Electronic – Understanding Working of High-Pass RC Filter

capacitorfilterhigh pass filterphaseresistors

I was trying to visualize what happens when frequency is swept from 1Hz to 1GHz. In case of capacitor, voltage lags behind current by 90 degree and capacitor reactance depends upon it's frequency.

I have tried simulating the RC filter using LTSpice. The picture is given below but my question is very general and for theoretical understanding.

enter image description here

I want to understand the phase response of the filter. Since, voltage lags behind the current in capacitor (which is connected in series with voltage source). Therefore, they will be some voltage across capacitor and it's intensity will depend upon the frequency (decreases with increase in frequency). I have tried to understand the RC filter working. I have tried to explain what happens in RC filter when frequency is swept from 1Hz to 1GHz.

  1. Stop-Band:
    The frequency will be low and reactance of the capacitor will be very high. All of the voltage will be across capacitor and will be delayed by 90 degrees. The output (voltage across resistance will be nearly zero).
  2. Transition-Band:
    The reactance of the capacitor will start to decrease relative to the resistance. The voltage at output (across resistance) will be subtraction of Voltage source's voltage and delayed version of voltage source (90 degree). Therefore, overall phase of the system will be less than 90 degrees.
  3. Pass-Band:
    In this band, reactance of capacitor will be nearly zero and hence, there will be 0 phase delay. Although there will be some voltage across capacitor which will appear as ripples in the pass-band.

Is my understanding correct?

Best Answer

In the transition band we can say this is typified when Xc = R but, it isn't a simple numerical addition of Vc + Vr = Vinput. We have to use Pythagorus: -

$$V_{input} = \sqrt{V_C^2+V_R^2}$$

This is because the voltage of the capacitor and voltage of the resistor are always 90 degrees relative to each other when they share a common current.

Another point to note is that when Xc = R in value, Vc and Vr will be equal in value (not phase of course) and, using the formula above and re-arranging, you can prove that the output voltage (Vr) is \$\sqrt2\$ lower than the input voltage. This of course is known as the 3 dB point of the filter because \$\dfrac{1}{\sqrt2}\$ is -3.01 dB.