Understanding Capacitors

I know this is not an electronics forum, so if you feel I should post to some electronics forum I will (and if you could recommend one, Id love to hear your suggestions).

So I “salvaged” a 10V-1000uF capacitor from some old electronics (a failed wall wart). Im wondering if it works as it should. So I charged it with a 9V battery which has itself a v of 8.15V. It charges the cap in about 10 secs, up to 7.90V.

As soon as its disconnected from the battery of course it begins to lose charge. Ive mapped how much it dropped over time and Im wondering if this is normal. Here is the chart, Q/1 It’s normal, right?:

Also I noticed that when I connect it to an LED thru a 1k Resistor, it lights the LED but its voltage drains down to about 1-1.5V instantly and of course the LED turns off. I understand this is what is supposed to happen, the energy stored in the cap quickly flows into the LED as the cap loses all charge. Q/2 How can I make the energy stored in the cap last a bit longer?

Finally, Q/3 how should I understand how the V and uF of a capacitor relate to how much power they can yield and for how long? In other words, if I want to power a small dc motor of about 5V and 300mA for about 1 minute, instead of having the cap loose its charge all at once?

Marciokoko:
In other words, if I want to power a small dc motor of about 5V and 300mA for about 1 minute, instead of having the cap loose its charge all at once?

That ain't gonna happen.

A capacitor big enough to store that much energy might fit in
the trunk of you car.
Some of the new super capacitors are not too bad on size.
Maybe a brief case size.
It would also deplete your cash reserves.
Dwight

how should I understand how the V and uF of a capacitor relate to how much power they can yield

Q = VC
where Q is the charge or energy in the capacitor in coulombs
V is the voltage across the capacitor in Volts
C is the capacitance value in Farads.

1 Amp is the flow of 1 coulomb in 1 second.
1 coulomb is the equivalent to the charge on 6.242×1018 electrons.

There is something called the [u]RC Time Constant[/u]. (It can be derived from the equation Mike posted.)

If you multiply the capacitance (in Farads) by the resistance (in Ohms) that's the RC time constant, and it's the amount of time it takes for the capacitor to discharge to 63% of it's initial charge. There's nothing "special" or "useful" about the 63% value... It's just the way the physics turn-out, and the time-constant is an easy thing to calculate.

A one Farad capacitor (one million microfarads) and a one Ohm resistor would discharge to 63% in one second. 1uF and 1 meghom will have the same one-second time constant.

1000uF and 1K is also 1 second.

In the chart you made, I assume the only resistance was the resistance of your multimeter. The capacitor also has some internal "leakage resistance" (so the capacitor won't stay charged forever, even with nothing connected).

From Ohm's Law, your motor at 5V and 300mA has a resistance of 16.7 Ohms. (It depnds on the motor-load and 300mA could be worst-case with the motor stalled, or at start-up... It may draw less current once it's running, and/or if it's running with no load.)

The RC time constant also applies when charging the capacitor, but we don't usually know the effective source resistance of the power supply or battery.

Ok so if my motor has a resistance of 16.7Ohms and my capacitor is rated at 1000uF or 0.001F, does this mean it would take 0.017 seconds to discharge?

So I would need a 1F capacitor to make it last 16.7 seconds, right?

I saw a video of graphene supercap that spun a motor for a few minutes. But of course it was graphene and it was larger than what I have. I was just wondering what the relation was.

I was looking through the caps I've collected from things I've taken apart recently and noticed a DVD recorder transformer that had a 400V 100uF cap which is the size of a strawberry whereas a 400V 4.7uF is the size of a grape. That's a 20x jump. To go from a 100uF to a 1F I would have to jump (100->1,000,000) 10,000x. Oh well, I dont know what 10,000x the size of a strawberry would be, but I guess it would be big, if it were strictly linear... :s

Ok so if my motor has a resistance of 16.7Ohms and my capacitor is rated at 1000uF or 0.001F, does this mean it would take 0.017 seconds to discharge?

So I would need a 1F capacitor to make it last 16.7 seconds, right?

Right… To 63%. If you start-out at 9V, 63% is about 5.6 Volts. After another 16 seconds, you’d be down 63% from there or about 3.5V.

There are [u]one Farad capacitors[/u] and there are small low-voltage [u]supercapacitors[/u].

The exponential [u]capacitor discharge curve[/u] is the opposite of what you want from a “battery”. The voltage drops rapidly at the beginning and then levels-out when the capacitor is almost completely discharged.

An ideal battery holds it’s voltage through most of it’s life, and then fall-off quickly at the end. Real batteries are not ideal or perfect, but they are better than capacitors. For example you wouldn’t want a capacitor-powered flashlight that starts getting dimmer as soon as you turn it on… You’d like the brightness to hold-up for most of the battery life. Or if you have an electric car, you wouldn’t want to be at half-power when the battery is half-discharged… Ideally, you’d like it to behave like a gasoline-powered car where you have full engine power until you suddenly run out of gas.

As Mike said, Q = V x C

From that we can derive the useful relation, I = C x dV/dt

or to put it another way, C = I / (dV/dt) which is very useful when designing power supplies.

So, in your case, if you take a constant 300 mA and allow say a 1 V drop in 60 s,

the capacitance you need is: C = 0.3 /(1/60) = 18 Farad!

Russell.

Also the energy stored in a capacitor is 0.5 C V^2, so for 10V and 1mF that is 50mJ

The RC timeconstant is for a fall to 37%, not 63%, BTW, i.e. to reduce by a factor of e (Euler's
constant) So the drop from 9V to 3.3V, not to 5.6V

The differential equation for an RC circuit solves to include a term e^(-t/RC)

I = C x dV/dt

differential equation

I prefer the form:
I.dt = C.dV

and for inductors
V.dt = L.dI