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1 KVA 240 v transformer output 6v,8v,12 v AC how to test if it can supply 83 amps?

Discussion in 'Electronic Basics' started by rob, Mar 4, 2005.

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  1. rob

    rob Guest


    I am a total novice in electrics and am not sure were to post this
    question?.. i hope some one can help me...I need a transformer that
    can supply aprrox 6 Volts Dc at 75 amps..
    I have just purchased a 240 v ac input 1 KVA rated transformer its AC
    output is 6v ,8v and 12 V..Now i understand that i need to rectify
    the current and have just got off ebay a dozen MBRP40030CTL 400A 30V
    schottky diodes with low Vf.. i assume that these will be able to
    supply a 75 amp current.. at 6 v?..I got them because there rated at
    400 amps..and were very cheap!

    My other question is that there is label on side of the transformer
    which is hand written saying that it is rated at 83 amps?..
    Label pic

    yet when i looked inside i found a large fuse holder with 63 amps at
    550v printed on it.?

    yet the fuse thats attached to it says 40 amps at 550VAC..

    Now i am confused ? if the fuse is 40 amps does that mean i have only
    transformer that can supply 40 max?..

    Inside the transformer there is a board with 1KVA printed on it

    i hope this makes some sense?

    Is there any way i can test to see if it can supply 75amps ?

  2. John Fields

    John Fields Guest

    First things first...

    1. If the transformer is rated at 1kVA with a 240V input, then that
    means it's designed to operate with 240V on its primary, and should be
    able to supply a total of 1kVA to a load connected to the secondary.
    From your photos it appears that the transformer has a single primary
    and a single, tapped secondary, so at 1kVa the currents which should
    be available from the secondary at the various taps will be:

    VA 1000
    I = ---- = ------ = 166.6A for the 6V case,
    E 6

    1000VA/8V = 125A for the 8V case, and 1000VA/12V = 83.3A for the 12V
    case, so it looks like the secondary was wound with wire designed to
    supply a maximum of 83A out of the secondary, no matter which tap you
    use, and that's what that "83" on the label means.

    2. The voltage and current written on the fuseholder is what the
    fuseholder is rated for, which is no more than 660V across it (at
    least it _looks_ like 660V to me) with no fuse in it, and no more
    than 63 amps through it with a fuse installed. That means you
    shouldn't put anything bigger than a 63 (60) amp fuse in it.

    3. The 40 amp fuse that's in there means that the last person who used
    the transformer fused it for 40 amps. If you want to fuse it for a
    different current you'll need to change the fuse, but if you want to
    run it at 75A you'll need to change the fuseholder as well.

    4. Since you're allowed 83 amps out of any tap on the secondary you'll
    be able to take 75 amps, no problem, but if you need 6VDC there might
    be a problem if you use the 6V tap, since you'll be dropping some of
    the voltage available from the secondary in the diodes when you
    rectify the AC to get DC. On the other hand, if you used the 8V tap
    you could still get 75A out of it, but if you wanted 6V out you'd have
    to get rid of the extra volt or so with about a 100 watt regulator.

    Then there's the question of the filter capacitor. That is, whether
    you need one or not. What's your application?

    5. Finally, to test it at 75 amps, what you'll need to do is put a
    resistance across the 6V tap which will allow 75 amps to flow through
    it with 6V across it. That resistor will need to be:

    E 6V
    R = --- = ----- = 0.08 ohms
    I 75A

    and it'll dissipate:

    P = IE = 75A * 6V = 450 watts while it's connected, and you
    should measure the current and the voltage out of the transformer _at_
    the transformer while it's being exercised.
  3. Rob

    Rob Guest

    Your transformer is rated at 1000kVA, so at 12V this gives
    1000/12=83A, which fits with the label. As this transformer is second
    hand, the original owner only needed 40A and so installed a 40A fuse.
    After all, your AC mains outlet may be rated at 13A, but that doesn't
    stop you fitting a 1A fuse in your table lamp. Remember to mount your
    rectifier diodes on a heatsink.
  4. John Fields

    John Fields Guest

    Why? take a look at the data sheet and find the curves for Vf VS If.
    If he's trying to eke out that last little bit of voltage from the 6V
    secondary tap he might want to let the diodes run warm.

    BTW, if you had bothered to read the thread before you posted, you
    might have noticed that you were only echoing what had been posted
  5. Err... I know you are smarter than this Johnnie boy. At 75A DC output the
    diodes will start smoking very quickly without adequate heatsinking. At
    around 75A instantaneous forward current the voltage drop at maximum
    junction temp. is somewhere around 0.25V, or around 19W per element. Add to
    this something vaguely around say 2W for reverse leakage current, and a
    heatsink is definitely required.

