## Alternator Stator Outputs Shown as Center Tapped Transformers.

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• A question that comes up is the difference between a delta and wye configuration. These things are usually looked at with respect to how a 3 phase load is
Message 1 of 1 , Jan 12, 2004
A question that comes up is the difference between a delta and wye
configuration. These things are usually looked at with respect to how
a 3 phase load is configured with respect to the three available
wires of input. However the 3 phase source of emf itself has two
options of how it can be connected, and is always connected in WYE.
The WYE connection from the stator coils, is theoretically 1.7 times
the voltage present as if the voltages were instead extracted as
Delta. If those stator coils themselves were arranged as center
tapped trasformers typically used on the "outside" application from
power sources, then the resultant voltage of the ending coils would
be twice that found on each coil alone, because of the 180 phased
voltages on each coil, but here it assumed that because the three
phase motional emf voltages are 120 degrees out of phase, instead of
180, we then get only 1.7 times the series coil voltage instead of 2
times that individual phases voltage. As output forms then we
generally think of delta and wye as the only two possible expressions
of stator wiring output. There is however a third possibility where
diodes could be used to switch the stator output between a delta and
a wye expression, depending on the polarity or the currents found in
the AC. It gets a little hard to imagine how that signal would look?

Heres a letter I recieved some time ago on the subject.

> WYE stator
> windings. I am just an automotive student and our
> why a Delta stator carries more current
Right off the bat I would think it does NOT contain
more current, but;
These are somewhat complicated issues when both
voltage and amperage issues are mentioned, but I would
hypothesise that the output in delta could only
contain more current if the delta phase were obtained
totally in isolation, meaning that we would then have
6 terminals, where each phase would have two seperate
terminals obtained for its output. This is a confusing
issue I have not yet mentally resolved, and I could be
wrong on this specific point.
and a WYE
> stator carries more
> voltage.
The wye stator does contain more voltage because it is
the result of two stator coils in series, rather then
just one. This is why it puzzles me that your
instructor would say that the Delta output should
contain more current, which ordinarily I would think
is untrue, and the only possible way I can think to
say that it might be true is the above reservation
where that MIGHT occur if the phases were derived in
isolation, with no interphasal connections. If we had
three delta phases in isolation, each outputing one
amp, that would be 6 wires holding 6 amps outwards
from the source. If we allowed each of those delivery
wires to be joined with the adjacent delta phase then
we would instead have three wires, no longer holding 6
total amps, but 1.7 amps times 3 or 5.1 amps.

My instructor is a 30 year ASE certified
> technition and he
> did not know the answer. If your could help me, not
> to show my
> instructor up, but to educate our class with an
> sincerely appreciate it.
>
> Thank You,
>
> Mike
>
Its rather simple, they simply dont make the stator
windings in Delta, they make them in Wye instead. If
they made then in Delta, each coil encompassing the
flux from the motional magnetic field would be
recieving that quantity, and a measurement of that
quantity would show a lesser amount of total flux
change converted to voltage. Say we have 1 volts emf
for a single delta stator winding. We have three of
them doing this. Now we take those same windings and
reconvert them to WYE. Formerly we only had one stator
winding per phase for output. Picture the delta
triangle being pushed inwards from the middle of each
delta pathway, we can only push it so far inwards
until the collapsed triangle becomes a WYE. But when
we have done this we have allowed one delta branch to
share the pathways with each of its neighbors delta
pathways. This however may only be a mental picture to
understand the analogy later on. If the analogy
applied then now the currents available on those new
pathways would be the vector sum of each two delta
stator currents, which is also what occurs when we
join 6 wires in delta isolation to delivery with only
three wires. If we look at the problem from the
reverse direction, from the currents found on the
delta load vs the line that carries those shared
a vector subtraction with respect to those currents.
Each action carries the inverse aspect; depending if
we are working from the inside-out, the more common
circumstance, or the outside-in, in which case the
laws are reversed.

Now picture instead the actual change of stator
connections. Now each output phase; (OBTAINED AS A
DELTA OUTPUT, THIS IS AN IMPORTANT ASPECT AS TO HOW WE
ARE VIEWING THE PROBLEM); it now has TWO stator coils
in series to make that output, instead of merely one
obtained in the Delta to Delta case. But just because
we formerly had one volt in the delta to delta case,
we do not expect that now because now we have two
sources of motional emf in series, that we would now
obtain 2 volts output. Rather we obtain the VECTOR SUM
of those two motional emf's, which is 1.7 volts,
because the AC waveforms on each of these stator
windings are 120 degrees out of phase. So there is a
trade off here, even though we are not recieving
double the amount by using twice the amount of stator
coils for delivery, we are still getting more then if
we drew on merely one stator coil alone.

