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Re: Voltage Multiplication Circuits and DC Powered Tesla Coils
All:
This is a longish informational posting about DC multipliers
and DC Teslas and related. I've spent several years designing, testing,
and destroying multipliers and so feel that I can shorten a lot of the
back-and forth questions by covering it all here, all at once.
(My apologies Terry!).
<><><><><><><><><><><><><><><><><>
Cockroft-Walton Multiplier bridges are in actuality AC-coupled adders,
not multipliers (in spite of the name). They can be half-wave or full wave
or even multi-phase in design. In more common, modern-day literature,
they are called "charge pumps".
In essence, the simplest implementation is a transformer with one
side grounded and the other side coupled through a capacitor to a diode.
This is the first half of the stage, which is called the peak detector.
The second half is another capacitor and diode on the other side of the
"ladder" ('cause it kinda looks and acts like climbing a ladder). This
part is called a "level shifter".
The two parts together comprise a complete "stage" that will provide
1X (that's right, one times) the AC peak-to-peak input voltage.
Each successive stage adds almost exactly the same amount, minus the
successive diode drops and other minor losses.
Here is an Ascii picture (you need to use Courier New font):
Stage #1 Stage #2
Peak Detector Peak Detector
| / | /
AC
O------||------.--------------.---------||------.---------------.------O
etc
| \ | | | \ | |
------- \-----/ ------- \-----/
/ \ \ / / \ \ /
/ \ \ / / \ \ /
/-----\ ------- /-----\ -------
| | | |
| | / | | | / |
GND
O--------------.------||------.-----------------.------||-------.-------O
etc
| \ | \
Level Shifter Level Shifter
Stage #1 Stage #2
As you can see, the AC input gets rectified and then stored. That DC level
then becomes the effective "ground level" for the following stage. The
second
stage then rectifies the AC again minus minor losses caused by putting two
capacitors in series and adds that amount to the DC level generated by the
first stage. Voila! 2X the AC PP input. (It helps a lot here to think of
capacitors as "AC resistors").
Each component for each stage MUST be able to withstand the AC Peak-to-Peak
voltage that is driving the first stage. Because of the design, this is an
"equal stress" architecture that allows all components to have the same
ratings.
If you have 15KV AC PP on the input, then the diodes and the capacitors
must be
rated for 15KV AC (not DC). Although the rectification seems to indicate
that
you could get away with 15KV DC, the reality is that there is an AC signal
present (as well as DC) at every point of the ladder except at the ground
input
of the first stage. It might also be necessary to put a higher voltage
diode
or string of diodes in the final diode position ... under a full "short to
ground" transient, the last diode could see the full swing).
For use with something like a Neon Sign Transformer (or NST), the basic
problem
is that the input frequency is 60Hz. The frequency is totally critical
when
it comes to efficiency in a charge pump. The lower the frequency, the
larger
the coupling capacitors have to be in order to couple and integrate the AC.
(Note: NSTs typically have center-grounded secondaries. You must use a
full
wave multiplier design in order to avoid a really massive ground loop).
Here is an example:
First, with 2 stages half-wave, 15KV input and 30ma load,
I'll use .015uFd caps (15KV AC rated):
Cockroft-Walton Multiplier Work Sheet
-------------------------------------
November 6,1998 K.Ottalini
-------------------------------------
Enter the number of stages:? 2
Enter the input voltage (volts):? 15000
Enter the input AC Frequency (Hz):? 60
Enter the capacitor values (Uf):? .015
Enter the load current (amps) or 'R':? .030
-------------------------------------------------------
The estimated ripple at 30 ma is: 100 KVolts <----- OUCH!
The output voltage is: 30 KVolts
The estimated ripple is: 166.6667 % maximum at this
current.
The estimated Vdrop is: 233.3333 KVolts
The real output voltage is: -203.3333 KVolts
The percent regulation is: 388.8889 %
The power requirement for 30 KVolts at 30 ma is: 900 Watts
This is equivalent to a load resistance of 1000000 Ohms
==========================================================================
Hmmm. Something is really wrong here. How can I get 100KVolts of ripple?
The problem is that the load is way too much for the size of the caps and
the frequency. Here is the same thing at 40KHZ:
Cockroft-Walton Multiplier Work Sheet
-------------------------------------
November 6,1998 K.Ottalini
-------------------------------------
Enter the number of stages:? 2
Enter the input voltage (volts):? 15000
Enter the input AC Frequency (Hz):? 40000
Enter the capacitor values (Uf):? .015
Enter the load current (amps) or 'R':? .030
-------------------------------------------------------
The estimated ripple at 30 ma is: .15 KVolts <--- good
The output voltage is: 30 KVolts
The estimated ripple is: .25 % maximum at this current.
The estimated Vdrop is: .35 KVolts <--- good
The real output voltage is: 29.65 KVolts
The percent regulation is: .5833333 %
The power requirement for 30 KVolts at 30 ma is: 900 Watts
This is equivalent to a load resistance of 1000000 Ohms
=====================================================================
This looks a lot better, only 150 volts of ripple and good regulation.
Here is the 60Hz worksheet again, but using 2.0Ufd coupling capacitors:
Cockroft-Walton Multiplier Work Sheet
-------------------------------------
November 6,1998 K.Ottalini
-------------------------------------
Enter the number of stages:? 2
Enter the input voltage (volts):? 15000
Enter the input AC Frequency (Hz):? 60
Enter the capacitor values (Uf):? 2
Enter the load current (amps) or 'R':? .030
-------------------------------------------------------
The estimated ripple at 30 ma is: .75 KVolts <---- not so good
The output voltage is: 30 KVolts
The estimated ripple is: 1.25 % maximum at this current.
The estimated Vdrop is: 1.75 KVolts <---- not so good
The real output voltage is: 28.25 KVolts
The percent regulation is: 2.916667 %
The power requirement for 30 KVolts at 30 ma is: 900 Watts
This is equivalent to a load resistance of 1000000 Ohms
=====================================================================
Not bad, but we still see a loss of almost 2KV. It would probably
take 5.0uFd caps to get a reliable and efficient multiplier.
Summary:
As you can see, the bottom line is that the caps are actually a liability
unless you are really desperate to get more than 15KV. For 15KVDC and
below,
you actually only need a full-wave rectifier, and use the energy storage
caps as the "filter". Just be sure to put a large resistor or inductor
in series with the output to limit the current surges.
For DC-powered TC's, it is unlikely that you'll need more than 15KVDC
since it is the stored energy in the caps that is critical. There
is certainly a voltage related efficiency profile (E=0.5CV^2 ... the energy
rises 2X the voltage but only 1x the capacitance), but it turns out (after
lots of experimenting on DC TC's) that attempting higher voltages only
slows
down the rep rate, and the performance is much more impressive with faster
rep rates. You can easily run from 1 (yes one!)BPS up to 600BPS or more
(BPS=beats per second) and literally get full power out at each beat.
No sync problems, no safety gaps needed, just great sparks.
With only 15KV, and 0.03Ufd storage caps (no multiplier, just a full wave
rectifier), a reasonably well tuned DC TC can reliably produce 42" sparks
with less than 1.5kw. Just changing the caps to 0.5uFd can result in
as much as a 30% increase in output at 1KV to 2KV LOWER input voltage!
(of course, if this is continued, at some point your spark gap will go
into total melt down).
(Safely ...) Enjoy!
*---------------------------*
* Kevin Ottalini *
* WhoSys / Who Systems *
* High Voltage with Style! *
* ottalini-at-mindspring-dot-com *
* Often in Another Reality *
*---------------------------*