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I'm doing a super simple Arduino led project. I followed a tutorial online that wired up the LEDs like this:

enter image description here

I'm wondering why this capacitor doesn't just short the power supply.

Can anyone provide me with some insight?

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    \$\begingroup\$ Have you considered the perspective that everything else is also connected across the power supply and yet none of it causes a short? I mean...the Arduino is connected across the power supply just like the capacitor is. If you provided a basis for why you think it might short, you would get better answers because we can correct misunderstandings. Is there some particular characteristic you know about capacitors that leads you to think it should short? \$\endgroup\$ – DKNguyen May 26 '20 at 20:54
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    \$\begingroup\$ Have you plugged arduino to AC? \$\endgroup\$ – Gregory Kornblum May 26 '20 at 23:48
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    \$\begingroup\$ It does short the supply, but very quickly reaches the same voltage and no longer shorts. If you use an array of (4?) 3-volt 400-Farad supercapacitors in series, it would almost definitely short the supply, until the supercaps were charged, if the power supply could even handle them. But I'm not sure that clears it up for you. Why do you think it would short the power supply? What is your understanding of what is going on? \$\endgroup\$ – MicroservicesOnDDD May 27 '20 at 1:24
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    \$\begingroup\$ This boils down to "what is a capacitor", meaning a downvote from me for "does not show any research effort". \$\endgroup\$ – pipe May 27 '20 at 13:05
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    \$\begingroup\$ sorry I haven't been able to reply till now. After reading the other responses it's pretty much cleared up for me. I know capacitors store charge. My misunderstanding was that a capacitor when charged would provide a 0 resistance path for electrons to flow shorting the circuit. I feel like I have a bit of a better understanding of electronics in general now. \$\endgroup\$ – WavePhaser May 27 '20 at 20:15
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The capacitor is in fact a short circuit, however only temporarily.

When you first turn on the power supply, the capacitor will act like a short circuit during this initial transient phase. There will be a large inrush current as the capacitor charges up (*).

Similarly, if you take a fully discharged capacitor, and connect it to a power supply which is already at some DC voltage, you will get a sudden inrush current into the capacitor which if large enough (e.g. very large capacitor) can be enough to overload the power supply. This is especially problematic if switching a large capacitive load into a circuit.


Once the power supply reaches its steady-state DC voltage, and the capacitor charged up to the same voltage, the capacitor will no longer act as a short circuit on the supply because it cannot conduct a DC current as explained by the other answers.

If however you have some noise or ripple signal, or a voltage transient, the capacitor will again act like a low impedance or short circuit (frequency dependent) for this noise signal or voltage ripple. This is how the capacitor has the effect of smoothing out the power supply.


(*) The power supply may provide some form of inrush current limiting or controlled voltage ramp-up, in which case the current flow will be limited.

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  • \$\begingroup\$ Of course, this assumes that you have the electrolytic capacitor wired in the proper orientation. \$\endgroup\$ – Hot Licks May 27 '20 at 16:25
  • \$\begingroup\$ @HotLicks which in the illustration show, it is. \$\endgroup\$ – Tom Carpenter May 27 '20 at 17:07
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There is no conductive path from one terminal to another in a capacitor. The symbol shows this.

schematic

simulate this circuit – Schematic created using CircuitLab

Figure 1. Non-polarised, polarised and variable capacitors. The symbol represents two parallel plates separated by an air-gap or a dielectric (insulator).

enter image description here

Figure 2. An unwound electrolytic capacitor showing the two layers of foil and one of the insulation layers. Source: Translators Cafe.

The capacitor can't pass DC current but it can accept charge and if the voltage across it varies current will flow in and out of both terminals. In this manner it is able to pass an AC current through it.

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It is possible, at least theoretically, to damage a power supply by connecting a discharged capacitor across its terminals. However, in real life the characteristics of the power supply and the capacitor usually prevent this from happening.

When you first connect the capacitor to the supply a large current will flow to charge the capacitor. The magnitude of this current is limited by the output resistance of the supply and the effective series resistance (ESR) of the capacitor. Neither of these resistances are shown on your cartoon schematic, but they are there nonetheless. As long as these resistances limit the current to a safe level no damage will occur. Another important consideration is that this surge current lasts for only a very short period of time...once the capacitor is charged to the supply voltage you will see relatively smaller currents.

