# Re: PFC Question

• To: tesla@xxxxxxxxxx
• Subject: Re: PFC Question
• From: "Tesla list" <tesla@xxxxxxxxxx>
• Date: Sun, 06 Nov 2005 11:11:16 -0700
• Delivered-to: chip@pupman.com
• Delivered-to: tesla@pupman.com
• Old-return-path: <vardin@twfpowerelectronics.com>
• Resent-date: Sun, 6 Nov 2005 11:14:04 -0700 (MST)
• Resent-from: tesla@xxxxxxxxxx
• Resent-message-id: <pDL8yB.A.cCH.CfkbDB@poodle>
• Resent-sender: tesla-request@xxxxxxxxxx

`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      |
|      |
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   |
|      |
o---------------|------|```

`M=L1=L2=0.25L, so:`

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

```                        -jw0.1875L
----------------uuuu---
|           |         |
3           3         |
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```