# Beginner doubt on capacitor+LED

Discussion in 'General Electronics Chat' started by J0ker, Apr 27, 2012.

1. ### J0ker Thread Starter New Member

Apr 18, 2012
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0
Hi!

So in this circuit, current flows first towards the capacitor as it charges but not through LED. When the capacitor is fully charged, current then flows through LED.

Why current doesn't also flow through LED while the capacitor is charing? Can't some current go to the capacitor and some to the LED? I've i have never tried it I would have thought LED is always on... but it's not

2. ### crutschow Expert

Mar 14, 2008
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You need to post a circuit diagram.

3. ### Jony130 AAC Fanatic!

Feb 17, 2009
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1,381
Empty capacitor act just like a short circuit. Notice also that in you circuit LED is connect parallel to the capacitor. So the voltage across capacitor is equal voltage across the LED.
Empty capacitor has 0V across his plates (act like a short circuit). As capacitor charge up the voltage across capacitor and LED rise.
When voltage across LED reach ON voltage (forward voltage) the LED start to to emit light and capacitor stop charging. Because the LED don't allow the capacitor voltage to be greater than LED forward voltage.

Dec 26, 2010
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A slightly more detailed picture of the currents in the capacitor and the LED would show some progressive rise in the LED current as the capacitor was charged, but the LED current would nevertheless be quite negligible until a fairly big fraction of its normal operating forward voltage was reached.

The capacitor charging would therefore be hardly affected by the LED until a fairly well defined voltage, beyond which the LED conduction would take current away from the capacitor, preventing the voltage rising further.

5. ### J0ker Thread Starter New Member

Apr 18, 2012
17
0
That's weird, circuit disappeared. I re uploaded to a different server:

Ok... I understand a little bit more. In the very first instants of powering on the circuit, isn't there a different potential in the LED from +5V to ground? Or better put, why isn't there different potential on the LED in the first moments? It's directly wired to a battery..

Also, regarding the shortcircuit concept... isn't the resistor evening both capacitor and LED out? I mean both end up facing the resistor.

Thanks for the patience.

6. ### crutschow Expert

Mar 14, 2008
23,091
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At the instant you turn on the voltage, the capacitor voltage is 0V and thus so is the LED voltage, since they are in parallel. Until the capacitor charges to the ON voltage of the LED, the LED will stay off.

7. ### WBahn Moderator

Mar 31, 2012
24,553
7,691
No, the LED is not connected directly to a battery. That would require that one terminal be tied to one side of the battery and the other terminal be tied to the other. However, in this case, the other terminal is tied to the resistor before going back to the battery.

If you are going to talk about 'ground', then you need to specify what node in the circuit that is. Normally, circuits are drawn with the bottom node being 'ground'. However, you also talk about +5V and since your battery is shown with the positive terminal on the bottom node, that implies that the top node is 'ground' and the bottom node is +5V (assuming that the battery is 5V, which is not stated in the diagram).

In general, there will be two different currents flowing, one in the capacitor and one in the LED. The sum of these two currents will be flowing in the resistor. The current flowing into a capacitor is proportional to the rate at which the voltage across it is changing. The current flowing in the LED, as a function of the voltage across it, is highly nonlinear and, for voltage more than a couple hundred millivolts below the nominal forward voltage for that LED (typically ~2V), the current flow will be negligible. Above that voltage, the current increased very, very rapidly..
As other posters have stated, initially the capacitor voltage is zero and all of the current (which is limited by the resistor) flows into the capacitor and starts charging it. As the voltage across the cap starts approaching 2V (or whatever for that LED), then some current starts to flow in the LED while the rest continues to charge the capacitor (with the sum of the two still limited by the resistor). Since the capacitor is still charging, albeit more slowing, the current in the LED rises quickly but, in doing so, this further robs the capacitor of current resulting in it charging rate going down even more. The end result is that the LED quickly reaches a current level that matches the amount allowed by the resistor and no more current flows in the capacitor which means that it's voltage stops changing and remains constant.

8. ### J0ker Thread Starter New Member

Apr 18, 2012
17
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This may sound dumb but would you happen to have a snippet of theoretical explanation behind that?

9. ### WBahn Moderator

Mar 31, 2012
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7,691
The explanation is in my post.

