# is this a short circuit????

Hi, i got come leds from Adafruit and their web tutorial recommend placing a 1000uF between positive and negative.

isnt it a short circuit?

i attach a photo from the tutorial.

thanks

Yes, it is a short for AC voltages. It is an open circuit for DC voltages because a capacitor contains an insulator between the two metallic plates. So, in your case, the capacitor is an insulator for DC, but would be a short or partial short circuit for AC or DC that varies rapidly in voltage value.

Paul

Paul_KD7HB:

Electrolytic capacitor

Paul_KD7HB:
but would be a short or partial short circuit for ... DC that varies rapidly in voltage value.

That doesn't sound right to me. Why would the impedance (resistance, in fact) be any less for rapidly changing DC than for constant or slowly changing DC? It's still DC: no polarity change and thus no return of the stored charge on opposite cycles.

Xc the captivate reactance = 1 / (2 pi f C)

Even if the DC varies rapidly, since the polarity never changes, f=0 and thus Xc is still infinite.

kenwood120s:
That doesn't sound right to me. Why would the impedance (resistance, in fact) be any less for rapidly changing DC than for constant or slowly changing DC? It's still DC: no polarity change and thus no return of the stored charge on opposite cycles.

Xc the captivate reactance = 1 / (2 pi f C)

Even if the DC varies rapidly, since the polarity never changes, f=0 and thus Xc is still infinite.

We are dealing with someone who doesn't know what a capacitor is or how it works an is unwilling to use Google, so I tried my best.

Paul

I'm not a huge fan of analogies, since people often push them too far and they break down but I like the analogy of the capacitor as a diaphragm in a pipe.

By that analogy, it's easy to see that with the current in one direction there's current for a moment while the diaphragm flexes and bulges, that's the capacitor charging. But once that stops, that's it. The pressure's still there, that's the voltage, but no current.

(Unless it's AC in which case the diaphragm returns the water in the bulge (the charge in the cap) whizzing back whence it came, rinse and repeat.)

So, if I put a DC offset on an AC signal so that it never crosses zero, it should not be affected by the capacitor?

KeithRB:
So, if I put a DC offset on an AC signal so that it never crosses zero, it should not be affected by the capacitor?

That's what I'm picturing yes, since f=0, but I'm not stating it as a fact, rather just my understanding. Certainly if f=0 then Xc if infinite from the equation. So I'll restate my understanding as a question: is f=0 for rapidly changing voltage that never crosses 0?

But that all said-maybe we should halt this discussion for OP's sake, and the answer to his question is "No, it's not a short circuit, since for the DC case in the pic, the capacitor charges up until the charge on the plates balances the applied voltage after which nothing moves. A cap is open circuit to DC."

(But I'd like to know if my thinking's flawed and maybe I'll open a thread on that.)

Edit: started this thread to discuss the varying DC stuff.

Any AC, without any regard to DC level, is affected by the capacitor. That is how power supply filters work.

Oh, and "rapidly changing" is not "f=0".

kenwood120s:
That's what I'm picturing yes, since f=0, but I'm not stating it as a fact, rather just my understanding. Certainly if f=0 then Xc if infinite from the equation. So I'll restate my understanding as a question: is f=0 for rapidly changing voltage that never crosses 0?

It appears that you don't know about Fourier analysis. Electrical signals do not possess just one single frequency, they are a sum of sinusoids with different amplitudes, frequencies, and/or phases.

A lot of people use "AC" not to refer to a signal that crosses 0, but a signal with non-zero frequency. I know it's not technically correct by a literal meaning of "alternating current", but that's just how it is.

kenwood120s:
That doesn't sound right to me. Why would the impedance (resistance, in fact) be any less for rapidly changing DC than for constant or slowly changing DC? It's still DC: no polarity change and thus no return of the stored charge on opposite cycles.

Xc the captivate reactance = 1 / (2 pi f C)

Even if the DC varies rapidly, since the polarity never changes, f=0 and thus Xc is still infinite.

I think DC means constant value. Physically hypothetically unchanging with time. If hypothetically or theoretically or physically changing with time in a sinusoidal fashion.... at some particular frequency......... then that's AC.

Also....theoretically.... if mathematically considering a time changing signal that is not purely sinusoidal in behaviour.... then it could '"mathematically" be considered as a collection of sinusoidal signals plus a constant unchanging signal if there is one ...... aka Fourier theory.

The capacitive reactance Xc expression has a negative sign. Eg. -1/(2Πf.C). At relatively high frequency, the capacitor impedance becomes relatively small......and at high enough frequency.... the ideal capacitor can be approximately a short circuit. At theoretical infinite frequency..... which we can never reach... the capacitor approaches a short circuit.

kenwood120s:
That doesn't sound right to me. Why would the impedance (resistance, in fact) be any less for rapidly changing DC than for constant or slowly changing DC?

DC is constant by definition. If something is changing it has a AC component. The impedance
depends on frequency (DC is just frequency zero). The AC component can get through the capacitor.

Hi,
As I pointed out in another thread.

AC CURRENT affects IMPEDANCE, not voltage.

If you vary the voltage on a capacitor form 2V to 5V to 2V to 5V etc, you have AC CURRENT, even though the applied voltage does not change sign.

2V to 5V, CHARGING the cap, current goes into the cap at its postive terminal.
then
5V to 2V DISCHARGING the cap, current goes out of the cap at its positive terminal.

Hence AC current.

Tom…
That is how an analog decoupling cap works in an amplifier.
(Sorry I don’t think they teach analog electronics in school/UNI anymore.)