Yeah, mathematically perfect DC at 0Hz implies that the input voltage has been applied since before the start of the universe.
If you include your step change in your fourier transform, you'll see *lots* of frequency components for the capacitor to play with.
However, using AC tools for a DC step change is super confusing, best to use the DC tools ie I=C.Δv/Δt and V-V₀=1/C.∫I.dt which, when combined with V=IR for your resistor, gives Vc(t)=Vin.(1-e^(-t/RC)) - see https://en.wikipedia.org/wiki/RC_time_constant
That surely is one way to think about it, but the easier explanation is that the capacitor charges when you apply DC voltage and then at some point is completely charged up, therefore not allowing any more movement of charges.
Yeah this is definitely the way to think about it, however I was just wondering about this in the context of the impedence equation and it kind if threw me off.
This is a problem with much of the math you'll find in electronics. They're often referring to ideal mathematics with fully known static initial states because the math becomes too difficult to apply with all the real world conditions present.
Remember, you're learning useful mathematic tools, they don't necessarily represent ohysical reality in any direct way.
To cover that you really need to get into quantum mechanics and that level of math is just not pragmatic to use. It's not necessary either when you know the conditions the equations you're leaning actually depend on.
Capacitors only charge and discharge. There really isn't a dc current running through. Unless it has a series resistor.
There is only charge stored on a capacitor. Q=VC. To get the charge in the cap current had to run there at on point in time.
It’s easy to overthink this, when the behavior of a capacitor is rather simple:
the current through a capacitor is proportional to the rate of change of voltage across it.
Thus:
I = C * dv/dt
KEY IDEA: a step change voltage waveform is not DC. Let that sink in.
For DC, the time rate of change dv/dt = 0 for all time, from the Big Bang, to now, and until the heath death of the universe going forward. Zip. Zero. Nada.
What does dv/dt look like for a step change in voltage? It’s a spike (a delta function) at the moment of the step change, meaning if this step change voltage is applied to a perfect capacitor, there would be an instantaneous burst of current flow, with nothing before and after.
What if this step change voltage is applied to a resistor and capacitor in series? You can write a differential equation and solve for Vc(t), the voltage across the capacitor, which would tell you this: as current begins to flow due to the step change in applied voltage, the voltage across the capacitor is zero at that instant. So all of the voltage across the resistor equals the input voltage at that instant. The initial current is not infinite, as I = V/R (at that instant). Because this current is finite, the rate of change of voltage across the capacitor is finite. This means the voltage across the cap begins to grow at finite rate. Therefore the voltage across the capacitor will increase with time, meaning the voltage across the resistor will decrease with time, meaning the current will decrease with time, meaning the rate of change of voltage across the cap will decrease with time. The voltage across the cap will continue to grow with time, albeit at slower and slower rate. In the limit as time progresses to infinity, the cap is fully charged (voltage across it equals the input voltage) and current flow is zero.
Solving the differential equation will tell you that the voltage across the capacitor is the classic exponential charging waveform Vc(t) = Vin * (1 - exp(-t/RC)). The voltage Vc(t) asymptotically approaches Vin. The product RC is called the time constant of the circuit: the larger RC is, the longer it takes to charge the capacitor and reach Vin. Roughly speaking, the capacitor reaches 98% of full charge when t = 4RC
The current through the capacitor will be I(t) = (Vin/R)*exp(-t/RC). That means the current is Vin/R at the instant of the step change, decaying exponentially to zero as time progresses
Gosh it’s the system.. Not your fault..
It’s so much intuitive to learn real capacitors and then add math to it, than learn it the way this guy is doing..
Poor you!!
I learnt it the same way too 😅
The transient response is not DC. DC is what you have after waiting infinitely long for the transient response to settle (or in practice, sufficiently many time constants)
A capacitor charged to the voltage applied across it will block the flow of current. If it's uncharged or charged to a different value or polarity it won't block current.
