Design of a CW multiplier circuit

  • #1
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Summary:
What are the steps of designing a CW (cockcroft walton). As I understand I start finding number of stages and then choose a ripple voltage (not sure what is the range on it) from there I calculate capacitance and voltage drop
And what kind of designs in CW that have low ripple without requiring transformer. Like HSVD and where can I find the equations for these designs for ripple,stage,voltage drop.
Im trying to design a circuit that switch from 220V AC to 2kV DC using 1 voltage source (without use of transformers). So CW (Cockcroft-Walton) seemed like a good design to do that. Looking at the equations of halfwave CW it looked to me that the optimal number of stages is set if you know the input and output voltage (in my case it was 7 stages) however both the ripple and voltage drop equations seemed to have no limitation (increase the value of capacitors and the ripple voltage will drop for example) so im wondering is there any equations that shows how low I can go? ( The book im studying says that in the CW halfwave ripple voltage is between 3-5%). So based on that I picked 4% as my ripple voltage and calculated the capacitors and voltage drop values from there. Is this the right approach to design a CW multiplier circuit ? pick a voltage ripple and design your circuit based on that ?

In half wave the ripple seemed to be too high for my design(70 voltage in simulation) so i looked for other designs of CW and I managed to find HSVM (Hybrid symmetrical voltage multiplier) which doesnt use transformer but I cant seem to find the equations for it. Is there any other design of fullwave that uses one voltage source directly (without a transformer) ? if so where can I find the equations for ripple,drop, and number of stages?
 
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Answers and Replies

  • #3
berkeman
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Also, what is your experience level so far with AC Mains power (like 220Vac) and high voltage? What is the application?
 
  • #4
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Also, what is your experience level so far with AC Mains power (like 220Vac) and high voltage? What is the application?
My english is kinda bad sorry if that was confusing but yeah I meant cockcroft walton. I have done two designs (theory and simulations): half wave and full wave which have the same diagrams on the wikipedia page. While the full wave results were good (low ripple) but the problem with the full wave is that it requires a center tapped transformer to get the two ac inputs..

Im rather inexperienced on AC mains practically since I only studied the theory and done simple experiments on the uni, as for the application im assisting on a project on my uni (HVDC) my part is comparing the theory with the simulations (simulation done by NI multisim)
 
  • #5
berkeman
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Im trying to design a circuit that switch from 220V AC to 2000kV DC
CW is merely the first step on the project since after that I will have to find ways to increase the current to ampers.
2000V at several amps is a significant amount of high voltage power. Is there a faculty sponsor/mentor on this project? Somebody with experience needs to be monitoring everything to ensure the safety of the students working on this project.

And is that a typo in your post, and you mean 2kV = 2000V? Or do you really mean 2000kV = 2MV?
 
  • #6
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2000V at several amps is a significant amount of high voltage power. Is there a faculty sponsor/mentor on this project? Somebody with experience needs to be monitoring everything to ensure the safety of the students working on this project.

And is that a typo in your post, and you mean 2kV = 2000V? Or do you really mean 2000kV = 2MV?
Yeah its a typo I meant 2000V. And yeah im under two professors who are monitoring this project but as of right now this is merely theory and simulations, for the experiment we have high voltage lab in the uni. The experiment probably will be done there under them.
 
  • #7
Tom.G
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I think it is time for you to post your circuit, parts values and output requirements, and simulator output if convenient. A link to the design information you used would also be a great help.

That way we have a better idea of what you have and what you want.

For instance the minimum capacitor values change with each stage and their values also depend on the ripple requirements and the rated load current. You have not said what those are.

You must account for the peak line current which will be many time the load current, and also account for this in the ESR (Equivalent Series Resistance) of the capacitors.

So far there are too many unknowns for us to be of much help.

Cheers,
Tom
 
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  • #8
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I think it is time for you to post your circuit, parts values and output requirements, and simulator output if convenient. A link to the design information you used would also be a great help.

That way we have a better idea of what you have and what you want.

For instance the minimum capacitor values change with each stage and their values also depend on the ripple requirements and the rated load current. You have not said what those are.

You must account for the peak line current which will be many time the load current, and also account for this in the ESR (Equivalent Series Resistance) of the capacitors.

So far there are too many unknowns for us to be of much help.

