# What is the Real AC Voltage Phase Shift Across A Capacitor??

Discussion in 'General Electronics Chat' started by ozsavran, Jul 30, 2016.

## What is the Voltage Phase Shift Across A Capacitor in a series RC Circuit with AC Power.

1 vote(s)
20.0%

3 vote(s)
60.0%

0 vote(s)
0.0%

1 vote(s)
20.0%

0 vote(s)
0.0%

0 vote(s)
0.0%

0 vote(s)
0.0%

0 vote(s)
0.0%
1. ### ozsavran Thread Starter New Member

Jul 30, 2016
21
1
Dear Guys,

Here is a really fundamental and elementary question for you.

A- When you charge a capacitor with a DC voltage source, you see opposite charges accumulate on either side, and measure as such relative to your ground.

B- When you apply an ordinary AC sinusoidal signal with equal positive and negative peaks to the same capacitor, both scopes and simulators show simultaneously positive or negative voltages on both sides, with a slight voltage drop related to its reactive impedance Xc. Even when you take down frequency as low as 0.1Hz or 10 seconds each cycle, or change capacitor sizes between picos and micros same thing is observed.

Question:
=======
Shouldn't there be a 180 degree (pi radians, half a cycle) voltage phase difference across a capacitor when applied AC. When one side of the capacitor is at positive peak of the wave, shouldn't the other side be at negative peak?

No book, or web source I digged gave me a satisfactory explanation as to the Physics of what is going on here. Even if its related to charge discharge times, still both sides should not show same polarity simultaneously, and no difference is visible with same source with different size caps. Measurement direction is the same in both DC and AC applications too.

Can a few people take a stab at this glaring gap in my humble little electronics knowledge chain please? It is one of those painful ones that stops most progress in my small head.

2. ### #12 Expert

Nov 30, 2010
16,663
7,310
Which model do you think in, electron flow or conventional current flow?

3. ### #12 Expert

Nov 30, 2010
16,663
7,310
15 minutes...no answer. Does that mean you don't think about current flow?

We can't get anywhere without considering current flow. Maybe that's where you should start.
It's 3 am here and I'm going to bed if you don't answer.

Logged off? That was a short conversation.

Last edited: Jul 30, 2016
4. ### crutschow Expert

Mar 14, 2008
13,473
3,361
Your question is somewhat confusing, but I think you must be talking about a DC block capacitor which is feeding a high impedance load relative to the capacitor reactance at that frequency.
In that case there is only a small charge flow through the capacitor which means there is very little voltage change across the capacitor and thus both sides of the capacitor have essentially the same in-phase voltage.

But if you do lower the load impedance (or frequency) sufficiently so that the capacitive reactance becomes comparable to the load impedance then there is sufficient charge flow to change the voltage across the capacitor and you will see a voltage drop and phase shift across the capacitor.

An LTspice simulation of two RC circuits is shown below.
The reactance of the 1μF capacitors @ 1kHz is ≈159Ω

Out1 has a high output impedance load relative to that reactance so V(out1) is basically equal to V(in).
The current through and the voltage drop across C1 are low.

Out2 has a low output impedance load relative to that reactance so V(out2) has significant attenuation with a leading phase-shift of near 90°.
The current through and voltage drop across C2 are relatively high.

Last edited: Jul 30, 2016
5. ### Jony130 AAC Fanatic!

Feb 17, 2009
3,990
1,115
Why would the capacitor behave differently in AC ?

First the AC voltage

And the capacitor at AC voltage

For the positive half cycle (AC voltage change from 0V to its peak value let say 10Vpeak) the Capacitor is in charging phase (capacitor current flow into capacitor). Now AC voltage change from 10V to 0V and in this phase the capacitor is discharging (capacitor current flow out off the capacitor) with the AC signal change rate. And at the end of this positive half cycle AC and Vcap reach 0V.
Now we have a negative half cycle, the AC input change from 0V to -10V and this time the capacitor is again in charging phase but this time in opposite direction. Now AC input voltage change from -10V to 0V so the capacitor is discharging (current flow from capacitor into AC input source).

