# Simple capacitor circuit (slowly illuminating LED)

L

#### [email protected]

Jan 1, 1970
0
I recently breadboarded this simple circuit in an attempt to better
understand how capacitors work (fixed-width font required):

1K
+-------(switch)-------+----------VVVV--------+
| | |
| | |
---- 9V Battery --- 1 mF Cap |
-- --- (LED)
| | |
| | |
| | |
+----------VVVV--------+----------------------+
10K

Before running the circuit I hypothesized that when the switch closed
the LED would stay dark until the capacitor was fully charged [(5 time
constants)*CR = 5*10 = 50 seconds], then light up at full intensity. I
also predicted that the LED would fade away when the switch was opened
again. My prediction was only partially correct...

When I closed the switch, the LED stayed dark momentarily, then
gradually illuminated to full intensity and stayed lit until I opened
the switch, at which point it slowly faded away.

What I would like to know is: why did the LED slowly illuminate like
that? It seemed like there was some sort of change in "resistance" in
the capacitor, in that it **behaved** like a short at first (all the
current seemed to go down its branch), then started behaving more and
more like an open circuit (more and more current started going down the
LED's branch). Is it some sort of voltage/current relationship that I
have overlooked? I've seen the curves for voltage/current for
capacitors discharging and charging, and that seems to make sense, but
I guess I don't understand why the capacitor's current affects the
LED's current in the way that it does. Or does this have something to
do with reactance/impedance or something (I haven't gotten that far)?

Thank you very much for your help. This group is really awesome. Two
months ago I didn't know anything about electronics, but I've learned a
lot from the posts on here. I even managed to repair a control board
for our Maytag Neptune washer (which will only be used if the new
control board fries...and then only if I'm standing right beside the
machine with a fire extinguisher).

P

#### Pooh Bear

Jan 1, 1970
0
I recently breadboarded this simple circuit in an attempt to better
understand how capacitors work (fixed-width font required):

1K
+-------(switch)-------+----------VVVV--------+
| | |
| | |
---- 9V Battery --- 1 mF Cap |
-- --- (LED)
| | |
| | |
| | |
+----------VVVV--------+----------------------+
10K

What do you mean by 1mF btw , 1000 uF ? mF is a deprecated measure since
it's confused by the fact that Americans once used it to mean uF.
Before running the circuit I hypothesized that when the switch closed
the LED would stay dark until the capacitor was fully charged [(5 time
constants)*CR = 5*10 = 50 seconds],

No why so ?
then light up at full intensity.

No. It'll gradually illuminate as the cap charges. It won't 'suddenly come
on'. Why did you think so ?
I
also predicted that the LED would fade away when the switch was opened
again.
Yes.

My prediction was only partially correct...

When I closed the switch, the LED stayed dark momentarily, then
gradually illuminated to full intensity and stayed lit until I opened
the switch, at which point it slowly faded away.

That's what I'd expect.
What I would like to know is: why did the LED slowly illuminate like
that?

Beacuse the current through it gradually increased with the cap voltage (
as the cap charged via the battery's internal resistance ) of course.
It seemed like there was some sort of change in "resistance" in
the capacitor, in that it **behaved** like a short at first (all the
current seemed to go down its branch), then started behaving more and
more like an open circuit (more and more current started going down the
LED's branch).

Sort of. It's called reactance.
Is it some sort of voltage/current relationship that I
have overlooked? I've seen the curves for voltage/current for
capacitors discharging and charging, and that seems to make sense, but
I guess I don't understand why the capacitor's current affects the
LED's current in the way that it does. Or does this have something to
do with reactance/impedance or something (I haven't gotten that far)?

Yes. It'll be clearer perhaps if you included a resistor between the
battery and the cap representing the battery's internal resistance.

Graham

A

#### Andrew Holme

Jan 1, 1970
0
I recently breadboarded this simple circuit in an attempt to better
understand how capacitors work (fixed-width font required):

1K
+-------(switch)-------+----------VVVV--------+
| | |
| | |
---- 9V Battery --- 1 mF Cap |
-- --- (LED)
| | |
| | |
| | |
+----------VVVV--------+----------------------+
10K

Before running the circuit I hypothesized that when the switch closed
the LED would stay dark until the capacitor was fully charged [(5 time
constants)*CR = 5*10 = 50 seconds], then light up at full intensity. I
also predicted that the LED would fade away when the switch was opened
again. My prediction was only partially correct...

