Cockroft-Walton generator

C

cloudguitar

Guest
Hi all,

I am trying to build a simple half-wave CW generator (aka Villard
cascade) for my Physics class.

At the moment, I have 4 stages composed of 1.5 nF capacitors (rated
for 2kV) and diodes with an internal resistance of 500ohm, connected
to a 12V 50 Hz AC supply.

I simulated the circuit with Qucs and, as I was expecting from
calculations, the no-load output would have been around 100V DC. What
happens is that my multimeter reads 2.5V DC, instead.

I suspect I am doing something wrong with voltage probing. Referring
to this circuit schema:
http://upload.wikimedia.org/wikipedia/en/4/4c/Cockcroft_Walton_Voltage_multiplier.png
I connected the red probe to the output of the last stage (the circle
in the image) and the black probe to where the ground should be in the
schema. The circuit is open, and the voltmeter should close it by
connecting the last stage to the generator - also, there is no
connection with the AC supply ground.

Unfortunately, I am a newbie and I cannot figure out what is
happening.
Any idea?

Thank you very much for your time,

Claudio A.
 
cloudguitar wrote:

Hi all,

I am trying to build a simple half-wave CW generator (aka Villard
cascade) for my Physics class.

At the moment, I have 4 stages composed of 1.5 nF capacitors (rated
for 2kV) and diodes with an internal resistance of 500ohm, connected
to a 12V 50 Hz AC supply.

I simulated the circuit with Qucs and, as I was expecting from
calculations, the no-load output would have been around 100V DC. What
happens is that my multimeter reads 2.5V DC, instead.

I suspect I am doing something wrong with voltage probing. Referring
to this circuit schema:
http://upload.wikimedia.org/wikipedia/en/4/4c/Cockcroft_Walton_Voltage_multiplier.png
I connected the red probe to the output of the last stage (the circle
in the image) and the black probe to where the ground should be in the
schema. The circuit is open, and the voltmeter should close it by
connecting the last stage to the generator - also, there is no
connection with the AC supply ground.

Unfortunately, I am a newbie and I cannot figure out what is
happening.
Any idea?

Thank you very much for your time,

Claudio A.

Is your meter a 10 meg ohm input type ?
Click to expand...
also, 1.5 nf is small for 50 hz. you may want to
try larger caps..
Also, you don't need 2kv caps at the voltage you're
starting at.


http://webpages.charter.net/jamie_5"
 
cloudguitar wrote:

On May 24, 8:32 pm, Jamie
jamie_ka1lpa_not_valid_after_ka1l...@charter.net> wrote:

cloudguitar wrote:

Hi all,

I am trying to build a simple half-wave CW generator (aka Villard
cascade) for my Physics class.

At the moment, I have 4 stages composed of 1.5 nF capacitors (rated
for 2kV) and diodes with an internal resistance of 500ohm, connected
to a 12V 50 Hz AC supply.

I simulated the circuit with Qucs and, as I was expecting from
calculations, the no-load output would have been around 100V DC. What
happens is that my multimeter reads 2.5V DC, instead.

I suspect I am doing something wrong with voltage probing. Referring
to this circuit schema:
http://upload.wikimedia.org/wikipedia/en/4/4c/Cockcroft_Walton_Voltag...
I connected the red probe to the output of the last stage (the circle
in the image) and the black probe to where the ground should be in the
schema. The circuit is open, and the voltmeter should close it by
connecting the last stage to the generator - also, there is no
connection with the AC supply ground.

Unfortunately, I am a newbie and I cannot figure out what is
happening.
Any idea?

Thank you very much for your time,

Claudio A.

Is your meter a 10 meg ohm input type ?
also, 1.5 nf is small for 50 hz. you may want to
try larger caps..
Also, you don't need 2kv caps at the voltage you're
starting at.

http://webpages.charter.net/jamie_5"


Thanks for your reply.

yes, my meter is a 10MOhm multimeter. I was planning to add 4 stages
(reaching 8 stages) and to switch to a 230V 50Hz supply, so 2kV should
be barely enough.
Do you think that I should try with 230V right now?

I also have 15nF @ 3kV caps, do you think the would be better suited?
I would try the larger caps first.
Click to expand...



--
"I'd rather have a bottle in front of me than a frontal lobotomy"

http://webpages.charter.net/jamie_5"
 
On May 24, 8:32 pm, Jamie
<jamie_ka1lpa_not_valid_after_ka1l...@charter.net> wrote:
cloudguitar wrote:
Hi all,

I am trying to build a simple half-wave CW generator (aka Villard
cascade) for my Physics class.

At the moment, I have 4 stages composed of 1.5 nF capacitors (rated
for 2kV) and diodes with an internal resistance of 500ohm, connected
to a 12V 50 Hz AC supply.