6. ### ozsavran Thread Starter New Member

Jul 30, 2016
21
1
Dear Guys,

Many thanks for listening to me and replying so quickly. Thanks for not ridiculing me especially. I am a physicist by creation more than an electronics guy. Can't take anything at just face value.

#12 :
-------
To be perfectly physical and visual I am talking about electron flow.

Crutschow :
----------------
Not sure how a DC block capacitor is different from an ordinary one, or what it is exacty. But I will try to do your simulations in LTSpice myself to get to the bottom of this basic question.

Jony130 :
-------------
My thoughts exactly. Why should it or how it could behave any different? Yet both attached pictures of DoCircuits simulation and my exact same lab measurements with 2 different scopes with 2 different signal generators show the same result from 1khz to 0.1Hz. Changed resistors from 1 ohm to 10k ohm, changed caps from 10nF to 4700uF. Results are the same. Both sides of the capacitor show the same polarity simulatenously as shown below. Opposing sides should be at polar opposites and show 180d voltage phase difference in theory. Yet here it is.

File size:
56.6 KB
Views:
18
File size:
82.7 KB
Views:
16
7. ### ozsavran Thread Starter New Member

Jul 30, 2016
21
1
Dear Crutschow,

Thanks for your detailed mathematical answer. Tested your answer by putting a 160 ohm capacitor instead on DoCircuits. Voltage phase delay now occurs to some relief. But it is even more puzzling now that it is not an expected 180d difference, but only 90d or so. Can you take a stab at explaining how this is happenig on electron flow level please? Rather than just calculus? We are now seeing that opposing sides of a capacitor have different levels of opposite charges, and can even have same level of same polarity(!?) I have to visualize this event in terms of charge particals to fully get to the bottom of it in my head. I am sure there is a completely satisfactory depiction of this event, but just can't find it anywhere currently.

File size:
124 KB
Views:
12
8. ### crutschow Expert

Mar 14, 2008
13,473
3,361
The maximum phase shift in a reactive circuit is 90°, why do you keep saying it should be 180°?

To try to simplify the discussion, below is the simulation of a single R and C with the C going to ground.
I will talk current flow (assumed positive charges) since that is what I am used to.

When the resistor value is much higher than the capacitance reactance as here, the capacitor voltage is below the absolute value of the input voltage for most of the cycle.
This means the capacitor is charging in one direction for most of the positive cycle and charging in the other direction for most of the negative cycle.
This can be seen in the capacitor current I(C1).

Thus the voltage on the capacitor is going in the positive direction (slope) for most of the positive half of the cycle and in the negative direction for most of the negative half of the cycle.
This gives a near 90° phase shift between the input voltage and the capacitor voltage as can be seen in the V(out1) plot.

Here is an animated simulation that also shows the phase shift relation.

Hope this all helps.

9. ### #12 Expert

Nov 30, 2010
16,663
7,310
Let's start with, "ideal" components. The capacitors don't leak, wires have no resistance, battery voltages never sag, and we can see what's happening in a picosecond.
A battery on the left, negative terminal up, positive terminal down. From the negative post of the battery, a switch to a capacitor to a resistor then back to the positive pole with a wire. Turn the switch on and electrons flow from the battery to the left plate of the capacitor. Those electrons kick electrons off the right plate of the capacitor and those electrons flow through the resistor to the positive terminal of the battery. Presto! It's a circuit.

Now, look at the voltages. When the electrons flow through the switch, the left side of the capacitor has a negative voltage. That negative charge repels electrons from the right capacitor plate, so electron current is still flowing in the original direction and the left end of the resistor has a negative voltage inflicted on it. There is no difference in polarity. Zero degrees. Capacitors do not invert the polarity of an applied voltage, they just allow a charge of electrons on one plate to repel electrons off the other plate. In the instantaneous moment, the electrons are flowing as if nothing was interfering with them.