Since the voltage across the capacitor is a function of the charge stored on it, it can only change in response to a change in the stored charge. But you can't instananeously make charge appear or disappear, it has to enter or exit over time continuously and therefore the voltage has to change continuously with time, as well.

Since the two components, the capacitor and the LED, are in parallel, they have the same voltage. That's what it means for two things to be in parallel. Since the voltage across the capacitor can't change instananeously, neither can the voltage across the LED.

10. ### J0ker Thread Starter New Member

Apr 18, 2012
17
0
Ok. But here's how i'm imagining it in my head: in the very first moment, if we could take a snapshot right in the beginning, isn't there a potential difference = voltage accross the capacitor? That's why it starts charging right? Because there's a voltage, charges want to go from one side to another.

I'm really a dumb stone with this matter right now, i'm sorry. In the that snapshot, if we only take care to the LED...isn't it wired on one side to a +5V and on the other side to ground? Why can't it draw electrons right there? I can't really see the physical impediment for that to happen

11. ### WBahn Moderator

Mar 31, 2012
24,553
7,691
You still haven't defined what 'ground' is. Since you keep talking about +5V, I have to assume that the top node in the diagram is your ground and that the bottom node is therefore +5V.

How are you saying that the LED is tied to +5V on one side and to ground on the other? On one side it is tied to the positive terminal of 5V battery. On the other is it tied to a resistor that, in turn, is tied to the negative terminal of that same battery. The ONLY way for there to be 5V across the LED is for there to be ZERO current in the resistor and, if that's the case, where is the current flowing in the LED going? It is NOT going into the capacitor because it would be charging the capacitor in the other direction!

The capacitor starts charging not because there is a voltage across it, but because there is a current flowing into it. Consider a capacitor that has been charged to 10V and is just sitting there on a desk not connected to anything. How much current is flowing in it? None, despite the fact that it has a voltage across it. Voltage and current are two different concepts and the relationship between them is defined by the component in question. For a resistor, you can't have a current without a voltage and vice-versa. For a capacitor, you can't have a current without a changing voltage, but you can most definitely have a voltage without a current and also a current with no voltage, albeit it only momentarily before it changes (in a continuous fashion).

At the very beginning, since there is 0V across the capacitor, there is 5V across the 100 ohm resistor and therefore 50mA flows in the resistor. Because there is no voltage across the LED, no current flows in it. The full 50mA therefore flows into the capacitor and its voltage starts changing as a result.

$
\frac{dV}{dt} \ = \ \frac{I}{C} \ = \ \frac{50mA}{1mF} \ = \ 50V/s.
$

Even if the charge rate stayed the same (and it starts decreasing immediately), it would take 40ms to reach the 2V needed to turn on the LED.

12. ### J0ker Thread Starter New Member

Apr 18, 2012
17
0
Yes, ground (the negative pole of the battery?) is on the upper side. From now on i'll draw diagrams with the ground at the bottom.

I think this is what is making me struggle. The notion that when electrons are not flowing, there's a 5V across the LED but those 5V are gone when current is flowing through the resistor...

Are there any specific resources I can read or exercises I can do to go deep into that concept? (which i would say is the physical behavior of current and voltage in simple circuits)

Thank you.

PD: i'm teaching myself, and now i'm starting to think maybe i went too fast on the basics, even though i don't remember not understanding something... :\

13. ### crutschow Expert

Mar 14, 2008
23,091
6,849
The concept you seem to be struggling with is Ohm's Law, V = I*R. The voltage dropped across a resistance is proportional to to the current through the resistor times its resistance.

Thus when no current flows, there is no voltage drop across the resistance. But as soon as current starts to flow, there is a voltage drop across the resistor which subtracts from the total voltage available.

Say you have the resistor is series with the capacitor (remove the LED for the moment) and suddenly apply a voltage, then the full voltage will initially appear across the resistor and a charging current equal to V/R will start to flow. The capacitor voltage will start to increase as it charges, which reduces the voltage available across the resistor, thus reducing the current through the resistor. This continues in an exponential fashion until the capacitor is fully charged to the supply voltage (after many time constants of time) and no more current flows through the resistor. So we now have the full supply voltage across the capacitor and no voltage drop across the resistor.

Does that help?

