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Power supply question

Discussion in 'Electronic Repair' started by klem kedidelhopper, Feb 18, 2013.

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  1. I'm in the process of building a dual purpose power supply. The supply
    will have two outputs. One will be 24VAC rated at 4.0 A. The load on
    this supply will probably never exceed 2.0 A. The second supply will
    consist of a bridge rectifier off the 24V tap connected to a suitable
    filter cap of say 1000UF. This filtered DC will then be connected to a
    small surplus 12V regulator board and heat sink assembly which by the
    looks of it can handle 5A or better. I plan on mounting this regulator
    in the cabinet and providing a terminal strip for the 12V output. The
    DC load will probably never exceed .500 - .750 A.

    Now here is the dilemma. I haven't tried to put all this together yet,
    but I know that once I rectify and filter the 24VAC I'll probably end
    up with something like 30VDC out. The regulator uses a ua723, a TO3
    and a smaller TO transistor as well a whole bunch of discrete
    components. I have no specs on this regulator, but it is well built,
    appears to be commercial grade, and although it might handle it fine
    I'd feel a lot better hitting it with something like 18VDC instead of
    30. I would hate to blow it up trying to see if it would work on the
    higher input.

    I could add some series resistance either before or after the bridge,
    but the voltage drop across this resistance would vary depending on
    the total load and I'm not sure how well such a scheme would work. So
    I was thinking about employing a voltage divider at the output of the
    filter. The resistance ratio would be easy to figure out, however I'm
    just not sure how to determine the optimum resistor values. Does this
    seem like a viable plan, or perhaps someone my have other thoughts as
    to how to address this? If someone could please help me with this I
    would be very grateful. Thanks, Lenny
     
  2. mike

    mike Guest

    Not clear exactly what you're doing, but the devil is in the details.
    Depending on the line voltage and the design of the transformer, but
    the unloaded voltage of the transformer may result significantly higher
    than 30V.

    The word "24V tap" begs the question of how the transformer
    is configured to get the 30VDC.

    You really need to characterize the 12V board. It doesn't take a
    "whole bunch" of discretes to make a 723 supply. There may
    be other things going on. You don't know
    what input min/max it can tolerate. You're gonna have a wide
    range in input voltage depending on line voltage and load current.
    And a couple of volts of ripple. Once you figure out how much
    headroom you have, insert a zener in the power input.
    You can make one from a small zener and a power transistor.
    Doesn't have to be all that accurate.

    And there are a lot of potential gotchas using an unknown
    board. For example, I build supplies that sense output
    current in the negative lead. They run from floating transformer
    windings. If I were to try to use the AC for some other purpose,
    stuffing current into my floating winding relative to ground
    would make a real mess of things.

    Making a dual-purpose power supply where both purposes are known
    is much easier than building an dual-output supply where
    none of the "purposes" are known in advance.
     
  3. Guest

    The best aproach would be to use a transformer with both 12V and 24V
    windings. Feeding 24VAC into a bridge rectifier will give you about
    32 - 36 volts, depending on load. Using a rough approximation, your
    regulator would be dissipating 15 watts. I'd also boost the value of
    the main filter cap by at least a value of 10.

    An alternative approach may be to use a 7812 series regulator as an
    adjustable regulator. By using a voltage divider between the output,
    the reference (ground) pin, and true ground you could have an
    intermediate 18V regulated output to feed to your regulator. Check
    the 7812 data sheets for voltage readings and how to do this.

    If you have a variable power supply, or a variable transformer it may
    be possible to evaluate your regulator. Use a 100 ohm resistor
    between the power supply and the input pin of the regulator. Measure
    the voltage across the resistor while slowly increasing the voltage
    into the regulator. If a slight increase in input voltage results in
    a signiificant increase in voltage across the 100 ohm resistor you are
    above the maximum input voltage for the regulator.

    PlainBill
     
  4. Jamie

    Jamie Guest

    It can handle up to 40volts on the input..

    http://www.ti.com/lit/ds/symlink/lm723.pdf

    There is PDF with examples at the bottom for its use..

    Jamie
     
  5. Guest

    With a full-wave bridge rectifier, if you use an 8300 uF capacitor, the
    volts of ripple will equal the amps of load current. (Don Lancaster
    taught me this.) The proposed 1000 uF capacitor will have about 8 volts
    of ripple at a 1 amp load current. This is still above what the
    regulator probably needs as a minimum input, but you might want to
    consider a bigger capacitor.
    I have a USB power supply that plugs in to the 12 V in my car that does
    this. The 12 V DC goes through a power resistor (around 5 W rating) and
    then into the input of a 7805 regulator in a TO-220 case. I'm not sure
    if they did this to move some of the heat dissipation from the regulator
    to the resistor, or just wanted to provide some current limiting for
    some failure modes of the regulator, or what. I've never put a meter on
    it but it seems to work OK. (The USB socket provides 5 V DC at 0.5 A,
    or 2.5 W.)