I simulated the circuit with Qucs and, as I was expecting from
calculations, the no-load output would have been around 100V DC. What
happens is that my multimeter reads 2.5V DC, instead.

I suspect I am doing something wrong with voltage probing. Referring
to this circuit schema:
http://upload.wikimedia.org/wikipedia/en/4/4c/Cockcroft_Walton_Voltag...
I connected the red probe to the output of the last stage (the circle
in the image) and the black probe to where the ground should be in the
schema. The circuit is open, and the voltmeter should close it by
connecting the last stage to the generator - also, there is no
connection with the AC supply ground.

Unfortunately, I am a newbie and I cannot figure out what is
happening.
Any idea?

Thank you very much for your time,

Claudio A.

Is your meter a 10 meg ohm input type ?
also, 1.5 nf is small for 50 hz. you may want to
try larger caps..
Also, you don't need 2kv caps at the voltage you're
starting at.

http://webpages.charter.net/jamie_5"
Click to expand...
Thanks for your reply.

yes, my meter is a 10MOhm multimeter. I was planning to add 4 stages
(reaching 8 stages) and to switch to a 230V 50Hz supply, so 2kV should
be barely enough.
Do you think that I should try with 230V right now?

I also have 15nF @ 3kV caps, do you think the would be better suited?
 
On May 24, 9:58 pm, "Paul E. Schoen" <pst...@smart.net> wrote:
"Jamie" <jamie_ka1lpa_not_valid_after_ka1l...@charter.net> wrote in message

news:lkZZj.1748$IF3.151@newsfe07.lga...



cloudguitar wrote:

On May 24, 8:32 pm, Jamie
jamie_ka1lpa_not_valid_after_ka1l...@charter.net> wrote:

cloudguitar wrote:

Hi all,

I am trying to build a simple half-wave CW generator (aka Villard
cascade) for my Physics class.

At the moment, I have 4 stages composed of 1.5 nF capacitors (rated
for 2kV) and diodes with an internal resistance of 500ohm, connected
to a 12V 50 Hz AC supply.

I simulated the circuit with Qucs and, as I was expecting from
calculations, the no-load output would have been around 100V DC. What
happens is that my multimeter reads 2.5V DC, instead.

I suspect I am doing something wrong with voltage probing. Referring
to this circuit schema:
http://upload.wikimedia.org/wikipedia/en/4/4c/Cockcroft_Walton_Voltag...
I connected the red probe to the output of the last stage (the circle
in the image) and the black probe to where the ground should be in the
schema. The circuit is open, and the voltmeter should close it by
connecting the last stage to the generator - also, there is no
connection with the AC supply ground.

Unfortunately, I am a newbie and I cannot figure out what is
happening.
Any idea?

Thank you very much for your time,

Claudio A.

Is your meter a 10 meg ohm input type ?
also, 1.5 nf is small for 50 hz. you may want to
try larger caps..
Also, you don't need 2kv caps at the voltage you're
starting at.

http://webpages.charter.net/jamie_5"

Thanks for your reply.

yes, my meter is a 10MOhm multimeter. I was planning to add 4 stages
(reaching 8 stages) and to switch to a 230V 50Hz supply, so 2kV should
be barely enough.
Do you think that I should try with 230V right now?

I also have 15nF @ 3kV caps, do you think the would be better suited?
I would try the larger caps first.

Also, check the forward voltage drop of your diodes. The 500 ohm resistance
figure you give does not make sense. Maybe you mean 500 mV at 1 mA, but if
you have high voltage diodes they may be several junctions in series, which
would explain a low output with 12 VAC.

Paul
Click to expand...
Thank you for your time.

I am using 1N-5408 diodes.
Here's their specification for your ease:
www.fairchildsemi.com/ds/1N/1N5408.pdf

In your opinion, how much should I raise the input voltage to get a
noticeable effect?
 
cloudguitar wrote:

On May 24, 10:51 pm, whit3rd <whit...@gmail.com> wrote:

On May 24, 11:34 am, cloudguitar <cloudgui...@gmail.com> wrote:


I also have 15nF @ 3kV caps, do you think the would be better suited?

A Cockroft-Walton ladder uses only the peak/peak
input voltage stress across each capacitor, so
50V rating is OK for 12VAC, and 1kV rating is more
than adequate for 240VAC. The same holds
for the rectifiers, of course.

Two considerations may account for low output voltage:
the rectifier capacitance (forms a capacitive divider with the
ladder capacitors), and the voltmeter shunt resistance.
I'd use an ammeter instead of a voltmeter, just calculate
the output current of the CW rectifier into a short and see if
it matches. If the current is low, that means it's the
rectifier capacitance (and using larger ladder capacitors will
solve the problem).



Again, I appreciate all of your feedback.