One may wonder why in the stuff I do, that I never
wire the outputs in WYE. This is because then each of
those WYE based outputs would then be recieving a
lesser amount of voltage. Twice the amount of load
would be present, just like in the analogy where we
pressed the delta triangle into a wye formation, since
we now have two loads on that former delta path,
shouldnt that 1.7 volts be cut in half for the new
voltage division? If this were done for just a single
phase that would be true. But once we do it for two
delta phases, now each of those former phases is
sharing a half a pathway with its neighboring phase,
and when we do it for all three phases, the total
pathways on the wye become all pathways where the
current is being shared by the adjacent phase. Thus
the 1.7 volts doesnt get chopped into half, because
the adjacent phase also contributes some voltage,
albeit a voltage that is not in phase. Suppose I had
that outside 1.7 volts being extracted from the WYE to
input, (which never changes, because thats how they
are manufactured), to a WYE output then another
question comes into being, then we are back to the
original problem. For that output conversion,
ordinarily we use a fourth return wire, and this
fourth return wire connects back to the WYE stator
junction. We have returned to the original problem,
because a DELTA TO DELTA conversion is conceptially
the same thing as connecting the alternator output to
a WYE load, because that is a WYE to WYE conversion.
Both of these things ARE the same thing, with the
exception of how may lines are being employed that
share currents from different phases. So in the WYE to
WYE conversion, once again we are deriving a source of
motional emf from only a single stator coil, and in
this conversion, only the return fourth wire back to
the stator is the line that shares currents from all
three phases, and if all the currents on all the
phases are equal, that wire will hold complete
cancellation of current, equalling zero return
current, Thus for balanced WYE delivery, all of the
returning current returns on the adjacent phases. The
only difference then from a delta to delta and a wye
to wye is that for the delta case we do not have one
wire sharing current for all three phases, rather we
have three (output) wires that each share currents
with the adjacent phase, but in both cases the
motional emf is derived from only a single motional
emf obtained by only one stator phase. Since a wye to
wye is the same as a delta to delta, there is no
reason to specifically make the stator combinations
into exclusively a delta output, because practically
the same effect takes place in procuring the outputs
in wye, from the wye based stator.

I am still in confusion regarding the statement that
the delta output will be an output where MORE current
would be obtained. Again if we wish to be totally
specific and talk only about what is being done to to
the stator connections; we can say the following for
ALL cases of shared line formations.

The DELTA wired stator delivers LESS voltage, but it
does not deliver MORE current, it delivers LESS
current because it is being sourced from only a single
stator coil winding.

The WYE wired stator delivers both MORE current and
MORE voltage because then the motional emf stator
winds use two of these windings in series, but it does
not deliver twice the voltage or amperage found on
each stator winding because of a consequence of the
120 degree phasing differences.

A 1 amp delivery for these stator lines attached to a
the balanced case.

A 1 amp delivery for these stator lines to a load in
delta will contain 70% less current to the delta
phase, but it will be delivered at 70% more voltage.

I would suppose then if each of the loads were equal
for both wye and delta extractions, the 70% higher
voltage would then mean that actually more current was
extracted for the equal resistive load values placed
in delta rather then wye. Perhaps this is what is
meant by saying that more current is obtained IN A

This is where things get totally confusing if we do
not specifically state whether the aspect is internal,
as the question seems to be expounded; or external as
the load configuration. The laws are exactly opposite
for each case, where the voltages divide for a wye
of how the source itself is wired.

Perhaps we are making the problem appear needlessly
more complicated. Perhaps it needs to be rephrased
with the initial condition that we are using EQUAL
resistive loads for both cases. Since we obtain more
voltage in the delta output from an alternator, which
is always manufactured with the stator construed in
wye, when we obtain that output in Delta, it is a
higher voltage, so for equal loads in both cases, we
also obtain more current in that delta extraction. For
the Wye extraction, we obtain less voltage on each
individual load, thereby less current on those
branches when obtained in WYE.

Are we confused yet?

Take this into consideration;

When we take the outputs before the field is turned
on, (parametric), at my rpm I can draw 1.5 amps out
for a single Delta short. Adding the second short
reduces both of these values to slightly under 1.5
amps, a total just under three amps. In this
condition, two thirds of the stator outputs are
exclusively made for use on the two shorts, and one of
the stator outputs is shared by both phases. When we
add the third phase short, the total amperages on all
three phases drops to .75 A, and now all the stator
outputs are being shared with all three phases. But we
actually obtained more total current out when we drew
on just two phases. I do not think such a thing would
happen in real operating conditions where the field
was actually energized, but it is something to wonder
about for future comparisons. It would seem that the
three shorts has placed the machine in overload
operation, where the voltage drops down below the
effective values for maximum power transfer, or
something similar to that scenario.

Sincerely HDN
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