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.