# Understanding Working of High-Pass RC Filter

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. 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?

• In 2 I think you wrote "resistance" where you meant "frequency". In 3, "ripples" is not the best word. In fact, any (ideal) 1-order RC pair will have a smooth pass-band just like the one you've plotted. – Vicente Cunha Jun 18 '18 at 13:50
• Just a hint: if you switch the Y-axis to linear (instead of dB, default), you'll see the three bands more clearly. – a concerned citizen Jun 18 '18 at 14:57

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.

• I have one more question. The output voltage is due to common current? Then, the output voltage will be always in phase with input source. What I think it is due to phasor subtraction of input voltage and voltage across capacitor (which is 90 degree delayed and of lesser intensity than input voltage) – abhiarora Jun 18 '18 at 13:56
• The output voltage will be in phase with the common current that flows but, that common current is not in phase with the input voltage except when the frequency is very, very high. In the transition band (where Xc = R) the current will lead the input voltage by 45 degrees. – Andy aka Jun 18 '18 at 14:15
• Can you explain that common current won't be inphase with input voltage? – abhiarora Jun 18 '18 at 14:59
• It's because the combined impedance of C and R has both reactive and resistive elements. This means the impedance is a complex value that has a phase angle of -45 degrees (Xc magnitude = R). If you are unused to the idea of an impedance having a phase angle, then you might want to do some background research before I progress further. – Andy aka Jun 18 '18 at 16:27
• I do understand the impedance has magnitude and phase angle. But what I was asking is if current is flowing through the capacitor and resistor and current phase is not altered by capacitor and resistor, then it should be in-phase with applied input source. – abhiarora Jun 18 '18 at 16:30

Yes, you are correct. For better understanding I recommend this link:

https://www.electronics-tutorials.ws/filter/filter_3.html

Although I would like to add that Voltage across the capacitor is always a phasor perpendicular to the Voltage across the resistor. So you can imagine a right angled triangle with the base being the V across R and the perpendicular being V across 1/(frequency times C). Initially, at $\omega$ = 0 or small (in the stop band), the perpendicular is infinitely long hence the angle that the hypotenuse makes with base is close to 90 degrees. As omega takes a finite value that is not too large(in the transition band) the length of perpendicular decreases, and the angles reduces to an acute angle. Finally, as omega becomes large (in the pass band), the angle gets close to zero. The angle is what we call the phase and its variation is the phase response.