Help with placement of filter inductor

P

pawihte

Jan 1, 1970
0
I want to modify an ATX PSU for use with a project and am
thinking of using a classic full-wave voltage doubler to get
about -20V from the center-tapped 5V winding (I'll be using a PSU
with separate 5V and 12V windings and the 12V winding will be
used for another section of the project).

The rectified output will be further regulated with a linear
series regulator, and the load will be of the order of a few tens
of mAs. Please disregard other factors such as the voltage
feedback circuit.

Where I need help is with how to place the filter inductor(s).
Will the second of the two arrangements below be as effective as
the first? Please view with monospaced font.

_______
D1 | |
---------|<----L1-------|7915 |-- -15V
| | | | out
--CT | C1 -------
| |
------------------|
| |
| C2
| |
|-->|---L2--|------- GND
D2

_______
D1 | |
---------|<-------------|7915 |-- -15V
| | | | out
--CT | C1 -------
| |
-----------L----|
| |
| C2
| |
|-->|-----|------- GND
D2
 
P

pawihte

Jan 1, 1970
0
Tim said:
No, it's a double half wave:

Yes. But together it's effectively full-wave because they conduct
on both half cycles of the AC input and the output ripple has
twice the frequency of the source.
_______
D1 | |
---------|<--+--L--+--L--+----|7915 |--o -15V
| _|_ | | | | out
--CT | D3 /_\ C1 | -------
| | | | |
-------------+-----+ C3 |
| | | | |
| D4 _V_ C2 | |
| | | | |
|-->|--+--L--+-----+-------+---- GND
D2

All three L's have to be independent.
I think you drew D3 with the wrong polarity. D1 and D3 will just
short the transformer when the upper terminal is negative. But
why are D3 and D4 necessary (even if D3 is placed with the
correct polarity)?
 
P

pawihte

Jan 1, 1970
0
John said:

I understand that a second LC filter will further reduce ripple,
but that isn't exactly the point of my question. It's whether
placing an inductor in the common transformer terminal will
adequately serve the same function as separate inductors in each
half-wave section.
 
P

pawihte

Jan 1, 1970
0
Tim said:
Oops, good point. Yes, D3 has to point down. So D1-D4 could
be a
FWB, if they make them in high speed silicon.

D3 and D4 are required so the L's don't just flap in the wind.
You'll
see a waveform between -0.8V and +Vpk at the rectifier, if
inductor
current is continuous (which it should be).

BTW, you'll get 5V if you FWCT rectify both halves of the
waveform
together, which is how it's wired stock. Half wave gets half
voltage,
because the inductor's averaging it out. Half wave on the
whole
winding gets 5V again (at poor duty cycle), and FW doubler gets
10V
(at poor duty cycle). So you won't actually get 20V out. To
do that,
you have to split the winding's CT, which I suppose isn't
practical.

A oops from my side. I'm so much more used to capacitor-input
filtering that I forgot about how an L-C filter would behave with
half wave.
(Then again, you said your 12V winding is being used for
something
else? Is it seperate? They usually use one monster winding
with
"+5", 0V and "-5" taps along it.)
Yes, as mentioned in my opening post, I have one with separate 5V
and 12V windings.
At this point, the most economical (= lowest parts) solution
involves
stripping the 5V winding off, replacing it with a proper 20V
winding
and using a FWB, one choke and one filter cap (and maybe
another LC
filter for kicks).

I've been considering that option even before I realised my
oversight about inductive filters with half wave. Taking out the
5V winding is too much hassle with the two halves of the core
glued together. But since the turns/volt and the load are both
small,
it may be possible to just add a 20V winding.
Or maybe you can use the 12V winding instead.
Nope. It's committed. I want to use it to get 24V with a FW
bridge rectifier using Schottky or fast-recovery PN diodes. The
load will be about a couple of amps max.
Oh, and the conventional cap-input rectifier won't work because
it
will explode your transistors. You can get away with it for
small
currents (under 1A??), but regulation is poor, depending on
input
voltage, not duty cycle.
That much I know. Thanks for your interest.
 
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