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Re: PFC Question



Original poster: "Dmitry (father dest)" <dest@xxxxxxxxxxx>


"someone else can explain" - yeah, sure! but there is no any community here, so you can wait forever :-\

ok - setting at the midway point:

    o---------------|
                    |*
                    3
                  > 3 L1
                 /  3
                /   |
               |    |
 ~230v         M    --------
               |    |      |
                \   |*     |
                 \  3      |
                  > 3 L2  load
                    3      |
                    |      |
    o---------------|------|

M - mutual inductance
M=k*sqrt(L1*L2)

ASSuming k~1 (toroidal core, single layer winding):

M=sqrt(L1*L2)
L1 and L2 - each consists of a half of all variac turns, so
L1=L2=0.25L, L - full variac inductance.
now let`s decouple the magnetic couplings - as L1 and L2 are connected
in a phase (pay attention to the "*" signs at the upper circuit),
then the circuit would be:

    o---------------|
                    |
                    3
                    3 L1
                    3
                    |
                    3
                    3 jwM
                    3
                    |
 ~230v              --------
                    |      |
                    3      |
                    3 jwM  3
                    3      3 -jwM
                    |      3
                    3      |
                    3 L2   |
                    3     load
                    |      |
    o---------------|------|

M=L1=L2=0.25L, so:

    o---------------|
                    |
                    3
              0.5L  3
                    3
                    |
                    |
 ~230v              --------
                    |      |
                    |      3
                    3      3 -jw0.25L
              0.5L  3      3
                    3      |
                    |    load
    o---------------|------|

the internal resistance of the mains is very small - we can neglect it
- right? so mains is a dead short across variac input, then:

                        -jw0.25L
          ----------------uuuu---
          |           |         |
          3           3         |
  jw0.5L  3  jw0.5L   3        load
          3           3         |
          |           |         |
          -----------------------

you see ---> jw*(0.25-0.25)=0

ZERO inductance! :-P

if the voltage at the variac output is 3/4 from the voltage at its
input , then:

L1=0.0625L
L2=0.5625L
M=0.1875L


o---------------| | 3 0.25L 3 3 | | ~230v -------- | | | 3 3 3 -jw0.1875L 0.75L 3 3 3 | | load o---------------|------|


-jw0.1875L ----------------uuuu--- | | | 3 3 | jw0.25L 3 jw0.75L 3 load 3 3 | | | | -----------------------

you see ---> jw*(0.1875-0.1875)=0

again - ZERO inductance! :-P
is that hard to understand?


> Original poster: "Gerry Reynolds" <gerryreynolds@xxxxxxxxxxxxx>

> Hi Dmitry,

> I hope you're correct, however, the performance curves from powerstat
> may suggest otherwise.  I'm having a hard time understanding how
> there can't be any series inductance when, say, the setting as at the
> midway point.  Maybe you or someone else can explain this.

> Gerry R.

>>Original poster: "Dmitry (father dest)" <dest@xxxxxxxxxxx>
>>
>>you wanna say that if there`s a number of "toroid turns in series with
>>the load", then these turns  are "seen" by the load as some
>>inductance? it`s wrong too - variac isn`t an inductive divider, the
>>voltage on the load doesn`t depend from the current through it, so
>>load doesn`t "see" any series  inductance.
>>

-----
Neon transformers are doomed in Tesla service! I should know, I've
killed more of 'em than many pople have ever even seen in a life time.
25-03-96 (c) Richard Hull, TCBOR