Simple capacitor circuit (slowly illuminating LED)

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.

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.

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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 ?

Yes.

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.

Graham

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 reading.

Thanks for the rest of your reply!

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

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.

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

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

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.

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

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

or

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%.

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, AD

That's called a residual voltage btw.

Transients are something else *entirely*.

Graham

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

Cheers! Rich

That's not QUITE right. You shouldn't th 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.

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!

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

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