Wow. That was a really helpful response. Fantastic, really. Thank
you for taking the time to write that.
No problem.
After reading your post, I took my multimeter and measured the voltage
across the capacitor when the circuit was open, and sure enough, there
was a transient voltage of ~1.65V left in the capacitor from previous
charges.
Yep, thats what a capacitor(and inductors too) does... it stores charge. It
charges and discharges in a special way(not really but it looks "special").
Capacitors can be used as batteries in some sense too(but batteries tend to
have constant voltage for the majority of there life and last much longer).
One thing that struck me about what you were saying is that it seems
impossible for the capacitor to reach 9V of potential, because that
would mean the LED's branch would also have a 9V drop, leaving nothing
for the 10K resistor to drop. After thinking about that for a moment,
I measured the voltage across the capacitor when the switch was closed,
and I noticed that the voltage would rise quickly to about 2.53V, and
stay there. I left the circuit closed for two hours just to see if it
would rise any higher, and it stayed at 2.53V. I started wondering if
it would be possible to figure out the maximum voltage the capacitor
could reach in this setup.
Sure, its similar to the analysis I did for the simple circuits but involes
a little more algebra(the calculus part is pretty much identical but just
looks more complicated).
Anyway, this is what I was thinking: (this is long and could be totally
incorrect)
If you replaced the capacitor and LED branches in the original circuit
with a short, you would get a circuit that looked like this:
+-------------------+
| |
| |
| |
---- 9V Battery | Removed capacitor and LED branches
-- |
| |
| |
| |
+-------VVVVV-------+
10K
In this case, the current through the modified circuit would be 9 V/10
000 ohms = 0.0009A = 0.9 mA.
Now, at their lowest, the resistance (impedance?) of the capacitor and
LED branches in the original circuit would be 0 Ohms (a short). So the
0.9 mA could essentially be considered an upper bound on the current
flowing through the original circuit.
Impedence is the total restriction of flow. Resistance is basicaly what a
resistor does and reactive is like a resistance but it is not due to the
same mechanisms that resistors use and it usually depends on frequency(maybe
always). Impedence means exactly what it means... it impedes the flow of
current. Capacitors impead the flow too but in different way than
resistors... there "resistance" is actually changing over time(even the
ideal capacitor has this) and is also dependent on frequency(unlike an ideal
resistor). Basicaly impedence covers everything while reactance is used for
something that is "reactive"(capacitors and inductors) that has no
resistance... resistance is things like resistors. Its kinda circular the
way I gave it but the point I'm trying to make is impedence = reactance +
resistance. Ideal capacitors and inductors have reactance only and ideal
resistors have resistance only... you combine a circuit with them and you
get an impedence.... although you can refer to resistance and reactance as
impedence if you wish since impedence implies each of them too.
Yes, in the circuit above it is the maximum current that will ever flow when
you add more stuff(unless you add a power source)... adding passive elements
can only reduce the total current and stuff.
your 0.9mA is at t = 0.. at t = 0.0001, say, it might be 0.88998mA or
something like that.
(put capacitor back into circuit for what follows but leave out the other
branch with the diode)
the ideal is given then t = 0 here
Vc(t) = exp(-t/(RC))*(Vc0 - Vb) + Vb
Ic(t) = (Vb - Vc(t))/R
or with Vc0 = 0, Vb = 9v, R = 10k, and C = 1uF
hence R*C = 1/100 and
Vc(t) = 9v*(1 - exp(-100*t))
Ic(t) = 9v*exp(-100*t)/10k = 0.9mA*exp(-100*t)
these are the equations that tell you exactly(in the ideal case) what the
voltage and current are through the cap at any time t.
-------------------------
if you are not familiar with the exponential function it is not a difficult
concept. You just have to know a few properties of it.
exp(anything) >= 0
exp(0) = 1
exp(some really large positive number) ~= some much larger positive number
exp(some really large negative number) ~= 0
so
exp(0/192883 - 0*9843) = exp(0) = 1
exp(39849389289) = something really big and much larger than 39849389289
exp(-39849389289) ~= 0
exp(-t) starts at t = 0 with exp(0) = 1 and decays to exp(-10) in a special
way but basicaly you just need to know that as time increases the thing gets
smaller but does it at a slower and slower rate.
another thing to note is that exp(anything) ~= 2.7^(x)
check out
http://en.wikipedia.org/wiki/Exponential_function
or
http://ourworld.compuserve.com/homepages/g_knott/elect38.htm
for how it looks and a better explination.
