Current output from Cockcroft Walton Voltage Multiplier

It all depends on a few factors.

What operating frequency you are switching at.

What size capacitors get used in the multiplier stages.

What if any capacitance is there at the output to store energy.

You need to be sure your multiplier diodes can handle the currents involved as well.

What you need: A good front end moving fast enough to keep the output caps charged up.

Option: A transformer between your multiplier and your driver. It provides isolation, and highly selective voltage(s).

Huh?!

Cockcroft-Walton multipliers are used to produce multi-kV outputs. They work best when driven with a square-wave rather than a sine and they can easily supply many amps of current. You select the Cockcroft-Walton multiplier when you need to make plasma, for example, not to boost the gate drive voltage of the high side FETs?! Hence my confusion, here.

Try the IR2103 and similar for high side driving!

-Jeff

--
Message posted via http://www.electronicskb.com

You just need to bootstrap the output, ie a capacitor on the output, when the high side fet turns on it boosts its own gate supply. when the botom fet turns on it charges the cap from the supply via a diode.

IR and others make chips wich do the level shifting, al you need add is the cap and diode.

Colin =^.^=

I'm not looking to make kV outputs but I'm pretty sure I have the correct name for the configuration. The output of an oscillator feeds into a capacitor (C1). The other side of the cap(C1) is connected to the anode of one diode (D1) and the cathode of another diode (D2). This node also provides the input to the next stage or the output of the multiplier as a whole. The cathode of D1 and the anode of D2 are connected to each other via a cap (C2). Two diodes and two caps make up the stage. Isn't this a cockcroft-walton multiplier? I don't see why it should be limited to multi kV applications.

I looked at the data sheet for the IR2103 and it will do the job (along with many other chips from many vendors) but I'm trying to stick to a discrete solution and to understand the multiplier function.

Thanks,

Kevin

That's the basis of the question. How are the values of the operating frequency, capacitance, etc. determined?

The oscillator can easily be configured for anywhere from 400 Hz to

60KHz. I'm looking to get about 40 mA out of the multiplier so even small diodes (i.e. 1N4149) should be able to handle the current.

What considerations go into determining the optimal values for frequency and capacitance? This is the part I'm trying to understand.

Thanks,

Kevin

It isn't limited to high voltage applications, it just finds a lot of application in HV low current applications since it is cheaper than HV transformers.

Plate supplies in tube equipment frequently used a doubler - probably closer to what you intend.

I was unable to find a design brief on them but came across it not so long ago - as you'd expect the wave form and frequency have a lot to do with current as do the capacitors.

I was tinkering with the same problem myself today - turning on a Mosfet from a 3 volt supply - one thing I've always wanted to try was a small blocking oscillator for a gate source. It works very well open circuit it was outputting 80 volts so I added a pair of white leds to keep the voltage at 6V. Only uses 5 milliwatts of power and takes up less than a square inch. It would also find application as a high-side N channel switch with another isolated winding on the ferrite core.

--

----== Posted via Newsfeeds.Com - Unlimited-Unrestricted-Secure Usenet News==----
http://www.newsfeeds.com The #1 Newsgroup Service in the World! 120,000+
Newsgroups
----= East and West-Coast Server Farms - Total Privacy via Encryption =----

Found something that may be useful.

Search on: volt_mul.pdf

There's only four hits and one is the General Semiconductor application notes index

--

----== Posted via Newsfeeds.Com - Unlimited-Unrestricted-Secure Usenet News==----
http://www.newsfeeds.com The #1 Newsgroup Service in the World! 120,000+
Newsgroups
----= East and West-Coast Server Farms - Total Privacy via Encryption =----

limited to

That is, indeed, a Cockroft-Walton multiplier. It isn't *limited* to high voltage applications; that's just where it is most commonly (almost always) used.

Well, in that case... Think of the initial capacitor - and all of the ones

*directly in series with it* - as "coupling" capacitors. They convert any unipolar input into bipolar and apportion it to each stage down the line. The diodes steer current from each coupling capacitor to the appropriate terminals of the corresponding "storage" capacitors, which are also in series. Consider, now, what happens if you drive a C-W multiplier with an initial 10Vpp square wave. The negative 5V portion gets routed to one side of the storage cap while the positive 5V portion gets routed to the other; hence, there's 10V stored in the first storage cap. But this 10Vpp square wave gets coupled to the *next* set of steering diodes, which also apportion the negative and positive halves to another storage cap - at this point, there is 20V across the two storage caps (which are in series). Etc. and so on. Does this help?

