RC 90 deg phase shift

Question

What does 90 deg phase shift actually mean?

Like, consider this below RC circuit.

^{simulate this circuit – Schematic created using CircuitLab}

From what I understand, when I apply a sine wave at the input node, the output will be 90deg out of phase with the input. I read that capacitor takes some time to get charged through the resistor so as the voltage is built up. So, that time is what creates the phase difference. Am I correct?

So, can I understand like, whenever we add a capacitor, it will cause a phase shift of 90deg? Capacitor and inductor will cause the same 90deg phase shift right?

Answer 1

With a single stage RC lowpass filter, you'll get a phase shift that varies between the limits of 0 degrees (at DC) and 90 degrees (at infinite frequency), passing through 45 degrees at the frequency corresponding to the RC time constant. At this corner frequency, the amplitude of the output has fallen to 0.7071 times the low frequency value, which is roughly -3dB.

For the shown 10k and 1 uF values, the RC product, aka time constant, is 10 ms, giving you a corner frequency of 100 radian/s or roughly 16 Hz (100/2π).

For 'back of the envelope' calculations, the phase shift at half and twice the 3dB frequency can be approximated by 30 and 60 degrees, while the difference from the asymptotic 0 or 90 value halves for every further halving or doubling of the frequency (OK, the 30 is nearer 26 degrees, and it's not strictly halving, but the approximation is easy to remember, and is good enough for freehand Bode plots and for rough calculations on filters and PLLs)

With your 16 Hz 3dB frequency ...

freq (Hz)	approximate phase shift (deg)
1	3.75
2	7.5
4	15
8	30
16	45
32	60
64	75
128	82.5

With an RC filter, you can only approach, never reach, 90 degrees.

If you want nominally 90 degrees at all frequencies, then you need an integrator, using a current source into a capacitor, often realised as a fed-back opamp configuration.