Basic high pass filter phase calculation

[Qacer]

I apologize for asking this basic question. I just don't know what I'm doing wrong with my calculations.

I have a simple high pass filter: capacitor in series with a resistor. The output voltage is taken at the node shared by both components.

I plotted the phase angle (measurement point at the shared node) of this circuit with PSpice from 1 Hz to 10 MHz. The simulation shows that at 1 Hz the the phase angle is about +90 degrees.

However, my calculations give me the opposite:

Zc = 0 - j (1 / (w*C)) , w = 2*pi*f, R = 10k, C = 10pF
Zr = R - j0

Zt = Zc + Zr = -j(1/(w*C)) + R

So at 1 Hz, Zt = 10,000 - j 15915494309.2 = 15915494309.2 /_ -89.9 degrees ...

By looking at my calculation, it seems to make sense to me since the voltage at the shared node should be lagging the main source by 90 degrees. However, PSpice shows a different value.

What mistake am I making?

The reason why I'm trying to do this calculation is because I was reading an article about op-amp stability compensation. It mentioned that low pass filters subtract from the circuits total phase angle, while high pass filters add. I wanted to see the calculations for myself.

Thanks!

 

[BrownTown]

Well, whipping out my trusty old analog filter design book I do see that high pass filters appear to add to the phase, and low pass filters subtract, but your calculations also seem right, so give me a second on this. Also, obviously you know this, but 1Hz isn't gonna make it across a capacitor like that, a signal at that frequency is gonna be annhiliated by that filter.

EDIT: well jsut drawing he circuit diagram and doing KVL i see that that gain is RCs/(1+RCs) which indeed is a high pass circuit. And as we see, at low frequency you get ~90 degrees, and at high frequencies you get ~0 degrees. Just plug some random numbers into your calculator is easiest.

in other words, the mistake you are making is that you are just adding the comonents together, this gives you the total imedance, but what you want is the transfer function, so you have to do KVL, or jsut use a voltage divider to get Vo = (R/(R+1/(Cs)))V, than multiply by the Cs term to get RCs/(1+RCs), if you want then you can substitute s = jw and multiply by the actual angular frequencies.

 

[Qacer]

Ahhh! My brain goes numb at times. The plot that I was looking at is in terms of the voltage phase. What I have is just the impedance. Thanks! Now, I can sleep better. I tend to roll around my bed if I can't figure something out.

 

[tidehigh]

i miss electronics courses!