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How much current can my small transformer provide?

Discussion in 'Electronic Basics' started by ErikBaluba, Oct 22, 2005.

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  1. ErikBaluba

    ErikBaluba Guest

    Hi,

    I have a tiny isolated mains transformer with one primary winding described
    as 230V, and two secondary taps, both rated 6V and 0.175VA
    It doesn't say anything about power factor.

    What current can this transfomer provide? If I use it to power a DC circuit
    I guess it can provide 0.175 watt, or 0.175/6 = 30mA only? That's barely
    enough to drive a couple of leds? I also read somewhere that often the VA
    rating may be reduced to aproximately 2/3 to get the real power in Watts.

    E
     
  2. I suppose each secondary can provide .175/6=.029A AC RMS.

    If you rectify the outputs and filter with a capacitor, you can expect
    the transformer to get that hot when you use about half that current
    as DC. This is because the capacitor filter gets all its current in
    short pulses near the peak voltage, each half cycle. This makes the
    windings hotter than when the load current is a sinusoid of the same
    average (of absolute) current.

    They must be really tiny transformers. I don't think I have ever seen
    a 0.35 VA mains transformer.
     
  3. Kitchen Man

    Kitchen Man Guest

    Power factor is irrelevant to the transformer rating, as it is given
    in VA, or "Volt-Amps." A secondary rated at 6V @ 0.175VA will give,
    using standard mathematic grouping of terms [Volt-Amps divided by
    Volts], 0.175 / 6 = 29mA. Your calculation below is thus correct, and
    not really surprising because as you said, the transformer is tiny.
    6VAC is typical for powering a vacuum tube cathode heater, is that
    where this came from?
    If you want to find Watts, now you have to know the power factor of
    your load, done by calculating X vs. R at line frequency. Watts is
    the measure of actual work done, or volt-amps into the resistive part
    of the load. Volt-amps delivered into the reactive component of the
    load is wasted (no work is done), and is measured in Volt-Amps
    Reactive, or VARs. If you are only getting 2/3 of your energy
    converted to Watts, then you have a rather inefficient load.
     
  4. Bob Eldred

    Bob Eldred Guest


    Are you sure? I can't imagine a mains transformer that is only 0.175VA.
    That is vanishingly small. The smallest one I can find in my junk box is 5
    Watts, 6V @ 300 mA. I suspect the 0.175 VA is really current, 0.175A, 175 mA
    and it is mislabled or you are reading it wrong.

    Put a 30 ohm load on it and see what it does. What is the output voltage
    loaded and unloaded and how warm does it get after an hour or so? That
    should answer your question.
     
  5. ErikBaluba

    ErikBaluba Guest

    Thanks for the answers guys, I was out for a few days, so late follow-up. By
    the way, the transformer is really that small, its a mains transformer in a
    blue plastic shield, about the size of a grape. It was very cheap also,
    about $1. I can mail you a picture of it if you want :)
    I bought these small transformers in a local store. I plan to build my own
    home automation system and I want to place small RF controlled relays around
    my house, between the electrical-outlets and the appliances. I need to
    provide the RF receivers and the logic with appropriate voltage, so I
    thought a really tiny 6V transf. would be the thing. I guess 29mA is enough
    to power some simple RF receiver from Laipac etc..
    I just saw an article in an Electronics Magazine that described a extremely
    useful small circuit to step-down the mains voltage to TTL level by just
    using a X2 safety capictor, a FET and a few compos, no transformator.
    Perhaps i should just use that instead.

    Anyway, the concept of powerfactor confused me at first because I read that
    powersupply equipment is often given a powerfactor rating. So is such a
    rating typically given for a particular load or ampere usage? Let me follow
    up on that a bit so I get this right.
    When a load is purely resistive the powerfactor is 1, right? So the reason
    we have the concept of apparent power and real power is the
    inductive/capacitive characteristics of a load. I read that if my load has a
    capacitive reactance I should be able to add some coil/inductance to
    compensate/cancel out that, in theory at least.

    So, if I build a simple powersupply with my transformator, using a full
    rectifier and a filter cap (no regulator), can you comment on the following

    - Assuming the Cap is a big electrolyte, the powerfactor I get when
    connecting a resistive load would be a good indication of the quality of my
    transformer?
    - If the transformer and the cap where "ideal", I could select the cap to
    "match" my transformer so that my setup would have powerfactor 1?