    If you don't have any better specs on what the parts will do, I'd try it
    and see. Install the transformer, bridge rectifier, and capacitor, but
    leave out the regulator. Load the 24 VAC output to 4 A. (If you don't
    have some power resistors sitting around, car tail light bulbs in series
    are good for this and you probably already own some.) Load the output
    of the bridge rectifier to 1 A. Plug the transformer in to the lowest
    line voltage you expect it to work at (use a Variac if you have one).
    Measure the DC at the output of the rectifier - that is the lowest DC
    input to the regulator you can expect. If you want, you can then
    disconnect the load on the 24 VAC output and on the rectifier, plug into
    the highest line voltage you expect (Variac again), and measure the DC
    again, to get the highest DC input to the regulator that you can expect.

    Then, you need to know the minimum DC input you can have to your
    regulator board for it to still regulate. If you already have a
    variable DC power supply, this is easy to figure out. If you don't, you
    have to guess; if it was a 7812 the standard answer is that you need
    about 13.5 to 14 V DC minimum for 12 V DC out. Your board may be
    designed for more than this, though.

    Once you know the lowest DC voltage you can expect from the rectifier,
    and the lowest DC voltage you can supply to the regulator, you can
    figure out how many volts maximum you have to drop in the resistor.

    Just on paper, if I guesstimate a minimum DC input to the regulator of
    14 V, and a 1 A load, I have...

    Peak secondary voltage 24 V * 1.414 = 33.9 V
    Minus two diode drops 33.9 V - 2 V = 31.9 V
    Ripple with 1000 uF @ 1 A 31.9 V - 8 V = 23.9 V
    Minus regulator minimum 23.9 V - 14 V = 9.9 V
    Resistor needed 9.9 V / 1 A = 9.9 ohms
    9.9 V * 1 A = 9.9 watts

    I would probably then look for something like an 8 ohm, 15 watt or
    better resistor. In perfect conditions, this would provide a little
    more DC to the input of the regulator, and in the real world, it would
    also account for getting less than 24 V DC from the transformer,
    slightly more diode drop, slightly worse ripple, etc.
    The series resistor *is* a voltage divider. The regulator is the
    bottom resistor in the divider.

    Matt Roberds
     
  6. Phil Allison

    Phil Allison Guest

    "klem kedidelhopper"
    ** Likely about 35VDC unloaded


    ** A string of say 20 x 3A diodes is the cheapest solution.

    Wired in series with the "+" terminal of the bridge, all mounted on a tag
    strip.

    Drops the DC level by about 12 to 15 volts, depending on the load.


    ..... Phil
     
  7. Phil Allison

    Phil Allison Guest


    ** The correct value is 6300uF.

    Such a cap will drop 1 volt in 6.3mS, the typical discharge time in a full
    wave rectifier at 60Hz.

    For 50Hz full wave supplies, the value is 7500uF for 1V ripple at 1 amp.



    ..... Phil
     
  8. This is some great advice. Thank you everyone for all your input. I
    especially like the diode and Zener ideas. They/re cheap and should
    work well. I've got lots of diodes around here and I'm going to
    experiment with them. However after all this I remembered that the
    transformer is center tapped! At the time, I simply wire nutted the
    tap, tucked it down under the transformer and forgot about it. How
    this simple fact and the possibilities it presents slipped my mind is
    anyone's guess. I've been building this thing in the evenings after my
    usual work and perhaps I've been tired. The transformer measures 25.2V
    at 120V line unloaded. Loading the transformer in this fashion will
    probably unbalance it somewhat and drop the 24V a bit as well but I
    don't see it as a real problem. I should be able to use one side of
    the secondary and the tap now get at least 15 -17 VDC out of the
    bridge and filter, and that will provide a healthier input to my
    regulator. Lenny
     
  9. amdx

    amdx Guest

    No need to unbalance the transformer, just use 2 diodes and use the
    center tap as negative. See Here,

    http://metroamp.com/wiki/index.php/Full_Wave_Center_Tap_Rectifier

    Mikek
     
  10. Phil Allison

    Phil Allison Guest

  11. Guest

    I looked over what I had previously written and perhaps I didn’t explain this properly. I have a 24VCT transformer. The transformer needs to power 24VAC equipment as well as 12VDC equipment. For the DC circuit I first triedusing a bridge directly off the 24V winding. As soon as I connected up thefilter cap the DC output went to 35V, which is probably too high to feed my 12V regulator.

    Several people came up with some great ideas to address this, and I then realized that I had a center tap that was not being used. So I connected my bridge across the center tap and one side of the secondary. This time the 13VAC when FW rectified using the bridge went to about 19VDC, which is a safe input to the regulator. Although this worked, I wasn’t happy about unbalancing the transformer this way so I posted my results.

    Unless I seriously misunderstood It was suggested here that I come off the full secondary output with two diodes, cathodes tied together, (typical FW rectifier), and use the center tap as my negative return. I didn't think that would make any difference in the output voltage from using a bridge without the center tap however I tried it anyway. As I suspected it would, the DC output again went to 35V when I connected the filter up.

