# Capacitors 101.

Discussion in 'Electronic Basics' started by Patrick O Shane, Oct 29, 2005.

1. ### Patrick O ShaneGuest

Hi,

I was wondering if anyone here could explain capacitors to me. I have tried
for a long time to understand exactly what they do. I 'think' I have a basic
understanding. I could be totally wrong here.

Basically they store a charge and release it and the Farad rating determines
how big of a charge it can hold and for how long. I've noticed that the
voltage ratings do not necessarily have to match what is being applied to
it.

What is there purpose in an energized circuit?

What do the Farad ratings mean?

What do the Voltage ratings mean?

How are the two related or are they related at all?

How does amperage effect capacitors?

Do I need more of an understanding of circuitry and circuit design to be
able to understand what a capacitor does and how it does it?

How would a 25v capacitor effect a 2.5v LED?

I asked for help regarding putting an LED on an HO scale train. Some very
generous people worked up a circuit for me that had a capacitor in it to
stabilize the light intensity when the train slowed down or hit a dead spot
on the tracks. They suggested a 25v @ 400uF capacitor assuming the maximum
input voltage to the circuit would be 16v.

Hope I'm not making myself out to be an idiot.

Thanks for any input.

Sincerely,

Shane B.

2. ### Tom BiasiGuest

Hi Shane,
You have many good questions. Rather than answer each one which would be a
lot of typing for me and a lot of reading for you, why don't you read at the
link that I provided and get back here with some questions.

Tom

Look Here:
http://www.pc-control.co.uk/capacitors.htm

3. ### John PopelishGuest

The capacitance tells you how much charge displacement you get for
each volt applied across the cap. A 1 farad cap displaces 1 coulomb
with a 1 volt potential difference across its terminals. a 1 uF cap
has 1 micro coulomb of charge displacement for each volt of potential
across its terminals.
I don't understand the question.
The farad is the unit of capacitance 1 farad equals 1 coulomb per
volt. 1 amp (1 coulomb per second) passing through 1 farad requires a
rate of voltage change of 1 volt per second.
The maximum voltage the capacitor can withstand without damage.
Higher voltage rating implys thicker dielectric between the plates of
the capacitor. This takes up space, so higher voltage capacitors are
generally larger than lower voltage rated capacitors of the same
capacitance.
I=C*(dv/dt)
Current requires a rate of change of voltage applied across the
capacitor. The larger the capacitor, the smaller the rate of change
of voltage is needed to achieve the same current. The current rating
of a capacitor involves the resistive losses internal to the capacitor
that get hot when current passes through them, and how well the
capacitor gets rid of that heat.
The definition of a capacitor is a simple concept that stands alone.
Learning all the useful ways a capacitor can be used in electronics
takes a lot of study and experience.
25 volts is the maximum rating for the capacitor. If you charged it
up all the way to 25 volts and connected it across an LED (that
normally operates at a few volts) the voltage across the capacitor
would drop to a few volts very quickly. a high rate of change of
voltage across a capacitor implys a large current for a short time.
either the LED would give off a bright flash and quickly fade to it
would give a bright flash and be destroyed. The size of the capacitor
(how many micro farads) would mostly determine which would happen.
You need a capacitor that is rated to survive the highest voltage the
track can charge it to. Since most train speed controllers operate of
rectified AC, they have a peak voltage that is higher than the average
you measure with a volt meter. To be on the safe side, I think you
should use a capacitor that is rated for twice the DC voltage you
measure between the rails with the controller set for maximum speed,

The capacitor will act something like a small rechargeable battery
that soaks up current when it is discharged and supplies current when
it is running down. The larger the capacitance (micro farads) the
bigger a surge it will take to charge it but the longer the LEDs will
run off it before it discharges too much to light them.

Most LEDs are rated for 20 milliamps of current, but give out almost
as much light at 10 milliamps (0.01 ampere).

Lets say you measure the DC voltage between the track at 15 volts.
This will charge the capacitor to a little more than 15 volts, because
the voltage has ripple, rather than being perfectly steady, but lets
ignore that for the moment and say the cap can reach a full charged
condition of 15 volts.
You pick a current limiting resistor that limits the LED current to
0.02 amperes at the highest voltage the capacitor can reach. So 15
volts minus the 3 or so needed by the LED leaves 12 volts to be burned
up by a resistor. Ohm's law tells us that the volts per ampere
through a resistor are equal to its resistance. So 12/.02= 600 ohms.
The nearest 5% standard resistor is 620, so lets use that. If the
track voltage should disappear for some reason, the capacitor voltage
will start running down by the I=C*(dv/dt) formula. If you use a 470
uF capacitor (a standard value) this means that its voltage will decay
at dv/dt=I/C and .02/470*10^-6 = 42 volts per second. This rate will
slow as the lower voltage lowers the current but that is still a
pretty fast light decay (visibly off in a fraction of a second.