    Also, can I combine the two taps in some way to get one 6V output with
    double VA rating, thus 2x0.175VA ?
    Wouldn't that mean that I = Vrms/30, or more than 100mA? That should surely
    make it hot yes...
    What is the likely senario when overloading a transformator like this? I
    guess the heat can melt the isolation on the copper wires and create a short
    circuit in the windings?
    I think I need to set up a few things in my work environment before I start
    fooling around with mains stuff. Rubber gloves and mat comes to mind :)


    cheers,
    erik
     
  6. ErikBaluba wrote:
    (snip)
    That is one application of power factor. Nonlinear loads like
    rectifiers that generate harmonic currents are another.
    This applies only to AC circuits, not DC.
    No. The power factor of the rectifier with a capacitor filter and
    resister load will be a low power factor load due to the harmonics the
    rectifiers generate when they charge the cap up with narrow pulses at
    the voltage peaks.
    That doesn't work, once you put a rectifier in there.
    You can just parallel the two windings, as long as both windings put
    out the same polarity. Test this by putting them in series, first.
    if they add up to about zero volts this way, just connect the free
    ends together and you are done. If they put out about 12 volts, break
    the connection between them, and remake the connection using the other
    end of one winding and test again to see if you have achieved the
    first case.

    When fully loaded, the output of the windings should be 6 volts, if
    the line is at the rated voltage.
    If the parallel combination is rated for only about 58 ma, yes, it
    will get hot, and the voltage will be less than 6.
    Yes. The time it takes to reach an over temperature condition depends
    on the thermal mass of the transformer and the amount of overload.
    Make sure there are no grounded surfaces within reach, is a good idea,
    also.
     
  7. ErikBaluba

    ErikBaluba Guest

    No. The power factor of the rectifier with a capacitor filter and
    Ok, I partly understand that. Bear with me if the following are lame
    questions and assumptions.

    So the cap is only being charged when the input voltage is near peak,
    because the cap only have time to discharge a little bit between peeks.
    But can you give a more "layman's" explanation of what is meant when
    harmonic currents are generated?
    Does it mean that the current delivered by the cap has a sinus-derived form?
    When something generates harmonics, is that generally understood as noise in
    the circuit?
    In this example, is it because diodes are not perfect that you get
    harmonics, or is it because of the inductance in the transformer trying to
    resist the increasing current to the cap? I am just trying to understand
    what it actually means.

    Also, if the frequency of the input AC was higher, what would that mean for
    the harmonics? I mean the the cap would not have time to discharge as much
    in each recharging cycle...But then again, the higher frequency would
    generate a higher impedance in the transformer...

    thanks for your help...

    erik
     
  8. Not necessarily. It just means there are currents at frequencies that
    add into the line RMS current that do not add into the average DC
    output current. Here is my attempt to show the line current of a
    capacitor filtered full wave rectifier load (vies with fixed width
    font, like Courier):

    ^
    | |
    | |
    ----- ----- -----
    | |
    | |
    V

    That kind of waveform is made up of the line fundamental and many odd
    harmonics. The RMS current is the square root of the sum of the
    squares of the RMS current of each of those harmonic frequencies.
    But the average DC is just the average of the absolute magnitude of
    the waveform. For a given load resistance, the larger the capacitor,
    the shorter the pulses and the higher the harmonic content in the
    current waveform, even though the average DC current will be about the
    same. The higher RMS value of the large, narrow pulses also have an
    increased heating effect on the transformer winding resistances.

    All those harmonic currents have to be supplied from the AC source
    (through the distribution system), even though they do not relate
    directly to making the DC.
    Neither. It is because the capacitor receives current in narrow
    pulses twice per cycle. Switching supplies with power factor
    correction circuits are attempting to spread the line current out over
    the entire cycle, making the load look more like a resistor to the
    distribution system.
    Everything just scales up in frequency, as long as the diode
    conduction time remains the same fraction of the cycle time. Changing
    to an inductor input filter makes the current almost constant, that
    produces a different mix of line harmonics, but a mix that contains
    more fundamental and less harmonics, so the power factor is better.
     
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