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Estimating transfomer current rating?

Discussion in 'Electronic Basics' started by Terry Pinnell, Aug 9, 2004.

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  1. I have a salvaged transformer and I'd like to estimate its max current
    rating. I've just installed it in a home-brew power supply whose
    transformer I burnt out the other day (after about 20 years usage).
    This 0-30V supply includes a current limiter with various settings,
    including 2A and 4A. So I need the estimate in order to be confident I
    can retain one or both of those top end settings.

    Physically the transformer is a bit smaller than its predecessor, with
    dimensions of about 7 x 3.5 x 6 cm (LxWxH of metal outer) and weighing
    about 1.0 kg.

    Unloaded
    Sec pair Ohms AC V
    -------- ---- --------
    Grey 8.0 9.6 ~ 10
    Blue 2.8 17.6 ~ 18
    Red 0.6 10.6 ~ 11
    Orange 0.9 26.2 ~ 26

    I assume that the grey winding is much thinner wire than the other
    three pairs, and that the orange pair, with 0.03 ohms/V, is the
    heaviest duty? That's the one I've used for the DC power supply.

    I've also made the red and blue secondaries accessible, to give me
    roughly 10, 18 and 28 V AC. The red pair gave me this data:

    Load current A AC V
    -------------- ----
    0 10.6
    1.45 9.3
    1.60 9.1
    1.77 9.0
    2.58 8.1
    2.7 8.0

    So presumably I could also use the red secondary at a fairly high
    current. But what - 1.5A, 2A, 3A...?

    For the main orange secondary (disconnected from the bridge and all
    subsequent DC circuitry), I measured these:

    Load current A AC V
    -------------- ----
    0 25.9
    0.84 24.8
    0.88 24.6
    1.76 23.0
    2.52 22.2
    3.70 20.1

    Does that provide an estimate of the 'max safe current' I can use over
    a long period (assuming the DC circuitry remains robust, which it has
    appeared to be for the last couple of decades)?
     
  2. John Larkin

    John Larkin Guest

    Figure about 15 watts per pound for conventional transformers, maybe
    30 or so for torroids. That assumes all the windings are properly
    loaded, and resistive loads.

    John
     
  3. I read in sci.electronics.design that Terry Pinnell <[email protected]
    As a rough guide, for that mass of transformer, you can take the load
    regulation to be between 5% and 10%. Using just one secondary at once,
    you can probably go for 10%. So the load current that causes the voltage
    to drop by 10% from the no-load voltage is the maximum permissible
    current.

    But don't forget that the AC current from the transformer is much larger
    than the DC output current - 1.6 to 1.8 times for a bridge rectifier.

    The final test is 'how hot does it get?', which depends on where you
    have put it as well as its own properties. It should not get hotter than
    you can touch and stay touching, as, again, a rough guide.
     

  4. Thanks all. On the basis of that advice I'll be cautious about
    anything over say 2.5A until I've used it enough to assess long term
    temperature rise.
     
  5. Robert Baer

    Robert Baer Guest

    A transformer is rated by power, which is directly related to the
    cross-sectional area, core material, and to some degree, frequency.
    In general, if your new line power transformer has about the same core
    cross-sectional area, then the power rating is essentially the same.
    Therefore, given a power rating (say of 100 watts), the current
    capability can be calculated by a simple equation P = I * E (all RMS); a
    30 V winding could be made to handle about 3.3 Amps - provided the wire
    size used gives almost zero I*R loss.
    Other secondaries can eat up so much space that a smaller than
    adequate (for max current) wire size would be used, thereby limiting the
    rating for that winding.
     
  6. Those gross figures suggest an overall output of
    about 60VA. Stretch it to say 70VA tops, which
    is probably pushing a bit.
    Assume that every winding was designed with the same
    ratio of power output to copper loss. A big assumption
    I know, but it allows a reasonable guesstimate of the
    power output of each winding. As below.

    Winding Current VA-out Copper Loss
    ~~~~~~~ ~~~~~~~ ~~~~~~ ~~~~~~~~~~~

    Grey 0.08A 0.80W 0.05W

    Blue 0.42 7.56 0.49

    Red 1.19 13.00 0.84

    Orange 1.87 48.60 3.15
    ------- ------
    Totals= 70 4.53
    ------- ------

    Those max output currents are for a resistive load,
    and should be appropriately reduced for a rectifier
    load and capacitive input filter.

    That total of 4.5W copper loss on the secondary side
    should be matched by another 4.5W in the primary.

    4.5+4.5 plus say another 2W of core loss suggests that
    the transformer would take about 81VA off the 240V mains
    at full load, ie about 0.337A full load current.

    You didn't give the primary resistance, but the sums
    above suggests that it should be about 40 ohms. That
    would be a useful cross check to see if the sums are
    anywhere in the right ballpark.
     
  7. Robert, Tony: Thanks, both. Reckon I'll downrate that 2.5A guess to
    1.5A for time being then.

    You're very close, Tony. The three primary windings measured 4.5, 42.5
    and 46.9 ohms respectively. I used the last.

    BTW, I was thinking of sticking my temperature sensor flush against
    the transformer and measuring temp with my DVM while using heavyish
    currents. But the heat dissipated from the 2N3716 mounted onto the
    rear of the case would grossly distort the result. To use that
    approach I'd have to disconnect the secondary, which of course means I
    can't use the unit for normal bench work.
     
