I'm looking for some insight on whether an 1000 Farads at 2.7v or 5.5v would be more effective than say 10,000uF at 400v-450v when creating a strong magnetic coil. The intent is something similar to what I find when examining Coil Gun experiments. Some of the equations I have referenced are :
The amount of potential energy stored in a capacitor depends on the voltage:
PE = ½ C V^2
Where C is capacitance in farads, and V is voltage
And
v = V * sqrt(C/m) / 10
Where v is velocity in m/s, V is voltage, C is capacitance in farads, and m is mass in kilograms
I feel like the Cost/Benefit seems like having loads of amps at low voltage would be better than lots of amps at high voltage merely because supercaps pack soo many more Farads. I am given pause because of the v^2 in the energy equation, and the sqrt in the velocity equation.
Additional Information to Consider: Materials cost as of 5/oct/14-
Small Aluminum electrolytic capacitors 450V 1000UF – $1.50 each
SAMWHA super capacitor 2.7V 500F – $5.67 each
Large Electrolytic capacitor 2200uf 450v – $13.20 each
This Last example is what I see in most coilgun examples use and may be more available than the quick search I did for my initial search.
Please also consider the loss of capacitance when run in series vs addition of capacitance when in parallel. As well as the use of the capacitors in other types of experiments if the end result is very similar to each other.
Your thoughts and insights would be appreciated.
Best Answer
If we consider only ideal components, then the only number you need to care about is maximum stored energy per unit cost. As you already found, the energy stored in a capacitor is given by
$$ 1/2 \cdot C V^2 $$
The SAMWHA supercapacitor can thus store a maximum of:
$$ 1/2 \cdot 500 \cdot 2.7^2 = 1823 \:\mathrm J $$
The large electrolytic can store up to:
$$ 1/2 \cdot 0.0022 \cdot 450^2 = 222.75 \:\mathrm J $$
In this perspective, only the energy storage matters. That can come from high capacitance or high maximum voltage. It doesn't matter which, because you can shift the balance between current and voltage by changing the design of your coil. A coil with a large number of turns will develop a strong magnetic field without much current. However, it will have a large inductance and will require a high voltage to develop that current quickly. A coil with fewer turns will require more current to develop the same magnetic field, but the inductance will be lower so less voltage is required to get the voltage to rise rapidly.
If you were purchasing ideal capacitors, and you could fabricate an ideal coil, then this would be all you needed to make your decision. However, the real world is not so simple.
Real capacitors have non-ideal effects. It's common to model a capacitor as a network of other components, like this:
simulate this circuit – Schematic created using CircuitLab
Leakage determines how quickly the capacitor self-discharges. Ideally, this value is infinitely high. It is probably not much of an issue in your application, since you are not storing a charge for a long time.
ESL is the equivalent series inductance. Ideally this value is 0. This inductance places an upper bound on how quickly the current through the capacitor can change. You will want to make sure that the combined ESL of all your capacitors is at least an order of magnitude lower than the inductance of your coil, otherwise you will not get current, and thus the magnetic field, to rise quickly.
ESR is the equivalent series resistance. Ideally this value is 0. Here is the real problem for your application. You will be, at peak, drawing a very high current through your capacitors. This current must also flow through the ESR, and it is subject to all the physical laws of resistance. That includes dropping voltage, according to Ohm's law:
$$ E = IR $$
And converting electrical energy to heat, according to Joule heating:
$$ P = I^2 R $$
Any voltage dropped across ESR is voltage not available to drive your coil. If current is high enough relative to the ESR, then the voltage your coil sees will be essentially nothing. What you have is essentially this:
simulate this circuit
For the current to increase, voltage across R1 must increase. Since the voltage across R1 plus the voltage across L1 must equal the capacitor's voltage, as the current increases, the voltage across L1 must decrease. That's bad for you, because it means you can ramp up the current in L1 less rapidly.
Furthermore, the heating caused by losses in the capacitor can damage the capacitor. That's also bad for you.
The capacitor datasheets will elaborate on these details. However, based on general characteristics of supercapacitors and electrolytics, I can tell you that the electrolytics will have a much lower ESR on average. Supercapacitors are intended for lower current applications, sitting somewhere between a battery and a capacitor. They will typically be damaged by high current, or at least their ESR will be so high they will not perform well in your application.
There are also particular electrolytic capacitors designed to have an especially low ESR. These are called simply "low ESR" capacitors. They are also more expensive, but you might look into them. They typically find application as ripple filters in power supplies, where a higher ESR in the constant charge-discharging cycle they do would be a problem for efficiency, performance, or reliability.