Capacitors have me crazy. I can't comprehend them even after lots of research. I know that capacitors deal with resistance, voltage, current and capacitance (ohms, volts, amps and farad) but I cant connect them. I am making a plane and want to use a capacitor as battery. The motor uses 3.7V at 100mAh. I have a 6.3V 1F capacitor. And 39ohms resistor and I can't make the motor run for more than 0.5 seconds. If anyone could explain me and help me I would be so grateful. Thanks
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3 Answers
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See How do I calculate how fast a capacitor will discharge? for the relationship between capacitance, voltage, and discharge current, and how to calculate the discharge time.
How are you charging the capacitor? The 6.3V rating on your capacitor only indicates the maximum voltage that can be applied to it without being damaged.
What is the 39 ohm resistor you have? Please draw a schematic of your circuit. Also note that the motor will have an inrush current at startup that is many times higher than its rated current. This will affect the runtime considerably when the total stored power is so low.
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Some fundamentals:
Volts are the amount of "Pressure" thats available to drive an electric circuit. Amps are the amount of Current or "Strength" of the ciruit.
If (in DC Circuits), you multiply the current (I) of a circuit times the amount of voltage (V), you get the total POWER measured in Watts.
The Capacitor gets charged up from a voltage source somewhere?
A magneto attached to the back of your bicycle tyre is a good example of a generating source.
In small motors the current is in the range of 100 to 500 millamps or 0.1 to 0.5 amps.
Assuming you have a 6 volt battery as the source to start with, and the capacitor is in the range of 1000 to 5000 micro-farads ( 0.1 to 0.5 Farads), it would only drive the motor for a second or two.
6 volts x 0.1 = 0.6 Watts (or 600 mW).
If you want to drive the motor for 60 seconds you would need a few Joules of power.
Joules are the measure of Power over time.
In this case you would need a "High capacity" Capacitor to store all of that energy.
While it is possible to have one that big it would be practically a large size in relation to the size of a bunch of "AA" Batteries, and what they can hold as a charge.
A capacitor has not been found of high enough storage capacity to store a lot of energy for a long duration (yet).
Even some of the High-Power Capacitors found for high end Auto Stereo systems only smooth out the power flows into the power amplifiers for that deep bass effect.
Once you get past filtering and short time frame (1-2 secornds) now you are talking a large amount of energy stored in the capacitor, (a big physical size for commercial varieties) then the risks of accidental shorts or sparks are very real, so beware of what you are working with!
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Simplified model of your load as a constant current source for a conservative bound. In that case, if your capacitor of capacitance \$C\$ is charged to \$V_c\$, and your constant current draw is \$I\$, then:
$$V(t) = V_c - I \cdot t / C$$
What is t when your voltage reaches zero?
$$t = V_c \cdot C / I$$
Let's say your current draw is 100mA, \$V_c\$ is 6.3V and \$C\$ is 1F, then \$t\$ is 6.3 * 1 / 0.1 = 63 seconds.
Problem with the constant current source model is your load actually stops drawing power when your voltage drops below some level (or has some more complex dependence on voltage), which significantly reduced the "run time" of your capacitor. That is to say, your real load is probably not a constant current source.
Also capacitors have leakage, but I'd expect that to be a secondary effect. Also my maths should be community checked.
If instead of a constant current model, you imagine a purely resistive load, then the (slower) voltage decay follows the perhaps more familiar form:
$$V(t) = V_c \cdot exp^{-t/(RC)}$$