    I've been playing with power supplies all my life and maybe my age is catching up with me. I guess I just don't understand the explanation of how to get a lower voltage to my DC regulator by what I thought sounded like just configuring the transformer properly. So at the risk of sounding like a complete idiot here can someone please explain this further? Thanks, Lenny
     
  12. Phil Allison

    Phil Allison Guest



    I looked over what I had previously written and perhaps I didn’t explain
    this properly. I have a 24VCT transformer. The transformer needs to power
    24VAC equipment as well as 12VDC equipment. For the DC circuit I first tried
    using a bridge directly off the 24V winding. As soon as I connected up the
    filter cap the DC output went to 35V, which is probably too high to feed my
    12V regulator.

    Several people came up with some great ideas to address this, and I then
    realized that I had a center tap that was not being used. So I connected my
    bridge across the center tap and one side of the secondary. This time the
    13VAC when FW rectified using the bridge went to about 19VDC, which is a
    safe input to the regulator. Although this worked, I wasn’t happy about
    unbalancing the transformer this way so I posted my results.

    Unless I seriously misunderstood It was suggested here that I come off the
    full secondary output with two diodes, cathodes tied together, (typical FW
    rectifier), and use the center tap as my negative return. I didn't think
    that would make any difference in the output voltage from using a bridge
    without the center tap however I tried it anyway. As I suspected it would,
    the DC output again went to 35V when I connected the filter up.


    ** Center tap transformers are ALWAYS double the AC voltage of bridge
    rectifier ones for the same final DC voltage.

    You have made a dopey wiring error.


    ..... Phil
     
  13. I'm grateful for everyone's input, and although this power supply as
    well as the equipment that it will power is to be used in a family
    business, never the less I'm trying to build this as "technically
    correct" as possible. I found some 6.8V 10W stud mount zeners in the
    junk box last night. I'm thinking that they may be a bit light for
    this but I might try stringing two or possibly three of them together
    on a heat sink as Arfa suggested or perhaps try the series diode
    trick, (from Phil), both as previously mentioned on the output of the
    filter and see what happens. Even though given the load it may work
    fine it would be nice as Arfa said to not have to "unbalance the
    transformer. Lenny
     
  14. Jamie

    Jamie Guest

    Your problem is simple

    If the 24V AC devices does not come in contact with the common or
    ground of this supply you have, you can use a 2 diode full wave config.

    The center tap would be your common for the 12 volt supply, one diode
    from each outter leg joined together to form a full wave. The diode
    output alogn with using the center tap as your common will give you
    12 * 1.414 volts unloaded. If you need to use a 12 Volt reg, I guess you
    could do that. Use a LDO type.

    As for the 24 v AC, the outter legs will supply that to the out board
    device. Make sure the AC legs do not come in contact with either grd or
    DC out..




    o o
    | |
    | | to your 24V AC device
    | |
    | |
    | |
    | |
    | |
    | |
    + | |
    --------------+. ,-o---------|----->|-+--+
    )|( | | 18DC
    )|( | +------++
    Line Voltage -. ,-+----+| | | ===
    )|( === | | /-\
    )|( GND | | |
    +-------------+-' '+----------o----->|+----+ ===
    GND
    (created by AACircuit v1.28.6 beta 04/19/05 www.tech-chat.de)


    Unless I missed something, I don't see a problem?
    Jamie
     
  15.  
  16. I know it's difficult when we're only using text but I just can't
    understand, (read that is) your schematic. From your description
    though it sounds like it might be the FW arrangement I've already
    tried and got 35V out of. Is there any way to make that appear any
    more readable? Lenny
     
  17. I'm not sure if my last reply worked so I'll try it again. I know that
    it's very difficult when all we can use here is text but I'm afraid
    that I just can't understand, (read) your schematic. From you
    description however it appears to be the FW arrangement that I've
    already tried and got 35 V out of after I connected up the filter. Is
    there any way to make your diagram any more readable in this medium?
    Thanks, Lenny
     
  18. Jamie

    Jamie Guest

    THe FW you tried was a 4 diode bridge, this is a 2 diode full wave config.
    THe difference being is, the CT (Center tap) is your common, the
    negative terminal and where the Cathode of the 2 diodes come together
    are the (+) terminal which will be around 18VDC with Cap attached

    If you truly do have a 24 volt transformer with CT, this means you'll
    get 12 Volts from each outer leg with respect to the CT. Because they
    are out of phase with each other (180), this is a split phase
    configuration, like seen in residential pole pigs in the USA.

    If you have done this and gotten 36 volts, then you actually have a
    48V transformer with CT>

    Jamie
     
  19. What you're referring to is a 12VCT transformer. Connecting either a
    bridge and not using a center tap or just two diodes across the full
    secondary with cathodes tied together using CT as common connected to
    an output filter will yield 18VDC. A 24 V transformer under the same
    circumstances will produce 36VDC. I've tried it and it does. Lenny.
     
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