Calculate super capacitance

Thread Starter

k1ng 1337

Joined Sep 11, 2020
435
Hi,

I have ordered a 2.7V 500F double layer capacitor and as it makes its way to my workshop, I wish to prepare myself with some mathematics.

If charging to 1 time constant can be approximated as

Power * Time = (Capacitance * Voltage^2) / 2

And if 5 time constants the capacitor is nearly "full" (do real capacitors ever fully charge or deplete?)

How would I go about controlling the time parameter to find an unknown capacitance, providing that I am able to control and measure the other parameters.

Thank you and I hope this day finds you well.

Regards,
Mark
 

Thread Starter

k1ng 1337

Joined Sep 11, 2020
435
I would appreciate an example as I don't think I am applying the variables correctly.

Assuming I am using a 2.7 DC source in series with a 10 ohm resistor the power is 0.729W, do I set the voltage parameter to 2.7v, time how long it takes to get to 2.7v (in seconds), and then solve for capacitance?
 

BobTPH

Joined Jun 5, 2013
4,770
You cannot use the full voltage as the threshold because, mathematically, the time is infinite. Time how long it takes the get together the voltage of one time constant.

Bob
 

MisterBill2

Joined Jan 23, 2018
11,612
For recovering energy from a discharging capacitor there is a basic problem in that as the energy is removed the voltage drops, and since many applications require a specific voltage, the operation of that function changes a lot.
Are you intending to measure the value of capacitance using the formula? Or is this a more scholastic learning sort of question? Or is there a practical application involved?
 

Thread Starter

k1ng 1337

Joined Sep 11, 2020
435
@Bob What is the voltage of 1 time constant? 2.7 / 5 = 0.54v?

Furthermore what is the relationship of 1 time constant being set what it is to 5 being "full"? Why was this number chosen? There must be some expression that relates the charging curve in more detail..?

@MisterBill

My interest in these calculations is to relate energy to actual charge stored at the electrodes so when I go to build my own I can understand what is happening. I built a foil capacitor and was amazed at the simplicity of formula compared to other devices. I am now investigating how much deviation the formula goes from reality.
 

MrChips

Joined Oct 2, 2009
25,929
One time constant is R x C.
For example 10Ω x 500F = 5000 seconds = 83 minutes.
This is the time it would take to reach 63% of the charging voltage,
i.e. 0.63 x 2.7V = 1.7V

The time taken to reach 50% of the charging voltage is 0.69 x R x C = 3450s = 57.5 minutes
 

crutschow

Joined Mar 14, 2008
29,514
What is the voltage of 1 time constant? 2.7 / 5 = 0.54v?
When charging it is approximately 0.632 of the charging voltage.
When discharging it is approximately 1-0.632 = 0.368 of the charged voltage.
what is the relationship of 1 time constant being set what it is to 5 being "full"? Why was this number chosen?
The formula is Vc = V(1-e^(T/t)) where Vc is the capacitor voltage, V is the charging voltage, T is time and t is the circuit time-constant (R*C).
If you plug-in the value of 5t for T in the equation you will get a value for Vc that is about 99% of the value of V, which is considered sufficiently fully-charged for most engineering purposes.
If you don't want to wait for 5 time-constants, 3 time-constants gives a value of 95% of V.
 

MisterBill2

Joined Jan 23, 2018
11,612
@Bob What is the voltage of 1 time constant? 2.7 / 5 = 0.54v?

Furthermore what is the relationship of 1 time constant being set what it is to 5 being "full"? Why was this number chosen? There must be some expression that relates the charging curve in more detail..?

@MisterBill

My interest in these calculations is to relate energy to actual charge stored at the electrodes so when I go to build my own I can understand what is happening. I built a foil capacitor and was amazed at the simplicity of formula compared to other devices. I am now investigating how much deviation the formula goes from reality.
Ok, but building an ultracapacitor is going to be a lot more challenging because of the surface, which is a layer of conductor with space inside for the energy storage material, (electrolyte) whatever it may be. That is why they are also referenced as "double layer" capacitors. I had been considering a project that would require a capacitor charged to 10,000 volts to be able to deliver a few milliamps for over a second. The alternative would be a really high powered supply able to deliver that current constantly, which gets to be quite a big project and use a lot of input power from 12 volts DC.
 

Thread Starter

k1ng 1337

Joined Sep 11, 2020
435
Thanks everyone for the replies, the information provided was helpful. It would seem the power characteristics of a capacitor are governed by the dielectric breakdown voltage, the electrode material, spatial geometry and for a double layer capacitor, the voltage at which an undesirable chemical reaction takes place; whereas the charge and discharge profiles are inherit to electron mobility and are basically the same for all capacitors. It is now meaningful to me to look at other devices such as a BJT in terms of energy and charge rather than the novice approach of voltage and current.

My goal was to predict the state of a capacitor between two extremes which can be made useful to an analog device or an ADC.

Live long and prosper,
Mark
 
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