Re: Rebuilding an OBIT / saturation

["Tesla list" <tesla-at-pupman-dot-com>]

Original poster: "Dr. Duncan Cadd by way of Terry Fritz <twftesla-at-uswest-dot-net>" <dunckx-at-freeuk-dot-com>

Evening All!


In the words of Monty Python's Flying Circus ("Flying Sheep" Sketch)

'A fair question, and one that in recent weeks has been much on my mind.'

>> >
>> > TRANSFORMERS DO NOT SATURATE DUE TO TOO MUCH CURRENT BEING DRAWN !
>> >
>>
>> Then how come that you need bigger cores for higer power?
>
>Jochen, Richie, all!
>
>Interesting question, which has kept me pondering for years.
>
>I think the answer is this: you don`t _need_ a bigger core for higher
>power,

Power = volts x amps.

More power = more (volts x amps).        :-)

Volts = turns.

More (volts x amps) = more (turns x amps).

Flux = turns x amps.

More power = more flux.

More flux in same size core = saturation.

More power = bigger core.

Primary (amps x turns) determines total flux.

On no load, the primary current is in two parts:

1) A magnetising component at 90 degrees lag behind E(pri) which produces the
working flux.
2) An in-phase component which accounts for real heating losses in iron
hysteresis and eddy currents.  We hope this is small.

Under no load conditions, there is no current in the secondary and hence no
flux
is generated by the secondary winding.  The flux produced by (1) induces a
voltage E(sec) in the secondary because the secondary turns are cut by the
alternating primary flux.  The voltage ratio is equal to the turns ratio.

As a current I(sec) is drawn, the secondary ampere-turns will tend to produce a
secondary flux, which if it were allowed to exist would disturb the flux
conditions existing in the core at no load and thereby change the voltage
ratio.
This secondary flux is therefore counteracted by an induced  increase in
current
I(pri).  The current ratio is the inverse of the turns ratio.  The increase in
I(pri) supplies the power now taken from the secondary.

Since the induced primary flux and the secondary flux always cancel each other,
the total flux at no load is therefore approximately the same as the total flux
at full load and hence the power capability of the transformer is set by the
flux which can be set up in the core under no load conditions!  Strange, but
true.

All of this assumes that the windings have neither resistance nor reactance.

Resistance in the wire leads to copper losses which leads to a drop in
secondary
volts as the current draw increases, because the current through the resistance
sets up a voltage drop in the winding (E = iR), whilst reactance produces
leakage inductance and the flux from this stray reactance does not link the two
coils, i.e. it does not contribute to mutual inductive coupling.  Since this
flux is not linked, it is not counteracted, and also leads to a drop in output.
If the current circulating is low, the leakage flux will be low.  As the
load on
the transformer increases, the current in both windings increases, and the
leakage flux becomes noticeable, hence attracting screwdrivers etc to the core.

Under short circuit conditions, it may be possible for the heating losses
in the
secondary to destroy the winding, since when E(sec) = 0 the whole output
voltage
must be dropped across the resistance of the winding, in which case the power
dissipated in the secondary wire is roughly equal to E^2 / R, which will be
equal to the power rating of the transformer (though the leakage inductance
will
reduce this).

Anyone smell bacon? (incendiary pigs)

The effects of resistance and reactance on the load characteristics will be the
same as if the transformer was perfect but connected to an appropriate
resistance and reactance in series with the primary and secondary windings.
Both will cause a drop in output as power is drawn from the secondary.

However, the *net* flux in the core cannot be increased by increasing the
secondary current because it is always cancelled by the induced primary
current.

Richie is right!

Dunckx