  8. That doesn't sit well Terry..... the volume
    allowed for the primary winding is the most
    valuable real estate on the bobbin. Having
    primary windings that do nothing doesn't sound
    right.

    Is that how they are marked, or could there be
    some combination of windings that allow those
    idlers to be used?
     
  9. Gary Lecomte

    Gary Lecomte Guest

    Why not put some Thermal Switches in the Circuit. So if anything
    overheats, It Shuts off. Much cheaper than New Transformers and other
    parts. An Old Computer fan to blow air through it is also a good idea.

    Take care.......Gary
     
  10. Rich Grise

    Rich Grise Guest

    These numbers indirectly tell you the internal resistance of the
    supply. It's the slope of the graph, of course. Use that to figure
    out the dissipation at various current levels for both of these
    windings, add them up, and decide how much you want the transformer
    to dissipate.

    Or, just put a load on it and watch the temp. rise. I think one of
    the gurus beat me to this one, however.

    Have Fun!
    Rich
     
  11. Rich Grise

    Rich Grise Guest

    42.5
    4.5
    -----
    47.0, which is pretty flippin' close to 46.9. I'd say, put the
    two small ones in series, and that string in parallel with the
    other one.

    Maybe for a first smoke test, put a, say, .1R resistor in
    series with each to see how well they current-share.

    Have Fun!
    Rich
     
  12. Get a variable load like headlight bulbs and keep loading it up till it gets
    hot after several hours so the temp will stabilize.. If you can comfortably
    hold your hand on it you've found a load that it'll be happy with and should
    work for years. Remember that half wave vs full wave recitifiers will have
    double utilization factor.
     
  13. Robert Baer

    Robert Baer Guest

    I have hand-wound hundreds of power transformers for personal use, and
    *NEVER* had to worry or compensate for "copper loss". Never ran out of
    window space, either.
    Now core loss is someting that cannot be avoided, but can be safely
    ignored for 50W and higher power transformers (small percentage of core
    rating).
     
  14. Sorry, that's my careless description. There's only *one* primary,
    tapped in the familiar way (presumably to accommodate mains in the
    range 220-240V, although there are no markings). I simply measured the
    three combinations, and used the full winding.

    orange
    o---------o
    |
    C|
    C| 4.5 ohms
    brown C|
    o---------o
    C| Total winding 47 ohms
    C|
    C|
    C|42.5 ohms
    C|
    C|
    C|
    white |
    o---------o

    created by Andy´s ASCII-Circuit v1.23.080803 Beta www.tech-chat.de
     
  15. Thanks, but I mentioned what IMO is the snag with that up-thread in
    Message-ID: <>

    I've since implemented the alternative I suggested. Currently (sorry
    <g>) I have the secondary delivering 2.2A (an arbitrary first choice),
    while I monitor transformer case temperature with my DVM.

    But I still don't see how even a protracted series of such tests is
    going to tell me with any accuracy what I can expect using DC loads at
    various voltages in the range of my supply.
     
  16. Ah! Ok now, thank you.
     
  17. I read in sci.electronics.design that Robert Baer
    Your windings have zero resistance? Maybe you are using a different
    definition of copper loss than I^2R?
    This depends on the ratio of window height to limb width. For scrapless
    and even semi-scrapless laminations, the copper loss considerably
    exceeds the iron loss if proper allowance on maximum induction is made
    fro high mains voltage. With modern silicon iron, an iron (hysteresis)
    loss of 5 W/kg at 1.5 T is typical.
     
  18. I read in sci.electronics.design that Terry Pinnell <[email protected]
    If you are using a series regulator following a bridge rectifier with a
    large filter/reservoir capacitor, the a.c. secondary current will be
    between 1.6 and 1.8 times the d.c. load current. It has a peaky
    waveform, so you need a true r.m.s. meter to measure it, or use your
    scope. The waveform is close to repeated half-cycles of a higher
    frequency than 50 Hz, maybe 150 Hz, interleaved by zero-current periods.
    So you can take the r.m.s. value as the peak value divided by sqrt(2)
    and then divided by the ratio of the duration of the pulse to the whole
    half-period, i.e. 3, if the pulse looks like 150 Hz.

    The load voltage is irrelevant, because the rectifier always produces
    the full voltage across the capacitor.

    All you need to check is that the transformer doesn't get too hot with
    the maximum current you want to draw from it.
     
  19. legg

    legg Guest

    The rms current values in the secondary winding are directly
    substitutable. The 1.8 relationship of RMSin/DCout is typical of
    capacitive rectifier filters with 15% ripple.

    Theres no reason why you couldn't bodge a capacitive rectifier-filter
    onto your winding as a test load.

    When shopping for suitable junk transformers, it's often wise to take
    a meter or two and scrap paper with you.

    L values give turns ratios.
    R values give first-order current-generated rises.

    Otherwise, why not think of other uses for those extra windings? While
    neccessity may be the mother of invention, opportunity or incident is
    it's father.

    RL
     
  20. I've had time for testing the isolated secondary at only two loads so
    far, but here are the results. The temperature measurement was with
    the miniature sensor that comes with my DVM, taped to top of
    transformer. Case was open though.

    Time Load (A) Temp (C) V (AC RMS)
    ---- -------- -------- ----------
    11:03 0 25 25.4

    11:03 2.20 25 22.0
    12:00 2.18 42 21.9 OK to hold.

    12:03 2.92 42 20.6
    12:50 2.87 53 20.4 Too hot to hold
     
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