On Wed, 08 Feb 2006 06:09:44 -0600, John Fields
Winding resistence doesn't depend ONLY on input voltage range. You can
get dual 115V's with higher resistence then dual 230V's
as far as I know neither does regulation ... unless you have some good
reasoning to prove that.
For the same core, to get the same flux density with a 240V input
would require twice the number of turns as for a 120V input.
The voltage on the secondary is related to the voltage on the
primary by:
Es Ns
---- = ---- (1)
Ep Np
Where Es and Ep are the primary and secondary voltages,
respectively, and Ns and Np are the number of turns wound on the
secondary and on the primary, respectively.
Since the transformer is tranferring power, the current in the
secondary has to be related to the current in the primary by:
Is Np
---- = ---- (2)
Ip Ns
So, for, say:
120V>--+ || +-->120V>--+
P || S |
R || E [120R]
I || C |
120V>--+ || +-->120V>--+
With 120V out of the secondary and a 120 ohm load, we'll have a load
current (in the secondary) of:
Es 120V
Is = ---- = ------ = 1 ampere
Rl 120R
Now, since the voltage across the primary and the voltage across the
primary, and:
Np Ep 120V
---- = ---- = ------,
Ns Es 120V
then there must be the same number of turns on the primary as there
are on the secondary.
Rearranging (2) to solve for the primary current will give us:
Ns
Ip = Is ---- = 1A * 1 = 1.0 ampere
Np
Now, with that behind us, let's take a look at a transformer with
dual 120V primaries and connect them in series so we can use it as
an autotransformer with a 240V input and a 120V output and see what
happens:
240V>------+
|
||P
||R
||1
|
+-->120V>--+
| |
||P |
||R [120R]
||2 |
| |
240V>------+-->120V>--+
Since we have twice the voltage across the inductance of the
primary, we'll need twice the number of turns to keep the flux
density in the core the same as it was for the 120V case, and that
criterion is satisfied with the two windings in series.
Now, just assume, for the sake of the argument, that our first
transformer:
120V>--+ || +-->120V>--+
P || S |
R || E [120R]
I || C |
120V>--+ || +-->120V>--+
Really had two 120V primaries which were connected in parallel:
120V>--+---+ ||
| P ||
| R ||
| 1 ||
120V>--|-+-+ || +-->120V>--+
| | || S |
| | || E [120R]
| | || C |
+-|-+ || +-->120V>--+
| P ||
| R ||
| 2 ||
+-+ ||
In which case each primary would be rated for 500mA and, in
parallel, the combination could carry 1 amp.
Now, though, since we're not using the secondary and taking
advantage of the current sharing we'd get with both primaries in
parallel feeding the load on the secondary, we're forcing one of the
primaries to supply the entire load, which is going to cause twice
the voltage drop across it than would occur with with the windings
in parallel. That, in and of itself, will cause the regulation to
be poorer than it would be with a conventional transformer with the
same amount of iron in the core.
Finally, consider another autotransformer wound on the same core,
but with two _240V_ primaries wired in series with 240V across the
ends. 120V will still appear from the center tap to either end, but
because the primaries were wound with smaller diameter (higher
resistance) wire, and exhibit a higher resistance than the other
autotransformer, the drop across the 240V primary with the 120 ohm
load connected across it will be even greater than across the other
autotransformer, resulting in even poorer regulation.