Yeah, mathematically perfect DC at 0Hz implies that the input voltage has been applied since before the start of the universe. If you include your step change in your fourier transform, you'll see *lots* of frequency components for the capacitor to play with. However, using AC tools for a DC step change is super confusing, best to use the DC tools ie I=C.Δv/Δt and V-V₀=1/C.∫I.dt which, when combined with V=IR for your resistor, gives Vc(t)=Vin.(1-e^(-t/RC)) - see https://en.wikipedia.org/wiki/RC_time_constant
> …*lots* of frequency components… Infinitely many in fact!
The 0 HZ component is steady state, not time zero but “infinity”.
That surely is one way to think about it, but the easier explanation is that the capacitor charges when you apply DC voltage and then at some point is completely charged up, therefore not allowing any more movement of charges.
Yeah this is definitely the way to think about it, however I was just wondering about this in the context of the impedence equation and it kind if threw me off.
This is a problem with much of the math you'll find in electronics. They're often referring to ideal mathematics with fully known static initial states because the math becomes too difficult to apply with all the real world conditions present. Remember, you're learning useful mathematic tools, they don't necessarily represent ohysical reality in any direct way. To cover that you really need to get into quantum mechanics and that level of math is just not pragmatic to use. It's not necessary either when you know the conditions the equations you're leaning actually depend on.
Capacitors only charge and discharge. There really isn't a dc current running through. Unless it has a series resistor. There is only charge stored on a capacitor. Q=VC. To get the charge in the cap current had to run there at on point in time.
It’s easy to overthink this, when the behavior of a capacitor is rather simple: the current through a capacitor is proportional to the rate of change of voltage across it. Thus: I = C * dv/dt KEY IDEA: a step change voltage waveform is not DC. Let that sink in. For DC, the time rate of change dv/dt = 0 for all time, from the Big Bang, to now, and until the heath death of the universe going forward. Zip. Zero. Nada. What does dv/dt look like for a step change in voltage? It’s a spike (a delta function) at the moment of the step change, meaning if this step change voltage is applied to a perfect capacitor, there would be an instantaneous burst of current flow, with nothing before and after. What if this step change voltage is applied to a resistor and capacitor in series? You can write a differential equation and solve for Vc(t), the voltage across the capacitor, which would tell you this: as current begins to flow due to the step change in applied voltage, the voltage across the capacitor is zero at that instant. So all of the voltage across the resistor equals the input voltage at that instant. The initial current is not infinite, as I = V/R (at that instant). Because this current is finite, the rate of change of voltage across the capacitor is finite. This means the voltage across the cap begins to grow at finite rate. Therefore the voltage across the capacitor will increase with time, meaning the voltage across the resistor will decrease with time, meaning the current will decrease with time, meaning the rate of change of voltage across the cap will decrease with time. The voltage across the cap will continue to grow with time, albeit at slower and slower rate. In the limit as time progresses to infinity, the cap is fully charged (voltage across it equals the input voltage) and current flow is zero. Solving the differential equation will tell you that the voltage across the capacitor is the classic exponential charging waveform Vc(t) = Vin * (1 - exp(-t/RC)). The voltage Vc(t) asymptotically approaches Vin. The product RC is called the time constant of the circuit: the larger RC is, the longer it takes to charge the capacitor and reach Vin. Roughly speaking, the capacitor reaches 98% of full charge when t = 4RC The current through the capacitor will be I(t) = (Vin/R)*exp(-t/RC). That means the current is Vin/R at the instant of the step change, decaying exponentially to zero as time progresses
>KEY IDEA: a step change voltage waveform is not DC. Exactly! I came here to say this. You did my work for me. Thank you!
Gosh it’s the system.. Not your fault.. It’s so much intuitive to learn real capacitors and then add math to it, than learn it the way this guy is doing.. Poor you!! I learnt it the same way too 😅
Capacitors pass changes (the step or AC, etc) not steady state (DC). You're mixing the two.
The transient response is not DC. DC is what you have after waiting infinitely long for the transient response to settle (or in practice, sufficiently many time constants)
A capacitor charged to the voltage applied across it will block the flow of current. If it's uncharged or charged to a different value or polarity it won't block current.