One Farad will put out 1Volt @ 1Amp for 1 Second.
No, that's not right. It doesn't even make sense. The formula is:
dV / dT = I / C where
dV = change in voltage across the capacitor over a period of time (in volts)
dT = the period of time (in seconds)
I = current into, or out of, the capacitor (in amps)
C = the capacitance (in farads).
This refers to the rate of change of the voltage across the capacitor.
It tells you that for a 1 farad capacitor, if you feed 1A of current in, or draw 1A of current out, the voltage across that capacitor will change at a rate of 1V per second.
A capacitor is like a reservoir of voltage. Imagine it as a straight-sided container that can be filled with water. The analogy is:
- V: The level of the water represents the voltage across the capacitor's terminals;
- I: The flow (measured in terms of volume per unit of time - for example, litres or gallons per minute) of water into, or out of, the reservoir represents the charge or discharge current;
- C: The floor area of the container represents the capacitance;
- T: Time is the same in the capacitor and the container.
Charging a capacitor is like pouring water into the container. When it reaches a certain level (voltage), if you stop pouring, the voltage will stop changing, and will remain constant forever (ignoring real-world imperfections like leakage - which affects a capacitor the way a leak would affect a container).
If you pour water into the container, or suck water out of it, the water level will rise or fall at a rate determined by the analogy's equivalent of the formula I gave earlier. A higher inflow or outflow rate (current) will make the level (voltage) rise or fall more quickly. Also, for a given in/outflow rate (current), the level (voltage) will change more quickly for a narrower container (lower capacitance).
Unlike a container, a capacitor (a non-polarised capacitor, at least) can be charged with either polarity. It's more accurate to say that current flow into one plate of the capacitor will make that plate more positive, or less negative, relative to the other plate, according to the formula. (Conventional current; positive to negative flow.)
Really 1 Amp for 1 second which would mean a 1R load. Are you sure it's not 36.8% left of both current and voltage after 1 second?
Adam is talking here about the specific case of a capacitor discharging into a resistor. This produces a voltage vs. time graph with a distinctive shape:
(image credit:
http://msc-ks4technology.wikispaces.com/file/view/discharge.png/40743385/discharge.png)
There is also a flipped version of this graph showing a capacitor charging (instead of discharging) through a resistor.
The graph shows the voltage across the capacitor, graphed against time. At the start of the graph, the switch is closed, and the capacitor starts to discharge into the resistor. The voltage drops according to the graph.
The formula at the start of this post implies that the voltage (across the capacitor) should change linearly over time, but the graph above is clearly not linear. This is because the discharge current is not constant. From Ohm's Law, for a constant resistance, current is proportional to voltage, so as the capacitor discharges, and its voltage drops, the resistor draws less current. This results in the curve above, with the voltage trailing off towards zero more and more slowly as the current drawn by the resistor gets smaller and smaller.
There is a specific point in this graph where the voltage reaches about 37% of the initial voltage. In that drawing, it's about 3.3V, because the initial voltage was 9V. This is marked with a big dot which lines up with the "1RC" tick on the X axis. This refers to the "time constant" of the capacitor and the resistor. The time constant is defined by the following formula:
T = R C
where T is time, in seconds;
R is resistance, in ohms;
C is capacitance, in farads.
So if you multiply the capacitance (in farads) by the resistance (in ohms), and take the answer as a number of seconds, it tells you the amount of time from the start of the graph to the "1RC" tick mark. This is the point where the capacitor's voltage will have dropped to about 37% of its initial value.
So to get back to the original question...
The voltage value on the capacitor indicates the maximum voltage that can be connected to the capacitor
Right.
Generally a safety factor of 20~50% is recommended. So if a capacitor is normally going to have 20V across it, I would specify a 35V capacitor, not a 20V capacitor; not even a 25V capacitor. This applies more to electrolytics, which are available in closely spaced voltage ratings; for small capacitors, cost and size are lower, and it's common to see a 50V ceramic capacitor used in a circuit where it will only ever have a few volts across it.
