Simple question about capacitor power banks

#12

Joined Nov 30, 2010
18,224
I feel like adding...the example I did will find the voltage on each capacitor in series if you don't use equalizing resistors. Only after you predict the first surge voltage does the voltage rating of the capacitor become a useful number. Look back at the example and see that the 12 uf capacitor will survive the first surge if it's only rated at 200 volts, but the 8 uf capacitor should be rated more like 300 volts. If you add equalizing resistors, that theory becomes useless, and you should use equalizing resistors because all capacitors leak in the DC range. If their DC leakage is very different. one will eventually be holding 60%, 80%, or 90% of the DC voltage.

See how complicated this can get if you just reach into a bag and grab a handful of capacitors? You can spend more time on math than it would take for a trip to the store!
 

Papabravo

Joined Feb 24, 2006
22,084
Impedance has both a real part and an imaginary part. Reactance is the real number used to construct the imaginary part of an impedance. These terms are very precise, but maybe it is OK not to care.
 

#12

Joined Nov 30, 2010
18,224
Would you like to explain how the imaginary part of the impedance of a capacitor is important for 2 capacitors in series as a DC filter?
 

#12

Joined Nov 30, 2010
18,224
Let's try something like the example I did. Two 10 uf capacitors in series, each rated at 250 volts, the power comes on and the rectifier charges the capacitors to 450 VDC, except this time the capacitors are equal in size. The power line frequency is 60 Hz. Use any capacitor available for sale today and demonstrate how the imaginary part of the reactance changes the voltage across the capacitors to something different from 50% each.
 

Thread Starter

Natron999

Joined Feb 1, 2016
14
Now you've got me thinking about the difference between impedance and reactance... I'll more carefully study this page, and then get back here if I have further questions.

I will also look at the page you have noted and see if I can gleen any information that will help in understanding the original question.

I’m sure it’s there, I just don’t know if I will understand it.

I did discover some information relating to this question and I believe it is a continuation on what was posted in reply to the question by #12;

where #12 said: ↑

I don't remember someplace to point to, but I can do the job.

4 example: 2 capacitors in series, both rated at 10 uf, 250 V and you want to hit them with 450 VDC.
(This is an old problem from the Fender guitar amplifiers.)
Theoretically, you have 5 uf at 500 volts.

In reality, if one capacitor is 20% over its labeled size and the other is 20% below its labeled size, you have an 8 uf in series with a 12 uf.

Pick a frequency, any frequency, and you will see that the impedance of each capacitor is the inverse of its capacitance. The 8 uf resembles 12.5 ohms and the 12 uf resembles 8 and 1/3 ohm. Now we're in a basic Ohm's Law situation. The voltage is distributed in a 60/40 proportion. Flip the standby switch to, "on" and the 8.333 ohm resistor charges to 180 volts in a quick pulse. The 12.5 ohm resistor charges to [450V x (12.5/20.83)] and that is 270 volts. Bang. It's outside its voltage rating.

This is the kind of worst case calculation you have to do.

I've seen ratings as bad as +80%/-20%. If you want to play mix&match with that kind of sloppy tolerances, you better do the math or wear a full face shield, a leather apron, and dragon hide gauntlets up to your shoulders.


and is as follows...

I think the condition of capacitors after an attempt to utilize various electrolytic capacitors with different voltages and capasitance without doing the proper math and allowing for the proper combination thereof might be also described as the The Burstein–Moss effect:

The Burstein–Moss effect:

Is the phenomenon of which the apparent band gap of a semiconductor is increased as the absorption edge is pushed to higher energies as a result of all states close to the conduction band being populated.

This is observed for a degenerate electron distribution such as that found in some Degenerate semiconductors and is known as a Burstein–Moss shift.

The effect occurs when the electron carrier concentration exceeds the conduction band edge density of states, which corresponds to degenerate doping in semiconductors.

In nominally doped semiconductors, the Fermi level lies between the conduction and valence bands.

As the doping concentration is increased, electrons populate states within the conduction band which pushes the Fermi level higher in energy and in the case of degenerate level of doping, the Fermi level lies inside the conduction band.

The "apparent" band gap of a semiconductor can be measured using transmission/reflection spectroscopy.

In the case of a degenerate semiconductor, an electron from the top of the valence band can only be excited into conduction band above the Fermi level (which now lies in conduction band) since all the states below the Fermi level are occupied states.

Pauli's exclusion principle forbids excitation into these occupied states.

Thus an increase in the apparent band gap can be observed.

Apparent band gap = Actual band gap + Moss-Burstein shift.

Also there are negative Burstein shifts which are due to the interactions terms created by adding the extra charges through doping.

Basically, I understand both #12's reply and this new to me information to mean, (paraphrase) without proper calculations and application of same the capacitors at best can and probably will be either ruined or severely distorted.

If I find a simple and easy way to calculate an acceptable solution to the question I will post it.
 

Thread Starter

Natron999

Joined Feb 1, 2016
14
Don't fry your brain over it. I am famous for not being precise all the time. Witness the, "Gold Standard" for blood pressure. A million physicians have believed it for over 100 years, but WBahn thinks they used the wrong word.:rolleyes:
I might say a million chemists know how dense mercury is (and a few dozen other elements) at any particular temperature, but I don't want to continue that conversation.

Impedance, reactance, I don't see much difference for the first surge of current into a DC filter....but I'm sure somebody could educate me.
Unfortunately, I don't care.

