# Low pass RC filters and phase

## Main Question or Discussion Point

Hi!

Bear with me:
When the frequency is low in an RC series circuit and we take the voltage across the capacitor, the capacitors reactance is high and thus most voltage is across it. But when the frequency is high, the reactance goes down, and little to no voltage is dropped across the capacitor.

I'm pretty much fine with the above. But, when the frequency is low, and most voltage is shared across the capacitor, the phase angle of the system is also low (towards 0 degrees). And when the frequency is high, the phase shift tends towards 90 degrees. This is what I don't get. Isn't there always a phase shift of 90 degrees across a capacitor? Or are we then talking about a 90 degree phase shift between current and voltage in a capacitor, while in this context (RC filters), we're talking about a phase shift of the output voltage vs the input voltage, and don't really care about the phase shift between output voltage and current? Are these two separate phase shifts??

Lastly: isn't it problematic that the phase voltage shifts as frequency shifts? Won't that "distort" the output in some way?

Confusing! :P

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vk6kro
There is always a 90 degree phase difference between the voltages across the capacitor and the resistor, if they get the same current.

However, the input voltage is across the resistor and the capacitor, not just one of them.

It is like this:

[PLAIN]http://dl.dropbox.com/u/4222062/low%20pass%20filter%20vector%20diagram.PNG [Broken]

The input voltage is kept constant. The angle between input and output is shown as theta.

As the frequency increases, the voltage across the capacitor decreases and the resistor voltage increases. Also, theta gets larger, so there is a greater phase shift between input and output.

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Thank you vk6kro, that makes sense!

Follow-up question: How is it that in when the frequency is low, and most voltage is across the capacitor, the phase shift is approaching zero and we effectively have a real voltage (while imaginary if the frequency is high and most voltage is across the resistor). My intuition tells me it should be otherwise..

Think of how long it takes the capacitor to charge. If the signal frequency is low it is changing slowly and a (relatively) small capacitor will charge (relatively) quickly. So the phase change is minimal because the capacitor's effect on the signal is minimal.

vk6kro
Thank you vk6kro, that makes sense!

Follow-up question: How is it that in when the frequency is low, and most voltage is across the capacitor, the phase shift is approaching zero and we effectively have a real voltage (while imaginary if the frequency is high and most voltage is across the resistor). My intuition tells me it should be otherwise..
Don't forget this is only about voltages.

Take the extreme case where there is no resistor and the input is directly across the capacitor.
Obviously, the output voltage and the input voltage are the same and they must be in phase.

Now introduce a little bit of resistance. The output will be almost the same as the input in voltage and phase.

So, can you see that as R increases, the output voltage would drop and the phase shift would get greater?

It is the same if R increases relative to the reactance of C due to frequency shift.
At low frequencies, there is little phase shift and it increases at higher frequencies.

That makes sense, yes!

But in the extreme case with no resistor, and source voltage and output voltage across capacitor in phase, is the current then 90 degrees out of phase with both? (since it must be 90 degrees out of phase with the voltage across the capacitor) ?

vk6kro
Yes, that is right.

The current in a capacitor is 90° out of phase with the voltage across it.

It is like this:

[PLAIN]http://dl.dropbox.com/u/4222062/current%20in%20a%20capacitor.PNG [Broken]

The purple line is the voltage and the yellow sinewave is the current.
Note that the current peak occurs 90° before the voltage peak, so it is said to be "leading" the voltage.

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Thank you! That made things very clear.

While we're on it, is there by the way a simple explanation as to why the current leads the voltage by exactly 90 degrees, and not any at any other arbitrary phase angle?

vk6kro