# 9.5 questions on capacitor, LED, resistor

Discussion in 'Electronic Basics' started by [email protected], Oct 21, 2005.

1. ### Guest

1. Fast charging capacitors-Possible? Is the charging speed of a
capacitor determined by the input voltage and current?

2. I don't understand the differences between these myriad of capacitor
chemistries. But I'm looking at aerogel caps. Does output voltage stay
constant while current varies?

3. Also, do capacitors accept charge from an a/c source or will it need
a diode?

4. LEDs require a resistor, but how do you calculate the required
resistance if you connect the LEDs in series?

4.5. Then the voltage has dropped across each LED, yes?

5. What determines the capacity of a resistors resistance, i.e., what
is the max voltage/amperage input of a resistor with impedance x. Or do
I have it all wrong and is impedance rated at the specified input
power?

6. Is the power sent through a resistor merely attenuated or burnt off
as heat?

7. Will a resistor pass proportionally more power through if less than
the power the resistor is rated at is experienced?

8. Can caps handle overvoltage/overcharge (unlike a battery)?

9. Are cap chemistries what determine a caps discharge rate? Are carbon
aerogel caps "fast?" What do they use for railguns?

And because I'm too lazy to look it up, is there a simple way to figure
how many watts/hour a cap of capacitance xF will put out before
draining?

2. ### John GGuest

Too Many questions Toomany mistaken ideas

Start here. http://www.ibiblio.org/obp/electricCircuits/index.htm

3. ### Rheilly PhoullGuest

LOTS OF LUCK WITH THE COURSE !!
With your attitude I don't think ya gonna make it Then again you're natural talent as a troll could be an alternative
occupation.

5. ### Tim WilliamsGuest

Uh...yeah...
I = C * dV/dt.
Huh!? By definition, a capacitor's voltage will always vary in proportion
to the current!
If you connect a capacitor, say 0.1uF, to the 120V line, the voltage on it
will be 120V AC, which I think even you can comprehend. If you disconnect
it at any random moment, the voltage left on its terminals will be equal to
the voltage at the instant you disconnected it. You have a random chance of
getting peak, i.e., 160V left on the capacitor, but you also have the same
chance of getting zero or a negative value.

A diode, however, will ensure that current is always in the same direction,
not random.
No. First you assume each LED has the same terminal voltage, independent of
current. This is a "bad" assumption, but it works remarkably well for LEDs
(and other diodes). Since voltage is independent of current, you merely
subtract the total LED voltage drop from the supply voltage and divide the
remainder by the desired current to find your resistor.
Not impedance, resistance.

There are three limits on all components: voltage, current and power. Too
much voltage can cause electrons to go where they shouldn't, for example
high voltage loves to jump across air between wires that you might want
insulated. Too much power (P = I*R) causes things to burn. Too much
current causes too much power to be spent somewhere it doesn't belong, wires
for instance.

The voltage limit of a resistor is V = sqrt(P*R), and the respective maximum
current is I = V/R = sqrt(P*R)/R = sqrt(P/R), which you'd know if you looked
at a damn Ohms Law table.
I don't know what that was supposed to mean.
Conservation of energy, it "wastes" it as heat. It also can waste ONLY as
much power as it is sent, i.e., volts times amps.
Almost all capacitors are rated for some sort of overvoltage; electrolytics
might be rated for a mere 10% while film capacitors are often rated for 50%
extra, or more. Self-healing caps can withstand even more, if you don't
mind weeding them out first!
If chemistry determines ESR (equivalent series resistance) and ESL
(inductance), then yes...

But it doesn't.
I don't know, but I doubt it. Faster than a battery but slower than say, an
electrolytic.
High voltage film capacitors. They build them specially to handle the
extreme currents required in a railgun. Likewise, such capacitors can
discharge very quickly, under a microsecond.

Oh, and learn to count too- that was two questions there.
I don't know what the hell watts per hour are. Did you mean watt-hours?

The equation for energy stored in a cap is E = 1/2*C*V^2, where E is in
joules, C is in farads and V is in volts.

There are 3,600 joules in a "watt-hour" (i.e., one watt consumption for a
duration of one hour).

Tim

6. ### John PopelishGuest

The relationship between currnt and voltage for any device that acts
like a capacitor is I=C*(dV/dt) or current equals capacitance times
the rate of change of voltage. The larger the current, the faster the
voltage changes. The large the capacitance, the slower the voltage
changes.
No. The voltage changes at a rate proportion to the current passing
through the capacitor. The voltage may also change instantaneously in
proportion to the magnitude of the current, because the capacitor will
also include some internal resistance.
If you want the capacitor to charge up in one direction, you will need
a diode or other means to keep it from swinging its voltage back and
forth in response to an AC current.
LEDs drop a roughly constant voltage over the normal operating range
of current. You select the current you want, subtract the normal
operating voltage from the available voltage, and divide that excess
voltage by the desired current. This will give you a value of
resistance that will consume all the extra voltage when the desired
current is passing through both the LED and the resistor.
I think you are confused about the meanings of voltage and current.
Voltage is applied across a pair of nodes, and is a force that moves
charge through the paths between those two nodes. Current is the
moving charge.
Resistors have a power limit, based on how hot a given power
dissipation causes them to get and the thermal limit of the material
they are made of. They also have a voltage limit based on the
resistor element arching over or other nonlinear things happening.
Current is sent through, power is dissipated within. It is released
as heat into the surroundings.
The essence of resistance is that it passes a current proportional to
the voltage applied across it. Ohms are a unit that just means volts
per ampere. As long as you keep the product of volts and amperes
lower than the rated power (and briefly if you exceed this) you can
expect this proportionality to be quite constant, but not perfectly so.
Caps can be destroyed by over voltage, caused by internal arching when
the dielectric layer gets a hole blown in it.
Not really. The resistance of the electrode system that coats the
dielectric layer determines how high its resistance is, so how much
heat is produced by charging or discharging current, and how much
voltage is used up to create that heat.
That's easy. The answer is zero. ;-)
At least if you define draining as any drop in voltage. The capacitor
voltage must change in order for current to pass through it.

The formula for energy in a capacitor is
watt seconds = 1/2 * C * V^2
With the C in farads and V in volts.

Pick a capacitance and a starting voltage and find the starting energy.
Pick and ending voltage and find the energy still left in the
capacitor when the voltage is unusable. The difference between those
two energies is the usable energy you can get from that capacitance as
the voltage falls over that amount.

Be prepared for disappointment.

You will soon find out why chemical batteries have not been replaced
by capacitors except for a few very short duration or very low power
applications.

7. ### Jasen BettsGuest

The charge in a perfect capacitor is proportional to the voltage across it.

real capacitise have a finite but small internal resistance so there is a
slight lag and the charge never reaches 100% (but it gets close enough for
all practical purposes)
not in any capacitor I've seen, I've not looked at aerogel capacitors.
if you want it to stick you'll need a diode.
add the voltages of the LEDs and then use the same formula.
its wattage rating.

voltage squared divided by resitance shouldn't exceed that figuure
it's convertted to heat.
no from the same supply equak resistances will pass the same current
overcharged capacitors may fail spectaacularly.
They do handle undervoltage much better.

a 12V capacitor with 9V of charge in it still holds over half of it's
fully charged energy.

a 12V battery in the same condition holds less than 1 percent
2
E=0.5V C

or in words

E = half V squared times C

E in joules , V in volts C in Farads.

since a joule is a watt-second
and there are 3600 seconds in an hour
divide E by 3600 to get watt-hours

(watts/hour is meaningless in this context)  