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I have been trying to understand Bode Phase Plot of RC High pass filter.

enter image description here

I know that from my engineering that voltage lags behind the current passing through the capacitor. So, I assume the current through capacitor will be in-phase with the input voltage waveform and that same current will pass through the resistor. So, output voltage should be in-phase with input waveform.

Can someone explain me the Bode phase plot of the filter when frequency is swept from zero to cut-off frequency via time domain analysis?

enter image description here

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2 Answers 2

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Note that current is actually common to both C and R, not really as shown in the schematic.

If you compare the input voltage (Vin) with the output voltage (Vout), the input amplitude is always >= output amplitude. At low frequencies, output is much smaller than input.
And at low frequencies, phase of Vout leads phase of Vin:
at frequency below cutoff (in stop-band)
At high frequency, well into the passband, Vout amplitude is nearly equal Vin amplitude, and phase of Vout approaches that of Vin:
at frequencies above cutoff (in pass-band)

vectors of circuit voltages

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  • \$\begingroup\$ I can understand that. But if only a capacitor is connected across the voltage source, then current and voltage will be out-of-phase by 90 degree. If a capacitor and resistor are connected in series and phase of the circuit changes by 90 degree gradually, then it means the current through capacitor changes the phase as frequency is swept. I am not able to understand how current through capacitor changes it's phase as it should be always out-of-phase by 90. \$\endgroup\$
    – abhiarora
    Nov 3, 2018 at 9:02
  • \$\begingroup\$ @abhiarora Seems that you should become familiar with vector diagrams. Kirchhoff's rules must be extended to a 2-dimensioned space instead of 1-dimensioned space allowed for resistor-only circuits. (see edit). Remember that the voltage source (Vin) must supply any current demanded by its load. This current can have any phase with-respect-to its well-controlled voltage. Your graph of "phase" applies to the angle of "i" w.r.t "Vin". \$\endgroup\$
    – glen_geek
    Nov 3, 2018 at 14:35
  • \$\begingroup\$ I do understand the vector diagrams. But I am getting confused because I have read the current and voltage through the capacitor will be 90 degree apart no matter what is the frequency of the source. But when we add the resistor, I am unable to understand why this doesn't hold true. \$\endgroup\$
    – abhiarora
    Nov 3, 2018 at 15:26
  • \$\begingroup\$ @abhiarora Notice that the vector Vr IS 90 degrees from Vc - at any frequency. It might help to move the ground point to the junction of C & R (making Vin a floating source). The current common to R & C forces Vr and Vc to take the form shown in the red vector diagram. Current magnitude is proportional to Vr. \$\endgroup\$
    – glen_geek
    Nov 3, 2018 at 16:00
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I know that from my engineering that voltage lags behind the current passing through the capacitor. So, I assume the current through capacitor will be in-phase with the input voltage waveform

I'm not clear how you went from the first sentence to the second one here.

If the voltage lags behind the current, then the current leads the voltage. In any case they are 90 degrees out of phase with each other, so assuming they are in phase is not what you should do.

Only at high frequencies, where the capacitor has very little effect on the circuit, will the current become very nearly in phase with the input voltage (which is applied across the RC combination, not just across the capacitor). And this is exactly what your Bode plot shows.

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  • \$\begingroup\$ I can understand that. But if only a capacitor is connected across the voltage source, then current and voltage will be out-of-phase by 90 degree. If a capacitor and resistor are connected in series and phase of the circuit changes by 90 degree gradually, then it means the current through capacitor changes the phase as frequency is swept. I am not able to understand how current through capacitor changes it's phase as it should be always out-of-phase by 90. \$\endgroup\$
    – abhiarora
    Nov 3, 2018 at 9:03
  • \$\begingroup\$ @abhiarora You can't just say "the voltage". You need to be specific about which voltage. The voltage across the capacitor is 90 degrees out of phase with the current through the capacitor at any frequency. The voltage across the RC combination is not the same as the voltage across the capacitor at all frequencies. The voltage across the RC combination is not 90 degrees out of phase with the current through the RC combination at all frequencies. \$\endgroup\$
    – The Photon
    Nov 3, 2018 at 16:24
  • \$\begingroup\$ And your \$V_{out}\$ is the voltage across the resistor, not the voltage across the capacitor or the voltage across the combination. \$\endgroup\$
    – The Photon
    Nov 3, 2018 at 16:24

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