    It appears Rob posted from Google Groups which has a large delay between
    when people post and when messages appear. It is possible he never even saw
    your post before he posted his own message. Cut him a little slack.

  6. These diodes are well suited to the task at hand.

    If you don't need any kind of filtering and are perfectly content with
    really lumpy DC power direct from the bridge rectifier, then yes this
    transformer should be adequately capable of producing your 6V(ish) at 75A
    DC. Naturally you would have to remove or replace the fuse with something
    greater than 75A. Fusing on the primary side is somewhat easier and cheaper
    since 250V fuses at a few amps are much cheaper and more common than giant
    75A+ fuses (although at your low voltages 30V automotive blade fuses should
    suffice). On the other hand a fuse on the primary side would need to have
    greater overrating to avoid nuisance fusing since it must contend with
    transformer saturation surge currents at initial power up.

    If you need 6V DC at 75A DC well filtered, then the situation changes
    significantly. The transformer outputs vaguely 6V AC, but when filtered
    yields a peak voltage of roughly 8.5V less two schottky diode drops. When
    well filtered with a large capacitor, the power factor will be substantially
    reduced. Capacitive input filters fed from full wave rectified sinusoidal
    sources can have typical power factors of around 0.6. In other words, if
    your load still draws 75A DC (at around or a bit less than 8V DC with the 6V
    AC tap) with the capacitive filter, the effective heating in the transformer
    secondary will be more than what a 6V 75A RMS AC load would produce.

    Suppose your load draws 75A DC at 8V but with a powerfactor of 0.6 (due to
    input filter capacitor). This means your load uses 75*8 = 600W, but
    600W/0.6 = 1000VA which is the maximum rating of your transformer. It isn't
    totally clear if your transformer is rated for 1000VA load from any of the
    taps, or if it is rated for 1000VA from the 12V tap only, but somewhat less
    on the lower voltage taps. The 6V tap has half the turns of the 12V tap, so
    if one assumes the length and size of the wire used for the 6V tap is half
    that of the 12V tap, then 1000VA at half the voltage is twice the current.
    However, P=I^2*R, so for half the resistance (half the length of wire for
    the full 12V tap), but twice the current, the power dissipated is double.
    On the other hand the 6V winding is probably located closer to the center of
    the transformer where the diameter of each loop of wire is smaller, so the
    6V tap probably has somewhat less than half the resistance of the 12V tap,
    so the heating situation isn't quite so bad.

    All and all, it is hard to say for certain if 75A DC is overdoing it a
    little bit if you use a capacitive input filter. Without the filter a load
    current of 75A DC is fine, but with the filter 75A might be pushing it
    slightly over its limits. Long term reliability may be somewhat
    compromised, so for peace of mind improved cooling using forced air and/or
    reducing the load current to something a bit less than 75A might be

    What exactly is your application?
  7. John Fields

    John Fields Guest


  8. Before someone calls me on this oversight, I should call myself on it. In a
    full wave bridge rectifier each diode only conducts the load current part of
    the time, and only blocks voltage part of the time, so the above theoretical
    dissipation figures would probably be closer to half that stated or
    somewhere around 10W total. 10W per device is still probably too much for
    use without proper heatsinking.
  9. rob

    rob Guest

    Thanks John for that fantastic reply..I am much more confident now
    that i have more information on my PSU...

    I am using the PSU to power an electrochemical cell for making
    Chlorate/Perchlorate.. Unsmoothed Dc current will work ok for this
    purpose....The voltage is not critical ..

    You mention that i will need a 0.08 Ohms resistor to produce a 75 amp
    I have purchased some 18 gauge Nikrothal 80 wire with a .040"
    diameter, 0.4062 ohms per foot..

    Am i correct in thinking that i can divide 0.4062 (foot ) by 12 to
    get the Ohms per inch?.. which = 0.03385 Ohms (per inch) so i would
    need 2.37 inches ?...I intend to use this as a variable resistor to
    increase/decrease the current flowing in to the cell..


  10. Lord Garth

    Lord Garth Guest

    Take care with this is a contact explosive.
  11. Rich Grise

    Rich Grise Guest

    Sure. Just put a 0.08 ohm, 450 watt resistor on the secondary, and meter
    the volts and amps. Or 75 ea. 6V @ 6W light bulbs. ;-)

    Fuse the primary at NO MORE THAN 4 amps. At 75 amps on the output, it
    should only be drawing - oh, I'll leave that arithmetic as an exercise
    for the reader. ;-)


  12. Rich Grise

    Rich Grise Guest

    People who post from google groups don't deserve slack.