I think you will want a capacitor quite a bit larger than 470 uF.
10 times larger, perhaps. If you use a higher resistance to limit the
initial current to a lower value (and the light to a dimmer peak
value) the capacitor will run down at a proportionately slower rate.
Double the resistor, and the light will start dimmer but fade over
twice as long an interval.

4. ### ErikBalubaGuest

Hi,

It can take some time to really understand how a capacitor works. In schools
capacitors are usually explained by starting with the physics, thus Q=CV.
And since current is defined as the change of charge (Q) over time you will
end up with a differential equation. This equation is not that complicated
for a simple RC circuit, but for many it wont give an intuitive
understanding of a capacitor.

Personally I found this article a good intuitive explanation of a capacitor,
http://amasci.com/emotor/cap1.html

In your model train situation the capacitor is best understood as a
reservoir for your powersupply during short blackouts. A typical LED
requires about 20mA to yield good luminosity, and to achieve that typically
about 1.7 volt is required across the LED (unless you have a high intensity
LED). Your capacitor of 400uF will only be able to provide this kind of
current in very short bursts. To determine this you can check something
called the time constant (R*C) for discharging your RC circuit. After R*C
seconds the voltage across your cap will be reduced to about 37% of original
value.
Example: if your train runs of 15V, then Ohms law yields that your resistor
is probably around (15-1.7)/20, or 650 ohm. That gives a time constant of
about 0.2 seconds when your capacitor is 400uF. So after 0.2 seconds the
current through your LED will only be about 6mA. But since the blackout
zones probably lasts significantly shorter than this your LED looks nice and
stable. During train slowdown the blackout zones will obviously last longer,
possibly discharging your cap and turning off your LED in brief periods.

Other prevalent ways of using caps are in filters, amplifiers and radio
circuits, where they are used to couple/bypass signals within a certain
frequency domain.

regards,
erik

5. ### RyanGuest

What do the Farad ratings mean?

This may be a terrible analogy, but it is the way I understand it.

I think of a capacitor as analagous to a water pressure tank in a well
water system. If you live in the country and use a well, then you
realize that the pump in the ground supplies water pressure, but only
when the pump is running. When we open a faucet, we don't want to have
to wait on the water pump to get started and develop pressure. To solve
this problem, we insert into the line a holding tank. This tank acts as
a buffer to smooth out the transitions when our source stops and starts
again. This way the output is held in a slower-to-change state rather
than quickly going off and then back on again.

The size of this tank (it's capacitance) can be varied from large to
small. If we make is smaller then it charges more quickly but
discharges more quickly too.

This container is rated for a certain amount of pressure. Too much
pressure and a wall will burst and it will not hold pressure. In a
capacitor, too much voltage will pierce through the insulation between
the many layers inside and it will not hold a charge. My example stops
here as a capacitor is not a big empty tank, it is a bunch of
alternating layers of insulation and electrolytic.

I would say farads and voltage aren't really related, but the two have
an influence on the size of a capacitor. This electrical insulation
takes space. It take mores insulation to prevent an arc when a higher
voltage is used, so this means that a lot of the volume of a capacitor
is occupied by insulation. Conversely, you could have the same volume
of capacitor which has a significantly larger farad rating, but can only
be used at just a few volts, or else it becomes damaged. All else
equal, a low capacitance, low voltage unit will be smaller.

As I understand it, you could use a 25v, a 200v, or a 2000v capacitor
and as long as it was 400uF, you should expect the same result in a
circuit. (unless a 2000v cap introduces some non-conforming properties)

Whoever helped you probably calculated the capacitor based on knowing
the voltage which you provided, and knowing how much current an LED will
use (and then using a resistor to limit the current), and then assuming
what duration of time this LED will need to be powered by the capacitor
(thereby deciding how many farads based on the current draw while also
remembering not to let the voltage drop so low that the LED goes out).

6. ### Jasen BettsGuest

they behave like springs in a mechanical device

- soaking up bumps,
- thec can also be used to block DC - because a capacitor won't pass DC but
will pass AC.
the softness of the spring, how much electtricity will flow into (or outof)
the capacitor for every volt of charge.
how many volts they can survive.
only in that more volts or more farads means a physically larger device
(also capable of storing more energy)
the time it takes to charge them.
it depends on how they are connected to the rest of the circuit.
I think they also put something between the capacitor and the led to protect
it from the 16V track voltage.

Bye.
Jasen  