It is only in an AC circuit where the Xc of the capacitor and the resistance of the load resistor interact to cause a phase shift. The capacitor is still what I call a stupid part. Other people call it passive. The capacitor inflicts no energy and makes no decisions, it just allows charges on one plate to attract or repel charges on the other plate. I'm trying to take the magic out of the capacitor here. Did it work?

10. ### DGElder Member

Apr 3, 2016
347
87
ozsavran,

Understand that you are not measuring the charge on the capacitor plates with your o-scope. You are measuring the voltage of the capacitor terminals with respect to ground. The voltages on the capacitor plates do not have to be of equal and opposite values, however the charge on the plates are of equal and opposite values. The voltage on the plates depend on that to which they are connected. If at some instant in time I disconnected the circuit from ground and raised the new "ground" by 30V then both plates on the cap would increase by 30V but the voltage difference between the plates would remain the same, the charges on the plates would remain the same.

Before you changed the circuit you had so little current flowing through the capacitor, with respect to the size of the capacitor, that it had no time to charge enough to create a significant voltage across its plates, so of course the voltage on both sides of the cap were nearly the same.

I suggest setting the scope to look at the difference between CH1 and CH2 so you can see what is going on with the cap voltage. Compare that to CH1 which is the source voltage and CH2 which is the resistor voltage. Then you can see the 90 degree phase difference between the voltage across the cap and the current through the cap (which is in phase with the voltage across the resistor).

Last edited: Jul 30, 2016
11. ### ozsavran Thread Starter New Member

Jul 30, 2016
21
1
Dear Guys,

Yes, DG Elder's diagnosis of my pesky confusionary illness here is about on correct spot. Again my idea here is not to dispute measurement devices, God forbid, just to pave way to better understanding of electrical concepts for slow thinkers like myself. I will have to develop my own version of physical/visual explanation of what exactly happens in a capacitor during AC excitation, and will be happy to share it here for fellow slow thinkers.

For one thing charge accumulation on capacitor plates is not instantaneous, slowed by whatever series resistor is there, hence your point. Still even when the capacitor is charged to +10% on one side, it has to be charged to -10% on the other side, and has to show -10% voltage relative to isolated absolute ground.

But probably not relative to voltage source ground. Assuming, both sides of the capacitor is at absolute 0 volts before connection to our fixed AC voltage source, negative side of the capacitor is charged only -10% at that point, its still at a positive voltage potential value relative to the negative side of the AC source, or the circuit "ground", but at absolute -10% voltage relative to real ground. That could be the explanation I am hungry for.

Writing back here from 1500m high mountain retreat with a dingy samsung windoze8 tablet, and my DoCircuits online simulator account is deaf. After I do little experimentation in the city, and do some simulations I will be happy to share my ground-breaking, world shattering results back here with you guys.

Thanks again to every responder for taking this enormous challenge with me! Everybody's kind help is really appreciated. Lets see if I produce even funnier misunderstandings with my humble explanation of AC capacitors.

12. ### MrAl Distinguished Member

Jun 17, 2014
2,548
514
Hello there,

Being a physicist that means you probably think deeply and try to come up with general ideas on what EXACTLY is happening in a given experiment.

In this case with the capacitor and voltage, you are absolutely right. The two voltages are 180 degrees out of phase. But can you figure out how to come to this conclusion?
The answer may not be very obvious, but it's there. I thought i would give you a little time to think about it again now that you have the confirmation that it really is out of phase and it must be out of phase.

Just to note, the other issue which is different than that one, the current is 90 degrees out of phase with the voltage, but yes, that's another issue. Your issue is also interesting and not thought about as much. The basic concept comes up in other areas too but i dont want to give too much away just yet.
Think about it just a little more first

I found this diagram which should clear this up quick:

Last edited: Aug 1, 2016
ozsavran likes this.
13. ### ci139 Member

Jul 11, 2016
341
39
it is really how you set up your capacitor in an ~AC sine or SW circuit . . .