14. ### WBahn Moderator

Mar 31, 2012
24,553
7,691
Read what I wrote again more carefully. You claimed that there was 5V across the LED. I pointed out that the only way for that to be true was for there to be no current flowing in the resistor (since, if 5V is across the LED there is no voltage left to appear across the resistor). This results in a contradiction because if there are 5V across the (forward biased) LED, there will be current flowing in it, and there can't be because there is no place for it to go. Hence, there is NOT 5V across the LED.

You need to start with the fundamentals such as what is charge, what is voltage, what is current, what is resistance, what is capacitance, what is inductance, and what properties have to be conserved. These will lead directly to the fundamental laws that are used in circuit analysis. I would recommend getting an introductory physics book. If you do not have at least a semester's worth of calculus, be sure to get a non-calculus based text (which is generally a high school text). If you do have some calculus, get a college text.

15. ### J0ker Thread Starter New Member

Apr 18, 2012
17
0
Ok. No voltage drop because there's no current flowing, but it's still at a higher potential than ground isn't it? So its electrons, even though they are quite, they "want" to go to the negative pole of the battery right?

I'm going to tell you how I'm imaging it, even though I'll sound silly, but for the sake of detecting what I got wrong, I imagine (in the original diagram): electrons come out from the positive pole (conventional) and they run towards the junction. Now, they can choose to go through the capacitor or through the LED. As far as I know, they feel the same attraction to any of those (in this point electrons are +5V, they want to go to ground, and both ways will take them there). They could even split and more electrons would go through the less resistive path, and fewer would go through the other one (like i've seen with resistances etc.). YET they ALL choose to go through the capacitor, and none through the LED (yes, until enough voltage is in the capacitor...but the same voltage that the battery is using to make the electrons go through the capacitor should be able to push electrons through LED..is what I can make up in my mind).

I understand the relationship between the voltage through the capacitor and the current flowing through the LED (i just did like a dozen similar tests on a sim to try to understand this thing), and it makes sense: the more charged the capacitor is (the more difference in potential there is between its legs) the more current flows through the LED. Just as I'd expect to happen with a battery.

Where am I mistaken in the way I try to imagine the electrons making a decision in the junction?

Thank you for the patience, I promise I'm smart with other stuff lol

PD: i just went through Ohm's law again and also did some kirchoff/thevenin just to refresh the math: nothing wrong there. It's just the physical need of an electron in a parallel scheme with a capacitor that i don't seem to grasp.

Last edited: Apr 29, 2012
16. ### #12 Expert

Nov 30, 2010
18,076
9,691
When the electrons start flowing through the resistor they don't have a choice at the LED/capacitor junction because they haven't built up enough voltage on the capacitor to break through the 2 volt threshold of the LED. Only after enough current has arrived to charge the capacitor up to 2 volts do they have a choice. At that point, the threshold voltage of the LED is defeated and the electrons only flow through the LED. All 30 milliamps flow through the LED and the capacitor never gets to 5 volts because the LED is acting like a zener diode.

Sorry if this doesn't work for you. It's just another way of saying what everybody else has said, but that's the magic of a public forum. One person says things in a different way than other people, and sometimes that is what makes the LED in your head light up.

17. ### WBahn Moderator

Mar 31, 2012
24,553
7,691
This might help a bit, but it's hard to make the analogy work really well.

Imagine that the capacitor, the LED, and the resistor are pieces of pipe and are connected by flexible tubing just like the wires. The water enters the tubing from the output of the pumping station (the postive side of the battery) and eventually ends up down at the intake to the pumping station (the negative side of the battery). Now, a pipe that is sitting horizonatal has no pressure difference across it, while a pipe that is rotate vertical has a significant pressure differential across it. The pressure difference represents the voltage. The pipe that represents the resistor is stuffed with a porous sponge or cloth such that if you rotate the pipe a little bit you get a little bit of current flowing through it and if you rotate it more you get more current. The pipe that representa the LED has a little wall inside the pipe such that if you only rotate it a little bit, the water backs up behind the wall but can't make it over it. If you rotate it far enough, then the water at the inlet of the pipe has sufficient potential to make it over the wall and, as you rotate it just a tiny bit more, you get a huge increase in current flowing over the wall. The capacitor is a pipe with a spring-loaded piston inside. As you tip this pipe, water flows into it and the piston is pushed down the pipe, but the spring pushes back so it only goes so far. If you tip the pipe a bit more, there is more pressure across the piston so it moves a bit further before reaching a balance point. Now, instead of us tipping the pipe, suppose there is a mechanism that rotates the pipe based on the position of the piston -- the further down the pipe the piston is, the more it gets rotated. Note that there is water on both sides of the piston - as water flows into one side, the piston moves and pushes water out the other side. Furthermore, it doesn't matter if we think of water being pushed into one side or pulled out of the other side, the same thing happens in either case; namely the piston moves and whatever water flows into one side and equal amount flows out the other and the pipe is rotates to match the piston position.

