0
\$\begingroup\$

I'm struggling to understand the phase response of a high pass RC filter.

At low frequencies, does the capacitor charge/discharge so quickly that there is a phase lead? But then we should have seen a similar behaviour in the low pass filter.

At high frequencies, in a low pass filter, the capacitor is unable to keep up with the frequency of VIN and charges/discharges at a slower rate resulting in a phase lag. Should we not see a similar behaviour in the high pass filter? On the contrary, VOUT actually gets more and more in phase in the high Pass counterpart which I don't understand.

\$\endgroup\$
  • \$\begingroup\$ The sinusoidal voltage across a capacitor will always be \$90^o\$ out of phase with the current through it - the frequency of the sinusoid is irrelevant in this case. The overall phase shift for a RC network is dependent on the values of R and C, and on the frequency (since the reactance of the capacitor is frequency dependent, and it is the relative magnitudes of resistance and reactance that determine the actual phase shift). \$\endgroup\$ – Chu Jul 15 '17 at 12:27
1
\$\begingroup\$

If you want to understand phase relationships of passive filters you have to buy into the right terminology else you will forever be going back to first principles and that first principal is that current flow is determined by capacitance multiplied by the rate of change of applied voltage

If you then apply sinewaves to that first principle you start to see and understand the phase relationships.

So, for ac applications we just remember the simple acronym CIVIL. It stands for Capacitor:I leads V and V leads I in an inductor (L).

We also remember that the impedance of a capacitor reduces as frequency increases and that the impedance of an inductor increase with frequency.

All proven from the basic formulas for capacitors and inductors but conveniently converted to different pictures when thinking about AC analysis and, if you talk about phase shift you are buying into AC analysis.

To talk about charging and discharging in the samesentance as phase shift is missing the point.

\$\endgroup\$
1
\$\begingroup\$

The current through a capacitor charges and discharges it. A positive current charges the capacitor, the voltage across its terminals rises.

It can be shown that a sinusoidal current produces a sinusoidal voltage across a capacitor. The positive half wave charges the capacitor and produces the rising edge of the resulting voltage, going from the minimum to the maximum. So, when the current reaches its peak the voltage is in the middle of the rising edge. It can be seen that for a sinusoidal waveform the current leads by 90 degrees.

The same behavior can be seen when analyzing a high pass filter. One decade below the cutoff frequency most of the input voltage drops across the capacitor and only a small fraction of it across the resistor. The current through the capacitor is leading its voltage. The output voltage is just the current through the capacitor times the resistor and therefore has the same phase as the current through the capacitor. Therefore the (small) output voltage is leading the input voltage.

For frequencies higher than the cutoff frequency output and input voltage are almost in phase since the capacitor is nearly a short circuit.

\$\endgroup\$
0
\$\begingroup\$

It is easier to understand the phase shift using complex numbers. By Ohm's law: \$ V = Z \times I \$, Z is the impedance, I the current and V the voltage.

The impedance of an inductor is noted \$ j \omega L \$ , with L the inductance value, \$ \omega \$ the frequency in radians per second (i.e. \$ 2 \pi f \$), j is engineering notation for imaginary number \$ i \$ and is equal to \$ 0 + 1i \$, i.e. a vector of amplitude 1 and 90° phase shift. Therefore voltage is ahead of current in an inductor.

The impedance of a capacitor is noted \$ \frac 1 {j \omega C} \$, which equals \$ -j \frac 1 {\omega C} \$ (*). Pay attention to \$ -j \$ here, which is equal to \$ 0 - 1i \$, i.e. a vector of amplitude 1 and -90° phase shift. Therefore voltage lags behind current in a capacitor.


(*) \$ \frac 1 i \$ equals \$ -i \$, just multiply both the numerator and the denominator by i.

\$\endgroup\$
0
\$\begingroup\$

AC current always lags voltage in a capacitor regardless of its use in series or parallel. THe impedance (f) always determines the current flowing thru it with an AC voltage across it.

But the impedance ratio as a function of frequency determines both transfer function and the phase shift relative to say a resistance where the breakpoint is 45 deg when the R=Zc(f) with lagging current then approaches 90deg depending on the position of the cap Series or Shunt and of course freq.

It is easy to visualize with Phasors and math if you understand it.

Perhaps harder is the intuitive understanding.

Transfer function of XY out vs in is shown for both HPF,LPF and Transfer function of each part is shown XY=VI where a resistor is always a 45 degree linear line for V/I slope , but Caps always 90 deg lagging current.

Thus the position of the Cap determines if the phase of the output is partially leading or lagging when the voltage across the cap reduces near 0 above the same breakpoint. We consider Caps as short circuits intuitively when Zc<< R and thus when attenuating in a LPF it is always 90 deg. at high f and in a HPF it is always 0 deg shifted or out=in for a transfer slope depending on scales for XY looking like ~45 deg more or less IF they were equal scales.

So in summary each R and C the XY plots or Y/X= I/V on the Y/X plots and transfer function X/Y= Out/in

So I made a simulation here.

enter image description here

Although sweeping both filters in slow motion, they must span almost 3 decades (50Hz to 20kHz) to show the entire 90 phase shift of each filter, where it shifts most rapid at the break point f-3dB or 0.707 of input.

Note only the HPF transfer function reaches vertical or 90 deg phase shift at max f to match the phase shift of I/V in the cap when it has near 0 voltage across is.

Thus the load V(R) shows the Cap current I(C) going thru it leading by 90 deg relative to the input X or + 90deg. because the cap current is ALWAYS lagging current or Leading Voltage by 90 deg.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.