# voltage multiplier + voltage divider charging capacitors bank

Discussion in 'General Electronics Chat' started by electricmaniac, Jul 25, 2011.

1. ### electricmaniac Thread Starter Member

Jul 25, 2011
15
0
Hello everybody,

I'm planning to use a half wave voltage tripler (stages n = 1.5, line connected 230VAC 50Hz 325Vpk through an isolation transformer, V_out = 975VDC with no load) to charge a capacitors bank (electrolytic capacitors 1000uF) to a voltage level of about 850V. This voltage is sampled from a voltage divider (made of three appropriate value resistors in series). A current limiting resistor (R = 1000 Ohm) is placed between the voltage divider and the capacitors bank.

We know that V_out_load = (V_out - V_drop) = 2n*Vpk - (I_load/6fC)*(4n^3 + 3n^2 - n), but according to the Ohm's law we must also have: I_load = (V_out/R) which is the maximal charging current for the capacitors bank once the main switch is closed (the capacitors bank is the load during the charging phase, right?). Now, we see that I_load depends on V_out, which in turn depends on I_load.

Simply substituting (V_out/R) for I_load in the first equation does not provide sensible results (e.g: with tripler capacitors of 1000uF = 0.001F each, I obtain V_drop = 61V, but with capacitors of 1000pF = 0.000000001F each, the numbers do not have sense to me any more...).
So:

1) How can I calculate V_drop?
Will the load reduce output voltage also of the voltage divider by some amount?
How I should take also this amount into account?

2) On which basis would you select the capacitance for each multiplier capacitors?
I've found no source that specifies the rules for this choice.
(I was thinking 3 capacitors of about 1000pF each or so with a breakdown voltage rating of 350 or above...)

3) Do you think that for capacitor charging purpose a full wave voltage tripler would work better than a half wave one?
If so, how do you draw a circuit scheme for such a full wave voltage tripler?

Thank you for your time.

2. ### Adjuster Well-Known Member

Dec 26, 2010
2,147
301
Please post a schematic showing you are planning to do: this will make it very much easier for somebody to advise you.

An accurate solution to this kind of problem might best obtained with a circuit simulation package: even a freebie like LTSpice should be able to give you a pretty good idea of what to expect.

I would think that the tripler capacitor value will depend partly on how long a charge-up time was acceptable, as well as any steady load current (including the divider current) and how far you can accept the output voltage to be down from the peak value.

P.S. I suppose that you realise that a 1000μF capacitor bank charged to 850V is a very dangerous thing?

3. ### #12 Expert

Nov 30, 2010
17,831
9,162
The capacitors in the voltage tripler actually represent an impedance in the power supply. Trying to charge 1000uf by using 1000pf capacitors will be rather slow.

A drawing is necessary.

Last edited: Jul 25, 2011
4. ### Adjuster Well-Known Member

Dec 26, 2010
2,147
301
The OP wants to charge 1000μF, so it really would be pretty slow with 50Hz input. Actually I doubt that even the maximum voltage will be much good, depending on his voltage divider current, capacitor leakage etc.

Note also that a transient simulation on a circuit with very small coupling capacitors and a big final reservoir will take a relatively long time to run.

5. ### electricmaniac Thread Starter Member

Jul 25, 2011
15
0
Hi,

Thanks.
I do realize that my questions can be a little confused(-ing) so this time I post the circuit scheme. I've tried to simulate the circuit by the circuitmaker software, but the simulation is limited only to 10ms due to the student version limits, and I also suppose there be a design mistake somewhere...

P.S: Please understand that I'm only an amateur, so forgive the mistakes and the poor quality schematic.

thanks again

6. ### Adjuster Well-Known Member

Dec 26, 2010
2,147
301
This circuit cannot work in its present form. For a start, the charge equalising resistors across the capacitors in the reservoir are 250Ω each, which times four and divided by three comes to 333Ω total. This is a low resistance compared to the 1kΩ and 130Ω series resistances, let alone the tripler effective resistance, whatever that comes to. This will prevent the circuit charging up very much at all, which is probably just as well because if it did get to 850V said resistors would be dissipating over 2kW, as much as an electric room heater!

Next, you have put the secondary of a trigger transformer directly in parallel with the capacitor bank. That effectively shorts out the capacitors (so they will not charge at all) , and also prevents the trigger transformer from developing any output pulse voltage. The potential divider feeding the capacitor at the input of the trigger transformer looks good to drain about an Ampere, if such current were actually available.

Your tripler is also arranged wrongly, with AC current being shunted into the output capacitor bank...

Perhaps it would be better to keep this as a strictly theoretical project for now. It would be better to learn to be able to "see" problems of this kind for yourself before trying to building a 300 Joule plus flash lamp.

7. ### electricmaniac Thread Starter Member

Jul 25, 2011
15
0
Hi Adjuster,

Your remarks are very useful to me. I intend to keep this project strictly theoretical till I fully understand the circuit's behaviour. So, in your opinion, where I should start modifying the scheme?
...what do you mean exactly? (I thought the tripler was the only right thing...)

Thank you.

8. ### Adjuster Well-Known Member

Dec 26, 2010
2,147
301
The way that you have your tripler grounded will result in it having a large amount of AC voltage on its output. This will lead to problems if you try to terminate it in a low impedance load like a capacitor bank. For some reason a good many web references show the circuit in this form, but it's not right for what you are trying to do.

I have made a couple of simple simulations to illustrate the problem - but note the diodes shown not very safe for this job - only 600V PIV rated, so they would not be a good choice in practice.

As you will see, moving the ground connection to the other end of the transformer secondary removes the large AC voltage from the output, and gives a much higher voltage into an external capacitor.

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9. ### electricmaniac Thread Starter Member

Jul 25, 2011
15
0
Hi Adjuster,

Your simulations were stimulating, thank you for the time spent on them, they represented also a starting point for further analyses.

I've increased the ohmic values of the balancing resistors in the capacitors bank: each of them now dissipates about 0.18 watts at full voltage. The simulation indicates also a satisfying cycle time (charging the main capacitors) of about 57 seconds thanks to the increased capacitance in the voltage tripler. Now everything seems working fine during the simulation at least, but some doubts remain:

1) The electrolytic capacitors are polarized but I'm not sure of their proper connection in the tripler circuit (the positive terminal of one capacitor must be connected to the negative one of the other capacitor...right?).
Must the tripler capacitors withstand a stage voltage defined as RMS 220 V or as the peak-to-peak value of 325 V?

2) The first resistor of the voltage divider (R1) must withstand a 96mA peak when the system is powered on. The current then exponentially decays to 12mA after about 63 seconds, therefore the average current from 0 to 30 seconds amounts to about 40mA. Selecting the wattage rating of the resistor would you take into consideration the peak value (96mA then about 100 W) or the average value (40mA then about 16 W)?

3) Can a frequency multplier circuit increase the charging time of the tripler capacitors as well as that of the capacitors bank?

Thanks again.

10. ### electricmaniac Thread Starter Member

Jul 25, 2011
15
0
some other more accurate diagrams...