# Capacitors

Discussion in 'General Electronics Discussion' started by biferi, Oct 12, 2016.

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1. ### biferi

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Oct 12, 2016
Hello everybody I am new to the WebSite and I am a Computer Graphics Tech. and a Computer Repair Tech.

I am Back in school to be an Electronics Repair Tech. and need some help.

I am at the point of Capacitors and I am having a hard time understanding how to use them.

I know if you Hookup a 25 microfarad Capacitor to say a 9 Volt Battery it will charge to the same 9 Volts as the Battery.

And yes I know how you use a Resistor with it to get your Time Constant TC of how long it will take for the Capacitor to Charge.

And yes I know it takes 5. TCs to Fully Charge a Capacitor.

But when you get a Capacitor you get it at the Charge you want it to hold say 25 microF or 12 picoF or what have you.

But I do not understand what a Farad is.

Lets say I have a 9 Volt Battery and a LED that takes 3 V at 20 mA to run.

So I connect a 900k Resistor in Series with the 80 microF Capacitor and the 9 Volt Battery.

Then I know to get my Time Constant I do this
R x C = 72 then I do this
72 x 5 = 360 TC.

So this tells me when the switch is closed the Capacitor will take 3 Minutes and 60 SEC. to Fully charge.

Ok so I understand the Capacitor will have 9 Volts and it will have a charge of 80 microF but if I take this Capacitor and Hook it up to the LED that needs 3 Volts and 20 mA I know I use a Resistor to bring the Voltage down but how do I set the Current?

How many Amps are in 80 microF?

Last edited by a moderator: Oct 13, 2016
2. ### davennModerator

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Sep 5, 2009
Hi
welcome to EP

well you should be able to discover that very easily

The current is set by the resistor
have a read of our LED's resource

none. current is the FLOW of charge, and since the capacitor is storing a charge (nothing flowing)
there is no current until it is connected to a load. Then the current flowing is determined by the resistance of the load

Dave

3. ### dorke

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Jun 20, 2015
Welcome to EP.
You are making a fruit salad of it all...

It is the unit to measure capacitance.
It tells us what is the maximum electric charge that can be stored in a capacitor.

It can be imagined to be the size of a "reservoir"
i.e the larger the capacitance(in Farad ) the larger is the electric charge that can be stored .

You can look at it to be equivalent to the volume of a container in cc.
The larger is the container volume in litters/gallons, the more liquid you can store in it.

2. "I know it takes 5. TCs to Fully Charge a Capacitor".
Not an accurate statement.
This is only a practical approximation.
To fully charge an ideal capacitor it will take infinity(that is conceptually important to understand)

3. "will take 3 Minutes and 60 SEC. to Fully charge."
It will take about 6 minutes.

4 ."it will have a charge of 80 microF"
No, the capacitance is not the charge stored !!!
The relation between charge stored Q in a capacitor of value C and voltage of V is
Q= C * V

5. "How many Amps are in 80 microF?"
Meaning less question.

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Jun 21, 2012
5. ### biferi

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Oct 12, 2016
Ok I do understand that 1. Amp is telling us how many Electrons are moving past One point in one SEC.
So yes a Capacitor is just holding a Charge.

Say I want to use a Capacitor latter and I charge up a 25 microF Capacitor.
And then I disconnect it I know it will keep the charge till I use it.

And yes I know the Resistor I use with it and the Load will tell me how long the Capacitor will Discharge.

But how do I know if I will be able to get 32 mA out of the Capacitor or 10 mA out of the Capacitor?

I know 25 microF can hold more then a 13 microF Capacitor but how do I know witch one will give me more mA?

I would think the 25 microF will give me more mA but how do I know how many?

I hope this helps.
Thanks again.

6. ### dorke

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Jun 20, 2015
The relation is Q= I *t or I=Q/t
And since we also have Q=C*V
we get I=C*V/t

So for a fixed time interval and voltage
I1=C1*K ; I2=C2*K
Thus I1/I2 =C1/C2

So if C1=25 ,C2=13
the 25 uF may give you 25/13 "more current" than the 13 one.

7. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

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Jan 21, 2010
In theory there is no limit to the magnitude of the current you can draw from any capacitor. The capacitance asking with the voltage across it when you start week tell you how long you can draw that current as the voltage drops to zero.

