Uriah said:
Believe it or not, I have been into electronics for a long time but
for some reason some things don't sink in. Like what exactly is
capacitance? I know it is measured in ohms.
Capacitors have impedance, a frequency dependent
relationship between the voltage across them and the current
through them. Impedance is two dimensional, and can be
expressed as magnitude (of ohms) and phase shift (polar
form), versus frequency, or in artesian form as real and
imaginary parts of the total.
Capacitance, itself, is measured in farads. A farad of
capacitance experiences a 1 volt change across its terminals
for each coulomb of charge that passes through it. Since a
coulomb per second is called an ampere, we can say that an
ampere passing through a Farad of capacitance changes the
voltage across it by a volt, every second.
As a formula, this is I=C*(dv/dt), where I is current in
amperes, C is capacitance in farads, and dv/dt is the rate
of change of voltage in volts per second.
I know it is like a
frequency dependent variable resistor.
Sort of, except that capacitance does not consume energy,
but stores it and gives it back, depending on whether the
voltage is rising or falling in magnitude. The energy
stored in a cap is energy=1/2 * C * V^2, where energy is in
watt seconds or joules, C is capacitance in farads, and V is
the volts across the terminals.
But when the capacitance goes
up in uF what is going on with the electrons? Does it hold more
Electrons?
It holds more electrons shifted from one plate to the other
with less applied voltage. Think of the capacitor as a
rigid tank with a rubber membrane stretched across the
middle, and a pipe into the vessel on each side of the
membrane. In this analogy, the pressure difference
(voltage) between the two pipes (terminals) is proportional
to the amount of fluid removed from the volume on one side
of the membrane and forced into the volume on the other side
of the membrane. The deformation of the membrane is what
stores the energy. In a capacitor, it is the electric field
between the plates that stores energy as the dielectric is
stressed with voltage.
Or is the field between the plates holding a higher
charge?
The charge is localized to the surface of the plates, but
the energy is stored in the electric field between the
plates these surface charges produce.
If you have a 1uF cap and a 1pF cap and you send a sign wave
through it of a particular frequency of 1k. What is happening that is
different between the two?
If the voltage applied between their terminals is similar,
as it would be if they were connected in parallel, the 1uF
cap would be passing a million times more current than the 1
pF cap. In other words, the impedance of the 1 uF cap is a
million times lower than the impedance of the 1 pF
capacitor. The energy stored at any voltage is also a
million times higher.
But if you connect these two capacitors in series, and pass
the same current through each, this is like connecting a
very low and a very high ohm resistor in series. Most of
the total voltage drop occurs across the higher value
resistor, and across the very high impedance capacitor,
which is the smaller one. The magnitude of capacitive
impedance is ohms = 1/(2*pi*f*C), where f is frequency in
hertz, and C is capacitance in farads. Ohms is a word that
means volts per ampere.
They both charge to the same potential? Do
the discharge at a different rate?
If they are forced to discharge their voltage at the same
rate (volts per second) the 1 uF cap will deliver a current
a million times larger than the 1 pF cap does during this
discharge. If they are drained at the same current, the 1
uF cap will have a discharge time a million times longer
than the 1 pF cap does.
I can't see the forest for the
trees, I think, what ever that means.
The main thing that complicates capacitors that is missing
from the understanding of resistors is time. You cannot
think clearly about what capacitors do in a circuit or in
isolation, if you don't include time. The only thing you
can say without referring to time is the instantaneous
energy stored in the cap, that is purely a function of
capacitance and voltage.
When you are testing something with your DVM and you have it set on
the diode setting. You usually get a .400 number and when reversing
the leads you get an open. What is the meter doing?
(snip)
It is delivering a regulated current and displaying the
millivolts of drop that current produces as it passes
through the leads and the component they are connected to.
The reading usually goes off scale if the leads are open
circuited. You can use a separate volt meter to measure how
high this voltage goes, open circuit. Something around 2
volts is common (2000 on the meter). Silicon diodes may
drop, between .3 and .6 (300 to 600 millivolts) when
passing this small (and also measurable with a second
milliamp meter) current.
If you test a large capacitor, you should see something near
zero (uncharged) that rises smoothly to the meter overload
(cutoff) as the current ramps up the capacitor voltage, as
charge is stored in the cap. Not a good idea to test the
charge capacitor in reverse, because the meter is not
expecting its voltage to go negative. Touch the capacitor
leads together to dump the stored charge, to repeat the test.
Once you calibrate this capacitor test (zero to cut off time
for a given capacitance) you can roughly estimate other
capacitances by their charge time, relative to this one.
For instance, if it takes 2 seconds for a 100 uF cap to
charge to meter cut off, it should take about 20 seconds for
a 1000 uF cap to do the same thing.
