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how do I regulate a 25A supply?

Discussion in 'Electronic Basics' started by Johan Wagener, Oct 10, 2003.

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  1. I have a 12V secondary 300VA transformer that I want to use for 12V
    applications. I bought a 25A rectifier bridge. Measuring the voltage after
    rectf. I get about 11.8V. When I connecect some smoothing caps I get 17V. I
    want to regulate this to stay at 12V. The highest amp regulator I can find
    is 1.5A. How do I go about doing this? Use a lot of these regulators?
     
  2. You will have to use a regulator together with some parallel coupled
    power transistors . The limiting factor will be power handling
    capabilities and as a consequence cooling of these transistors.
    Do a search for power supply schematic on Google and I am sure you will
    find some schematics that will help you.

    /John
     
  3. I

    Try putting some load across the thing; you'll find that the voltage is
    around 12V. Transformers without load will have a higher voltage across the
    secondary than transformers with a load. Put a 2W 100 ohm resistor across
    the secondary and test it.
    Unfortunately, your transformer won't give you 12V regulated. It has a
    dropout of 3V or so.

    If you get another transformer, you can use a power transistor to overcome
    the limitatioins of the regulator:

    IRFI1010G

    ---------------------------------+|+---+-----+
    | | |V| | |
    | | === | |
    | | | | |
    | | | | |
    | | | | |
    | |\| | | .-.
    | +------|-\ | | | |
    | +------+ | | >--------- | | | 0.8 ohms
    +-+--+---| 7812 |-----------|+/ | '-'
    A A | | | |/| | |
    +--+ | | | | | | |
    |+-(-+ | | | | | |
    || A A +--+---+ | | | |
    || +-+--+ | +--------|-------------+ |
    || | | | |
    || | | | |
    A/C in | | | |
    | | | |
    | | | |
    +------+---------+-------+-------------------+
    |
    ===
    GND


    A/C should be at least ___20V___

    created by Andy´s ASCII-Circuit v1.24.140803 Beta www.tech-chat.de

    You can also use a switching regulator, which will certainly be a more
    efficient solution. They are cheap when bought from one of the the various
    electronics surplus places.

    Regards,
    Bob Monsen
     
  4. When connecting a load to the circuit the volts drop to only about 16.3. ?
     
  5. John Fields

    John Fields Guest

    ---
    Unfortunately, that doesn't say much. If you want to determine what the
    output voltage of the transformer is under a 25 amp load, then you've
    got to pull 25A out of it at the same time that you're measuring the
    voltage across it.

    That means that if the transformer is rated to put out 12VRMS while it's
    delivering 25A, you'll need to put R = E/I = 12V/25A = 0.48 ohms across
    it. 1/2 ohm is pretty close, and during the time it's connected it'll
    be dissipating about P = IE = 25A*12V = 300W, so you'll need a pretty
    big resistor even if you make the measurement quickly!
     
  6. John Fields is right, you can't test it the way I recommended, you need a
    larger (er, smaller) load. 12V at 100 ohms is only 120mA, so that is about
    1/200 of the current you are going to ask it for eventually.

    Again, a switching power supply will be far more efficient. You will easily
    be able to find a big PC power supply that puts out 300W at 12VDC. Look
    here:

    http://www.supermicro.com/PRODUCT/Accessories/Power_Supply.htm

    Regards,
    Bob Monsen
     
  7. default

    default Guest

    If the smoothing cap is large enough, the voltage to the valley of the
    ripple will allow you plenty of room for a 4 volt regulator drop.

    A rule of thumb for filter caps is 2,000 micro farads for every amp of
    current you pull, for about 10% drop to the valley of the ripple
    (where it starts dropping out of regulation). It is only a rule of
    thumb and will also depend on transformer regulation, current carrying
    conductors, quality of the caps used, rectifiers, etc.

    You can decrease the 4 volt drop on the regulator by using something
    other than the conventional self biased NPN (collector served with raw
    DC and emitter output circuit). Running the regulator bias from a
    higher source than the same 17 volts you are using for the collector
    input, is a way to use the NPN circuit and still achieve low dropout
    and/or higher efficiency. Dropout with a small additional 4 volt bias
    source drops to ~.6-.7 volts.

    You can also regulate by using a phase controlled Thyristor on the
    transformer output and rectify and control in one operation - if
    output ripple isn't too critical. That way the rectifier drop of .6-1
    volt (for a center tapped transformer, 1.2-2V for non tapped) is all
    the drop you will have, and efficiency will be higher. It is a good
    option for brute force power supplies, but not ripple sensitive
    electronics (good for motor speed controls, bench supply for testing
    car radios and electronics, magnet drives, welding, etc.) If the
    output cap is large enough, and the current drain small enough - you
    may never notice the ripple.

    With the thyristor controller or regulator, all the voltage dropped
    isn't wasted as heat. With a 25 amp conventional linear regulator
    dropping 5 volts across a bank of NPN transistors and rectifiers
    represents 130 watts of waste heat. (and your 300 VA transformer
    won't put out anywhere near 25 amps for very long) Thyristor reg is 15
    to 50 watts wasted, depending on the transformer tap.

    The reason switching regulators are replacing linear regulators . . .
     
  8. If this was my project, I would try to keep the heat under control by
    using schottky diodes for the bridge (about half the heat compared to
    silicon junction rectifiers and about .6 volt extra output voltage).
    And I would design a switching buck regulator, probably a two phase
    design, if you could tolerate the bit of increased ripple current,
    compared to a linear design. With these methods, you might actually
    get 25 amps of regulated output.
     
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