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24Vdc Power supply

Discussion in 'Electronic Basics' started by Ian Tedridge, Jul 29, 2006.

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  1. Ian Tedridge

    Ian Tedridge Guest

    Could somebody please help me with a very simple 24Vdc power supply, capable
    of delivering 20amps.

    Does not need to be a very clean supply, Just to drive 3 of 6amp dc motors.

    Hoping to just use 240ac to 24ac transformers, some single phase rectifiers,
    and capacitors.

    My problem is not knowing what size components I require.

    Ideally, something that I can build on at a later date for smoothing out the
    output without reducing its load capability.

    Do need to achieve 23.5 to 24Vdc at 20amps plus

    Thanks in anticipation
     
  2. Phil Allison

    Phil Allison Guest

    "Ian Tedridge"

    ** No need for capacitors.


    ** Really ?

    So you don't imagine a 24 V ( ideally 26.5 V ) 20 amp tranny would do?

    Just add one 40 amp bridge rectifier on a modest heatsink.

    If there is any chance one of the motors could stall, then add a 20 amp
    breaker in series as well.





    ....... Phil
     
  3. Ian Tedridge

    Ian Tedridge Guest

    Thanks Phil,

    My thoughts on the capacitors, was just to clean up and stabilise the output
    a little. Something I've seen used before.

    Must admit, had not thought about a breaker for protection. Think I need to
    slow down a little and take stock of what I am trying to do.

    Cheers
     
  4. Chris

    Chris Guest

    Hi, Ian. Ditto on everything Phil said. If you'd like some control of
    the motor speed, though, you can drive the primary of the transformer
    with the output of a Variac like this (view in fixed font or M$
    Notepad):

    |
    | _ T1
    | o--o_/ \o--.
    | FU1 )
    | )0-240VAC .-----.
    | )<-----. ,---o~ +o-------.
    | 240VAC ) )|( | BR1 | |
    | ) )|( | | +|
    | ) .--' '---o~ -o--. / \
    | ) | T2 '-----' | ( M )
    | ) | 0-24VAC | \ /
    | o-----o----' | | -|
    | | | | |
    | '--------' '----'
    |
    (created by AACircuit v1.28.6 beta 04/19/05 www.tech-chat.de)


    But plain ol' rectified 24VAC should do the trick just fine.

    Cheers
    Chris
     
  5. default

    default Guest

    Some more to think about . . . if you filter it you raise the voltage
    because with raw RMS DC 24V=24V, with filtered DC you approach the
    peak voltage which is 1.4142 X the RMS value, or about 36 volts
    (unloaded - the transformer will drop a little voltage as will the
    rectifier diodes) Transformer ratings are for some specified current
    and at some nominal voltage - Voltage can be plus or minus 10%. off
    nominal.

    And, you are using 20 amps. each diode in a four diode bridge will
    drop .6 volts for 1.2 volts in each half cycle at 20 amps that's 24
    watts wasted power and requires a good heatsink. Use a center-tapped
    transformer and you cut the waste power in half - ditto the heatsink
    and size, and likely as not cost as well.

    I use an adjustable 24 VDC 10 amp to run some fans - the SCR's doing
    the controlling also rectify the AC to save on wasted power and keep
    the heatsink small.
     
  6. Phil Allison

    Phil Allison Guest

    "default"

    ** Shame how you then *waste* the output capacity of the transformer
    instead - by loading only half of the secondary copper at a time.

    Very inefficient .




    ......... Phil
     
  7. John  Larkin

    John Larkin Guest

    Correct.

    John
     
  8. Eeyore

    Eeyore Guest

    Incorrect. A half bridge configuration wastes transfomer utilisation. The I^2R losses
    increase far more than anything you'll save in a bridge.

    Graham
     
  9. default

    default Guest

    No way.
     
  10. default

    default Guest

    I see the errors of my thinking. Intrinsic lossses in the transformer
    cause the I sq R loss ot increase faster than the diode drop losses
    in that (24V) voltage range. At lower voltages the diodes waste more
    power than the transformer and a center tap makes more sense.
     
  11. Eeyore

    Eeyore Guest

    I'm afraid you're utterly wrong.

    Graham
     
  12. Phil Allison

    Phil Allison Guest

    "default"

    ** Not really.


    ** Drivel.

    The "full wave, center tap" ( two diode) arrangement REQUIRES that a
    transformer's VA be DE- RATED compared to using full wave bridge.The reason
    is as I posted earlier - only HALF the secondary copper is being used at
    any moment.

    This applies to a transformer of any size.