Cheers,
Tom
Hello Tom, Yeah I think my post was little confusing but since then I have figured some stuff out so basically here the situation. The final design goal is to use a 220V 60hz AC source to generate a result in the range of 1.5-2kV DC with around 1-2A load current. So for the voltage part I figured that I will be using a cockcroft walton multiplier. The half wave has too much oscillations I was told so its not viable option. I tried to use a full wave cockcroft walton (Picture of circuit attached) Im not sure if I can remove the coupling capacitor here (C1,C3) based on the simulations I didnt see much difference on the output if removed.

As a start I was simply asked to get a result of the voltage (1.5-2kV with low ripple) with 1mA current which I think I have achieved in my design (Though the number of stages is too high so if I can reduce it to 4-5 stages with another design that will be better).

Based on that given data I have reached out that the optimal number of stages to achieve 1.5kV is 8. Then using this ripple equation for equal capacitor values : Eripple = [Iload/(2fC)]*n at 0.5% ripple (around 7V) I have calculated the capacitor to be 9uF. for the voltage drop VDROP = [Iload/(6fC)] * (n^3 + 2n) = 163V which gets me around 1.6kV output voltage which results in 1.6Mohm load resistance. My simulations results (Using NI multisim reading results with multimeter and Oscope) is similar to this 1.62kV with around 6V ripple and 1.01mA.

The next step would be increasing the current to around 2A. The diodes shouldn't be big problem (probably gonna need to pick other diode since this one is designed for less than 1A). but diodes aside what would be the method to increase the current to that range ? An obvious method would be reducing the load resistance which will lead to higher load current which means I would need to reduce the capacitor to mF range which I think might cause some problems. Anything I can add to the circuit to increase the current ? or is it just playing around with parameter if so what parameters I should look into ? And is there a way to reduce number of stages ?
 

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  • #9
Tom.G
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Im not sure if I can remove the coupling capacitor here (C1,C3) based on the simulations I didnt see much difference on the output if removed.
Sure looks like they can be removed. I don't see that they serve any useful purpose in this circuit.

Anything I can add to the circuit to increase the current ?
Increase the capacitor values, and don't forget the diode current ratings.

I suggest you look at the current waveform on the supply and on C2. The peak current will be many times the output current with a short duty cycle. It is also informative to look at the current traces of the other main capacitors, C2,5,8,11...etc.

If you look at the voltage waveform across C2 (and C5,8,11...) you will probably see a large voltage sag between the source-voltage peaks. That's safe to look at with the simulator, but don't try it on the actual hardware without the approval of your instructor. Depending on the 220V supply configuration where you are, that could be... spectacular, expensive & dangerous.

As usual, there are more details that come up as the design progresses, which we will get to as they arise.


Cheers,
Tom

Addendum:
Eripple = [Iload/(2fC)]*n at 0.5% ripple (around 7V) I have calculated the capacitor to be 9uF.
I don't have a definitive reference at hand, but formula on the website below shows about 13V ripple, closer to what I would expect.

http://home.earthlink.net/~jimlux/hv/cw1.htm
 
  • #10
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Sure looks like they can be removed. I don't see that they serve any useful purpose in this circuit.


Increase the capacitor values, and don't forget the diode current ratings.

I suggest you look at the current waveform on the supply and on C2. The peak current will be many times the output current with a short duty cycle. It is also informative to look at the current traces of the other main capacitors, C2,5,8,11...etc.

If you look at the voltage waveform across C2 (and C5,8,11...) you will probably see a large voltage sag between the source-voltage peaks. That's safe to look at with the simulator, but don't try it on the actual hardware without the approval of your instructor. Depending on the 220V supply configuration where you are, that could be... spectacular, expensive & dangerous.

As usual, there are more details that come up as the design progresses, which we will get to as they arise.


Cheers,
Tom

Addendum:

I don't have a definitive reference at hand, but formula on the website below shows about 13V ripple, closer to what I would expect.

http://home.earthlink.net/~jimlux/hv/cw1.htm
Thanks for the help, I did increase the capacitance (My calculations shows 15mF for 0.5% ripple which is around 8.5V) and I changed the diodes to a virtual ones for now (Just for simulation still need to find a diode that have high PIV and current).

But I noticed a problem on my circuit (attached) while im getting 2A on this resistance, however if a load is connected to the circuit, the current will decrease from 2A which is really bad since the main objective of this design is to test for high dc high current so a load will probably be connected to the circuit.