    The _only_ thing google beta has going for it is posts show up right away.
    But it's still the person's responsibility to read the thread, and copy/
    paste context material.

    What they really _should_ do, of course, is use a real newsreader and
    their ISP's news server. I've never heard of an ISP that doesn't have
    a news server.

  13. Rich Grise

    Rich Grise Guest

    Yeah, use 2 3/8" of wire if you want to start a fire.

    6V * 75A = 450 watts. Your wire will disappear, and splash molten
    Nikrothal all over the bench and your lap and your face.

    Get some fat wire with a R of about 0.01 ohms per foot, and use 8 feet
    of it.

    Or shop around for a 6V, 450 watt electric heating element.

    And fuse the primary at about 2 amps, for the 75 amp test.

    Good Luck!
  14. Rich Grise

    Rich Grise Guest

    According to a current thread on you don't even
    need an electrolytic cell - with the right mixture of reagants, the KClO4
    precipitates out of solution on its own.

    And, on its own, it's not explosive. It's a very strong oxidizer, but it
    still needs fuel. You might be thinking of ammonium perchlorate - that's
    pretty nasty stuff.

    Good Luck!
  15. John Fields

    John Fields Guest

    I mentioned using a 0.08 ohm load to test your transformer because if
    you place a 0.08 ohm load across a 6 volt source, from Ohm's law
    you'll get:

    E 6V
    I = --- = ------- = 75 amperes
    R 0.08R

    flowing through the resistance if the source is capable of supplying
    that current. Also, I mentioned that the resistor will be dissipating
    450 watts, which is a HUGE amount of power for a 2.37 inch long piece
    of Nichrome to handle.

    There's also another problem, and that's that if you're planning on
    using the Nichrome wire as a current limiting resistor, it'll be in
    series with the cell, so if you want to get 75 amps into the cell the
    sum of the resistances of the cell _and_ the resistor will have to be
    equal to 0.08 ohms, which will make the resistance of the resistor
    substantially less that 0.08 ohms.

    Here's your proposed circuit:

    +------+ +------+ +------+
    MAINS-----|P S|--[FUSE]--|~ +|---[R1]<--|--+ |
    | | | | | | |
    | | | | | [R2] |
    | | | | | | |
    MAINS>----|P S|----------|~ -|----------|--+ |
    +------+ +------+ +------+

    Notice that if the output of the bridge is 6VRMS, in order for 75 amps
    to flow through the cell the _sum_ of R1 (the nichrome resistor) and
    R2 (the cell itself) must be 0.08 ohms. That means that you'll
    probably be dealing with _very_ short lengths of white-hot Nichrome
    wire to get 75A into the cell. Not a good idea, for two reasons: The
    first is that white-hot Nichrome is dangerous, and the second is that
    as Nichrome heats, its resistance increases. You could make things
    better by getting longer lengths of Nichrome and connecting them in
    parallel, but that's still a messy way way to do it, and if you need
    to be changing resistances in the middle of a run it's going to turn
    into a nightmare.

    What I _strongly_ suggest you do is get a Variac (240V 50Hz in, 0 to
    240V out at 5 or 10 amps. Five would work and would be less expensive
    than ten, but ten wouldn't have to work as hard.) and use it to
    control the voltage _into_ the transformer. Neat, clean, and you can
    monitor the input current and get a pretty good idea of what's going
    into the cell, although mneasuring the output current would still be
    best. Also, using the Variac would allow you to use the 8V tap (or
    even the 12V tap) and get away from too low a voltage out of the
    bridge because of the diode drops and too low a voltage into the cell
    because of wiring.
  16. You would have been right 10 years ago, but today very few ISP's have a
    newsserver, and even fewer run it in a good way, if they have one.

    Today most participators in usenet have to find newsservers on their own.

    Up till a few weeks ago the individual net university server in Berlin
    was the best, but they decided to start taking a small fee, so a lot of
    people have to find new free newsservers, or pay.

    More and more people are forced to use some web interface, because they
    do not know where to find a free newsserver. And many youngsters have no
    idea that there is a thing called usenet to begin with. They think
    newsgroups is the same as web forums.
  17. The Phantom

    The Phantom Guest

    Do you have any instruments that can measure currents of ~75 amps
    AC? Of 5 amps AC?

    Do you have an AC voltmeter?

    The tests that others have proposed so far require you to dissipate
    power on the order of .5 kw.

