# Voltage Multiplier Minimum Frequency

The shown Capacitor-Diode (Cockroft-Walton) voltage multiplier, with $n$ capacitors stages (from ground to output) has a AC input with amplitude $v_{in}$ and a DC output $v_{out}$, with diodes with junction voltage $v_{j}$ which in steady state happens to be equal to:

$$v_{out}=2n(v_{in}-v_j)$$

This depends on the frequency of the input, which need to be some minimum value for converging to the previous approximated value.

Below is the simulation, with $v_{in}=5V$, $v_{j}=0.5V$ approx., $n=4$, $C=1uF$, $f=10Hz$, for the voltages through each component, plus the output voltage in magenta (greater value), estimating the value of $v_{out}=36V$.

But i am unclear on how to get an estimate for other values of $C$ and $f$. Do somebody have a reference for this kind of circuit, and for calculating|estimating an expression for this frequency?

• I have never seen a good one. From experience the output impedance and internal losses increase surprisingly fast as you add more stages, but I've not seen (and failed in my attempts to create) a satisfactory math model that explains why. – Brian Drummond May 29 '18 at 21:52
• It doesn't really make sense talking about the frequency and capacitance without any load. – Harry Svensson May 30 '18 at 8:13
• For source frequencies below some point, the capacitors do not get fully loaded, and the given estimate is not even closely reached. Only the natural PN junction is participating on this. – Brethlosze May 30 '18 at 23:14
• @HarrySvensson Check comments in answer below. – Brethlosze May 30 '18 at 23:32

The capacitance or frequency does not influence the open-circuit output voltage. It remains

$$v_{out} \approx 2n(v_{in}-v_j)$$

The influence of the frequency and capacitance is only visible when loading the Cockroft-Walton multiplier. The total capacitance on which the output voltage is stored is roughly

$$C_{out,eq} \approx \frac{C}{n}$$

This capacitance will discharge when the input is low, proportional to the current the load is sinking. Using big capacitors will make the output more stable. Note that this equivalent capacitance approximation only holds if the voltage drop at the output is less than $v_{in}-v_j$.

If you increase the frequency, you can build up the voltage more quickly, and the output voltage will be pushed up. Of course, provided the top capacitors can be charged quickly enough.

The balance of these two effects will influence the ripple and average output voltage, but I don't think it is very obvious in what mathematical way though.

[EDIT] I worked through the references you posted in the comments, and found the following (relevant) scientific publications.

1. J.S. brugler, Theoretical Performance of Voltage Multiplier Circuits, IEEE Journal of Solid-State Circuits, June 1971

2. P.M. Lin and Leon O. Chua, Topological Generation and Analysis of Voltage Multiplier Circuits, IEEE Transactions on Circuits and Systems, vol. CAS-24, no. 10, october 1977

One difference between their multiplier and yours, is that they connect the final capacitor to ground ($C_o$). All other capacitors have the same capacitance of $C$.

$$\begin{cases} V_{o,even} = N\cdot V_i - \frac{N}{6} \left( \frac{N^2}{2}+1 \right)\frac{I_o}{C\cdot f} \\ V_{o,odd} = N\cdot V_i - \frac{N}{12}(N^2-1)\frac{I_o}{C\cdot f} \end{cases}$$

Your circuit is one with an even multiplication $N = 2\cdot n$. And so the output voltage would be

$$V_o = 2n\cdot V_i - \frac{n}{3} \left( 2n^2 + 1 \right) \frac{I_o}{C\cdot f}$$

The ripple (in percents) was determined, quite logically to be:

$$r = \frac{I_o}{C_oV_of}$$

$$\Delta V_o = \frac{I_o}{C_of}$$

However, then I came across this paper:

1. S. Iqbal, Elimination of odd harmonics in symmetrical voltage multipliers, Journal of Instrumentation, vol. 7, April 2012

And this one opens his paper with the statement

(...) The operation of half-wave multiplier is very well described in the literature. The on-load average output voltage of this multiplier is

Followed by the formula without reference

\begin{align} V_o &= 2n\cdot V_i - \frac{I_o}{f\cdot C}\left(\frac{2n^3}{3} + \frac{n^2}{2} + \frac{n}{3} \right) \\ &= 2n\cdot V_i - \frac{n}{3}\left(2n^2 + 1 + \frac{3n}{2} \right)\frac{I_o}{f\cdot C} \end{align}

$$\delta V_o = \frac{I_o(n+1)n}{f\cdot C}$$

After searching through more recent papers, it seems that the most common reference is

1. M. Khalifa, “High-voltage engineering, theory and practice,” in Electrical Engineering and Electronics, A Series of Reference Books and Textbooks, vol. 63. New York: Marcel Decker, Mar. 1990, ch. 16.

However, the formula attributed to this reference is again different, although the circuit is identical to yours.

$$\Delta V_o = \frac{I_o}{f_sC}\left(\frac{2n^3}{3} + \frac{n^2}{2} - \frac{n}{6} \right)$$

$$\delta V_{pp} = \frac{I_o}{f_sC}\frac{n(n+1)}{2}$$

I have no access to the reference book, so I can't say for sure which of the two it is.

• Because of the PN junction between parallel diodes, the capacitor dont get never fully charged below a given frequency. When i simulate the circuit at frequencies below 1Hz the approximate is not even close. Maybe this is too harsh to model and thus the question would be then unfeasible to answer without a proper analysis. Or perhaps a good idea would link that frequency and the capacitances in a very approximate way. – Brethlosze May 30 '18 at 23:19
• Indeed, thanks to your reference to the name of the circuit (Cockroft Walton) i found this and this, which indeed express the drop and ripple of the circuit as inversely proportionaly to the frequency. Though i am not sure if the drop is the same that i think. I think this completes the answer, though i will validate them. – Brethlosze May 30 '18 at 23:31
• Thank you for that piece of information. I believe the derivation was done here. Feel free to take a look. I will edit my response with this information as well. – Sven B May 31 '18 at 8:50
• I have edited my answer to include some papers that you can use to find what you're looking for. – Sven B May 31 '18 at 17:31
• I guess your answer deserves a "super like". Being legally borderline with the ESE rules, I am pretty sure when you read this you'll discover what it is... – Brethlosze Jun 1 '18 at 1:44