Author Topic: Calculate needed capacitance for capacitive divider  (Read 497 times)

Offline Zipdox

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Calculate needed capacitance for capacitive divider
« on: September 07, 2020, 11:34:37 AM »
I've built the kaizer SSTC III and I've tested it at 85V. I've chosen a larger secondary and top load, meaning the resonant frequency is quite a bit lower. This leads me to speculate that I might need larger caps in the capacitive divider or risk discharging them before a full resonant period completes.

So how do I calculate the required capacitance? I know that resonant frequency, primary current and perhaps bus voltage are relevant parameters in the equation.

Offline Zipdox

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Re: Calculate needed capacitance for capacitive divider
« Reply #1 on: September 07, 2020, 11:52:12 AM »
After consulting wikipedia I came to this conclusion. Is this correct?

Offline Mads Barnkob

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Re: Calculate needed capacitance for capacitive divider
« Reply #2 on: September 07, 2020, 09:17:40 PM »
I would just calculate for the required ripple current according to the SSTC RMS current: http://kaizerpowerelectronics.dk/tesla-coils/sstc-design-guide/

I also wrote about half-bridge voltage doubler / voltage divider capacitors, voltage sharing etc here : http://kaizerpowerelectronics.dk/tesla-coils/drsstc-design-guide/dc-bus-capacitor/
https://kaizerpowerelectronics.dk - Tesla coils, high voltage, pulse power, audio and general electronics
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Offline klugesmith

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Re: Calculate needed capacitance for capacitive divider
« Reply #3 on: September 07, 2020, 09:36:02 PM »
The formula with 1/f, A, and V is obviously wrong, or at least poorly drafted.  Can we see the wikipedia link, for context?
Variables are not defined, but we can guess.
Where f stands for frequency, it's customary to use i to represent current.
A is the symbol for a unit of measurement (of current), analogous to Hz for frequency and F for capacitance.

Ohm's law is traditionally stated as E=IR, V=IR, or U=IR depending on when and where.   
Not V=AΩ, except for humor (see photo).
Formulas made from units of measurement may be dimensionally correct but numerically wrong.
For example, energy in a capacitor is e = C V^2 / 2.
If we substitute the units of measurement, we get J = F V^2 / 2.   That's not only silly, it is wrong by a factor of two, because 1 joule = 1 farad x 1 volt^2.

« Last Edit: September 07, 2020, 09:42:48 PM by klugesmith »

Online davekni

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Re: Calculate needed capacitance for capacitive divider
« Reply #4 on: September 08, 2020, 02:58:25 AM »
Kludgesmith:  The final lines in your post confuse me:
"For example, energy in a capacitor is e = C V^2 / 2.
If we substitute the units of measurement, we get J = F V^2 / 2.   That's not only silly, it is wrong by a factor of two, because 1 joule = 1 farad x 1 volt^2."

Agree that it's not the normal way to write the formula, but I'm confused by your "factor of two" comment.  A 1 farad capacitor charged to 1 volt stores 0.5 Joules of energy, not 1 Joule.
David Knierim

Offline klugesmith

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Re: Calculate needed capacitance for capacitive divider
« Reply #5 on: September 08, 2020, 05:09:26 AM »
Yup, it's ambiguous.
The formula for energy in a capacitor is e = 1/2 C V^2.
I was making a case for never expressing it as J = 1/2 F V^2, because that can be confused with a statement about units of measurement.

1 joule of energy is a unit of measurement, identical to the product of
 1 farad and 1 volt squared,
 or 1 watt and 1 second,
 or 1 henry and 1 ampere squared,
 or 1 newton and 1 meter,
 or 1 kilogram and (1 meter per second) squared.
The third and last examples have no factors of two,
in spite of the fact that inductor energy e = 1/2 L i^2, and kinetic energy e = 1/2 m v^2.

Fourth example is a good place to dig deeper.
Nobody can argue that joule is another name for newton meter (of work, not of torque).
Consider a Hookian spring whose force is proportional to deflection. 
If the force reaches 1 newton at deflection of 1 meter, the stored energy is 1/2 joule.
Work done by force that does not change with displacement: e = f x.
Work done by spring force that's proportional to displacement: e = 1/2 f x = 1/2 k x^2.
It gets sketchy when the last formula is dumbed down to "joules = 1/2 times newtons times meters".

« Last Edit: September 08, 2020, 05:31:55 AM by klugesmith »

Online davekni

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Re: Calculate needed capacitance for capacitive divider
« Reply #6 on: September 08, 2020, 06:01:39 AM »
Kludgesmith,

Thank you, now I understand your point - statements about definitions or equivalency of units is implied when using unit names in equations.
I'm used the unit newton meter being written "newton-meter" or N-m or Nm, rather than "newton * meter" or "N*m".  As a unit label, the "-" is a dash, not subtract.  That's why I associated the equations as applying to values rather than units, since it used "*" instead of "-".
« Last Edit: September 08, 2020, 06:12:39 AM by davekni »
David Knierim

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Re: Calculate needed capacitance for capacitive divider
« Reply #6 on: September 08, 2020, 06:01:39 AM »

 


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