File size:
38.8 KB
Views:
18
14. ### ozsavran Thread Starter New Member

Jul 30, 2016
21
1
Hello Mr. AI,

Well, yes to kindly prove my point to Crutschow and gentelman who removed his comment from here about how a single capacitor cannot be a voltage inverter, I setup the proteus experiment below in attached picture. Voltage phase difference between opposite capacitor plates has to be always 180d, not 90d. The term "phase shift" became a cliche to automatically refer to voltage-current phase difference, which is not my interest here.

Well, lo and behold that when applied AC, opposite sides of a capacitor is 180d apart relative to absolute ground at zero volts. Created a zero volt ground by using two synchronized AC sources instead and using middle point as such.

The 2 DC voltage meters from top show opposite sides of the capacitor at +4.04V and -4.05V relative to real ground simultaneously. However, that simulator was "unsuccessful" at showing voltage between bottom lead of lower AC voltage source and lower capacitor lead. There has to be voltage difference there as there is a resistor in between. But that meter keeps flickering between +0.00 and -0.00 for some reason, I am not sure.

In my humble poor lab, I do not have 2 synch-capable AC sources like this simulation. However I am looking to ground connection isolation circuits, or using a naked re-bar on my buiding roof as a real ground connection path instead.

The problem with the initial circuit was bottom of the AC source was connected to the bottom of the capacitor via a resistor, not to a real absolute 0V ground connection.

Thanks for taking my question seriously and please do let me know if I am still missing something. Couldn't quite get the significance of charge fields depicted in your picture in this case.

Cheers!

File size:
376.9 KB
Views:
20
15. ### nsaspook AAC Fanatic!

Aug 27, 2009
3,008
2,350
The first principle is that a capacitor is a energy storage element not a charge storage element so all potentials across it must be generated from the capacitors field energy. The electric field energy is the result of charge separation in a electronics capacitor (in a electronics capacitor the coupling between plates is extremely high as opposed to most physics sphere capacitor examples where the 'earth' is a important factor as a circuit path). The overall number of charges in the plates doesn't change, for every extra charge on one plate a charge is removed from the other plate in the circuit.

So at frequencies where the Capacitive Reactance is low IRT other elements it acts much like a resistor (SEEMS to be a kind of conductor) in a circuit with a small voltage drop across the capacitor element. To say this capacitor voltage drop is a 180 degree phase shift is misleading in the same way as calling the voltage drop across a resistor a 180 degree phase shift.

Now if you need a specific 180 degree phase shift with single ended sources using a capacitor then the typical RC network is used.

If the source is floating then a resistive divider can be used to provide a 180 degree phase shift IRT the signal ground.

Last edited: Aug 1, 2016
16. ### DGElder Member

Apr 3, 2016
347
87
But probably not relative to voltage source ground. Assuming, both sides of the capacitor is at absolute 0 volts before connection to our fixed AC voltage source, negative side of the capacitor is charged only -10% at that point, its still at a positive voltage potential value relative to the negative side of the AC source, or the circuit "ground", but at absolute -10% voltage relative to real ground. That could be the explanation I am hungry for.

You are using undefined terms that make it difficult to understand what you are trying to say. We are, I assume, still talking about capacitor charging in the circuit you posted. In this circuit, which is your only defined system in this thread, there is only one ground. There is not a "source ground", "absolute ground", "real ground" or "circuit ground". The only defined ground Is that which is defined by the ground symbol in your circuit and by definition all those points are connected and at the same voltage and all other voltages in the circuit are defined as being relative to that point. Anything outside this defined system is irrelevant and undefined.

An AC source does not have a negative and positive side, by definition they change polarity every half cycle. However, the AC source in your diagram does, for some reason, show a positive and negative terminal and it is the "positive" terminal that is connected to ground. So that terminal is at zero volts - always. Assuming the AC source has a 1V peak to peak voltage then the so called "negative" terminal swings from +1V to -1V. and so does the left hand side of the capacitor.

