
    the positive feedback transient and ensure a minimum PWM frequency in
    order to keep the headlight from flickering, a problem I had. The
    original circuit uses hysteresis, so the switching rate depends on
    external factors -- charging system, wiring harness, loads.
    f = 1/(2RC)?
    If so, then adding 0.01 uF in series with R4 at 120 k would ensure
    predictable switching at a bit over 400 Hz.
    Did I get this right?
     
  2. gearhead

    gearhead Guest

    The coil is a problem and there's nowehere else in the circuit to get
    feedback, so I guess one has to add another transitor in parallel with
    T2, and instead of connecting to the coil put a resistor in the
    collector. And take the feedback from there. I suppose that would
    work.
     
  3. readeraz

    readeraz Guest

    -----T2 is connected to ground via R3, and T2 is conductive: current
    flows through the rotor winding L, and the magnetic field in the
    alternator increases, which in turn increases its output voltage. When
    the voltage is high enough to bring D1 in its conductive state, T1
    will also become conductive and will connect the base of T2 against
    the positive line.---

    I dont think the description is right. magnetic field goes up can
    increases voltage output?
    no!
    and the voltage increase can make D1 conduction?
    no.
    the correct description should be,
    current saturates gradually, which makes potential at DF reduce, this
    will cause T1 conduction, then positive beedback takes place. the
    oscillator occurs.
     
  4. gearhead

    gearhead Guest

    How about

    ,----+-------------------,
    | | |
    | | ,----|--100K--,
    --- | | | |
    - | | |\| .01 =
    --- \ Vref--100---+--|+\ |
    - / | >-------+---to mosfet
    | \<----------------|-/
    | / |/|
    | \ TLC3702 |
    | | |
    '----+-------------------'

    The TLC3702 comparator has a push-pull output and uses less than
    100uA, to preserve battery life when the bike is parked.
    For Vref I'd use a series regulator with a very low ground pin
    current, for the same reason.
     
  5. gearhead

    gearhead Guest

    A rectifier prevents any flow from the battery back into the
    generator. Inside the generator, the high end of the field connects
    to the high end of the armature. The whole thing gets its start from
    remanent magnetism in the pole shoes, and builds up from the there.
    The battery doesn't have anything to do with it.
     
  6. gearhead

    gearhead Guest


    Yes, I can run a comparator (or whatever) from the anode side of the
    rectifier. But the bike has to be revving around a couple thousand
    rpm to get anything out of the generator -- at idle the voltage
    drops. Then the comparator inputs will sit at higher voltage than the
    comparator's supply, a big no-no.
    Powering off the ignition circuit would be impractical. People have
    these antique bikes and they want to take the guts out of the old
    relay box and pop in a circuit board. The bikes don't have a separate
    wire that runs from the ignition switch down to the relay box.

    Actually one of the guys came up with a circuit and installed it in
    his bike already:

    http://www.hydra-glide.com/phpBB2/download.php?id=1173

    Guess I ought to just leave well enough alone.
     
  7. gearhead

    gearhead Guest

    I just had a little bit of a Eureka moment about why that circuit
    oscillates.
    When I put that circuit together I tested it on a bench with a
    resistive load in the power mosfet drain to simmulate a field coil.
    When I used a 2 or 3 amp load, it oscillated. So inductance didn't
    cause it.
    But the top rail has some impedance, and when the load switches on it
    pulls the top rail down a fraction of a volt, which feeds back to the
    base of the small transistor that controls the power mosfet.
    As you increase the load, at some point the magnitude of this voltage
    droop negative feedback approches the magnitude of the positive
    feedback or exceeds it, and you get oscillation.

    To solve the problem I could try
    1) increasing the positive feedback, or
    2) bypassing the negative feedback

    but if I just hang a cap off the transistor base for bypass it will
    slow down the switching.
    So:

    Vbatt
    |
    +---------------------------+----,
    | | |
    | 1K field
    | .01 | |
    diode ,--100k--||--|----+
    string | | _|
    | | ,-------+-||_
    | | / |
    | | | |
    +---1K--+--1K--+--| |
    | | | |
    1K 10 = > |
    | uF | | |
    '-------+-----------+------------+
    |
    gnd

    The negative feedback has a time constant of 10 mS, much longer than
    the
    1 mS time constant of the postive feedback. Should work?
     
  8. gearhead

    gearhead Guest

    for practical and theoretical reasons.
    I just had a little bit of a Eureka moment about why this circuit
    oscillates.
    When I put that circuit together I tested it on a bench with a
    resistive load in the power mosfet drain to simulate a field coil. At
    light loads it didn't oscillate but when I used a 2 or 3 amp load, it
    oscillated. Resistive loads, so inductance didn't cause the
    oscillation.
    But the top rail has some impedance, and when the load switches on it
    pulls the top rail down a fraction of a volt, which feeds back to the
    base of the small transistor that controls the power mosfet.
    As you increase the load, at some point this increasing negative
    feedback approches the magnitude of the positive feedback or exceeds
    it, and you get oscillation.

    To solve the problem I could try
    1) increasing the positive feedback, or
    2) bypassing the negative feedback

    but if I just hang a cap off the transistor base for bypass it will
    slow down the switching.
    So:

    Vbatt
    |
    +---------------------------+----,
    | | |
    | 1K field
    | .01 | |
    diode ,--100k--||--|----+
    string | | _|
    | | ,-------+-||_
    | | / |
    | | | |
    +---1K--+--1K--+--| |
    | | | |
    1K 10 = > |
    | uF | | |
    '-------+-----------+------------+
    |
    gnd

    The negative feedback has a time constant of 10 mS, much longer than
    the
    1 mS time constant of the postive feedback.

    On the right track?
     
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