Negative phase shift in low pass RC filter but positive in high pass? In the context of pure capacitance circuit

Question

That there is a 90° phase shift where "current leads the voltage" in a "pure capacitance" circuit such as in schematic below, is easy to understand. It also fits 100% with the hydraulic analogy of it, an elastic rubber sheet. In both hydraulic and electric circuits, the current will flow when the voltage changes, and not at the peaks. It is the derivative of the voltage.

So, I am capable of understanding that.

But going from pure capacitance to RC filters, I read that:

"In the low-pass case, the output of the filter lags the input (negative phase shift); in the high-pass case the output leads the input (positive phase shift). " (source)

This to me makes no sense, and I do not know where to start to make sense of it. There must be a very easy way to make sense of it, and so I figured I would ask. I understand high pass and low pass filters effect on the amplitude, e.g., why they are actual filters, and I understand them at a circuit level too. But, not why they would have reverse phase shift.

The phase shift curves used in examples seem to mix what parameters they show, but they all show the same thing. Some use Vs, voltage source, others Vc, the voltage at the capacitor, but those are equivalent in the context in my understanding (the capacitor reaches its fully charged state when Vs is at its peak. ) To simplify, I will use Vs, and then Io, the current measured at the output. In the low pass RC filter, and the "pure capacitance" circuit, the effect is the same. But, in the high pass RC filter, the measured current is in the reverse direction. But it must still be flowing from capacitor plate closest to Vs, and away from Vs, when Vs starts to decrease from its peak.

Based on the circuits for low pass and high pass RC filter, I do not understand why current measured at output is reverse in one case.

Based on my assumption that the current must be reversed since that is what it seems to measure, in a "pure inductance" circuit, the current is the opposite to "pure capacitance" because the inductor will oppose the current as the voltage is increasing towards positive peak, and then it will oppose decreasing the current. This makes the current “lag” behind the voltage. If a "high pass RC filter" has an output phase shift similar to an inductor, what is causing this? My source for the claim that it has is still this article.

Adding another image to explore why the current is reversed, I drew a little output circuit onto the schematic in the image above, for both low pass and high pass RC filters. And, specified a point t where a measurement is made. Unless I misunderstand something, it must be measuring the reverse current direction in either filter. Starting to get it now.

My interpretation is that Vc and Vs will cause currents to flow in opposite directions along output as shown in the circuits below.

This seems to be equivalent to these circuits, that measure opposite voltages at the output.

Edit: I finally understood, and posted an answer. The reason it was not directly intuitive to me was that the causes for the phase shift are entirely different in the low pass filter.

Answer 1

When you have a circuit with R and C in series, as a whole it does not matter which way they are connected, there are identical currents flowing in both circuits and it also means that voltages over the matching components are identical.

It's just that the point of view is different.

So if capacitor voltage lags the capacitor current, it means that the capacitor current leads the capacitor voltage. The resistor simply works under Ohm's law and converts current into voltage. Thus if capacitor voltage lags resistor voltage, the resiator voltage leads the capacitor voltage.

In the RC low pass filter, the output voltage gives you the actual voltage waveform of the capacitor.

In the RC high pass filter, the output voltage is the voltage of resistor, but due to Ohm's law the resistor voltage is actually related to the current running through both components so output voltage gives you the actual current waveform of the capacitor converted into voltage.

Answer 2

To understand why in low pass filter the output voltage lags the input try read this:

RC Circuit Current

AC Circuit Having Only Capacitor

No, we can see what is going on in the high pass filter (HP). In this case, we again will have the same situation with the capacitor voltage. The voltage across the capacitor will lag the input voltage exactly the same way as in the low pass filter. Also notice that in the HP filter the instantaneous value of an output voltage is equal to:

\$V_{OUT}(t) = V_{IN}(t) - V_{CAP}(t)\$

And the problem to see what is going on is that the input voltage constantly changing its value. But the key point here is to see that the capacitor voltage lags the input voltage. And because of this the output voltage will "peak" early than Vin peak is.

For example, if Vin reaches 0V, and capacitor in a previous half-cycle manages to charge to some positive value. The output voltage will be negative earlier before Vin reaches the negative half. For example when Vi approaching 0 from positive side --> Vout = Vin - Vcap = 2V - 1.8V = 0.2V but when Vin = 1V, the output voltage is Vout = 1V - 1.9V = -0.9V and for Vin = 0 we have Vout = -1.8V. But now the Vin is also negative thus Vout will continue to grow in a negative direction. And we get the peak value at the output when Vcap =0V. But it will occur sooner than Vin negative peak. because when Vin is at a negative peak the output voltage will be lower due to voltage drop across the capacitor voltage (capacitor manages to charge to some negative value).
A hope that you see why output voltage leads the input voltage. As you can see it is not so easy to see in the time domain.

Here you have a better explanation about phase shift in low pass filter

How are current and voltage out of phase in capacitive circuit?

