0
\$\begingroup\$

I do not understand how this voltage multiplier circuit works:

enter image description here

wherever I have read (also Wikipedia), I have not found a step by step explanation. I do not understand if the voltage drop across the diodes in conduction is negligible or not, and in general I need to understand the temporal trends of the voltages at the ends of the capacitances. Moreover, I do not understand what happens when the input AC voltage is different from its maximum value.

Can you give me a detailed explanation on what happens?

\$\endgroup\$
2
\$\begingroup\$

As Bimpelrekkie explained, whether or not the voltage drop of the diodes is significant or not depends on what voltage you are working with.

Since I've been playing with these things for the last few days, I'll post some of the diagrams and measurements I've made.

I think the voltage traces explain fairly what happens to the output when the AC is not at its peak value.

Just FYI: The voltage I'm using (30 volts peak to peak) is kind of on the edge of making the forward voltage of the diodes I'm using not matter much. The forward voltage is about 3% of the available voltage, so noticeable but small enough to squint and and say "eh, close enough" most of the time.

For starters, the Cockcroft-Walton half wave multiplier you showed is built of a bunch Greinacher voltage doublers. The Greinacher is composed of a Villard doubler followed by a rectifier and a filter.

This is a Villard doubler:

enter image description here

It takes AC, and adds half the peak to peak voltage to the AC as DC. My transformer was putting out about 30 volts peak to peak AC. The Villard circuit added about 14V to that. I ended up with a peak voltage at about 30V. It isn't really DC. It is "fluctuating DC" with all of the voltage above 0.

The gray line across the middle of the image is 0 volts. As you can see, the AC was moved upwards and resulted in the "DC." The flat spots at the bottom of the DC come from the diodes. The diodes have a forward voltage of around 1 volt, so that's missing from the bottom of the curve.

enter image description here

Now here's the circuit for a Greinacher doubler:

enter image description here

This circuit is the basis of the Cockcroft-Walton multiplier. Look at your Cockcroft-Walton diagram, and kind of tilt your head to one side. I'm sure you'll recognize its "cells" as nothing more than a bunch of Greinacher doublers strung together.

This is what the output of a Greinacher doubler looks like:

enter image description here

It is a little bit lower than the Villard output, but it is much closer to DC. If I had used larger capacitors, then it would look exactly like DC.

Now, you string a bunch of Greinachers together, and you get a higher voltage. It would look a lot like that last picture, just with higher output voltage.

I'm getting about 24V out of a single Greinacher. If I put a bunch of them together to get a Cockcroft-Walton multiplier, then I get 24V for every stage I add to it.

1 stage 24 V

2 stages 48 V

3 stages 72 V

and so on.

You take the peak to peak voltage of your AC (30 V in my case,) subtract the diode forward voltage, and that's approximately the voltage you can expect in DC for a single stage. Multiply by the number of stages, and you have your (approximate) DC output voltage.

The next trick is figuring out the size of the capacitors to use. I used 100nF capacitors for my experiments (that was way too small when working with 50 hertz.) That makes my doubler have a very high resistance. A "load" of 1 megaohm noticeably drops the output voltage. I've been trying to figure out how to express the impedance of a Cockcroft-Walton multiplier as a function of the capacitance. I'm playing hooky from that right now. My plan was to build a couple of more multipliers with different capacitors and see how they respond to loading.

\$\endgroup\$
1
\$\begingroup\$

I do not understand if the voltage drop across the diodes in conduction is negligible or not

It is not a case of "negligible or not", as with many things in electronics: it depends.

The diodes have a certain forward voltage, typically between 0.5 V to 1.0 V depending on how much current flows though the diode (more current => more voltage drop).

If I multiply an AC input voltage of 5 V then a drop of 0.5 V per diode is significant, 0.5 V is 10% of 5 V.

If I multiply an AC input voltage of 50 V then a drop of 0.5 V per diode is less significant, 0.5 V is 1% of 50 V.

If I multiply an AC input voltage of 500 V then a drop of 0.5 V per diode is practically insignificant, 0.5 V is only 0.1% of 500 V.

Do you see what I'm doing here? The diode drop voltage remains constant but relative to the voltages at which the circuit is working, that constant drop become less significant as that voltage increases. So: it depends, on the voltages at which the multiplier operates.

Moreover, I do not understand what happens when the input AC voltage is different from its maximum value

An AC voltage has a peak value which is its maximum value. We generally don't talk about an AC voltage's maximum value, we say peak value.

An AC voltage is generally described using an RMS value and/or peak value and/or peak-to-peak value. It depends on your application what is most convenient. These voltage multiplier work on the peak value of the signal as they are rectifiers.

\$\endgroup\$
1
\$\begingroup\$

The Vf drop of the diode does have a significant, cumulative effect. Each stage reduces the output voltage by one Vf drop.

In the example below, there are two stages. With a 10Vp-p input, the output voltage is 18.66V instead of 20. The difference is two Vf diode drops of 0.66V each (1.33V total.)

schematic

simulate this circuit – Schematic created using CircuitLab

You can reduce this drop by using Schottky diodes, but you will still have some drop.

To analyze this in detail, the key is to understand the current flow and how the diodes are steering it. Building a simulation helps with this immensely. I built this sim with the CircuitLab tool (available as an option on the edit toolbar); or use a free Spice sim like LT-Spice.

\$\endgroup\$
0
\$\begingroup\$

Remove all components except for D1 and C1. Analyse the circuit behaviour.

Now add in D2 and C2. What's different? What's the same?

Now add D3 and C3. What's the same and what's different? Do you notice a pattern?

Generalise to adding an arbitrary number of Ds and Cs in the same pattern.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.