Hello everybody!

Since i'm new to the forum I guess it's a good idea to introduce myself. I'm 27 year old & from Germany. I'm a hardware designer and mostly work with digital circuits. I'm currently reading "The Art of Electronics" by Horowitz & Hill to freshen up my analog skills.

I stumbled upon a phasor diagram which has been bugging me for hours now, maybe someone can shed some light.

I attached the figures from the book. It shows a

• a phase splitter circuits
• a circuit to use the splitter as a phase shifter
• a phasor diagram to show how and why the shifter works

So \(v_1\) is the voltage in phase, \(v_2\) is 180° out of phase. This can be seen as the two vectors pointing right and left.

To the question(s):

• The vector direction \(V_R\) seems to be chosen arbitrarily. Why is that poosible? In the phasor diagrams I know, a resistor vector is 0° -> along the real axis.
• The vector \(V_C\), by defintion, needs to form a 90° angle with the \(V_R\) vector (Z = -xC). It does that, but in the wrong direction I believe (+90°). To my understanding, a capacitive impedance should be -90°

This may seem like a detail, but that's just the kind of stuff that makes me question my knowledge on the subject. Any help appreciated!

 

Your understanding is correct (at least the drawing you have posted seems fine to me). If you want to make sure you can check out the materials in the VOL. II -AC material in this website.

 

You are trying to compare a 90 degree system to a 180 degree system. Your diagram shows a 90 degree relation between resistance and reactance. The book diagram is showing the relation between opposite voltage, which is a straight angle, 180 degrees as shown in the book diagram. As you adjust resistor.....that operating arch can swing the entire 180 degrees. Also....there is no ground. The signals are compared to each other. Hope that helps.

 

Hello BR-549,
thanks for your reply, but i'm afraid that still doesn't answer my question. You pointed out the opposite voltages forming a 180° angle. I do understand that part, it's the other two vectors VR and VC that are bugging me, specifically why VR is pointing upwards (that could be arbitrarily) and VC forming an angle of PLUS 90 degrees instead of MINUS 90 degrees.

Your explanation is very similar to the one in the book, so I guess you are correct. Could you please try to exlain it in further detail, because i just can't seem to get the approach.

 

Hi,

When you look at the phase shift of a capacitor alone you would see that the current and voltage are out of phase by 90 degrees.
If you put a resistor in series with the cap, if we call the voltage drop across the resistor at zero degrees then the voltage drop across the cap will be at -90 degrees. This makes sense, but notice we are looking across each element, not at the output of the circuit.

When we look at the output of the circuit, the phase could be say -80 degrees. The difference now is that we are referencing the phase shift to the driving voltage itself. When we do that with all of the measurements, we get different phase shifts.

For example, with a series RC circuit driven by a voltage source and R=1 and C=1 and w=6, and with one end of C tied to ground and one end of R being driven by the source, we get a phase shift at the output (referenced to ground) of about:
-80.54 degrees.
Since this also happens to be the voltage across the cap, the voltage across the resistor would be at about:
9.46 degrees.

With two voltage sources it gets a little more complicated, but luckily the original problem it looks like V2 is the negative of V1 so it's a little simpler.
We would calculate the phase shift at the output of the network to start with.
With different circuit arrangements the results could be very different.
However, it looks like they may have just drawn a generalized diagram because the phase shift at the output looks like it should be negative when they have drawn it as positive.
So this means a more exact drawing might be that one flipped over, or rotated (mirrored) about the x axis so that everything is negative. We could do some calculations if you like.

 

For standard phasor convention with CCW +ve phase rotation, the diagram is "incorrect" to the extent that the capacitor voltage phasor appears to lead the resistor voltage phasor - as the OP observes.

Perhaps it depends on the phasor drawing convention being adopted by the author of the source document from which the diagrams are copied.

This may be readily "corrected" by redrawing the phasor diagram to reflect the correct phase displacement between VR and VC, whilst maintaining the intent of the diagram which shows how a phase shift of the output signal at constant amplitude is achieved.

 

Hi,

When you look at the phase shift of a capacitor alone you would see that the current and voltage are out of phase by 90 degrees.
If you put a resistor in series with the cap, if we call the voltage drop across the resistor at zero degrees then the voltage drop across the cap will be at -90 degrees. This makes sense, but notice we are looking across each element, not at the output of the circuit.

When we look at the output of the circuit, the phase could be say -80 degrees. The difference now is that we are referencing the phase shift to the driving voltage itself. When we do that with all of the measurements, we get different phase shifts.

For example, with a series RC circuit driven by a voltage source and R=1 and C=1 and w=6, and with one end of C tied to ground and one end of R being driven by the source, we get a phase shift at the output (referenced to ground) of about:
-80.54 degrees.
Since this also happens to be the voltage across the cap, the voltage across the resistor would be at about:
9.46 degrees.

With two voltage sources it gets a little more complicated, but luckily the original problem it looks like V2 is the negative of V1 so it's a little simpler.
We would calculate the phase shift at the output of the network to start with.
With different circuit arrangements the results could be very different.
However, it looks like they may have just drawn a generalized diagram because the phase shift at the output looks like it should be negative when they have drawn it as positive.
So this means a more exact drawing might be that one flipped over, or rotated (mirrored) about the x axis so that everything is negative. We could do some calculations if you like.

Hey MrAl,
thanks for joining the discussion and getting into the problem. I can follow your example and get the same results. As you and t_n_k pointed out, the diagram is probably flipped. I just couldn't believe a book with a reputation like this would let that slide. As I said in the original post, this is just the kind of stuff that makes me question who's wrong, the author or me. Well, glad it turned out the author is, or at least he follows a convention that I don't know of.

Thanks to everybody who took the time to answer!

 

Hi again,

Yes, and as i think about it a little more, is it possible that the phase diagram was not even intended to show the phase shifts of the other diagram with the circuit with the R and C and two voltage sources? Perhaps it was a generalized diagram for a phase shifter but did not refer to any specific circuit.

In any case, i decided to do a couple calculations on the diagram with the phase shifter circuit with the two voltage sources, and using the other diagram just to set the two voltage source values so that V2=-V1, and for reference V1=1 volt. With values of R=1, C=1, and w=1/sqrt(3) i get the following results:
Vout=1 at -60 degrees
Vr=1 at -120 degrees
Vc=1.732 at +150 degrees

From this i can see that the phase shift between Vr (the voltage across the resistor) and Vc (the voltage across the cap) is 90 degrees.
All the voltages are taken across the respective component from top to bottom, so the bottom is the reference for that measurement.
For the output voltage of course the reference is ground.

I explained all this because i am not sure if the diagram i produced will show up with the new site software. That diagram shows the phases and amplitudes as the original diagram was supposed to do, for those set values.
NOTE: Yeah, the 'new' software will not allow an upload of the diagram yet.
LATER: Ok it works now

 

Hi MrAl,
i am pretty sure the diagram does belong to the circuit, it is a very short subchapter of maybe half a page of text and the images.
When I rearrange you diagram a little, we end up with what we already suggested: A flipped version of the diagram in the book.