When I closed the switch, the LED stayed dark momentarily, then
gradually illuminated to full intensity and stayed lit until I opened
the switch, at which point it slowly faded away.

What I would like to know is: why did the LED slowly illuminate like
that?

The brightness of the LED depends on the current flowing through it,
but current doesn't start flowing until the voltage across it reaches a
threshold. Look at the current versus voltage characteristic curves of
LEDs e.g. http://www.telatomic.com/art/pepc5.jpg

As soon as the minimum threshold is exceeded, LED current will begin to
flow "stealing" current from the capacitor. The capacitor continues to
charge, and the voltage continues to rise, but not as fast as it would
if the LED wasn't there.

The threshold voltage for LEDs is typically between 1 and 2 volts,
whereas for silicon PN junction diodes (e.g. 1N4148) it is around 0.7V.

N

#### Nick

Jan 1, 1970
0
Yes. It'll be clearer perhaps if you included a resistor between the
battery and the cap representing the battery's internal resistance.

Compared to the 10K resistor already in-circuit, I doubt it'd have any
significant effect...

P

#### Pooh Bear

Jan 1, 1970
0
Nick said:
Compared to the 10K resistor already in-circuit, I doubt it'd have any
significant effect...

Ahh - ok - for some reason ( the usual trouble with ASCII circuits ) That
10k wasn't clear.

Graham

L

#### Lacy

Jan 1, 1970
0
You are correct in your observation and explaination. The Cap acts a a very
low resistance at first and as it charges up the resistance increase
allowing more current and Voltage across the LED. Eventually the Cap will
reach a point where it will no longer allow DC to flow in its branch.
Therefore, all the volatage and current will pass to the LED. When you turn
off the swith the Cap will continue providing voltage to the led until the
voltage bleeds off the capcitor below the foward voltage requirement to
light the LED.

Before I purchased a Cap Meter, I used a modified version of this to check
for a defective Capacitor. Put the LED in Series with the Cap and it will
remain lit until the Cap charges and then slowly dim until it goes out if
the capcitor isn't bad and leaking. If it is leaking, the LED will remain on
dimly and not go out..

I tried to explain this in normal Human English. I hope I didn't confuse you
and hope this helps. You are on the right track to learning and seem to have
good observation skills. I hope you good luck in your studies.

L

#### [email protected]

Jan 1, 1970
0
Pooh said:
What do you mean by 1mF btw , 1000 uF ? mF is a deprecated measure since
it's confused by the fact that Americans once used it to mean uF.

Yeah, I meant 1000 uF. Sorry about that. I remembered that uF was the
preferred unit after I had posted the original message.
No. It'll gradually illuminate as the cap charges. It won't 'suddenly come
on'. Why did you think so ?

I really didn't know what was going to happen... I thought I had read
somewhere that a capacitor behaved like a short, and I figured it was
good enough for a prediction. I must have misread whatever I had been

A

#### Abstract Dissonance

Jan 1, 1970
0
I recently breadboarded this simple circuit in an attempt to better
understand how capacitors work (fixed-width font required):

1K
+-------(switch)-------+----------VVVV--------+
| | |
| | |
---- 9V Battery --- 1 mF Cap |
-- --- (LED)
| | |
| | |
| | |
+----------VVVV--------+----------------------+
10K

Before running the circuit I hypothesized that when the switch closed
the LED would stay dark until the capacitor was fully charged [(5 time
constants)*CR = 5*10 = 50 seconds], then light up at full intensity. I
also predicted that the LED would fade away when the switch was opened
again. My prediction was only partially correct...

When I closed the switch, the LED stayed dark momentarily, then
gradually illuminated to full intensity and stayed lit until I opened
the switch, at which point it slowly faded away.

What I would like to know is: why did the LED slowly illuminate like
that? It seemed like there was some sort of change in "resistance" in
the capacitor, in that it **behaved** like a short at first (all the
current seemed to go down its branch), then started behaving more and
more like an open circuit (more and more current started going down the
LED's branch). Is it some sort of voltage/current relationship that I
have overlooked? I've seen the curves for voltage/current for
capacitors discharging and charging, and that seems to make sense, but
I guess I don't understand why the capacitor's current affects the
LED's current in the way that it does. Or does this have something to
do with reactance/impedance or something (I haven't gotten that far)?