So basically any idea on how to be able to get a constant current (1-2A) no matter what load is connected ?
as for the site you linked correct me if I'm wrong but I think these equations are for half wave CW.
 

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  • #11
Baluncore
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So basically any idea on how to be able to get a constant current (1-2A) no matter what load is connected ?
The CW multiplier is not the most efficient solution for high currents below about 5kV. It is best suited to very high voltages with very low current.

Input power is limited by the voltage and frequency of the driving waveform, multiplied by the capacitance. Output power; W = Iout · Vout; cannot exceed input power.

Continuous high output current requires high frequency three phase input.
 
  • #12
Tom.G
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But I noticed a problem on my circuit (attached) while im getting 2A on this resistance, however if a load is connected to the circuit, the current will decrease from 2A which is really bad since the main objective of this design is to test for high dc high current so a load will probably be connected to the circuit.
That 788Ω resistor IS the 2Amp "load". Or it at least represents the load for the final product.

as for the site you linked correct me if I'm wrong but I think these equations are for half wave CW.
Yes, it is.
But the circuit you are using seems to be a half wave circuit, with some extra (harmful?) components that reduce the effectiveness. Try simulating a 'normal' half wave circuit with the same part values and compare the outputs for equal output voltage, regulation and ripple.

For even more understanding, simulate a true fullwave circuit; that is one with two 110V line inputs with their common point connected to the CW "Ground". I think you will find the output ripple is different. (get the source polarity/phasing right!)

Continuous high output current requires high frequency three phase input.
Or very large capacitor values. 😁

Cheers,
Tom
 
  • #13
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1
Yes, it is.
But the circuit you are using seems to be a half wave circuit, with some extra (harmful?) components that reduce the effectiveness. Try simulating a 'normal' half wave circuit with the same part values and compare the outputs for equal output voltage, regulation and ripple.

For even more understanding, simulate a true fullwave circuit; that is one with two 110V line inputs with their common point connected to the CW "Ground". I think you will find the output ripple frequency is different. (get the source polarity/phasing right!)


Or very large capacitor values. 😁

Cheers,
Tom
I did have a normal half wave circuit it could reach the same output with 4 stages, but the problem it had too much ripple (No matter how much I change on capacitance there will always be too much ripple). So I used a normal full wave with two sources (2 110V) I only needed 4 stages but then I was told that they cant get me these sources on the lab and that I would need to use a full wave circuit with 1 220V source.

So I made the adjustment and reached the circuit I attached before (which gets me 220V on each stage well not 220V since it drops a little on each stage) I think it's called HSVM (Hybrid symmetrical voltage multiplier) but the equations for it I couldn't find so I used the same ones on full wave but I put half the values for input voltage. From the simulations im getting similar results, maybe 50V difference in output and 1-2 voltage in ripple so i'm satisfied with my equations I think ?

As for the current im still not sure what is the solution here to get a constant 2A DC. since I want the voltage and current to be constant always but resistance would be changing in different loads which from ohm laws means that the current will change. The only thing I studied in theory that fits this case is current sources ? but im not even sure how to implement it here and not sure if it would affect the voltage. Any idea on how I should approach this ?
 
  • #14
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The CW multiplier is not the most efficient solution for high currents below about 5kV. It is best suited to very high voltages with very low current.

Input power is limited by the voltage and frequency of the driving waveform, multiplied by the capacitance. Output power; W = Iout · Vout; cannot exceed input power.

Continuous high output current requires high frequency three phase input.
I do think it's doable since the professor i'm under told me there is a solution for this problem, but Im not limited to CW so if there is a better solution to achieve 1.5-2kV 2A DC using 220V 60Hz AC through using other circuits I would look into it (Can't use transformers). But I would much rather to solve this using CW. any idea how to approach this ?
 
  • #15
Tom.G
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As for the current im still not sure what is the solution here to get a constant 2A DC. since I want the voltage and current to be constant always but resistance would be changing in different loads which from ohm laws means that the current will change.
Yup, Ohms Law rules, always, full stop. Pick any two of the three and the third will be what is needed to satisfy the other two. That's just the way the Universe works.

solution to achieve 1.5-2kV 2A DC

From what little I could find in your posts about the exact requirements, this one seems to be the most complete:
The final design goal is to use a 220V 60hz AC source to generate a result in the range of 1.5-2kV DC with around 1-2A load current.
My 'Engineering Conservative' reading of that statement says that as long as the DC voltage is between 1.5-2kV when the output current ranges between 1A and 2A, you have met the requirements.

