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transformer theory questions

Discussion in 'Electronic Basics' started by [email protected], Dec 29, 2003.

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  1. Guest

    I already understand how the turns ratio between primary and secondary
    of a transformer translates into the voltage ratio. What I would like
    to understand is why this works this way.

    Consider a simple 1:1 transformer. I'll get the same voltage out on
    the secondary as I input on the primary. Now if I add some turns to
    the secondary, I get more voltage. But if I add some turns to the
    primary, I get less voltage out on the secondary. This can obviously
    be seen by turning the transformer around and applying power to what
    was the secondary, and getting more voltage out on the primary when
    the primary has more turns.

    What I want to know is how it accomplishes this. I would think that
    if I add more turns on the primary, I'm going to get more magnetic flux
    in the core. Or is this simply not the case?

    When answering this, please try to avoid falling back to turns ratio
    itself, since I am accepting that already, though on faith (because it
    can be observed to work). I'm trying to understand _why_ it works.

    Along these lines I have other questions, and the answer to these may
    even help understanding the above.

    Turns ratio is one thing (establishes the voltage ratio). But how is
    the actual number of turns to be determined (e.g. what factors will it
    depend on)? I'm sure that if I build a transformer with a nice solid
    iron core and wrap 2 turns on the primary with some nice thick 0 gauge
    wire, and wrap 4 turns on the secondary, and plug it in, I'm going to
    get a blown fuse or tripped breaker, instead of double the voltage out
    on the secondary (well, maybe it will be for an instant).

    Suppose I have a given transformer with room to add more windings to
    both primary and secondary, and I add an equal proportion to both so
    the ratio stays the same. What does that do to the capacity of the
    transformer?

    I've seen the internal windings on a 1000 KVA transformer, and it just
    didn't look like much. They were thick, but not that many. It just
    seemed like it would "nearly short out" whatever power was applied to
    either primary or secondary. Or is it the case that this sizing is
    also going to translate into the waste factor of a transformer (e.g.
    the waste in a 1000 KVA transformer with a 120/240 volt primary would
    be more than a normal home 15/20 amp circuit would take)?
     
  2. Something like faraday's law of induced voltage is probably the easiest way
    to describe it. The induced voltage is equal to the number of turns times
    the rate of change of the magnetic flux through the loops. Since the loops
    of the transformer are around the same core, the rate of change of flux is
    equal for both sides (ignoring various issues). Thus, the induced voltage
    divided by the the number of turns is equal. This means that

    Vin/Npri = Vout/Nsec

    or

    Vin/Vout = Nsec/Npri

    So, it really is the ratio...

    Adding more turns on the primary does increase the flux through the core.
    Faraday's law, however, causes the change in flux term to fall out of the
    equation. Within certain limits, the number of turns isn't important, its
    the ratio of primary to secondary that is important.

    Note that you can't increase the flux indefinitely by increasing the turns
    of the primary, since the permeability of the core decreases with current
    through the turns. Other such issues (like resistance in the turns) conspire
    to cause the secondary voltage to diverge from the ideal.

    I'll let somebody who has actually built transformers answer the rest of the
    questions.

    Regards
    Bob Monsen
    ---
     
  3. John Larkin

    John Larkin Guest

    Nope: decreases it. That's why the secondary voltage will go down when
    you add primary turns.

    John
     
  4. Guest

    | Something like faraday's law of induced voltage is probably the easiest way
    | to describe it. The induced voltage is equal to the number of turns times
    | the rate of change of the magnetic flux through the loops. Since the loops
    | of the transformer are around the same core, the rate of change of flux is
    | equal for both sides (ignoring various issues). Thus, the induced voltage
    | divided by the the number of turns is equal. This means that
    |
    | Vin/Npri = Vout/Nsec
    |
    | or
    |
    | Vin/Vout = Nsec/Npri
    |
    | So, it really is the ratio...

    Well, I know it is. I'm looking for an understanding of how things work
    to make it so, not just reducing some equation to see what falls out.


    | Adding more turns on the primary does increase the flux through the core.
    | Faraday's law, however, causes the change in flux term to fall out of the
    | equation. Within certain limits, the number of turns isn't important, its
    | the ratio of primary to secondary that is important.

    Right. But what changes when you add more primary turns? How does the
    secondary end up getting a lower induced voltage as a result?


    | Note that you can't increase the flux indefinitely by increasing the turns
    | of the primary, since the permeability of the core decreases with current
    | through the turns. Other such issues (like resistance in the turns) conspire
    | to cause the secondary voltage to diverge from the ideal.