-------------------------
Back to the circuit and the equations we can see that
Ic(t) = 0.9mA*exp(-t/100)
then Ic(0) = 0.9mA*exp(-0100) = 0.9mA*exp(0) = 0.9mA
then at t = 100s
Ic(100) = 0.9mA*exp(-100/100) = 0.9mA*exp(-1) = 0.9mA/e ~= 0.9mA/2.71 ~=
0.3mA
so after 100 seconds with a 1uF capacitor and 100k resistor and the 9V
battery there will be a 0.3mA current going through the capacitor
you can also see that if t = n*RC then
Ic(n*RC) = 0.9mA*exp(-n*RC/RC) = 0.9mA*exp(-n) = 0.9mA/exp(n)
so after one time constant(the time constant though depends on the circuit
and isn't always R*C but could be much more complicated) one has
Ic(RC) = 0.9mA*exp(-1) = 0.9mA/e ~= .331
so we can find out the percentage it dropped in 1 time constant by taking
the ratio of Ic before and after
Ic(RC)/Ic(0) = (0.9mA/e)/(0.9mA) = 1/e ~= 36% hence it dropped by 1 - 36% =
64%.
there for the current is 64% smaller after 1 time constant... you can do the
math to get what it is at after n time constants and its just 1/e^n. If you
want it at 50%, say, then you have to do a little more math though and you
would be dealing with fractions of time constants and it wouldn't be so
nice. The reason time constants are important is that they are independent
of the actual circuit... if you know the time constant of the circuit then
you know after one time constant whatever you are measuring will be 64%
smaller.
just so you know,
1/e ~= 36%
1/e^2 ~= 14%
1/e^3 ~= 5%
1/e^4 ~= 2%
1/e^5 ~= 0.7%
so you can see how the rate gets slower and slower. between the t = 0 and t
= RC it drops 64% and between the next time constant t = RC and t = 2RC it
drops only about 22% more. This is why, I suppose, they have the 5 time
constant rule thing... you would actually, theoretically, have to wait
forever to get a drop of 100%.
Back in the original circuit, by Kirchoff's law, if (I <= 0.9mA) then
(I_c + I_led <= 0.9mA). So at most, the current through the LED's
branch can be 0.9mA. Here I'm assuming I_c isn't negative... Maybe
that's not a safe assumption.
right, because the cap acts like short at t = 0 we have the simple circuit
with just Vb and R in series and nothing else... so it will be supplying
0.9mA's at that moment you flip the switch. I_c will never be negative cause
it will never be able to out do that battery(else it would charge it and
hence it would be supplying current... that is though only if the cap
doesn't have a larger voltage across it from being previously charged... but
its still ok long as the voltage across it isn't larger than the
battery(else it would be stronger and be able to force current through it).
So you wouldn't want to charge the cap up with a 100V battery say then stick
it in the circuit with the 9v battery cause it might screw up that battery).
Normally one can assume current is going in the wrong direction without
causing to many problems as long as one is consistant. If you do this, say,
mathematically and assume the wrong direction you always will get the a
negative in front of your current when you solve for it... i.e., if you
assume I goes one way you will eventually get I = -|I| which means you
guessed wrong. Even if you measure the current by putting the leads in the
wrong way you will get just a - of what you thought. While its important to
get the right sign in a qualitative analysis so it makes sense it always
happens to come out right in the math unless you make a mistake somewhere.
In this case though you just have to think how the 'electrons' are flowing
to see that the cap will not be able to change the direction of current when
the battery is hooked up. Remember, just think about the two extreme cases
of a capacitor when it is completely discharged(a short) to when it is
completely charged(open)... all the stuff inbetween is some a smooth
transition from one to the other.
i.e., say I have some complicated circuit with tons of resistors and
capacitors. If the circuit is in equalibrium(for DC) then it means that all
the capacitors are either acting as a short or an open circuit. Its just a
simple matter of figuring out which and then simplifying the circuit to get
the new circuit. Note that not all circuits will be in equalibrium because
say there will be a switch like you have that will keep on switching on and
off causing the capacitor to charge and discharge... or who knows what else
could be happening.
So if the maximum 0.9mA was running through the LED's branch (the
branch includes both the LED and the 1 Kohm resistor), it that would
mean that, at most, 0.9mA is running through the 1 Kohm resistor.
Using this you could calculate the maximum voltage drop across the
resistor:
V_s : resistor voltage drop
R_s : resistor resistance
I_s : resistor current
I_s <= 0.9mA
V_s/R_s <= 0.9mA
V_s <= 0.9mA * R_s
V_s <= 0.9mA * (1000ohm)
V_s <= 0.9V
Now if you add the voltage drop of the LED, or about 2.0V, you get 2.0V
+ 0.9V = 2.9V dropped on the LED's branch. This could be considered
the maximum possible voltage drop, because the LED's voltage drop
wouldn't change much for this current range.
