Transformer theory for computer hackers.

From: Philip.Belben_at_powertech.co.uk <(Philip.Belben_at_powertech.co.uk)>
Date: Mon Jan 19 12:30:15 1998

This would be easier with some diagrams, but I'll do my best. This is
not strictly on topic but I think I'd better clear up some of the
confusion since I caused it!

The equivalent circuit of a transformer is an "ideal transformer" in
series with an impedance and in parallel with a magnetising impedance:


O---------------+----------UUUUUU----/\/\/\/----)||(-------------O
                | )||(
                / Xs Rs )||( O
I \ Rm )||( U
N / )||( T
P | )||( P
U ) )||( U
T ) Xm )||( T
                ) )||(
                | )||(
O---------------+-------------------------------)||(-------------O


The ideal transformer merely changes voltage.

Xs = series reactance (inductive)
Rs = series resistance (copper loss - due to resistance of windings)
Xm = magnetising reactance
Rm = magnetising resistance (iron loss - due to hysteresis of core)

There are two things limiting the power you can shove through a
transformer. First (and quite easy) is the thermal limit - the current
flowing in Rs. In reality, the thermal limit is the current flowing in
_any_ winding, which might burn out that winding.

Next is the flux in the core. The two windings of the ideal transformer
carry equal and opposite ampere-turns, and their flux cancels out. The
flux in the core is determined by the current in Xm, which depends on
voltage and frequency (remember it's an inductor - more current at lower
frequencies)

Usually the transformer has a primary winding with a rated voltage (so
you don't over-flux it. It also has a VA rating which determines the
current (so you don't burn out the windings)

If you are trying to kludge things, there are some special cases to
worry about.

1. 50Hz MAINS.

If your transformer is designed for 60Hz mains and you want to use it at
50Hz, the lower frequency will meaan that Xm is lower. (X=2*PI*f*L).
The magnetising impedance will draw more current, and the flux in the
core will be higher. From a practical point of view, this means that
the rated VOLTAGE must be lower. For example if it is designed for
240V, 60Hz, at 50Hz the rating is 200V. This is called DERATING.

Since the rated voltage is lower, and the rated current is the same
(excessive current will melt copper whatever the voltage), the VA rating
is also less.

Most small commercial transformers are designed for 50Hz anyway, so this
shouldn't be a problem. However (following the discussion of mainframes
and things) most [North American] medium to large power transformers are
not! (Medium > 5kVA I suppose. Large probably > 50 kVA. Hard to say -
in my job I regularly deal (at least on paper) with 500000kVA
transformers)

2. HOME MADE AUTOTRANSFORMERS.

An autotransformer has the low voltage end of one winding connected to
the high voltage end of another. This saves copper, reduces core size,
etc. It is often used for converting between mains voltages and is at
its best around the 2:1 voltage ratio.

Many small transformers are made with two primary windings, each rated
at 120V or thereabouts. The idea is to connect them in series for 240V
operation and in parallel for 120V. A trick I mentioned a few days ago
is to wire them together as an autotransformer (Windings are W1 and W2
in the diagram:

                  +---------------O
                  |
                  )||
               W1 )||
                  )||
O-----------------+||
                  )||
IN- W2 )||
PUT )||
                  |
O-----------------+---------------O

Call the VA rating of the transformer V*I, where V is the voltage rating
of the windings in series (usually 240V) and I is the current that flows
in each winding.

On the HV (higher voltage) side, we want V and I. On the LV side we
must therefore provide V/2 and 2I. (Assuming an ideal transformer for
the moment)

Now each winding is rated at V/2, so energising one at this voltage is
not a problem. But each winding is only rated at I, so what do you do
with the 2I you have to supply?

2I must flow in at the junction of the two windings, and I must flow out
at the HV terminal. It is easy to see that I will flow through W1; the
rest of the current entering at the junction must flow through W2.
Simple arithmetic tells you that each winding carries a current of I;
the ampere turns are equal and opposite, so the transformer will work.

However, once you add the imperfections, you do start to get a few
problems. I'd have to do some serious calcs to find out exactly how Zs
(= Rs + i*Xs) and Zm affect it, but you may have to take a VA rating 10%
or so lower than nameplate. Which is not much of a loss.

You also have a regulation problem. Most small transformers have extra
turns on the secondary to compensate for the voltage drop across Zs.
Here you are only using the primary, so you don't get this. Expect a
voltage drop (proportional to load current, of course) of about 10% to
20% (less for larger transformers) at full load.

So, in response to Tony's comment, the transformer won't mind if you use
it almost up to nameplate rating - but you may not get as many volts out
as you expected. Using a transformer at 30% of nameplate rating is as
good a way as any to avoid this!

In response to (I forget your name - sorry!) I hope this has explained
what I was getting at.

Finally,

SCOTT CONNECTION etc.

Someone yesterday(?) tried to perpetuate one of the myths about the
Scott transformer, which is used in converting balanced 2-phase to
3-phase power.

If you have two phases at 90 deg to each other, you have a balanced
2-phase supply. This is very rare! There will usually be 3 wires - two
phases and neutral - and the phase to phase voltage will be about 1.41
times the phase to neutral voltage. The scott transformer (actually two
single phase transformers) converts this to and from 3 phase.

Most US domestic supplies (and a few UK rural supplies to farms and
things) have two phases at 180 deg to each other - phase to phase
voltage is twice phase to neutral voltage.

THERE IS NO TRANSFORMER THAT WILL GIVE YOU THREE PHASE FROM SUCH A
SUPPLY. You need preferably a motor-generator or dynamotor
(motor-generator sharing a common rotor winding). As mentioned
previously, an induction motor with suitable capacitors and things will
do this quite well up to a point. Electronic converters exist and are
getting cheaper all the time...

Finally, you may have two of the phases of a three-phase system. If you
have three wires - two phases and neutral - there is a transformer that
will derive balanced 3-phase from this. If not, you're in the same
position as with a single phase or unbalanced 2-phase (which is actually
single phase centre tapped to earth) supply.

If anyone STILL doesn't understand what's going on, ask privately. Who
knows? If I get very many questions, I may be posting "3 phase ac
theory for computer hackers" before long...

Philip.

<><><><><><><><><><><><><><> Philip Belben <><><><><><><><><><><><><><>
      Bloedem Volke unverstaendlich treiben wir des Lebens Spiel.
      Grade das, was unabwendlich fruchtet unserm Spott als Ziel.
      Magst es Kinder-Rache nennen an des Daseins tiefem Ernst;
      Wirst das Leben besser kennen, wenn du uns verstehen lernst.

Poem by Christian Morgenstern - Message by Philip.Belben_at_powertech.co.uk
Received on Mon Jan 19 1998 - 12:30:15 GMT