    And flux saturation. But we could ignore those issues and look at just the
    number of turns. It's the ratio. But what I want to know is what fundamental
    physics are involved to get that.
     
  5. Guest

    | On Mon, 29 Dec 2003 21:51:44 GMT, "Robert Monsen"
    |
    |
    |>Adding more turns on the primary does increase the flux through the core.
    |
    | Nope: decreases it. That's why the secondary voltage will go down when
    | you add primary turns.

    How/why does the flux go down with more primary turns? Is the coil reactance
    a culprit in that?
     
  6. The flux in the core is proportional to sum of the amp turns of all
    two windings (total amps going around the core). If we deal with an
    open circuit secondary, so we don't have to deal with any amp turns in
    the secondary (and just look at the voltage) then all the amps come
    from the primary. The primary RMS current is the applied voltage
    divided by the impedance of the primary, which is proportional to the
    primary inductance, which varies with the square of the primary
    turns. So doubling the primary turns, multiplies the primary
    inductance by 4, lowering the primary current to 1/4, lowering the
    peak flux in each direction to (1/4)*2=1/2 (amp turns, remember?).
    The volts per turn is proportional to the rate of change of the flux.
    Have you seen the V=L*(di/dt) differential definition of inductance?
    Keeping in mind that flux is proportional to amp turns, voltage must
    also be proportional to df/dt (f being flux). So doubling the primary
    turns also lowers the df/dt to (1/4)*2=1/2 also. So the volts per
    turn goes to 1/2, and we are back to another way of seeing that
    doubling the number of primary turns halves the volts per turn for all
    windings without assuming it.
    It is limited by the peak flux in each direction during a cycle. The
    lower the number of turns, the faster the flux has to change to
    produce the higher volts per turn, so the farther it ends up swinging
    by the end of each half cycle. If we could come up with a low loss
    transformer core that could handle twice the peak flux without
    saturation, we could halve the number of turns in all windings for the
    same integral volt seconds per half cycle. Integral volt seconds is
    proportional to peak flux.
    Because the low inductance of the primary will allow large current to
    pass, saturating the core very quickly. If the frequency were high
    enough that the high volts per turn did not have to be supported for
    so long, you might get away with it. High frequency transformers tend
    to have low numbers of turns compared to low frequency transformers.
    It shifts the losses from core (because of lower hysterisis (which are
    driven by peak flux swing) and eddy current (which are driven by volts
    per turn) losses, while increasing the copper I*R losses (adding turns
    adds resistance).
    A big transformer has a lot of core cross sectional area, so it can
    carry a lot of total flux through the windings. Big rate of change of
    flux means big volts per turn, means low number of turns. So there is
    room for fat wire to get low loss at high current.

    In general, the bigger the transformer, the smaller the waste compared
    to the power carried through the transformer. This is not something
    you get for free, but is a design necessity. You have to under
    utilize the flux capability of a very big transformer core, compared
    to a very small one, because the big one has a lower surface to core
    volume ratio. The heat is produced in all the volume, but has only
    the surface area through which to escape. Same for the copper
    volume. So you have to design a big transformer to carry slightly
    less power per pound of iron and copper than a small one. Stressing
    that pound less produces less waste heat per watt carried in the big
    one because that is the only way to keep the temperature the same.

    So a wall wart transformer may be only need to be 80% efficient at
    rated power to keep from self destructing, while a substation
    transformer may have to be designed to be 95% efficient at full load
    to have long term reliability (and this while being cooled with
    circulating oil). It is all about temperature rise.
     
  7. John Larkin

    John Larkin Guest

    Yes; John P explains it pretty well below.

    Another way to look at it: assuming the wire resistance is pretty low,
    disappear the core and apply 120 VAC. The current will be huge and
    fuses will blow.

    Now reinstall the core. The current goes down because there is a 'back
    emf' effectively in series with the winding that cancels out most of
    the applied 120 AC. This voltage is generated by the AC magnetic flux
    within the core passing through the turns. Where does this flux come
    from? From the applied AC! Current in the primary magnetizes the core,
    the AC magnetic field induces voltage into the primary turns, and that
    voltage semi-magically cancels out most of the applied 120V input,
    leaving only enough residual current to magnetize the core!

    More primary turns need less AC magnetic flux to induce the almost-120
    volts back EMF.

    Turns out that - first order approx - flux density is proportional to
    volts-per-turn.

    Another myth: loading the secondary can saturate a transformer. In
    reality, AC loading on the secondary *decreases* core flux density.