Practically there is a limit to the current due to practical issues such as the resistance of the connecting wires and many other factors.

If you had a 1F capacitor that was charged up to 1V and you wanted to draw 1A from it, you could do that for 1 second (this is essentially the definition of 1F).

If you wanted to draw 0.1mA from a 1000uF capacitor charged to 25V, you could do it for:

(1/.0001) * (1000 * 10^-6/1) * (25/1) seconds

That's 250 seconds

8. ### davennModerator

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Sep 5, 2009
not really, real world capacitors are not perfect nor idea, they leak charge between the plates
The capacitor WILL NOT keep its charge for a long period

9. ### Ratch

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Mar 10, 2013
You are using technical slang when you or anyone else says a capacitor is "charged". The correct description is energized. When a voltage is applied across the cap, electrons accumulate on one plate and deplete by the same amount on the opposite plate for a net change of zero. So even when a cap is energized to a high voltage, it contains the same number of electrons that it had at zero volts. It takes energy to unbalance the charges on the plates, and energy is what the cap stores, not charge.

A farad is a measure of how much energy a cap can hold at a defined voltage.

Asking how many amps there are in a cap is like asking how many watts there is in a battery.

The words volt ,amp ,farad, battery, capacitor, resistor, current, minutes, time constant, and series are not proper nouns, and therefore are not capitalized in the English language.

Ratch

10. ### dorke

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Jun 20, 2015
@biferi,
I hope you are not discouraged by the "bombardment" of the forum members(including myself).
Keep asking,that is the only way to learn...
anything you need to make it as clear as possible for you.

Here is a water analogy:

Larger container volume is equivalent to larger capacitance.
Water level is equivalent to voltage.
for the same water level(voltage) the larger container will hold more water(charge).

Now,
Let's see a "discharge" or draining of water analogy.
Using a "straw" of the same diameter at the bottom is equivalent to the same current discharge.

For the same initial water level(voltage) ,
the larger the container the more time it will take to empty it(discharge the capacitor).

Hope this helps

Arouse1973 likes this.
11. ### hevans1944Hop - AC8NS

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Jun 21, 2012
I like this pressurized air-tank analogy better. But remember this: analogy is only an aid to understanding, relying on previously learned concepts to help describe new effects. The new effects, in this case electrical capacitance, are NOT the same effects described by the analogy.

The defining relationship for any capacitor is C = Q / V, where Q is the difference in the number of electrons stored on the plates of the capacitor, and V is the potential difference that results from this separation of charges over a finite distance. The potential V is a result of an electrical field, measured in volts per meter, in a dielectric between the two plates of the capacitor. This implies that you can change the capacitance if you can change the strength of this electrical field without moving any charges. There are several common ways to do this. The easiest is to simply vary the separation distance between the plates. Another is to vary the dielectric constant of the insulation between the plates. Still another way is to vary the effective area between the two plates. Doing any of these, when there is already a charge separation between the plates, will introduce additional forces that will affect the energy stored in the electrical field of the capacitor, either increasing or decreasing the amount of energy stored, the difference in energy being proportional to the force multiplied by the distance through which it moves the object responsible for the change in capacitance.

The rabbit hole leading to full understanding of capacitors and inductors is very deep. Simple RC circuits are just the first step, but before diving into that you need to review basic circuit concepts and learn how to analyze series and parallel resistor circuits with multiple voltage or current sources. The tools you need for this are Ohm's Law and Kirchoff's Voltage and Current Laws plus a little algebra. For example, Ohm's Law, applied to a resistor used to discharge a capacitor, will quickly tell you what the initial discharge current will be: I = V / R, where V is the initial voltage on the capacitor terminals, and R is the resistor value. Of course the initial value will decrease exponentially as the energy in the capacitor is converted to heat dissipated in the resistor, but understanding how that works is the beginning of your trip down the rabbit hole. When you get to the bottom, Alice, say hello to the Red Queen for me.

12. ### bushtech

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Sep 13, 2016
When you get to the bottom you might turn into the mad hatter

13. ### dorke

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Jun 20, 2015
Hop,
There is a little, but very important fact about capacitance:
It is a geometrical property !

In fact the unit of capacitance in the cgs unit system is centimeter [cm] !!!

hevans1944 likes this.
14. ### davennModerator

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Sep 5, 2009
thread is getting silly and going nowhere

time to close