A loose interpretation would be that as long as there is a single operating point that is within that range, say 1A at 1.5kV, you are good.

You seem to be worried about ripple voltage but nowhere is there a stated requirement in your posts. Please state any requirement, or forget about it.

In your present circuit:
  1. What is the voltage at 1A output?
  2. What is the voltage at 2A output?
  3. What is the voltage at 0A output, no load?

Cheers,
Tom
 
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  • #16
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A loose interpretation would be that as long as there is a single operating point that is within that range, say 1A at 1.5kV, you are good.

You seem to be worried about ripple voltage but nowhere is there a stated requirement in your posts. Please state any requirement, or forget about it.

Yeah 1.5kV is fine but 1A was just for simulation kinda. Final design should be any value between 1500-2000V DC and 2A DC. Ripple voltage also should also be low around 1%.

  1. What is the voltage at 1A output?
  2. What is the voltage at 2A output?
  3. What is the voltage at 0A output, no load?
At 2A im getting 1614V.
At 1A im getting 1680V.
At no load im getting 1748V.

Yup, Ohms Law rules, always, full stop. Pick any two of the three and the third will be what is needed to satisfy the other two. That's just the way the Universe works.

The problem is the load will be changing hence the resistance will be changing.
 
  • #17
dlgoff
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... still need to find a diode that have high PIV and current
Have you ever considered tubes? :approve:

I apply ~4kV (full wave rectified) when playing with my vacuum deposition system.

HVsupply_2.jpg
 
  • #18
Tom.G
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The problem is the load will be changing hence the resistance will be changing.

Your own measurement show that you have met the stated requirements:
QUOTE="Khalid, post: 6246613, member: 668350"]
The final design goal is to use a 220V 60hz AC source to generate a result in the range of 1.5-2kV DC with around 1-2A load current.
[/QUOTE]
At 2A im getting 1614V.
At 1A im getting 1680V.
At no load im getting 1748V.
Good! Your design works as needed and the values are well within the design requirements.

Please give a detailed explanation why you think there is a problem. Or perhaps follow the @dlgoff suggestion.

Cheers,
Tom
 
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  • #20
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Please give a detailed explanation why you think there is a problem. Or perhaps follow the @dlgoff suggestion.
Because I can only reach 2A with a specific load value (around 800Ohm) which is bad since this design is used for testing and will be connected to different loads with different resistance values. What should happen is: no matter what load I connect I should be getting 2A 1.5-2kV DC.
 
  • #21
Baluncore
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What should happen is: no matter what load I connect I should be getting 2A 1.5-2kV DC.
A CW generator will change it's duty cycle depending on the output voltage. So you cannot use a CW generator as a stable current source without an external current limiter. For a ±500 V output compliance, at 2 amps, you are dissipating maybe 1 kW in the current regulator.

I would use a switching current generator to produce 2 amps, operating at about 50 kHz. But you have specified “no transformer”, so you have precluded the most energy efficient solutions.
 
  • #22
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I would use a switching current generator to produce 2 amps, operating at about 50 kHz. But you have specified “no transformer”, so you have precluded the most energy efficient solutions.
If the transformer is the better solution (economically) then I suppose I can use it since main reason it was rejected before is that it was not economically efficient to use it for the desired voltage output. Can you explain a bit on your idea it's a bit unclear for me.
 
  • #23
Baluncore
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A step-up switching voltage converter feeds the voltage back to regulate the voltage.
https://en.wikipedia.org/wiki/Boost_converter

If the same circuit is controlled by the current flowing in the output circuit it becomes a switching current regulator. Just as a voltage boost converter needs a current limiter, so a high voltage, regulated current generator needs a voltage limit to protect it when the load is open circuit.

Rectify 230 VAC input to produce 320 VDC. You want 2 amp at say 2kV = 4 kW. The input current will need to be about 4000 / 325 = 12.3 amps. That is close to the limit of a single phase electrical circuit. Efficiency should be better than 95%.

See; https://www.st.com/en/applications/power-supplies-and-converters/ac-dc-converters.html

Resonant converters are used for higher power output. Here is a link to a resonant voltage converter;
https://www.st.com/content/st_com/e.../ac-dc-converters/llc-resonant-converter.html
 

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