    John
     
  8. Rich Grise

    Rich Grise Guest

    ....
    Strictly speaking, that's not faith, because it's observed. I think
    the fancy-schmancy term is "empirical." :)
    I had a sort of "flash" of inspiration here - when you stack up 2
    turns in the primary, each only gets half of the applied voltage,
    which is then transformed over to the other wire. But the current
    is applied in parallel, as it were, since it goes by twice.

    At least it sounds pretty cool to me, and is fun to visualize
    with some good bud. %-}

    Cheers!
    Rich
     
  9. Rich Grise

    Rich Grise Guest

    Actually, I rather like Robert Monsen's answer. :) At least it makes
    sense to me, if I try to see your POV.

    Cheers!
    Rich

     
  10. Guest

    | Another way to look at it: assuming the wire resistance is pretty low,
    | disappear the core and apply 120 VAC. The current will be huge and
    | fuses will blow.

    But there will be some inductance, just not very much? Could you wind
    enough turns over air to draw no more than 10 amps at 120v 60Hz? Surely
    I would think so (but it would be a lot of turns).


    | Now reinstall the core. The current goes down because there is a 'back
    | emf' effectively in series with the winding that cancels out most of
    | the applied 120 AC. This voltage is generated by the AC magnetic flux
    | within the core passing through the turns. Where does this flux come
    | from? From the applied AC! Current in the primary magnetizes the core,
    | the AC magnetic field induces voltage into the primary turns, and that
    | voltage semi-magically cancels out most of the applied 120V input,
    | leaving only enough residual current to magnetize the core!

    Somewhat like a reflection of power that would not only be induced into
    the primary, but also into any winding. So double the windings and you
    get double the induced voltage back, given a particular flux?


    | More primary turns need less AC magnetic flux to induce the almost-120
    | volts back EMF.
    |
    | Turns out that - first order approx - flux density is proportional to
    | volts-per-turn.
    |
    | Another myth: loading the secondary can saturate a transformer. In
    | reality, AC loading on the secondary *decreases* core flux density.

    Actually, that would make sense to me. The secondary would be taking
    power out of the system.
     
  11. Guest

    | wrote:
    | (snip)
    |> This can obviously
    |> be seen by turning the transformer around and applying power to what
    |> was the secondary, and getting more voltage out on the primary when
    |> the primary has more turns.
    |>
    |> What I want to know is how it accomplishes this. I would think that
    |> if I add more turns on the primary, I'm going to get more magnetic flux
    |> in the core. Or is this simply not the case?
    |
    | The flux in the core is proportional to sum of the amp turns of all
    | two windings (total amps going around the core). If we deal with an
    | open circuit secondary, so we don't have to deal with any amp turns in
    | the secondary (and just look at the voltage) then all the amps come
    | from the primary. The primary RMS current is the applied voltage
    | divided by the impedance of the primary, which is proportional to the
    | primary inductance, which varies with the square of the primary
    | turns. So doubling the primary turns, multiplies the primary
    | inductance by 4, lowering the primary current to 1/4, lowering the
    | peak flux in each direction to (1/4)*2=1/2 (amp turns, remember?).
    | The volts per turn is proportional to the rate of change of the flux.
    | Have you seen the V=L*(di/dt) differential definition of inductance?
    | Keeping in mind that flux is proportional to amp turns, voltage must
    | also be proportional to df/dt (f being flux). So doubling the primary
    | turns also lowers the df/dt to (1/4)*2=1/2 also. So the volts per
    | turn goes to 1/2, and we are back to another way of seeing that
    | doubling the number of primary turns halves the volts per turn for all
    | windings without assuming it.

    So to build a higher KVA transformer at a given voltage means fewer
    turns because you have to have more current ... and a correspondingly
    larger core? That would explain why these big KVA transformers have
    so few turns.


    |> I'm sure that if I build a transformer with a nice solid
    |> iron core and wrap 2 turns on the primary with some nice thick 0 gauge
    |> wire, and wrap 4 turns on the secondary, and plug it in, I'm going to
    |> get a blown fuse or tripped breaker, instead of double the voltage out
    |> on the secondary (well, maybe it will be for an instant).
    |
    | Because the low inductance of the primary will allow large current to
    | pass, saturating the core very quickly. If the frequency were high
    | enough that the high volts per turn did not have to be supported for
    | so long, you might get away with it. High frequency transformers tend
    | to have low numbers of turns compared to low frequency transformers.
    |
    |> Suppose I have a given transformer with room to add more windings to
    |> both primary and secondary, and I add an equal proportion to both so
    |> the ratio stays the same. What does that do to the capacity of the
    |> transformer?
    |
    | It shifts the losses from core (because of lower hysterisis (which are
    | driven by peak flux swing) and eddy current (which are driven by volts
    | per turn) losses, while increasing the copper I*R losses (adding turns
    | adds resistance).