I connected the CW to a 230VAC 50Hz supply. My meter read 1080V (meter
scale: 0-1kV),
and I clearly heard the cracky sound of a sparkle when I closed the
circuit, meaning that arcing happened, I suppose.

I will test it again with a professional voltmeter to see the actual
output voltage, but it really seems to me that it works.
If I need to run it at a lower voltage, I will increase the
capacitance.

Thank you!

Claudio A.
Doing the calculations on the 15 nf you suggested yields ~212k Xc for
Click to expand...
each cap. That is a high value for what you're trying to do. If very low
current is what you need then I guess it's fine how ever, it makes it
very hard to measure using a conventional meter even at 10 Meg Z after
you do a few stages.
And remember that you'll get ~ 500--600 mv loss on each multiplier.

For 50 Hz, you really need large caps.!
when I say large caps, I'm talking in the range of Uf's not nf's

P.s.
We have vessel units at work that employ this technology that outputs
2 Mvolts at 60 ma's
the input frequency is 100 KHz being generated from a hot cathode osc
tube that is able to do 375Kwatts into a Pie xformer made from litz wire.
That goes into a full wave Cockroft Walton multiplier which is inside
a pressured vessel of SF6 gas.

In the vessel is a photo multiplier tube so sensitive that is can
detect a very small emission of photons which would indicate so arching
taking place and thus shut down the system!...
In our system, we half rings that form the capacitance's via the side
walls of the vessel when the unit is closed up..

http://www.easternisotopes.com/rdi/services.php
Look there, that may impress you! :)


http://webpages.charter.net/jamie_5"
 
On May 24, 11:34 am, cloudguitar <cloudgui...@gmail.com> wrote:

I also have 15nF @ 3kV caps, do you think the would be better suited?
Click to expand...
A Cockroft-Walton ladder uses only the peak/peak
input voltage stress across each capacitor, so
50V rating is OK for 12VAC, and 1kV rating is more
than adequate for 240VAC. The same holds
for the rectifiers, of course.

Two considerations may account for low output voltage:
the rectifier capacitance (forms a capacitive divider with the
ladder capacitors), and the voltmeter shunt resistance.
I'd use an ammeter instead of a voltmeter, just calculate
the output current of the CW rectifier into a short and see if
it matches. If the current is low, that means it's the
rectifier capacitance (and using larger ladder capacitors will
solve the problem).
 
On May 24, 10:51 pm, whit3rd <whit...@gmail.com> wrote:
On May 24, 11:34 am, cloudguitar <cloudgui...@gmail.com> wrote:

I also have 15nF @ 3kV caps, do you think the would be better suited?

A Cockroft-Walton ladder uses only the peak/peak
input voltage stress across each capacitor, so
50V rating is OK for 12VAC, and 1kV rating is more
than adequate for 240VAC. The same holds
for the rectifiers, of course.

Two considerations may account for low output voltage:
the rectifier capacitance (forms a capacitive divider with the
ladder capacitors), and the voltmeter shunt resistance.
I'd use an ammeter instead of a voltmeter, just calculate
the output current of the CW rectifier into a short and see if
it matches. If the current is low, that means it's the
rectifier capacitance (and using larger ladder capacitors will
solve the problem).
Click to expand...

Again, I appreciate all of your feedback.

I connected the CW to a 230VAC 50Hz supply. My meter read 1080V (meter
scale: 0-1kV),
and I clearly heard the cracky sound of a sparkle when I closed the
circuit, meaning that arcing happened, I suppose.

I will test it again with a professional voltmeter to see the actual
output voltage, but it really seems to me that it works.
If I need to run it at a lower voltage, I will increase the
capacitance.

Thank you!

Claudio A.
 
On May 25, 9:28 am, "Phil Allison" <philalli...@tpg.com.au> wrote:
"cloudguitar"

I am trying to build a simple half-wave CW generator (aka Villard
cascade) for my Physics class.

At the moment, I have 4 stages composed of 1.5 nF capacitors (rated
for 2kV) and diodes with an internal resistance of 500ohm, connected
to a 12V 50 Hz AC supply.

** What is the impedance of a 1.5nF cap at 50Hz ????

The answer is in megohms.

...... Phil
Click to expand...
I would use the CW to charge some other capacitors. Therefore, a low
current output shouldn't be great a problem.
Unfortunately, here in Italy components are difficult to find and
quite high priced. In fact, the most interesting cap I
can get here is a polystyrene 0.1uF rated for 3kV at 3.50€ each (5$),
way too expensive. And I cannot even get
more capacitance.
I would probably have to purchase from abroad, but I a moving from
Milan to Boston in a couple months so it isn't probably
worth it.

I'll try with 1uF if possible.
 
You must log in or register to reply here.

Welcome to EDABoard.com

Sponsor

Back
Top