Thank you very much for your help. This group is really awesome. Two
months ago I didn't know anything about electronics, but I've learned a
lot from the posts on here. I even managed to repair a control board
for our Maytag Neptune washer (which will only be used if the new
control board fries...and then only if I'm standing right beside the
machine with a fire extinguisher).

Two things to note about a cap that are important. First is that the voltage
across the capacitor is proportional to the charge it contains and second is
that the current through a capacitor is equal to the rate of change of
charge in the capacitor.

i.e.,

V = Q/C
I = dQ/dt = CdV/dt

(or I = CdV/dt and dQ/dt = I is equivlent to int(Idt) = Q (which is saying
the sum of the current I over time is equal to the total charge Q)

This is sorta like an ohms law for capacitors except the V-I relationship is
not simply linear but depends on the rate of change of something.

+-------(switch)-------+
| |
| |
---- 9V Battery --- 1 mF Cap
-- ---
| |
| |
| |
+----------VVVV--------+
10K

if we try and analyze this part of the circuit qualitatively what we can say
is when the switch is closed the battery supplies current to the cap at a
certain rate. Basicaly the electrons flow from the batter and get "stuck" on
one plate of the capacitor and electroncs on the opposite plate are "forced"
off and move to the other terminal of the battery.

So over time the more electrons get deposited on capacitor and create a
repulsion again more incoming electons. The battery is only so strong and
can only force so many. Eventually it gets harder and harder to add
electrons to the cap and the charge on the cap(amount of electroncs) are
simply the sum of all the electrons on that have entered the capacitor(cause
the capacitor is really an open circuit and the electrons couldn't have left
anywhere... hence they had to be stored).

Maybe an analogy is a bathtub filling with water. The charge is simply the
amount of water in bathtub and the current is the amount of water per second
comming out of the faucet. Now the knob that controls the water comming out
of the faucet is controled by how much water is in the tub. For a capacitor
its a very "special" relationship while for a human its much simpler one
where we just turn it off once it gets "full"... a cap would always be
slowly turning it off...

Anyways, we can find out quantitatively by analyzing the circuit above.

We see that the total voltage drop around the circuit must be 0, that the
capacitor "resists" the voltage of the battery (hence it will have opposite
sign), and the current I through the wire is the same throughout(cause there
are no branches for the current to "split" up).

i.e.(skip this if you don't understand... the final equations are whats
important),

Vb - CQ - IR = 0

Vb = voltage supplied by battery
CQ = voltage drop by capacitor
IR = voltage drop by resistor

but I = dQ/dt

so

Vb - Q/C - R*dQ/dt = 0

or

dQ/(Q - CVb) = -dt/RC

==> (solving the DE by integration)

Q(t) = exp(-t/(RC))*(Q0 - CVb) + CVb = exp(-t/(RC))*(Q0 - Qt) + Qt

where Q0 = initial charge(the stored charge on the capacitor from previous
charges)

CVb = total charge that the battery can put on the capacitor = Qt

or if we want the voltage across the cap then we just do(since Vc = Q/C)

Vc(t) = exp(-t/(RC))*(Vc0 - Vb) + Vb

and the current is(since I = dQ/dt)

Ic(t) = -exp(-t/(RC))*(Q0 - Qt)/RC = (Qt - Q(t))/(RC) = (Vb - V(t))/R

notice how IC(t) looks like ohms law except V(t) is changing with time
depends

Heres the equations again

Vc(t) = exp(-t/(RC))*(Vc0 - Vb) + Vb
Ic(t) = (Vb - V(t))/R

(note that exp(0) = 1 and exp(-infinity) = 0)
we can see at time t = 0

Vc(0) = Vc0
Ic(0) = (Vb - Vc0)/R

If there is no charge on the capacitor at t = 0 then we can see that Vc(0) =
0 and Ic(0) = Vb/R... this is exactly what one would have if C was shorted
act, although it only acts that way for an instant... We can also see that
if there is an inital charge(inital voltage drop) then this reduces the
inital current through it(which is what we would expect too).

as t -> infinity(or oo) one gets

Vc(oo) = Vb
Ic(oo) = 0

hence after a long time the voltage drop across the capacitor reaches Vb(the
voltage across the batter) and the current through ends up stopping(i.e.,
open circuit). The reason for this is that over that time period the battery
deposited its charge on the capacitor but eventually it cannot stick any
more electrons on it, it "turns out"(actually its defined that way) that the
number of electons it can put on there is due to the voltage it has(its
strength). It just gets harder and harder for the battery to do this in such
a way simply way(and exponentials are pretty simple when you consider it is
fundamental when considering dependencies on rates of change).

