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?

  • \$\begingroup\$ Well, the phase shift of 90 degrees doesn't take place until after the frequency gets high enough, with that circuit. It's 0 degrees near DC, then transitions (over about one decade of frequency change) around where the corner frequency is located (which is in the center of that decade transition span.) Does that much make any sense? The resistor dominating near DC and the capacitor dominating at HF? (Think about the currents relative to the voltages and how a capacitor is a derivative-function.) \$\endgroup\$
    – jonk
    May 25, 2022 at 5:46
  • \$\begingroup\$ Thank you. Yes, I understand that capacitor blocks DC and therefore at DC, there will be no phase shift. But I am still not clear on which frequency will the phase shift be 90deg? Can you explain at which frequency and why is it at that freq the phase shift will be 90deg? \$\endgroup\$
    – user220456
    May 25, 2022 at 5:51
  • 1
    \$\begingroup\$ Do you understand/accept this:$$I_{_\text{C}}=C\frac{\text{d}}{\text{d}t}V_{_\text{C}}$$? \$\endgroup\$
    – jonk
    May 25, 2022 at 6:05
  • \$\begingroup\$ Yes, I understand \$\endgroup\$
    – user220456
    May 25, 2022 at 6:08
  • \$\begingroup\$ The way to read that is that the current through the capacitor (I don't want to belabor the physics of how a current can be through a capacitor when there is a dielectric that prevents electrons from flowing through the dielectric) is proportional to the rate of change of the voltage across it. Right? \$\endgroup\$
    – jonk
    May 25, 2022 at 6:08

1 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.

  • \$\begingroup\$ Just to be exact: An opamp-based integrator (a real one, not a mathematical model) can produce a phase shift of "exactly 90 degrees" at one single frequency only. However, in the vicinity of this 90deg crossing we can expect a phase shift only slightly less or more than 90 deg (+- 2....3 deg or so). \$\endgroup\$
    – LvW
    May 25, 2022 at 10:22
  • \$\begingroup\$ @LvW My first thought was 'pedant!'. My second was 'yes, it could use some improvement, without cluttering the answer'. \$\endgroup\$
    – Neil_UK
    May 25, 2022 at 11:09
  • \$\begingroup\$ The background resp. reason for my comment was your wording "exactly 90 deg". As I can see, you have edited your contribution. Looks good to me. \$\endgroup\$
    – LvW
    May 25, 2022 at 18:25

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