    My original presumption of more turns = more combined flux was just all
    wrong. The flux is proportional to the current and more inductance means
    less current (as well as more copper resistance as a small side effect).


    |> I've seen the internal windings on a 1000 KVA transformer, and it just
    |> didn't look like much. They were thick, but not that many. It just
    |> seemed like it would "nearly short out" whatever power was applied to
    |> either primary or secondary. Or is it the case that this sizing is
    |> also going to translate into the waste factor of a transformer (e.g.
    |> the waste in a 1000 KVA transformer with a 120/240 volt primary would
    |> be more than a normal home 15/20 amp circuit would take)?
    |
    | A big transformer has a lot of core cross sectional area, so it can
    | carry a lot of total flux through the windings. Big rate of change of
    | flux means big volts per turn, means low number of turns. So there is
    | room for fat wire to get low loss at high current.
    |
    | In general, the bigger the transformer, the smaller the waste compared
    | to the power carried through the transformer. This is not something
    | you get for free, but is a design necessity. You have to under
    | utilize the flux capability of a very big transformer core, compared
    | to a very small one, because the big one has a lower surface to core
    | volume ratio. The heat is produced in all the volume, but has only
    | the surface area through which to escape. Same for the copper
    | volume. So you have to design a big transformer to carry slightly
    | less power per pound of iron and copper than a small one. Stressing
    | that pound less produces less waste heat per watt carried in the big
    | one because that is the only way to keep the temperature the same.

    That volume vs. surface issue sure shows up in a lot of places :)


    | So a wall wart transformer may be only need to be 80% efficient at
    | rated power to keep from self destructing, while a substation
    | transformer may have to be designed to be 95% efficient at full load
    | to have long term reliability (and this while being cooled with
    | circulating oil). It is all about temperature rise.

    Which is a critical rating on both oil and dry distribution transformers.
     
  12. Guest

    |
    | | ...
    |> When answering this, please try to avoid falling back to turns ratio
    |> itself, since I am accepting that already, though on faith (because it
    |> can be observed to work).
    |
    | Strictly speaking, that's not faith, because it's observed. I think
    | the fancy-schmancy term is "empirical." :)

    OK, sounds good to me.


    |> I'm trying to understand _why_ it works.
    |
    | I had a sort of "flash" of inspiration here - when you stack up 2
    | turns in the primary, each only gets half of the applied voltage,
    | which is then transformed over to the other wire. But the current
    | is applied in parallel, as it were, since it goes by twice.

    And there being 2 sets of turns in the primary, twice the inductance
    and half the current?


    | At least it sounds pretty cool to me, and is fun to visualize
    | with some good bud. %-}

    It's starting to resemble radio transmission antennas. If I can connect it
    all back to that, I think I'll have it (I do know more about that).
     
  13. John Larkin

    John Larkin Guest

    Yup, a lot of turns. The permeability of air is something on the order
    of 1/10,000 or so that of a transformer core, so the 'magnetizing
    current' (just the primary current into the primary inductance) is
    correspondingly higher if you remove the core. In most practical
    situations, an air-core coil will have more copper-wire resistance
    than inductive reactance at 50/60 Hz.

    You could download a coil-inductance calculator and try some cases.

    A single turn of #0 wire, about 2 miles in diameter, should be close.

    John
     
  14. Right, with fixed V, N and dPhi/dt are inversely proportional.

    V = N.dPhi/dt

    Thanks for pointing that out.

    Regards
    Bob Monsen
     
  15. Peter

    Peter Guest

    Great thread. I learned a lot.


    ---
     
  16. Don Kelly

    Don Kelly Guest

    ------------
    John, I am late jumping into this but one point that is generally forgotten
    or ignored by many is that, on the basis of e=N(d(flux)/dt) , for AC the
    relation Eave =4(frequency)*(Maximum flux) * Number of turns (For a
    sinusoid change the Bougerre factor to 4.44 ). Trade offs between flux,
    frequency and turns, as you indicate, then occur. The core material
    doesn't enter this. The core affects only the excitation current needed to
    get the required flux.

    By the way, I would not be at all interested in a substation transformer
    with only a 95% efficiency at full load. I would want what is readily
    obtainable, an efficiency in the order of 99.5% at full load. I have tested
    3KVA air cooled 120/240V transformers which had an efficiency over 95% from
    1/4 load to 1-1/4 times full load rating, peaking about 98%. :)
     
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