    ........ Phil
     
  13. John Fields

    John Fields Guest

    ---
    I disagree.

    For a resistive load and the same output voltage, current, and
    regulation, both transformers must use the same gauge of wire on
    their secondaries. However, since the CT arrangement requires that
    each half of its secondary conduct only half the time, it will
    _inherently_ have a higher VA rating than the FWB transformer, but
    some of its capacity will be unused. That is, it will run cooler
    than the FWB transformer for the same load.

    The same is true if the load includes a reservoir capacitor, as
    indicated by example 1 at:

    http://www.mcitransformer.com/i_notes.html
     
  14. Phil Allison

    Phil Allison Guest

    "John Fields"

    ** Dig you own grave deep as you like if you wish .....



    ** Your SILLY false assumption is now plain.

    The **context** here, is where the SAME transformer is used in one or the
    other rectifier mode.

    Do try to improve your reading skills - John.

    Unconscious context SHIFTING is such a GLARING indicator of your
    congenital AUTISM.





    ........ love, Phil
     
  15. default

    default Guest

    Why would a designer use the same size wire in the CT secondary? Each
    winding is working at a 50% duty cycle so the required ampacity of the
    wire is cut by half - things being what they are, the transformer
    maker would save copper by using smaller wire.

    AND Looking at transformer ratings where they provide two independant
    secondaries the output will be rated 12V @ 2A , and 24V @ 1A (same
    game I would think)
     
  16. John Fields

    John Fields Guest

    ---
    Well, let's see...

    If we want to use the _same_ transformer to feed a full-wave bridge
    or as a center-tapped transformer feeding a 2 diode full-wave
    rectifier, then we need to have a transformer with two secondaries
    which can be paralleled or connected in series.

    Just to nail things down, let's say that we want to feed the
    rectified output of the transformer into a 12 ohm resistive load and
    that we need 1 amp through that load.

    That means that we need a 12VA transformer, and in the case of the
    FWB, the entire secondary needs to pass 1ARMS continuously. For a
    regulation of 10%, that means that the no-load secondary voltage
    will be 13.2V, the secondary's resistance will be:

    13.2V - 12V
    R = ------------- = 1.2 ohm
    1A

    and it'll be dissipating:


    P = I²R = 1A² * 1.2R = 1.2W


    Since both secondaries will be wired in parallel, that means that
    each one will have a resistance of 2.4 ohms and will be dissipating
    0.6 watts.

    Now, if we wire the secondaries in series in order to use them
    center-tapped, when either side of the secondary is delivering
    current to the load its resistance will be 2.4 ohms,

    So, instead of:

    13.2V
    |
    1.2R
    |
    +-->12V
    |
    12R
    |
    0V

    the circuit now looks like this:

    13.2V
    |
    2.4R
    |
    +-->11V
    |
    12R
    |
    0V

    Notice that now, instead of 12V into the load, we'll have 11V, with
    one half of the secondary dropping the other volt.

    That means that we're now looking at about 18% regulation instead of
    10%, and to get the regulation down to 10% we'll have to double the
    area of the wire in the secondary, which means _increasing_ the VA
    rating of the transformer using the center-tapped full-wave
    arrangement.

    No only that, each half of the secondary will be dissipating:


    (13.2V - 11V)² 4.84
    P = ---------------- = ------- ~ 2 watts,
    2.4R 2.4R

    But for only 50% of the time, so that's about 1 watt.

    Since the paralleled secondaries will only be dissipating about 0.6
    watts each, that means the series-connected secondaries will be
    running hotter that the parallel connected ones, requiring the
    series connected arrangement to be rated for greater, not less,
    than 12VA.

    That is, unless I missed something...
     
  17. John Fields

    John Fields Guest

     
  18. John Fields

    John Fields Guest

    ---
    That is, by trying to get 2A out of each secondary for half the
    time.
     
  19. Phil Allison

    Phil Allison Guest

    "John Fields" = desperately needs a remedial reading class



    ** Which is IDENTICAL to my original comments.


    ** English class at school ??




    ........ Phil
     
  20. John Fields

    John Fields Guest

    ---
    If you meant that one would have to run the transformer at a lower
    VA rating than it was rated for if the center-tapped arrangement was
    used, then I agree with you.

    However, the argument you gave, that the reason for that was that
    only half the copper was being used at any given instant doesn't
    seem to support that, since half the copper is being used, but only
    for half the time.

    Had you delved into it a little deeper, as I did, (since this is
    seb) your meaning would have been clearer and this exchange would
    not have been necessary.
     
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