When there are no other components involved, like inductors or resistors, the combined or differential reactance with the other parts that don't exist (or are not significant) isn't important. You could study this down to the inductance in the power transformer, the inductance of the wires, the resistance of the rectifier at every level of current flow, the Equivalent Series resistance and inductance of the capacitors, but it won't make a percent of difference in the example I provided.

I apologize, my intention was not to burden anyone with the actual project I am currently working on but in light of some of the replies it might make better use of everyone's time if I just describe the project.

If I need to start another thread or post, I can and will do that.

If it is easier for a moderator to move this to another post please feel free to do so and link me, message or whatever.

I will briefly describe,

CURRENT PROJECT:

With a 1.5 volt aa d/c battery as the power supply connected by a simple on off switch used as the trigger to an EM35-1387-NCW Transformer and the switch being switched on and then off very quickly (1 second or even 1/16th of a second), including a fast switching high frequency diode as both a vehicle for the charge to pass as well as a protection method for the transformer and battery. I desire to capture the initial surge and voltage spike including flyback voltage of conventional energy into a capacitor that will store the charge for some period before discharging into a much larger capacitor.

My ultimate goal / purpose of this project is to excite the radiant / ambient energy that is made available at the zero point or the opening of the gate when the vaccum is present and gather, direct and or funnel in the radiant energy into a useful circuit consisting of a condensing coil and if the same capacitor / capacitor's of choice can be the same as the one's used for the conventional energy then that would be the way to go.


PREVIOUS METHOD AND PROJECT:

I have been able to collect radiant energy, ie (cold electricity) on a polished aluminum panel approx 4' X 4' at a 20' elevation, I ran a single copper wire from the elevated aluminum plate 20' to ground level but not grounded, then 90 degrees laterally approx 15' then 90 degrees up again for another 15' and finally laterally another 5' to the place where the circuit I used was placed.

Directly under the aluminum panel approx 4' X 4' at a 20' elevation, was a copper plate 1' X 2' buried at a depth of 1'. 90 degrees over and 15' away I had.set a copper ground rod only 3' deep. Both the rod and copper plate were connected together in line with one copper wire that followed the same route to the circuit as the wire that was connected to the aluminum plate.

The circuit consisted of a full wave bridge rectifier I built with 4 diodes.

The single positive wire from the aluminum plate to the + a/c in, the negative wire from the two grounding sources to the - a/c in, and the + d/c out to the - leg of an electrolytic capacitor and the + out of the capacitor to the two wires twisted together on a bifiller wound torrid of a joule thief and the - d/c out to the - leg of the JT transistor.

Then the led light + to + leg of transistor and the - to the - leg of transistor.

I was able to keep a single led light on 24/7 for 3 months that is when I disconnected it and I was able to light it again any time. I was able to run it on one wire while holding the other leg with me as the ground or on two wires.

I was also able to completely submerge under water both wires with the lighted bulb still lit and it stayed lit until I removed it with no shock to my hands which I left in the water with it the lit bulb for more than a minute.

I always got a 100% flashing and beeping from the wireless mains detector on anything metal within 20 feet + and many times on myself.

I had extremely high voltage readings on the voltage meter many times, the frequency was surely as the current was not.

The limitation I had with this circuit is that the only oscillation I had was the led and I now know it wasn't at the right place in the circuit perform the function of oscillation required to extract serious power from the atmosphere and it was also the wrong component for the job.

I was able to get several significant visible sparks across the the gap of the capacitor the few times I shorted it out on purpose but I did not have a spark gap in place to pump the circuit and extract any significant power.

I am reasonably certain I can achieve successful results of extracting significant power with a much smaller circuit and method but I am ruining my voltage meters and having trouble estimating the voltage and current I am currently getting with just a single manual and mechanical oscillation of the circuit I am currently using with a 1.5 volt d/c battery.

I'm sure that within some future time I will figure it out but I was hoping to present this situation to everyone and get some direction, learn and share information because this type of energy is certainly there, it is extremely powerful and there is as much as anyone might need, I am just seeking a safe, repeatable, scaleable, functional, (it's already affordable), fixed and portable form and method of extraction and utilization.

There are so many ways to do this very thing, these are just two that I have worked with.
 

Thread Starter

Natron999

Joined Feb 1, 2016
14
Thank you, but that quote is not originally mine. I found it in this place, and liked it so much that I decided to use it as my signature here.

I read the stories on that page and it was very uplifting and positive, thanks for sharing where it came from and it's a very good signature quote to have chosen. I will look for something to use for mine, it's a good idea.
 

Thread Starter

Natron999

Joined Feb 1, 2016
14
I feel like adding...the example I did will find the voltage on each capacitor in series if you don't use equalizing resistors. Only after you predict the first surge voltage does the voltage rating of the capacitor become a useful number. Look back at the example and see that the 12 uf capacitor will survive the first surge if it's only rated at 200 volts, but the 8 uf capacitor should be rated more like 300 volts. If you add equalizing resistors, that theory becomes useless, and you should use equalizing resistors because all capacitors leak in the DC range. If their DC leakage is very different. one will eventually be holding 60%, 80%, or 90% of the DC voltage.

See how complicated this can get if you just reach into a bag and grab a handful of capacitors? You can spend more time on math than it would take for a trip to the store!
True enough! I agree that it seems to be very complex, I think the store and matched capacitors are better.
 
Top