Capacitor question

[magnetman12003]

I have a 65 volt cap rated at 20,000 uf.  If I charged the cap with 12 volts would it be full or would I have to go to 65 volts to see that?   The current I would be charging the cap with is in the milliamp range.??

[ayeaye]

When you charge the capacitor with 12V, then the capacitor is full when the voltage on the capacitor reaches 12V. How fast that happens, depends on the resistance between the capacitor and the voltage source. The current first is is such as it were when the capacitor were short circuited, that is if there were a piece of wire instead of the capacitor. The current then goes down by an exponential curve, and the voltage rises until it reaches 12V.

If you want to know, when is the maximum energy stored in the capacitor that this capacitor can store. It is of course when you charge your capacitor with 60V, and the voltage on the capacitor rises to 60V. The energy in the capacitor is equal to C * V * V / 2 , where C is the capacitance and V is the voltage on the capacitor. When C is in farads and V is in volts, then the energy calculated that way is in joules (J), which is watt seconds (W * s). Which means that when you divide that by the time which it took to charge the capacitor to that energy from zero, you get the average power used during that time to charge the capacitor.

[TinselKoala]

I have a 65 volt cap rated at 20,000 uf.  If I charged the cap with 12 volts would it be full or would I have to go to 65 volts to see that?   The current I would be charging the cap with is in the milliamp range.??

The "65 volts" rating is the maximum voltage that the capacitor can safely handle before the internal dielectric is shorted and ruined.

Energy on a capacitor goes as E=(CV²)/2 where E is in Joules, C is in Farads and V is in volts.
So if you charge your capacitor to 12 volts, it will contain (0.020000 F x 12V x 12 V) / 2 = 1.44 Joules of energy.
If you charge the capacitor to 65 volts, it will contain (0.02 F x 65V x 65V) / 2 = 42.25 Joules of energy. (If I did the math right, that is.)

This would qualify as "fully charged" since going higher in voltage would risk blowing up or otherwise ruining the capacitor. The current controls the _rate_ at which the capacitor charges... less current = more time to reach the voltage level desired.

[magnetman12003]

The "65 volts" rating is the maximum voltage that the capacitor can safely handle before the internal dielectric is shorted and ruined.

Energy on a capacitor goes as E=(CV²)/2 where E is in Joules, C is in Farads and V is in volts.
So if you charge your capacitor to 12 volts, it will contain (0.020000 F x 12V x 12 V) / 2 = 1.44 Joules of energy.
If you charge the capacitor to 65 volts, it will contain (0.02 F x 65V x 65V) / 2 = 42.25 Joules of energy. (If I did the math right, that is.)

This would qualify as "fully charged" since going higher in voltage would risk blowing up or otherwise ruining the capacitor. The current controls the _rate_ at which the capacitor charges... less current = more time to reach the voltage level desired.

What I am planning to do is run say around 20 DC volts into the large cap and monitor how fast the current charge goes in.  When 12 volts are reached I plan to disconnect the charge and use what's in the cap at that time.

Thanks for your help with this.  I am currently involved with making a free energy from air setup and have successfully made a receiver that actually works.  Will post that when I try a few more parts for better results.  The large cap being one of the parts.

[sm0ky2]

 

What I am planning to do is run say around 20 DC volts into the large cap and monitor how fast the current charge goes in.  When 12 volts are reached I plan to disconnect the charge and use what's in the cap at that time.

Thanks for your help with this.  I am currently involved with making a free energy from air setup and have successfully made a receiver that actually works.  Will post that when I try a few more parts for better results.  The large cap being one of the parts.

20v areal, nice work. have you already charged smaller capacitors?
you can use a transistor with a 12v cut-off to switch the cap from charge to discharge when the voltage is at the appropriate level.
remember there are two times you have to deal with.
the time it takes to charge it, and also the time it takes to discharge it.
I think you stated you want to slowly discharge the cap? this is sometimes more complicated than just raising the resistance of your circuit.
while that will change the discharge current (avg), the time is so short, and not greatly affected
a larger cap might give you a few more microseconds
There are these things called "supercapacitors", basically caps with a built-in internal resistance that can be discharged slowly, like a battery.
but charge quickly like a capacitor. maybe worth looking into once you develop your design a bit more