So, in the case of a HP filter when Vin voltage is equal to the capacitor voltage, the current is 0A thus, the output voltage is 0V as well. So we get Vout = 0V sooner than Vin reaches 0V. Because Vin will reach Vcap when Vin decreases from the peak value. And we geting the peak value at the output when Vcap is at 0V.

Also, you can play with this simulation

Answer 3

Try using the picture you uploaded and add resistors. Now you can see that the lowpass has a series L and a highpass a series C, and adding a resistor as the output will mean there is no longer a purely imaginary part, but a real part, as well, thus the phase shift will be lessened, but in the same "direction".

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Edit: Given your recent edits and comments, I thought I'd expand a bit.

First, (purely) reactive elements have a reactance (XL for L, and XC for C), which is nothing but a resistance that varies with frequency: proportionally for the inductor, and inversely proportional for the capacitor. That means that the value of the reactance will increase with frequency for the inductor, and will drop for the capacitor.

Then, consider the modified image I've shown above: the two L+R and C+R circuits form a lowpass and a highpass filter, respectively, because the output resistance and the L or C reactance will form a voltage divider that varies with the frequency. Since XL increases, the output will be lower and lower as the frequency increases; the reverse for XC. Therefore the L+R circuit will let lower frequencies pass more, and higher frequencies pass less, so it's called a lowpass; similarly for the C+R.

Now that we established which one is a lowpass and which one is a highpass, think of what happens from the output's side. Those LR and CR circuits have their outputs taken from across the output resistor, both. That means that whatever current flows through the L or C will appear on the resistor, and the (ideal) resistor's voltage will have the same phase as the current through it. That menas that in a lowpass (L+R) the current will lag and, thus, the output voltage will lag; similarly, for the highpass (C+R), the current will lead, so the output voltage will lead. And the currents are lagging or leading due to the way the reactive elements affect it: the inductor applies an integration (for a sine voltage input the current will be a negative cosine), the capacitor a differentiation (for a sine voltage input the current will be a cosine).

Now consider your two approaches: R+C and C+R. For the C+R case, the output is taken from across the resistor, and that is what we discussed above. For the R+C case, the output is taken from across the capacitor, and this time the voltage will no longer have the same phase as the current: it will be the integral of it. But since the current first goes through R, it will have no phase alteration thus, for a sine input, the current will be a sine. And the integral of it will be a negative cosine, lagging, which is exactly the case as the L+R -- therefore it's a lowpass.

You can very easily verify these things using a simulator. There are plently of free choices, and even the most basic ones can deal with RLC elements. I won't post any pictures, instead, I'll leave you to find out for yourself. A picture may be worth a thousand words, but a simulation will provide the understanding -- this is what simulators are all about: insight.

Answer 4

there is a 90° phase shift where "current leads the voltage" in a "pure capacitance" circuit such as in schematic below

There is a phase shift between the the voltage and current. Look carefully at your schematic. That is the only interpretation that can be given of phase shift in that schematic.

But going from pure capacitance to RC filters, I read that:

"In the low-pass case, the output of the filter lags the input (negative phase shift); in the high-pass case the output leads the input (positive phase shift).

The phase shift that they are referring for filters is the phase shift between input voltage and output voltage. A filter has two ports, an input port and an output port. The input port consists of an input signal AND a reference voltage (usually ground). The output port consists of an output signal AND a reference voltage (usually ground). Thus, two voltage signals are being compared.

Note that in the single capacitor case in your schematic, it doesn't make sense to label one side of the capacitor "input" and the other side "output". The signal is applied to both sides of the capacitor. The capacitor has a single port.

Your schematic lacks a ground symbol. If you placed a ground symbol in your schematic, you might, at first, think you could take the voltage on the left side (relative to the ground) as your input, and the voltage on the right side relative to ground on as the output. But you would soon see that at least one of the sides of the capacitor is at ground, wherever you attached the ground symbol to your circuit in the schematic. Thus, one side, would always have zero voltage relative to ground.

Take away. It is important to distinguish cases where the voltage and current are compared from cases where an input voltage and an output voltage are compared.

Answer 5

In a high pass filter, the phase shift increases as the wave length of the AC generator increases, because the capacitor impedance increases. This is logical, and it is the same as in a “pure capacitance” example, current flows only when the voltage is changing, the derivative of the voltage, and leads the voltage. When impedance decreases with higher frequency, the capacitor does not have as much time to charge, and the current is not impeded since it never reaches the charged state, and the phase shift effect does not take place.

In the case of the low pass filter, the actual case of the phase shift is very different. Here, it is caused by a delay in charging the capacitor. The output is measuring the voltage or charge on the capacitor. Whereas in the high pass filter or “pure capacitance” circuit, it is caused by that current only flows when the capacitor is either charging or discharging, but not when it is charged. The phase shift is capped to at most 90° because the capacitor can at most reach its fully charged state at the point just before where the wave reaches 0.