This thread sparked some discussion, and I thought I'd start a new thread here since the OP's question there is answered.
In his answer to that OP's question about a cap being a DC short, Paul KD7HB said no it's open circuit, but also said:
the capacitor ... would be a short or partial short circuit for ... DC that varies rapidly in voltage value.
And that got me wondering.
The equation for impedance of a cap has f on the bottom, and f=0 for DC, so Xc is infinite for DC, hence open circuit.
Xc = 1 / (2pi f C)
Does it matter if that DC voltage is changing, while remaining on one side of 0, and never changing polarity so never becoming AC? To my thinking, f is still 0 if it never crosses 0V, so Xc is still infinite and it's still open circuit.
larryd:
A voltage does not have to cross 0V to become AC.
Well I always thought the crossing of 0V, the reversal of polarity, was the very definition of alternating current.
(Recent discussions on the current rating of switches for example, say that the AC current can be higher for the very reason that AC changes polarity and continually puts the fire out.)
Perhaps don't think of it as a straight DC voltage or AC, but as a change in charge. A capacitor doesn't necessarily know ground reference, only a difference in voltage whether it's from 0V to +10V or -5V to +5V, only that both have a 10V swing. DC isn't instantaneous, and any capacitor acts as a short even to DC until it plateaus (TC). Look at PWMing mosfet gates. Why use a resistor in series with the gate while using an Uno. It has 40mA's of current available, and the mosfet is voltage controlled, right?
kenwood120s:
Well I always thought the crossing of 0V, the reversal of polarity, was the very definition of alternating current.
(Recent discussions on the current rating of switches for example, say that the AC current can be higher for the very reason that AC changes polarity and continually puts the fire out.)
I think you need to realise when we talk about AC we refer to, AC CURRENT , so when you vary the voltage up and down on a capacitor, you are charging and discharging it.
This means the current flows IN and then OUT.
So even though the varying voltage is always one polarity the current is not.
Tom...
kenwood120s:
say that the AC current can be higher for the very reason that AC changes polarity and continually puts the fire out.)
Do you mean quenches the spark?
Yes because the direction of current flow is changed and so eliminates the ionised path that started as the contacts were just broken and a lot closer together then they are when the current changed polarity. This means the voltage is now insufficient to ionise the air inbetween and so the spark does not start with the current going in the opposite direction.
TomGeorge:
I think you need to realise when we talk about AC we refer to, AC CURRENT , so when you vary the voltage up and down on a capacitor, you are charging and discharging it.
This means the current flows IN and then OUT.
So even though the varying voltage is always one polarity the current is not.
Tom...
That's what I've been trying to visualise, especially that last part "even though the varying voltage is always one polarity the current is not.
I'm not seeing that if a cap is charging with DC from left to right so to speak, and that DC voltage decreases, then the cap discharges. In my mind it still charges, just not as quickly. With current by definition being the flow of charge, if the current is left to right, albeit at changing voltage, then the cap will carry on charging since charge is still arriving from the left (at least until the cap is fully charged, anyway). It will only discharge when the polarity of the supply changes, and the current arrives at the cap from the right now, meaning the charge that previously arrived from the left is leaving to its left and heading round the to the other side of the cap through the circuit. Is there something wrong with that scenario?
Oh wait maybe me writing that down made me see something else, and I get what tinman was about with the swing thing: let's say a "left to right" current has charged the cap to some degree (perhaps, but not necessarily, fully) and the cap now has a certain voltage. If the supply voltage drops, but still has that left to right polarity, then since voltage is force, the cap's voltage overcomes the now low supply voltage and the net voltage causes a discharge with a reverse current although the supply didn't change polarity.
larryd:
A voltage does not have to cross 0V to become AC..
MarkT and others agree with that, MarkT having said in the thread which spawned this one:
MarkT:
DC is constant by definition. If something is changing it has a AC component.
And TomGeorge said this over there:
TomGeorge:
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.
The consensus being, and which makes perfect sense to me belatedly, that if the direction of the current changes it's AC, even though the polarity of the supply didn't change. If the relative voltages are such that the current reverses, even though they may well both be positive, we have AC. (My original understanding of AC needing to cross zero probably coming from that's the way it is with house power, and an assumption that that's the way it has to be.)
I was happy with that and should have left well alone, but then went scouting around on stackexchange and found this snippet in a reply to a question:
someGuyOnStackexchange:
a time-varying signal is not necessarily AC, even if it is periodic. It must also have a zero mean value (i.e. zero DC component). In electronics sometimes we cut corners and say AC to mean any time-varying signal
TomGeorge:
No just remember
DC == DIRECT CURRENT
AC == ALTERNATING CURRENT
Current not voltage.....
Tom...
Yeah I'm remembering that indeed Tom...
But it seems I'm not alone in thinking that the applied voltage has to have changed polarity for it to be deemed AC. And I'm glad I'm not alone in that thought tbh, so I don't feel to bad about it.
Well it did just mean that in the 1800's, with the wars between DC and AC mains distribution but with the
advent of signal processing using electronics with the thermionic valve (vacuum tube) the meaning must
have changed to encompass varying signals in amplifiers (where DC offsets were commonplace).
Note that an ac signal with a dc voltage offset often has no dc current offset, so the current may be
pure ac whatever...
MarkT:
Note that an ac signal with a dc voltage offset often has no dc current offset, so the current may be
pure ac whatever...
Ok well I'm not even going to try to understand that bit; I have just thought it safe to get back in the water and I'm going to quit while I'm still in the game.
kenwood120s:
I was happy with that and should have left well alone, but then went scouting around on stackexchange and found this snippet in a reply to a question:
And so now I'm back where I started....
No that random guy was wrong. Their are a lot of people on the internet that think, in error, they are correct.
kenwood120s:
The equation for impedance of a cap has f on the bottom, and f=0 for DC, so Xc is infinite for DC, hence open circuit.
Xc = 1 / (2pi f C)
Does it matter if that DC voltage is changing, while remaining on one side of 0, and never changing polarity so never becoming AC? To my thinking, f is still 0 if it never crosses 0V, so Xc is still infinite and it's still open circuit.
The expression for Xc has a negative sign .... Xc = -1/(2.pi.f.C)
If talking about ideal (theoretical) conditions, then DC is associated with physically unchanging signal instantaneous amplitude.
If there's any physical time-changing amplitude, then the signal could 'mathematically' (hypothetically/theoretically - using a mathematical model) be 'considered' (treated) as 1 (or a family) of sinusoidal signals (plus a constant instantaneous amplitude signal - if it exists). For any particular 'AC' (sinusoidal) component having a particular frequency 'f', a capacitor's impedance defined at that particular frequency 'f' is -j/(2.pi.f.C). So, from the perspective of frequency analysis - the capacitor's impedance depends on what frequency we're studying (or focusing on).
For a DC signal, the ideal capacitor IS an open circuit as far as DC signals are concerned. For a purely sinusoidal signal at one particular frequency f, the capacitor will have a particular impedance. It is only when the frequency is considered (by somebody) to be high enough, where the capacitor is considered to be approximately a short circuit. For ideal capacitors, the capacitor is only a short circuit when the sinusoidal signal's frequency is infinite....but we can never reach infinity....(since infinity is infinite). But we usually get the picture. It's approximately a short circuit for AC signals when frequency is sufficiently high (or equivalently - when the impedance of the capacitor is deemed to be sufficiently low enough at the frequency being studied).