-Jeff

--
Message posted via http://www.electronicskb.com

The capacitance will determine how well the multiplier performs on a pulse to pulse basis, and it also is your energy storage location. In HV, one only makes what one uses. With your low voltage application, one would think that the same type of optimizations would be in order.

The thing is, you need to experiment at the bench for idealizing this kind of thing. Calculations get you into the ball park, but real circuit operation observations dial it in much better.

You could start with small multiplier element capacitance, and work toward larger elements, and even tail end storage (on the output) for the final design configuration.

I had an HV supply with only 3 Volts in, and we had to make a 6 V multiplier using 1N4148s so the op amps would run even after the source dropped below 2 Volts. That was a fun HV circuit to make.

You did not indicate your supply voltage, but if you need 4 stages it would presumably be around 2.5 volts. The diode drops will be a significant share of the input voltage and must be accounted for in the design of the system. As others have indicated, there are much better ways to accomplish your objective, but certainly a voltage multiplier could be an interesting thing to tinker with.

Look at this way; one charges N capacitors in parallel and discharges them in series. If there were no losses, the Q=C*V and Q=I*T equations can be used with the note that C*V is constant per capacitor. Charge (say) 4*C to input V and then discharge at 4*V at capacitance of C.

In article , seanacais wrote: [snip]

I found a few cooking sums in an old book. They are based on a sinewave stimulus, but probably good enough as a starting point. You need to switch to a fixed width font and no auto wrapping.

Iout ( 2n^3 n ) Vout = 2.n.Vpk - ----*( ---- - --- ) F.C ( 3 6 )

Iout ( n(n+1) ) Vripple,pk-pk = ----*( ------ ) F.C ( 4 )

n is the number of stages.

Vpk is the peak value of the stimulus assuming perfect diodes and a zero ohm source resistance.

For a dc supply, Vpk = Vsupply - 2xVdiode. So Schottky diodes would look to be very useful here, and it would be prudent to LTspice a circuit to see the effect of the driver source resistance.

--
Tony Williams.

Correcting a possible ambiguity.......

"2n^3" is "2 times n-cubed".

--
Tony Williams.

Iout ( 2n^3 n ) Vout = 2.n.Vpk - ----*( ---- - --- ) F.C ( 3 6 )

And we can see that load currents are murder in a high-N (many stage) Cockcroft-Walton generator.

Tony, does your old reference say anything about the case where all the capacitors are not equal?

Absolutely. Even in this case the OP may be better doing a single stage doubler and then using that dc output to power a second oscillator doing another single stage doubler.

Of course Win, they all do. :-)

The charge sums are a little complicated to put down, and understand (for me anyway). Basically they state that each stage has a reduction in voltage step (from Vpk) of dV, and dV increases as you go down the column, being largest at the lowest stage.

The situation can be improved by increasing the cap values going down the column.... If C1 is the top cap and Cn is any lower cap at position n, then Cn = n*C1.

I ( 1 2 3 n ) Similarly, Vripple = --- ( -- + -- + -- + ... -- ) F.C ( C1 C1 C3 Cn )

For High Voltage multipliers they state that it is only normally practical to make the bottom cap values equal to 2x the values of all others in the column. This is because the bottom caps only have to withstand Vpk, whereas all other have to withstand 2xVpk.

And here's where I thank you Win for asking that question, because I misread the Vout sum from the book..... the Vout sum I gave was that circumstance above of 2x cap values for the bottom caps in the column. The all-equal-caps sum is.

I ( 2.n^3 n^2 n ) Vout = 2n.Vpk - --- ( ---- + --- + --- ). F.C ( 3 2 6 )

--
Tony Williams.

Pray tell us what great old book you have?

High Voltage Engineering (Fundamentals). By: E. Kuffel and W.S. Zaengl. Pergammon Press, 1984.

Reading from the 'simple stuff' on pages 14-17.

--
Tony Williams.

Indeed!

...Jim Thompson

-- | James E.Thompson, P.E. | mens | | Analog Innovations, Inc. | et | | Analog/Mixed-Signal ASIC's and Discrete Systems | manus | | Phoenix, Arizona Voice:(480)460-2350 | | | E-mail Address at Website Fax:(480)460-2142 | Brass Rat | |

| 1962 | America: Land of the Free, Because of the Brave

1984 isn't that bloody long ago. Is it the first printing?

So it is a "Great recent book you have..."