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Solid States Devices => solid state devices => Topic started by: magnetman12003 on October 01, 2015, 12:12:49 AM

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.??

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.

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=(CV^{2})/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.

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=(CV^{2})/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.

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 cutoff 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 builtin 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

Hi,
As others have stated a cap can be charged up near the max voltage, I like to charge it to about 80% of it rated value.
Voltage is the level at which you want the cap and amps is the speed at which is charges.
I have built a pulse heater using solar. This is not overunity.
Currently using 600watts of solar, pulsing a 800 watt infrared heater element(pulsed only not constant) using 7x 2600F(2.5v) ultracaps in series(371F).
It's coded to charge to 14.4vdc, this is hooked to a 1000watt inverter which then discharges through to 13.5vdc then cycles again. The full cycle time is around 19seconds in full sun. In an enclosed/insulated room the heating is accumulative and works nicely.
This patent may help you, while it is based on solar any type of DC input may be substituted.
US 6367259  Batteryless solar power system
https://www.google.com/patents/US6367259

Nicely done!
But have you thought about eliminating the inverter? It's just wasting energy that could be going into the heating element.
But since your final product is heat, maybe the inverter losses (which are also going to heat) are acceptable.

Hi TinselKoala,
Nicely done!
Thank you.
But have you thought about eliminating the inverter?
Absolutely, I have a DC version also using a 480watt 12vdc@40 amp cab heater which works great. I can also use a DC dump load through the charge controller(small range though). The only issue I have with the DC is I want to run the cables about 25feet from the solar charger. Which means heavy duty wiring to minimize the voltage drop/cable heating with DC, with the inverter I can use a standard 12Awg extension cord but then the loss is with switching.
So there are pros and cons with each method.

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.
I just purchased 10 Super caps. Each one rated at 1.5 farad and 5 volts.
For me to see a total working voltage of 12 volts placed across 6 of those caps can you tell me if all the caps should be placed in series? What would the total Farad count be then? How long would a small 300 milliamp 12 volt motor run from that fully charged group of 6 caps?

Capacitors in series decreases capacitance. You may consider that it decreases the capacitance to less than the smallest capacitor in series. Capacitors in series are calculated so, 1 / C = 1 / C1 + 1 / C2 + 1 / C3 ... If you have only two capacitors in series, then you can derive from that, that C = C1 * C2 / (C1 + C2) .

I just purchased 10 Super caps. Each one rated at 1.5 farad and 5 volts.
For me to see a total working voltage of 12 volts placed across 6 of those caps can you tell me if all the caps should be placed in series? What would the total Farad count be then? How long would a small 300 milliamp 12 volt motor run from that fully charged group of 6 caps?
You could make a 3series, 3parallel array using 9 capacitors. (3 groups of (3 series) in parallel). This will take you back up to the 1.5 F total capacitance and give a 15 volts rating, which will give you a little safety margin over your 12 volt charging voltage.
It's not possible to say for certain how long your motor would run, because the voltage will begin dropping on the cap array while running the motor and we don't know the minimum voltage that will run the motor. Build the array and try it!

If the caps are all the same rating you can just divide by the number in series.
1.5F/6 in series = 0.25F or 250mF
1.5F/10 in series = 0.15F or 150mF
One other thing you need with larger farad capacitors is a balancing circuit. Sometimes there is a slight difference in charging between capacitors due to dielectric absorption. This can cause one cap to have a much higher voltage than the others and can actually exceed the rating of the cap.
This is not good so we need to balance the voltage across each cell. The caps I got came with a balance circuit but it is similar to the setup from LaserSaber's site.
BoostPack Balancer Circuit Tested
http://laserhacker.com/?p=292
The circuit is simple just a diode and led across each cell. There are other ways also like using opamp balancing but the circuitry becomes more involved.
The discharge time for constant current is:
TimeInSec = (CapInFarad * (ChargeVoltageDischargeVoltage))/IDischargeCurrent;
So for your motor example:
CapInFarad = 1.5F/6 in series = 250mF
ChargeVoltage = 12vdc
DischargeVoltage = 9vdc for example (This is the stopping point or whatever motor minimum)
IDischargeCurrent = 300ma
TimeInSec = (.250F*(12vdc9vdc))/.300a = (.250F*3)/.300a = 2.5seconds
The other way if you want to calculate the nearest capacitor for the load and runtime with constant current you can use this formula.
CapInFarad = IDischargeCurrent * TimeInSec / (ChargeVoltageDischargeVoltage)
So using your motor example lets say we want to target a 1minute(60sec) run time so you need to figure out what kind of cap you need.
IDischargeCurrent = 300ma
TimeInSec = 60
ChargeVoltage = 12vdc
DischargeVoltage = 9vdc
CapInFarad = .300a*60sec/(12v9v) = 18/3 = 6F
If anyone sees a mistake let me know.