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Because as far as DC is concerned, a capacitor is an open circuit.

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    \$\begingroup\$ Once it is charged to the DC voltage \$\endgroup\$ – relayman357 May 26 '20 at 20:50
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    \$\begingroup\$ @relayman357 If it's not charged to the DC voltage, that means there was a transient and thus it's not pure DC. \$\endgroup\$ – Hearth May 27 '20 at 1:13
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    \$\begingroup\$ If you put some 400-Farad supercaps in there, they will act like a short circuit until they are charged. So an uncharged capacitor suddenly connected does behave like a short-circuit (especially if it is big). 1000uH sounds pretty big, but it's actually pretty small. \$\endgroup\$ – MicroservicesOnDDD May 27 '20 at 1:30
  • \$\begingroup\$ @MicroservicesOnDDD that's a really long transient then :D \$\endgroup\$ – quetzalcoatl May 27 '20 at 7:41
  • \$\begingroup\$ @MicroservicesOnDDD "1000uH sounds pretty big", 1000uH? I thought we were discussing capacitors, not inductors. \$\endgroup\$ – Glen Yates May 27 '20 at 21:56
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You can imagine the capacitor as a battery. The charge on a capacitor is dictated by voltage and it's capacitance. Since capacitance will be static and charge doesn't teleport, we can see that as charge increases, the voltage increases. The voltage can only go as large as what's feeding it so there's a maximum amount of charges the capacitor can hold, which is again dictated by the voltage.

Its purpose here is to provide a current source for the LEDs. There's a few reasons for this, the top one is that large instantaneous current draws cause a droop in the voltage from the power supply, so the capacitor has a lower voltage across it, which makes it lose charges. So it turns into a current source for the short amount of time of the droopped voltage.

An additional fun question to ask yourself is why you had to add a resistor to the control signal going to the LEDs!

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A capacitor is two electrodes separated by an insulating layer. In some cases the insulating layer could be air or vacuum. In other cases it could be a dielectric (such as some type of plastic or ceramic). Unless there is some type of fault, such as dielectric failure, the capacitor is not a short-circuit at DC, by the definition of a capacitor.

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DC doesn't travel through a capacitor as long as its charged.

The capacitor on LED strips is there to dampen sudden Amp changes in the system, when your LED Strip suddenly needs more power etc. The thin cables on these PSUs are most often too high inductance to manage these fast, big shifts well.

This is a great answer to that.

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The capacitor was connected in parallel with the circuit. Current reaches both at the same time and it would be too late for it to protect the circuit. This would have been different if the connection was done in series inline with the circuit. The current would have to reach and pass through the capacitor before it reaches the circuit to protect it.

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  • \$\begingroup\$ question isn't asking about protection ... \$\endgroup\$ – VanGo May 27 '20 at 17:54
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It IS going to short the power supply. It's also going to cause the capacitor to explode, in all likelihood. The problem is that the capacitor is installed backwards. Electrolytic capacitors are polarized, and that dashed white line on the side of the capacitor is actually a series of arrows pointing to the negative terminal. Polarized capacitors really don't like being installed backwards and will typically "vent" their frustration in rather violent ways.

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    \$\begingroup\$ While you are correct that Alu-Elec caps are polarised, and reverse polarity will danage them, the white line is in fact the negative terminal, which is correctly connected to the negative terminal of the DC supply. \$\endgroup\$ – Tom Carpenter May 27 '20 at 17:04
  • \$\begingroup\$ "... that dashed white line on the side of the capacitor is actually a series of arrows pointing to the negative terminal." Actually the arrows are pointing away from the negative terminal. They're just supposed to mark the side of the capacitor with the negative pin. \$\endgroup\$ – Transistor May 27 '20 at 18:21
  • \$\begingroup\$ I've had this circuit running on my desk for hours at a time with no problem so this is very incorrect. \$\endgroup\$ – WavePhaser May 27 '20 at 20:02

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