LTSpice phase angle of RC circuit, why is phase angle increasing with frequency?

[Daniel Hamilton]

Hi I'm really new to electronics so sorry if my question has an obvious answer,

I started using LTSpice today and i've spent far to long trying to figure out why my circuit has a decreasing (increasingly negative) phase angle between Vs an I as i increase the frequency of the source?

I though RC circuits lose their capacitive reactance as frequency increases and become relatively more resistive, thus reducing the phase angle tan^-1(xc/r)
but the software seems to say otherwise
btw i'm not having this problem with RL circuits, the phase angle makes sense there. I've placed the positive end of the probe between the resistor and capacitor. also when i place the probe between the resistor and the source i get a flat line, is this because the two ends of the probe are shorting through the source?

Many thanks!! for any replies, this has me fairly frustrated

 

[Ratch]

Hi I'm really new to electronics so sorry if my question has an obvious answer,

I started using LTSpice today and i've spent far to long trying to figure out why my circuit has a decreasing (increasingly negative) phase angle between Vs an I as i increase the frequency of the source?View attachment 27156

I though RC circuits lose their capacitive reactance as frequency increases and become relatively more resistive, thus reducing the phase angle tan^-1(xc/r)
but the software seems to say otherwise
btw i'm not having this problem with RL circuits, the phase angle makes sense there. I've placed the positive end of the probe between the resistor and capacitor. also when i place the probe between the resistor and the source i get a flat line, is this because the two ends of the probe are shorting through the source?

Many thanks!! for any replies, this has me fairly frustrated

Sometimes, you got to look at the algebra of a circuit instead of a simulator to see what is going on.


As you can see, the imaginary (orthogonal) part of the voltage completely dominates the term when omega becomes large. Even if the magnitude of the voltage becomes smaller.

Ratch

 

[Daniel Hamilton]

thanks a million @Ratch , i guess there is more going on here than I thought, the text i had doesn't even use com numbers, it just uses tan and Pythagoras, I'll see if i can find a more mathematically inclined text tomorrow

with thanks,
Daniel

 

[LvW]

I though RC circuits lose their capacitive reactance as frequency increases and become relatively more resistive, thus reducing the phase angle tan^-1(xc/r)
but the software seems to say otherwise

Without any algebra:
You have plottet the output voltage across the capacitor versus the constant input voltage.
As you know, the voltage across a capacitor will lag the corresponding current through the device by 90 deg.
On the other hand, for rising frequencies the current through the whole circuit will be more and more dominated by the resistor (as you correctly wrote:...becomes relatively more resistive).
That means, the phase difference between the input voltrage and the current will continuously decrease (until both are in phase for very large frequencies).
As a consequence,
(a) the output voltage will continuously decrease (because of the decreasing capacitive impedance), and
(b) the phase difference between input and output voltage will approach -90 deg. for rising frequencies (because the phase difference between input voltage and total current becomes smaller). Of course, the phase difference between output voltage (across C) and current is always -90 deg.

 

[Daniel Hamilton]

This now makes complete sense. I kind of feel like an idiot for not seeing it, I don't know why but I had the idea in my head That phrase angle was only ever between a current and a voltage. Thanks so much @LvW!! At the same time I know this doesn't get me off the hook for not knowing all the relevant math, but it gives me back my confidence to breadboard while I catch up

Thanks guy! I never used forums before, but I'll definitely be using them more now

 

[LvW]

Daniel - here is another recommendation: Without any math, a so-called "vector diagram" can visualize the phase relations between all voltages and the current. Because in your example, both elements are in series, you start with the current vector and add the voltage vectors for the resistor (in phase with the current) and the capacitor (Vc lags by 90 deg). Then, both voltage vectors are geometrically added and the result is the driving voltage. The length of the vector for the capacitive voltage is inversely proportional to the frequency.
This method gives a nice impression on the situation for rising frequencies.

 

[Ratch]

Daniel - here is another recommendation: Without any math, a so-called "vector diagram" can visualize the phase relations between all voltages and the current. Because in your example, both elements are in series, you start with the current vector and add the voltage vectors for the resistor (in phase with the current) and the capacitor (Vc lags by 90 deg). Then, both voltage vectors are geometrically added and the result is the driving voltage. The length of the vector for the capacitive voltage is inversely proportional to the frequency.
This method gives a nice impression on the situation for rising frequencies.

Because voltage has magnitude but no direction, it should not be called a "voltage vector". Voltage has a phase difference with current, so it should be called a voltage phasor.

Ratch

 

[LvW]

Because voltage has magnitude but no direction, it should not be called a "voltage vector". Voltage has a phase difference with current, so it should be called a voltage phasor.
Ratch

OK - agreed.