Now, again keeping in mind that this is not a really good analogy, imagine that we insert a quick-disconnect coupling between the resistor and the tubing connecting it to the LED and capacitor pipes. We then position the LED and capacitor pipes so that they are both horizontal and even with the output of the pump. Since the are in parallel, we mount them both on a platform that can rotate is such a way that the ends of both on the 5V side are always even with the output of the pump house but that the other ends can rotate downward, noting that the lower ends will always be at the same level. The resistor, on the other hand is presently laying horizontal on the ground on a platform that is similarly hinges such that the side that is connected to the pump inlet will always be at that level but the other end can rotate up and, once we connect the coupling, will be forced to always be at the same height as the downhill side of the capacitor/LED platform.

So now we start the circuit by rotating the resistor up to the output side of the other platform and connect the coupling so that current can flow. The resistor now has a pressure difference across it and current starts flowing through it, but the only place the current can come from is by being pulled out of the capacitor (because of the wall in the LED pipe). As this happens, the capacitor/LED begins to tilt as the piston moves. Now, as it tilts, the uphill side of the resistor is getting lower, so there isn't as much pressure differential and the current flow in the resistor is reduced and, hence, the rate at which the piston is moving in the capacitor is reduced and the rate at which the platform is rotating is reduced. But they continue rotating, just getting progressively slower. At some point, the platform will have rotated just enough to start letting water flow over the wall in the LED pipe. At this point, the resistor still has a pressure differential across it and a certain amount of current still wants to flow through it, but it no longer had to pull the current through the capacitor because the current flowing over the wall is just enough to meet the needs. So the platform will have found the equilibrium point and will stay in this position.

Hopefully that will help you visualize one way of thinking about this stuff, just remember that you can't push any analogy to far and you can't push this one very far at all.

18. ### crutschow Expert

Mar 14, 2008
23,091
6,849
The electrons don't "want" or "choose" to go anywhere, they just are pushed around according to the relative potential between two points. Thus at the instant the capacitor starts charging its potential is at ground while the LED has a 2V potential and the battery has a 5V potential. So the electron is forced through the resistor by the potential difference to go into the capacitor and avoids the LED unit the capacitor potential exceeds 2V. Then the potential difference forces the electrons through the LED.

Remember once the electrons go through the resistor they no longer have a 5V potential pushing them. The lose an IR drop worth of potential.

Note: You mention electrons flowing out of the positive pole when, of course, they actually go into the pole. Normal current flow implies positive charges flowing out of the positive terminal.

19. ### djsfantasi AAC Fanatic!

Apr 11, 2010
5,434
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I've always envisioned an uncharged capacitor as a dead short. Then, in your diagram, it should be obvious why the LED doesn't light, as it is shorted. As a dead short, all the current will be absorbed by the capacitor. The next part I am unclear of. As the capacitor charges, the current will then split once the voltage on the capacitor is above what the LED needs.

20. ### bretm Member

Feb 6, 2012
152
24
No...the LED and capacitor are in parallel, so they will always have the same potential difference across their terminals. At time=0, the potential difference is 0V across the capacitor and 0V across the LED.

Because of the way diodes work, 0V means 0 current.

Because of the way capacitors work, the rate of voltage change of 0V/sec means 0 current. But once the circuit is switched on, the rate of voltage change is not zero, so the current through the capacitor is not zero.

The current through the diode depends (exponentially) on the voltage across the diode. This starts out at zero.

The current through the capacitor depends (linearly) on the rate of change of the voltage across the capacitor. This isn't zero right after switching on the circuit, so current will flow through the capacitor.

This current flow causes increasing polarization of the capacitor's plates, which means an increasing voltage drop over time. This also means an increasing voltage drop across the LED (since they're in parallel). This increasing voltage means the current through the LED increases over time.

Eventually the circuit reaches equilibrium when the current through the LED matches the current through the resistor, and no current is flowing through the capacitor.

Last edited: Apr 30, 2012