Patrick said:

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.

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.

What is there purpose in an energized circuit?

I don't understand the question.

What do the Farad ratings mean?

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.

What do the Voltage ratings mean?

The maximum voltage the capacitor can withstand without damage.

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

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.

How does amperage effect capacitors?

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.

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?

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.

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

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.

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.

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,

and no train loading the voltage down.

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.