Re: Simple capacitor circuit (slowly illuminating LED)


<longjohnstuartmill@xxxxxxxxxxx> wrote in message
news:1136372483.833860.324860@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
>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,

AD












.

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