Current Flow in an (AC) RC Circuit (Low Pass Filter)

Hi!

Ever since Veritasium’s video on current flow I have been re-examining and questioning everything i “know”. I was looking into Low Pass filters - I have always applied the formulas for the 3dB cutoff point I need, maybe used a simulator, got the result and called it a day! I am comfortable with the math if need be or to use a phasor diagram. However, I was thinking about the current flow and I fear I am missing a fundamental concept

let’s take this circuit

and running it through the simulator, we can see this

From t=0

  • Vin increases in the positive direction, as such current starts to flow, and the voltage across the capacitor increases as charge builds on the plates.
  • as long as Vin > Vc (Vc = Vout), the voltage on the capacitor will continue to rise, albeit after the peak of Vin this will slow
  • the current through the resistor is essentially in line with Vin
  • When Vin = Vc, (around t=6ms), the capacitor will begin to discharge. This will mean the current will reverse - and indeed, at 6ms Ir goes negative - this implies that despite Vin still being positive and pushing current into the circuit, it is being opposed by the voltage on the capacitor, as the voltage on the capacitor is higher it is forcing current back into the voltage source

my question here is, what is the real world implications of this? I have been thinking about various voltage sources, like an AC generator, and how that would handle current coming back in when it’s still generating a positive voltage and trying to drive current. This lead me to think about smoothing capacitors in rectifiers - however in this case, there are diodes that prevent reverse currents and so the smoothing capacitor will discharge into the load - RC filters don’t appear to do this.

I pulled out my Art of Electronics to see what they had to say on the topic, but didn’t find much on the implications of the current flow. I can follow the math and can draw the waveforms I see - however thinking about the actual current flow here has got me stumped i must admit!

I have seen examples where the RC filter is driving an op-amp; an op-amp having an extremely high (ideally infinite) input impedance, won’t accept current in, and so in this case again, the discharge current from the capacitor in the filter must go back toward the source.

When the capacitor discharges, for a period in time whilst Vin is still positive, we have two voltage sources in parallel with different voltages which is generally not considered a good idea. The resistor in between will create a voltage drop of IR - if we look at the point 7.8ms, the current through R1 is -1A, with a 30 ohm resistor we would expect to see -30v across R1 and indeed Vin - Vout is - 30v. However the current must still flow somewhere and as far as I can see it can only go back into the source

I had thought that as there was no load, it doesn’t really make sense - sure you can construct the circuit without a load (and again, the only place the discharge current can flow is back to the source; this would also be the same case as connecting an op-amp with high input impedance), but to be sure I can add a load; the voltage across is still Vout = Vc, but we can see a current through that follows Vc/Rload - which makes sense at face value, but again, not if I think about where the currents are flowing!
image

As the capacitor is discharging, we know there must be a current flowing! but where does it flow to? without a load, it can only go one way; with a load there is a discharge path, but then i’d expect the current through R1 to change accordingly…

I have been thinking back and forth on this for some time! I am really, really sure I have missed a fundamental concept here - I can follow the math to get the answer, but conceptually it seems to be messing with me!

I have been thinking about this and this is what I came up with; i’d appreciate any feedback on my analysis

  • as is always the case with voltage, it’s the difference in potential between two points that matters; if there is a potential difference, there will be an electric field and thus current (opposite the direction of electric field of course).

  • an electron doesn’t care what else is around, all it knows is “there is an electric field, i’d better move!”, and since Vcap > Vin, there will be a potential difference and the source just has to suck it up and deal with it!

  • if we remove the load from the circuit (for example we could be feeding an (ideal) op-amp), then there is only one way for the current to go, back through R1 an into the source. Anything else is impossible

  1. let’s consider an AC power source from a coil rotating in a magnetic field with slip rings
  • the rotating ring generates an EMF proportional to the number of turns, the magnetic field strength and the speed of rotation.

  • the generated EMF will exist across the coil, when connected to a load a current will be produced which is proportional to the circuit impedance as a function of time. The current that flows is only due to the fact that there is a potential difference across the coils and thus a field is generated in the wire that causes current to flow.

  • when Vcap > Vin, that current reverses due to the reversed potential difference, the fact that Vin is positive is actually irrelevant. the relative voltage is lower than Vcap… the fact that Vin is positive just means the relative potential difference is smaller.

  • being a simple coil of wire, the reversed current will just cause electrons to flow from the capacitor through the resistor, through the coil and into other side of the capacitor resulting in, over a few time constants, the capacitor returning to electrically neutral on both plates. this would be consistent with KCL

  • if we add a load, then we have have two loops for the capacitor to discharge through, back through R1 as above but also through the load. The discharge current will flow proportionally to the impedance of each loop. But I(R1) + I(R4) = I(cap) during the discharge part of the cycle. And the result is the same. And indeed if I look at the absolute value of the currents, I(cap) = I(R1) + I(R4)

  1. I looked at when the input is a square wave, i.e.: this circuit
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    for this we get waveforms as below (the current being I(R5) - and we clearly see the negative current spike when the 555 switches off; again, this has no where to go but back through the output pin

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looking at the circuit internals for a 555,

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we can see that when the output is 0v, any reverse current will be passed through to circuit ground harmlessly.

assuming I am right in my analysis (which I think I am! and feel free to correct me if i’m not) - my intuition was right all along. And you do need to consider what you are connecting capacitors to, they will generate negative currents and you have to deal with that. And in fact, this is exactly the same with inductors - we are used to putting flyback diodes in to snub the reverse voltage generated with the electric field collapses!

Your intuition is correct, though because of R1 (30 ohms) in your filter there never is a time during which two voltage sources are truly in parallel.

When the capacitor discharges, for a period in time whilst Vin is still positive, we have two voltage sources in parallel with different voltages which is generally not considered a good idea. The resistor in between will create a voltage drop of IR - if we look at the point 7.8ms, the current through R1 is -1A, with a 30 ohm resistor we would expect to see -30v across R1 and indeed Vin - Vout is - 30v. However the current must still flow somewhere and as far as I can see it can only go back into the source.

Indeed, connecting two voltage sources in parallel is generally not a good idea unless they have virtually identical characteristics, because any imbalance in voltage will create a current that generates heat in the conductors and in the internal battery resistances. These internal resistances are distributed and, in many cases, rather complex to model.