If you connect 12v battery to 50v capacitor, the capacitor will only get charged up to 12v.
Right.
The order of units uF > nF > pF (pF the smallest)
Right. The base unit is the farad (F) named after Michael Faraday. A farad is quite a lot of capacitance, and in practice, most capacitors are a tiny fraction of a farad.
Capacitance, like all quantities, can be expressed using the standard SI (Systeme International) unit multipliers and submultipliers. See
https://en.wikipedia.org/wiki/International_System_of_Units and specifically
https://en.wikipedia.org/wiki/International_System_of_Units#Prefixes. Since practical capacitances are mostly much smaller than a farad, we use the submultiplier prefixes.
farad (F) - a large amount of capacitance
millifarad (mF) - 0.001 farads or 10
-3 farads (see below)
microfarad (μF) - 0.000001 farads or 10
-6 farads
nanofarad (nF) - 0.000000001 farads or 10
-9 farads
picofarad (pF) - 0.000000000001 farads or 10
-12 farads - a small amount of capacitance.
The millifarad unit, mF, is not widely used; generally a value in the millifarad range is stated in microfarads; for example, a 15 mF capacitor would be described as 15,000 microfarads. There are other conventions too; older schematics will avoid the nanofarad (nF) unit, and give all values in microfarads (μF) and picofarads (pF), even though this requires extra zeros in the value. Very old schematics (over ~50 years old) sometimes used "μμF" to indicate pF!
How do you work out what farad value do you need, I know it depend on what you want to do, so I would appreciate if you could give me a simply example with explanation how it works. And also why when I charge 35v 47uF electrolytic capacitor using 9v 250mAh battery, the capacitor does not release its energy when I connect it to a small motor?
A thorough answer to your first question would take hours to write up. I may do that later. You could see whether it's answered elsewhere on the web first.
The second question is easier to answer. The capacitor actually does release its energy, but that energy is too small to do anything useful to a motor; even a small one.
Let's use the time constant formula from earlier this post, with a 47 μF capacitor charged to 9V, discharging into a 6V motor that draws, say, 0.1 amps, which is typical for a small DC motor.
A DC motor is not a resistor; it's much more complicated than a resistor, but it's still somewhat meaningful to pretend that it is a resistor for the purposes of this explanation. So this motor is designed to run from 6V and it draws 0.1A. So we can calculate a rough equivalent resistance using Ohm's Law, which rearranges to:
R = V / I
= 6 / 0.1
= 60 ohms.
Now we can use the time constant formula to work out the time constant of a 47 μF capacitor discharging into a 60Ω resistor:
T = R C
= 60 × 47×10
-6
= 2.82×10
-3
= 2.82 ms (milliseconds)
= 0.00282 seconds!
This tells us that after you connect your charged 47 μF capacitor to your motor, the voltage across the capacitor will have dropped to about 37% of its initial value (which is about 3.3V) after 0.00282 seconds.
This amount of time is so short that the motor won't even twitch. If you want it to do something useful, you need a lot more capacitance than 47 μF.
Since we know that 47 μF will run the motor for about 3 ms, and we know that T is proportional to C (for a constant R, which the motor is, roughly), we can say that if we use 1000 times as much capacitance, the motor will run 1000 times as long, i.e. about 3 seconds. That isn't very useful, but at least you can see it. So you would need a 47,000 μF capacitor.
As I said before, this value is in the millifarads (mF) range, but that unit isn't widely used for historical reasons to avoid confusion, so the capacitor will actually be specified, and marked, as 47,000 μF, not 47 mF.
I did a search on Digi-Key (
http://www.digikey.com) which is a huge electronic components e-tailer, and found a suitable capacitor:
http://www.digikey.com/product-detail/en/ESMH160VSN473MR50T/565-2620-ND/757833 47,000 μF, 16V electrolytic capacitor, USD 4.50.
Now, if you got one of those, charged it up to 9V, and discharged it into your motor, it would run for about three seconds