So what does the above example have to do with anything? Well you can use it
to assess qualitatively how your original circuit will behave. Notice that
if we hadd the new resistor and LED into it then what happens is we end up
allowing some current to flow through that branch. It will only do this
though if the LED is on(atleast idealy) which means there is a voltage drop
than the voltage rating of the LED + I*R where I is the current through
the added resistor R. But LED + I*R is the voltage drop across the capacitor
= Vc(t) (since its a parallel branch).

So when the switch is open we can see at first the extra branch(the LED and
resistor) will cause the cap to charge slower than it would without it...
eventually though the LED will turn on when the cap reaches a certain
voltage across it(enough to forward bias the LED) and then more current will
flow into the LED... this will farther cause the cap to slow its
charging(which is already happening because its a cap) and eventually the
cap will be completely charge and all the current will flow through the LED.
(and hence the total current through the LED would be from 0 to Vb/(R1+R2)
and change in an exponential way due to the capacitor).

You can analyze the original circuit quantitatively in the same I did but in
this cause you will get a slightly more complex problem due to two branches
and the extra voltage drop across the LED.

Note that when the switch is off though one has the simple equation for the
cap + R + LED

Vc - IR - VL = 0

or

Vc - dVc/dt/(RC) - VL = 0

or

dVc/(VL - Vc) = -t/(RC)

or

Vc(t) = -exp(-t/(RC))*(VL - Vc0) + VL

==> I(t) = (Vc(t) - VL)/R

We can then see when Vc(0) = Vc0 or that the cap has a voltage across it
just like a battery and I(t) = (Vc0 - VL)/R. as t -> oo one gets Vc(oo) = VL
and I(oo) = 0.

hence the capacitor will never discharge fully... (Also it will not
discharge at all if Vc(t) is never above VL).

Hope that helped in some way. The main point is to notice the way a
capacitor charges. The current throught he capacitor indicates in an
"inverse" way how much voltage is across it(inverse doesn't mean 1/x but a
sorta exponential inverse... i.e. if exp(-t) then its "inverse" is 1 -
exp(-t)).

You can also think, when the switch is closed in your original circuit, that
the capacitor is sorta acting as a time dependent drain on the current going
through the LED... at first the cap is draining away a lot of current but
then eventually it slows down and hence more will end up going through the
LED... enough for you to see it... over time more and more current will go
through the LED and it will get brighter and brighter(you'll need to do the
circuit analysis to get the "exact" times). When the switch is open then
the charge on the cap that was stored when the switch was closed will turn
into current... but this there will be more current(enough to power the LED)
and will eventually die down(as the charge equalizes among both plates of
the cap)... eventually there will not be enough current to charge the LED
and it will shut off(but there will still be some charge on the cap but it
just won't be enough to force itself through the LED).

Anyways,

L

#### [email protected]

Jan 1, 1970
0
Abstract said:
Two things to note about a cap that are important. First is that the voltage
across the capacitor is proportional to the charge it contains and second is
that the current through a capacitor is equal to the rate of change of
charge in the capacitor.

...

Anyways,

Wow. That was a really helpful response. Fantastic, really. Thank
you for taking the time to write that.

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.

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.

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.

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.

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.

D

#### Daniel Pitts

Jan 1, 1970
0
The way I think about it: A capacitor can only "hold" so much voltage,
which is basically electrical pressure. As the capacitor "fills" it
pushes against the current flow (adding resistance), which makes the
electricity want to flow through any other path it can.

Keep in mind this is just an analogy, but it helps me understand why a
capacitor does what it does.

P

#### Pooh Bear

Jan 1, 1970
0
and sure enough, there
was a transient voltage of ~1.65V left in the capacitor from previous
charges.

That's called a residual voltage btw.

Transients are something else *entirely*.

Graham

A

#### Abstract Dissonance

Jan 1, 1970
0
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
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)

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,

B

#### Bob Myers

Jan 1, 1970
0
Daniel Pitts said:
The way I think about it: A capacitor can only "hold" so much voltage,
which is basically electrical pressure. As the capacitor "fills" it
pushes against the current flow (adding resistance), which makes the
electricity want to flow through any other path it can.