Since the capacitor's branch and the LED's branch must drop the same
voltage, the capacitor has the same maximum voltage drop as the LED
branch, or 2.9V.
Anyway, 2.9V seems to be pretty close to the actual value of 2.53V.
Is anything I wrote plausible? At the moment, most other kinds of
analysis are way over my head for RC circuits.
yeah, it makes sense. You are looking at the maximum(final) case which only
occurs after a long time but it is useful to know because you need to make
sure you don't burn anything up.
that is, initially the current all goes through the "shorted' capacitor but
it will slowly become an open circuit causing more and more current to go
through the LED.. after an infinite amount of time all the current will go
through the LED and the capacitor will be an open circuit... to calculate
the current we need just to analyze the circuit without the capacitor(i.e.,
it opened),
I-> R1
+-------(switch)-------+----------VVVV--------+
| |
| |
---- V Battery |
-- (LED) VL
| |
| |
| |
+----------VVVV--------+----------------------+
R2
so V - I*R1 - I*R2 - VL = 0
or
I = (V - VL)/(R1 + R2)
R1 = 1k
R2 = 10k
VL = 2
V = 9
so
I = (9 - 2)/(11k) = 7/11k ~= .63mA
the voltage drop then across the R1 is
..636mA*1kOhm = .636V
the voltage drop across R2 is
..636mA*11kOhm = 6.36V
the voltage drop across the LED is 2V
hence the voltage drop across the cap is
2V + 0.63V = 2.63V
(the reason you got 2.9 is you took into account the wrong current through.
You took the initial current instead of the "final")
i.e., we have three circuits we are dealing with
for very large time t(say 5*RC)
+----------------------+----------VVVV--------+
| |
| |
---- V Battery |
-- (LED) VL
| |
| |
| |
+----------VVVV--------+----------------------+
R2
and for very short time t(t = 0)
+----------------------+
| |
| |
---- 9V Battery --- 1 mF Cap
-- ---
| |
| |
| |
+----------VVVV--------+
R2
(we can forget the switch as it is used in the discharge part which gives
you the circuit
R1
+----------VVVV--------+
| |
| |
--- 1 mF Cap |
--- (LED)
| |
| |
| |
+----------------------+
but this is an easy circuit to analyze to some degree and only depends on
the initial charge of the capacitor)
The "mistake" you made was to to take the current in the time t = 0 circuit
and use it in the time t = infinity one... the problem is that its not the
same current... its not close but. Its true that the current will never get
above 0.9mA in the LED but it doesn't even get close to 0.7mA(well, 0.63
might be close to 0.7). So your upper bound is a bit exaggerated but might
be fine if you just need an approximation.
i.e., as I mentioned before, basicaly what you do in situations like this is
that you short the cap and reduce the circuit and figure out the values of
the quantities you want to know then you open the cap and do the same and
get the new values. What you then know, atleast in a simple circuit like
this, that there is a "smooth" decay or growth from one value to the other.
If V(0) = 10 and V(100) ~= 0
then I know that V(t) smoothly(exponentially) decays from 10 to 0 over that
time between 0 and 100 seconds.. we could linearly say something like this
then as a first approximation
V(t) = 10 - 10*t/100 for t between 0 and 100 else v(t) = 0
ofcourse we already know it decays exponentially but this is just a quick
way to get some idea. in actuality it would be something like
V(t) = 10*exp(-t/TC) for TC = the time constant which we could find a upper
bound for(if its to large, say TC = 100 then we know it will not be 0 but
only 3.67 after t = 100 but its suppose to be 0, hence the TC for this
hypothetical circuit is probably at most about TC = 10 since exp(-100/10) ~=
0.0004 which is close enough to zero for me)
Thank you in advance for wading through that.
I hope that makes some sense and is helpful. It seems like you got the gist
of how it works which is probably good enough for most basic things. Just
remember that many circuits depend on the the way a capacitor works to do
some "cool" stuff. They usually use the way a capacitor charges and
discharges to accomplish this. Another aspect of this charging/discharging
ability is that it can act as a filter for high and low
frequencies(depending on how its used)... among other things.
Hence, if you wanted to know the exact time the LED would be "on" you would
need to know how the capacitor is charging and discharging. This can be done
relatively easy by finding the equations and and figuring out when the
voltage drop across it when the switch was opened and also when there was
enough voltage across it so that there would be enough current going through
the LED(here this is an simply relationship... the higher the voltage across
the capacitor means the higher the current going through the LED)... then we
would have to know when the switch was opened which means the capacitor will
discharge through the LED... it will take a certain length of time for hte
current to drop below the amount needed to power the LED. (the nice thing
about the mathematical way is that you would know it very precisely for any
R's and C's instead of having to guess how it depends on them).
Anyways,
AD