Electronic – Phase Shifter: why is simple RC circuit’s output 0 when phase is 90 deg

acfilterphasevoltage divider

I am rereading an electronics book (Fundamentals of Electric Circuits; Alexander) that says that the circuits

Lagging-Output RC Phase Shifter or Leading-Output RC Phase Shifter

have the total impedance

Z = R + jXc

and that the phase shift is given by

theta = arctan(Xc/R)

where Xc = -1/wC

It then says:

We should keep in mind that [these] simple RC circuits … also
act as voltage dividers. Therefore, as the phase shift θ approaches
90◦ , the output voltage Vo approaches zero.

I can see that as theta -> pi/2, Xc/R goes to infinity, meaning that C or R must be very small. In the second example, if Xc is so much larger than R, then I can see that eventually all the voltage will be across it and none at Vo. (Is that what he means by voltage divider?) However, in the first example, if C is very small then Xc is very large, and doesn't that mean that all the voltage will be across it and Vo will approach Vi? Or is C so small that it acts like a short? If the latter is the case, then why isn't C a short in the second example and all the voltage across R?

Best Answer

Let's start by pointing out that circuit 1 is the same as circuit 2 and that measuring an output across the resistor will look like a high pass filter whereas measuring it across the capacitor will look like a low pass filter. The filter will have a transitional point (or 3 dB point or half power point) at: -

f = \$\dfrac{1}{2\pi RC}\$

If you measure across the resistor and the capacitor at this transitional frequency you'll see that the RMS voltage is the same at: -

\$\dfrac{V_{IN}}{\sqrt2}\$

You'll also find that |Xc| = R at this frequency and that the phase shift of the signal across the resistor is exactly 90 degrees from the voltage across the capacitor. If you messed around a bit more you will find that this is always the situation because the current through both components is the same. If current is the same then the voltage across C always lags the voltage across R by 90 degrees.

It's also true that at very high frequencies Xc is very much smaller than R so the voltage across R closely resembles the input voltage whereas the voltage across C will be very small. At very low frequencies the opposite is true - the voltage across C is very similar to the voltage on the input and the voltage across R is tiny.

Only at these extremes will the phase shift produce close to 90 degrees and that will be seen across the component that has the lowest impedance.

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