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Re: Re: transformer question
Hi Grayson,
> Original Poster: Grayson B Dietrich <electrofire-at-juno-dot-com>
SNIP
> Just to be sure you (and Reinhard!!) get it, I am talking about
>LEAVING THE CASES UNGROUNDED. Just from a
>hypothetical, and not neccesarily practical, point of view. If the
>primaries are in series, then there would be no extra stress on
>the insulation for the seriesed secondaries,
> as the total volatge would be equal to that of one original
>transformer....and the current would be doubled, would it not?
If we leave out the "practicality" factor and do as you suggest, then
you are in so far correct that the total voltage of the two seriesed
xformers would equal that of one "normally" connected xformer. As
you are "putting in" one half of the total juice in each xformer, you
will obviously only get one half on the output side. But as you are
seriesing 2*0.5*Vnormal, you get the "normal" output voltage across
both xformers.
However, now to the current. Why should the current increase, when
you series the two xformers, above the rating of a single xformer
connected "normally"? This would be a sort of perpetium mobile.
Letīs, first, take a look at a normal (non HV) PSU (actually two):
2x 120V primary
2x 12V-at-10A secondary
Each PSU is capable of 12V*10A=120VA. Let us assume the PSU
are 100% efficient. In other words VAin = VAout. If I now
PARALLEL the two units (on both sides) I get:
on the input side:
2x 120V -at- 1A each (for a total of 120V*2A = 240VA)
on the output side:
2x 12V-at-10A each for a grand total of 12V and 20A or
also 240VA.
Now letīs look at your case of seriesing them. Here we will
further ignore impedance, etc and look at the xformer as a
pure resistor. This has nothing to do with the actual problem,
it just makes things easier:
We said that the PSUs pull 1A at 120V (primary) each, so
the "effective" resistance is U/I=R or 120V/1A = 120 ohms
On the secondary side they put out 12V-at-10A. Following
U/I=R this means 12V/10A or 1.2 ohms.
Now we series both primary and secondary windings. Letīs
look at the whole thing via Ohmīs Law:
Primary: 120 ohms + 120 ohms across 120V. This means
the total current that CAN flow is 120V/240 ohms or about
0.5A. Remember, we are taking the xformer for a perfect
resistor. In real life this isnīt true. As you have halved the
voltage and therefore halved current PER xformer, you
have just 1/4 the VA FLOW capability (0.5V*0.5A = 0.25VA).
Now letīs look at the secondary.............
Secondary: 1.2ohms+ 1.2 ohms = 2.4ohms at twice the
voltage (24V instead of 12V). Looking at this (and ignoring
the primary VA for the moment), we can see that each
12V xformer CAN only push a maximum of 10A across
itīs secondary (12V/1.2ohm = maximum current possible)
or looking at it from a different view: 24V across 2.4 ohms,
which, too, is 10A.
Now comes the "real life" problem:
TOTAL InputVA is 120V*0.5A (from above) or 60W.
TOTAL OutputVA is 24V*10A (from above) or 240W.
Obviously, here lies a problem. We cannot look at the
xformers as perfect resistors in real life. So, you will
get neither the 10A from the secondary nor the 0.5A
on the primary side (in the dual seriesed configuration).
The values will lie somewhere in between. The true
values are also dependant on xformer and core design.
If we now series the secondaries and PARALLEL the
primaries, we get:
Primary: 2 x 120ohms in parallel or 60 ohms, which
means our two PSUs will draw 2A equally distributed
or 1A each (as designed to) for a grand total of 120V
* 2A or 240W
Secondary: 2x 12V and 10A each. Or 24volts across
2.4 ohms for a total POSSIBLE current of 10A. Check:
24V*10A= 240VA = InputVA (we said the PSUs are
lossless in our example).
And now lastly, your case:
Input:
Letīs call the voltage VP, for each xformer
Current draw for each xformer; Letīs call it CP
Output:
Letīs call the nominal voltage VS (each xformer)
Letīs call the current CS (for each xformer)
Now letīs do a VA check (we will also assume the
lossless case):
InputVA: as both xformers are equal, we can say:
Total VA= 2x(VP*CP)
However, VP is NOT line voltage (seriesed), so CP
CANīT be the same as when the xformer is used
in "normal" (i.e. VP = 120V per xformer) operation.
Using Ohmīs Law, we find that:
RealVP= VP/2
RealCP= CP/2
(for one half the applied nominal voltage, we CANNOT
get the same current as in a single xformer case. Ohmīs
Law)
SO actual VA is really (VP/2)*(CP/2) or (VP*CP)/4
(All I did was pull the factors together)
Now you say that output current SHOULD be twice
that of normal operation, so we can say:
TotalVAout= VS (really: 2*0.5VSsingle)* 2*CS
So VAout is VS*CS*2
Lossless case: VAin= VAout or concrete in your case:
(VP*CP)/4 = VS*CS*2
VAout CAN NEVER BE larger than VAin (otherwise you would
be making energy), so this CANīT be true. As a further check,
we can say VP*CP and VS*CS are constant (for a certain
load "X") AND(!) VS*CS can NEVER be larger than VP*CP.
Letīs go for the best case (i.e. lossless case). If I now put this
in the above equation, I get:
(K1)/4= (K2)*2
or (lossless case for which K1 CAN ONLY EQUAL K2):
(K1)/4= (K1)*2
As K is constant and a real number, I can simply forget about
it for the moment (i.e. I set it to the value of one) and I get:
1/4 = 2
and this isnīt true. There are NO real "K" values (K2 can NEVER
be bigger than K1), which I could insert into the equation to get
a true result in the equation: (K1)/4 = (K2)*2. In other words,
the OutputVA (for your theory to hold true) would HAVE to be
bigger than the InputVA and the conservation law holds against
this.
V*A greets from Germany,
Reinhard