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Simple capacitor circuit (slowly illuminating LED)

Discussion in 'Electronic Basics' started by [email protected], Jan 4, 2006.

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  1. Guest

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

    | | |
    | | |
    ---- 9V Battery --- 1 mF Cap |
    -- --- (LED)
    | | |
    | | |
    | | |

    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). :)
  2. Pooh Bear

    Pooh Bear Guest

    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.
    No why so ?
    No. It'll gradually illuminate as the cap charges. It won't 'suddenly come
    on'. Why did you think so ?
    That's what I'd expect.
    Beacuse the current through it gradually increased with the cap voltage (
    as the cap charged via the battery's internal resistance ) of course.
    Sort of. It's called reactance.
    Yes. It'll be clearer perhaps if you included a resistor between the
    battery and the cap representing the battery's internal resistance.

  3. Andrew Holme

    Andrew Holme Guest

    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.

    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.
  4. Nick

    Nick Guest

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

    Pooh Bear Guest

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

  6. Lacy

    Lacy Guest

    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.
  7. Guest

    Yeah, I meant 1000 uF. Sorry about that. I remembered that uF was the
    preferred unit after I had posted the original message.
    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

    Thanks for the rest of your reply!
  8. 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.


    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.

    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

    Vb - CQ - IR = 0

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

    but I = dQ/dt


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


    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

    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

    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
    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


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


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


    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 -

    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).


  9. Guest

    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

    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

    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
    -- |
    | |
    | |
    | |

    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

    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.

    Thank you in advance for wading through that.
  10. Daniel Pitts

    Daniel Pitts Guest

    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.
  11. Pooh Bear

    Pooh Bear Guest

    That's called a residual voltage btw.

    Transients are something else *entirely*.

  12. No problem.

    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).
    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).

    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


    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


    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 ~=

    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% =

    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%

    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%.

    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.

    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
    | |
    | |
    ---- V Battery |
    -- (LED) VL
    | |
    | |
    | |

    so V - I*R1 - I*R2 - VL = 0


    I = (V - VL)/(R1 + R2)

    R1 = 1k
    R2 = 10k
    VL = 2
    V = 9


    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)

    | |
    | |
    ---- V Battery |
    -- (LED) VL
    | |
    | |
    | |

    and for very short time t(t = 0)

    | |
    | |
    ---- 9V Battery --- 1 mF Cap
    -- ---
    | |
    | |
    | |

    (we can forget the switch as it is used in the discharge part which gives
    you the circuit

    | |
    | |
    --- 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).

  13. Bob Myers

    Bob Myers Guest

    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
    load in parallel with it.

    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.
  14. Rich Grise

    Rich Grise Guest

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

  15. Jasen Betts

    Jasen Betts Guest

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

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

    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.

    | |
    |+ _|_
    --- \ /
    - -Y-\\
    : 9V | ``
    --- |
    - |
    | |

    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.

  16. Guest

    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
    an open circuit until after I had read your second post. That makes
    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!
  17. Rich Grise

    Rich Grise Guest

    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! :)

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