# How to find what value capacitor do you need?

Discussion in 'General Electronics Discussion' started by nazar1000, Jul 16, 2014.

1. ### nazar1000

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Jul 16, 2014
Hey, I know it is a noob questions, however, I was looking on many websites, reading all the information but I did not found the answer. I am new to electronic but I know some basics such as, how a circuit works e.g positive goes to negetive and also how resistors works and how to find the value I need.

Capacitors
• I know they are made of 2 conductive materials separated by non conductive material
• There are few types of them.
Tell me if I am wrong but..
1. The voltage value on the capacitor indicates the maximum voltage that can be connected to the capacitor
2. If you connect 12v battery to 50v capacitor, the capacitor will only get charged up to 12v.
3. The order of units uF > nF > pF (pF the smallest)
And now the part I do not understand the farads
1. 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.
2. 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?.
Thanks for response

2. ### Gryd3

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Jun 25, 2014
One Farad will put out 1Volt @ 1Amp for 1 Second.

http://physics.nist.gov/cuu/Units/prefixes.html

Take a look at what that μ really means... you will find that it is an incredibly small number.
Capacitors are rarely used to drive a load, and when they are, they are either rated much high in terms of Farads, or more are connected in parallel to each other.
They are commonly used as filters, and as a sort of timer or delay for circuits. (These timers/delays typically draw a miniscule amount of current.)
A motor is an inductive load, and requires a very large (compared to other components) current that cannot be provided nearly long enough by the capacitor.

One more thing to add to your brain about capacitors... They store a voltage potential, and can supply way more current than a batter... so connecting a capacitor directly across something may cause an incredibly excessive current to flow through if you do not have any form of current limiting. (Motors have a very low resistance when they are not moving, using a large capacitor directly on one, may not be a good idea )

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3. ### nazar1000

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Jul 16, 2014
Thanks for response
I kind of understand what you mean, but now I am not sure about the units, if one farad will put out 1volt, 1Amp for 1 second, so is 1pF = 1000 farads or what, sorry if it is simply but I just cannot simply get it.

4. ### Gryd3

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Jun 25, 2014
Pico is 10 ^ -12 This means that the decimal point moves to the left 12 spaces, not the right.

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5. ### nazar1000

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Jul 16, 2014
Ok so; if I would have a capacitor that has value of 1uF, this would mean that the capacitor can be charged at a rate of 1*10^-6 volts and 1*10^-6 amps per second which in order to produce 1 volt would take (produce a delay) of 1*10^6 seconds ? Or is it around the way; it will charge quckly but release the energy at the rate of 1*10^-6. Is either one close to the answer?

5,165
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Dec 18, 2013
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?

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7. ### Gryd3

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Jun 25, 2014
Getting closer... Think of it like a water balloon. (That is strong enough to not break)
If you pinch the neck, the current flow is less, so it will take longer to discharge.
The amount of water in the balloon is the capacitance. How fast the water shoots out is the voltage (Which is determined by the incoming voltage), and the width of the water stream is current.

The tricky thing about the units you had stated... is that you shrunk ALL of the items by 10^-6 ... You only need to shrink one of them
So 1V, for 1 Second, would be limited to 1*10^-6 Amps

Be careful with your math when you start using those units.
Think of shrinking the area inside a square... if you chop one side in half, the area is now half... if you chop both sides in half your area is now 1/4 (because 1/2 x 1/2 1/4)

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8. ### Gryd3

4,098
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Jun 25, 2014

5,165
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Dec 18, 2013
Generally the voltage source would be a fixed d.c in this case and yes you could charge it with 1uA constant current which would take a very long time. We generally fix the voltage and add a resistor which charges up the capacitor and work out the time constant from there. RxC is one time constant and will charge the capacitor to 63.2% of your applied voltage within this time frame. So for a 1F and 1R resistor it will reach 63.2% or (632mV) in one second.

5,165
1,087
Dec 18, 2013

Thanks

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11. ### Gryd3

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Jun 25, 2014
Ahh, that's right. I over simplified. The voltage from the capacitor would decrease, and thus so would current as the capacitor drains resulting in a longer discharge time... kinda like my balloon analogy. I was however, unaware of the math. Thank you

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5,165
1,087
Dec 18, 2013
Discharge time will be the same it's governed by R*C. But I know what you mean.

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13. ### nazar1000

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Jul 16, 2014
Now I feel like internet explorer, I don't know what links with what and how Thanks for explanation, but I just cannot put it in order.

14. ### Gryd3

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Jun 25, 2014
Sorry if we confused you. Probably due to my analogies.
You understand though that Farads is a measurement on how much charge can be stored in a capacitor?
1 μF is 1000 greater than 1 nF. > http://kaffee.50webs.com/Science/images/SI.Prefixes.Chart.gif
Capacitor will not charge above a voltage you put in. Larger capacity require longer times to fully charge. There is math involved to determine how long a capacitor will take to dis/charge depending on what you feed it

Connecting a capacitor to a load (resistor or motor) will discharge it. Faster if the resistance is low, dangerously fast if they resistance is near 0. (if you short it with a screwdriver)
Your motor did not do anything because the capacitor you used did not hold a big enough charge to be able to provide power to the motor for a decent length of time... (What happens to a ceiling fan if you turned it on, then immediately turned it off... the faster you do this, the less it will spin. If your fast enough, it wont move.)

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15. ### KrisBlueNZSadly passed away in 2015

8,393
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Nov 28, 2011
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.)

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;

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...
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.
Right.
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

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!
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

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16. ### Gryd3

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Jun 25, 2014
This is what I had misinterpreted. I did well in physics, and in electronics. I'm kicking myself for falling off the track so far.

17. ### nazar1000

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Jul 16, 2014
Thanks for writing this all, you really clarify lots of things for me
I would also like to thanks Gryd3 and Arouse1973 for trying to explain it to me.