That's not QUITE right. You shouldn't think of capacitors
as having a fixed voltage limit (except for the breakdown of the
dielectric in a practical capacitor, but we're presumably talking
about the purely-theoretical, blackboard-variety cap here).
Instead, it would be more accurate to say that a capacitor will
only hold a certain amount of CHARGE for a given voltage, and
that's exactly what the definition of capacitance says:

Q = CV or C = Q/V

You can ALWAYS put more charge into a capacitor by
increasing the voltage across it (again, this is the theoretical
"perfect" capacitor that never breaks down), but for a given
voltage there is a fixed amount of charge that a given cap
will accept, and so once that point is reached no more current
("charge flow") will enter the capacitor, and all will go to the

As the charge contained within the capacitor nears this maximum,
and the voltage across it increases, the rate at which additional
charge enters (which is the current) must be reduced, since there
is less voltage across any impedance which is in series with the
capacitor (i.e., the resistance which limited the peak available
current in the first place). Eventually, the capacitor is charged to
the same voltage as the source, and no more current can flow -
no charge can enter the cap - simply because there is no
voltage ("pressure") available to "push the charge in."

Bob M.

R

#### Rich Grise

Jan 1, 1970
0
That's called a residual voltage btw.

Is it really "residual"? Because that's about what a cap would discharge
to through an LED.

Cheers!
Rich

J

#### Jasen Betts

Jan 1, 1970
0
Wow. That was a really helpful response. Fantastic, really. Thank
you for taking the time to write that.

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.

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.

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:

no that's the wrong way to go about it...

_
+-----o-~ o--------+---[1K]----+
| | |
|+ | _|_
--- ----- \ /
- ----- -Y-\\
: 9V | | 
--- | |
- | |
| | |
+----[10K]---------+-----------+

when the capacitor is has charged fully (all it will charge)
no current will flow through it and so it will have
no effect on the rest of the circuit.

so it can be removed from the circuit and the remainder analised.

+-----o---o--------+---[1K]----+
| |
|+ _|_
--- \ /
- -Y-\\
: 9V | 
--- |
- |
| |
+----[10K]---------+-----------+

now we have 4 components in series conneted to the 9V battery
the switch, the 1K resistor, the LED and the 10K resistor.

to figure out the current throug resistors is easy, but LEDs are harder...

fortunately in the useful range the voltage drop on a LED is pretty
constant. for red LEDs it's about 1.8V

so subtract the 1.8V of the LED from the 9V and that leaves 7.2v pushing
the current round the circuit. there's 11K resistance in the rest of the
circuit so 7.2/1l000 gives .0000655 A (or 655uA misusing the lower case
u to represent the symbol for micro as is common in places like this)

now the capacitor connects across the 1K resistor and the LED

655uA through a 1K resistor is 655mV that plus the 1800mV for the LED
is 2.455V

you measured 2.53 so maybe your LED has a higher voltage drop than the 1.8
I used... or possibly your 9V is a little more than 9V.
or maybe your resistors weren't exactly their marked values.

in electronics often the parts are out by a few percent from their nominal
value, and where this is critical designs are made in ways that compensate
for this, or a premium is paid for high precision parts.

Bye.
Jasen

L

#### [email protected]

Jan 1, 1970
0
Abstract said:
... (lots of good stuff) ...

Anyways,

Thank you for another very helpful reply. I would also like to thank
everyone else who replied to my original message.

Heh, I didn't even notice the significance of treating the capacitor as
things a lot easier and it's more accurate. I'm starting to get a
better idea of how you go about analyzing RC circuits, and I'll try to
read some more about it. I checked out a book from the local college's
library called "Network Analysis" by M.E. Van Valkenburg -- hopefully I
can learn a few things before it completely loses me. =)

Thanks again!

R

#### Rich Grise

Jan 1, 1970
0
Thank you for another very helpful reply. I would also like to thank
everyone else who replied to my original message.

Heh, I didn't even notice the significance of treating the capacitor as
things a lot easier and it's more accurate. I'm starting to get a
better idea of how you go about analyzing RC circuits, and I'll try to
read some more about it. I checked out a book from the local college's
library called "Network Analysis" by M.E. Van Valkenburg -- hopefully I
can learn a few things before it completely loses me. =)

Thanks again!

Since this is "basics", I feel safe to reveal something one of my tech
school teachers said that opened new vistas of understanding: "A capacitor
opposes a change in voltage, an inductor opposes a change in current."

It's worked for me, ever since!

Cheers!
Rich

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