    The standard method for testing the capability of a transformer
    without actually dissipating a power equivalent to the rating of the
    transformer is the short circuit test.

    You need a variac for this. You dead short the 6 volt output of the
    transfomrer. You connect a variac capable of about 5 amps output to
    the power line. Turn the output of the variac all the way down to
    zero output (THIS IS VERY IMPORTANT). Connect the variac output to
    the 240 volt winding of the transformer with an ammeter in series to
    monitor the current. SLOWLY turn up the variac until the ammeter
    reads the rated current for the 240 volt winding (about 4 amps AC in
    this case). You may need to put a (power) resistor of a few ohms in
    series with the 240 volt winding of the transformer if the current
    comes up too fast (which it undoubtedly will) when you SLOWLY turn up
    the variac. (If you don't have any suitable power resistors, use an
    electric heater or some light bulbs as resistors). (An even better
    alternative would be to follow the variac with a small step-down
    transformer. Something with about 5 amps output at 6 or 12 volts.
    The variac would power the line-voltage winding and the low voltage
    winding would power the 240 volt winding of the 1kva transformer you
    are testing.)

    When you have the current into the 240 volt winding set to 4 amps AC,
    you just wait around and see how hot the transfomer gets. The current
    in the shorted 6 volt winding should be about 160 amps. Ideally you
    should short the 6 volt winding into a suitably rated shunt and then
    you could directly measure the current there.

    Without knowing the details of the transformer insulation system, I
    can't say what the maximum allowable temperature rise is, but if the
    transformer doesn't start stinking too much, you're probably ok. So
    far, we would be testing the transformer at its full rated power.

    Since you only want 75 amps from the 6 volt winding, you could set the
    primary current in this test to 1.875 amps AC. This would correspond
    to 75 amps in the 6 volt secondary. Just let it run for a few hours
    while monitoring the temperature rise. This will provide a test of
    the transformer at the actual current which you will be using.

    Rectifying the output of the transformer will cause the current to be
    non-sinusoidal which will increase the copper (I^2*R) losses.

    You could also connect your rectifiers in whatever configuration you
    will use, and short the output from the rectifiers. This will give
    you a method of testing the heatsinking of the rectifiers. As before,
    turn the variac which is supplying the 240 volt winding ALL THE WAY
    DOWN before you start the test. Turn up the variac SLOWLY while
    monitoring the current in the 240 volt winding until you reach 1.875
    amps AC. You should use a true RMS ammeter when the rectifiers are in
    circuit for best results. I wouldn't turn the current up to 1.875
    amps right away with the rectifiers in circuit; turn it up gradually
    while monitoring the temperature of the rectifiers. If the rectifiers
    sizzle when touched with a wet finger, you probably should use some
    more heat sinking.

    The advantage of the short circuit test method is that you only
    dissipate a power equal to the losses in the transformer which should
    be under 100 watts for a transformer of this size. (light bulbs or
    electric heaters extra) It does not include the core loss, but that
    shoud be a negligible part of total losses at nearly full power out of
    the transformer.
  18. Lord Garth

    Lord Garth Guest

    You're right Rich, I was thinking of ammonium perchlorate. Sorry!
  19. Are you sure? When designing a power transformer you get allot of variables
    to play with to try to optimize the design. It has been my understanding
    that often the most optimum design (for size/weight) makes the core losses
    approximately equal to the copper losses while operated under full load.
  20. John Fields

    John Fields Guest

    AFAIK, the short test is used to measure the copper losses, not the
    "capability" of a transformer, whatever that means.
    NO!!! If you do you will exceed the current rating of the secondary
    and, possibly, damage the transformer. The transformer is rated for
    1kVA out of the _entire_ secondary which, at 12VRMS out comes to the
    83.3 amps noted on the transformer's faceplate. That is, the
    transformer secondary is wound with wire which is designed to carry
    83.3 amps no matter which voltage tap is used.
    That's not really a valid criterion, since the core losses are only
    going to be a fraction of what they would normally be with 240VAC on
    the primary. Moreover, if you're going to do it properly you need to
    monitor the temperature rise of the transformer over ambient and make
    sure it doesn't exceed the spec.
    The shunt's an OK idea, but it needs to short out the _entire_
    secondary, for the reason given earlier.

    No, you'd only be exciting the copper, not the core. To determine the
    core losses, you'd need to do the open circuit test.
    This isn't a good test either, because the core losses will be even
    less than they were previously and won't reflect the conditions under
    which the transformer will be expected to function in real life.
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