It is not correct to say the voltages on the two sides of the capacitor are 180 degrees apart. When you refer to a phase difference you are talking about a difference in the relative position in two periodic or similar waveforms at the same instant in time. You can say that the plates at any instant in time have equal and opposite charges, but their voltages with respect to ground are not, in this circuit, equal and opposite at any instant in time and are not 180 degrees apart.

What you can say is that with respect to the point half way between the two plates of the capacitor the voltages on the capacitor plates are equal and opposite. And in an AC circuit their voltages with respect to that midpoint are 180 degrees apart. This can be inferred from the physics of capacitors and the o-scope waveforms, but you can not directly measure it in your circuit.

P.S. If God posted a comment saying that the ground in your circuit is actually at 100V with respect the the Universal Holy Ground, everything I said above would still be true - or at least not less true.

Last edited: Aug 1, 2016
17. ### ozsavran Thread Starter New Member

Jul 30, 2016
21
1
Dear Mr. Elder,

Just as I suspected this is in fact a fruitful refreshing discussion, not a redundant and boring one, as old and as simple the subject is.

We all agree that its a matter of measurement point. Maybe terminology and convention of notation. But we should be defending nothing.

I have to say there are repeating contradictions in your answer unfortunately. I am sure you will spot them when you re-read your last message. But its fun and fruitful to exercise our minds mutually. Your answers are much appreciated even if you judge mine.

So the terms I used were not pre-defined, that is correct. Just trying to use plain English to literally mean what I say. Sorry if that was confusing. Maybe I should have said "source reference" or "citcuit reference" instead.

Coming back to contradictions I notice, when two sides of something is equal in magnitude but opposite on polarity relative to its mid-point, that mid point has to be at zero. No way around it.

Now, if absolute, real, single and only ground also happens to be at zero, then those 2 sides has to be polar opposites relative to ground as well. Two things that has to stay at polar opposites at all times are 180d apart. Not 90d apart. Then you have to have less than the same amount of charge at one side. That is like mother of all wrongs for a capacitor isn't it?

One important benefit I am getting from these exchanges is that I was wrong in assuming that Electronics is an arm of Physics. Its really 70%+ mathematics with pre-definef part behaviour. 30% Physics just get in the way. Electronics could be defined as electrical programming discipline. Much like software programmers like to isolate their brains from what hardware really does when a command is executed.

Of course I am not making this observation/classification/definition in any demeaning or ridiculing way. Its a hard, involved, highly fruitful, deep impacting field obviously. But just observing what its practice really involves.

Electronics weaves graphs and networks to go further, while Physics dig down to go deeper and deeper to find real explosive stuff. That's my simple humble observation about thinking in Electronics domain.

Again, thanks and keep discussing with me please. I feel truly honored and enriched.

18. ### crutschow Expert

Mar 14, 2008
13,473
3,361
Yes, there is a way around it.
That appears to be a source of your confusion.
For a capacitor, the two sides are not "equal in magnitude but opposite in polarity" relative to a "mid-point".
One side of the capacitor is opposite in polarity relative to the other side of some given magnitude but this is no zero voltage mid-point for that.
Voltages are always relative to some actual point, not some imaginary "mid-point".
If you can't measure the voltage with a meter then it has no real meaning.
A meter has two terminals that must be connected to two nodes and one of those nodes can't be imaginary.
I think that's where your confusion about the two sides of a capacitor being 180° out of phase comes from.

19. ### ozsavran Thread Starter New Member

Jul 30, 2016
21
1
Aw please, c'mon now Cruts... Don't turn this into a one upmanship debate here please. How can you have different amount or charges on 2 sides???

DG Elder said it himself above:

> What you can say is that with respect to the point half way between the two plates of the capacitor the voltages on the capacitor plates are equal
> and opposite. And in an AC circuit their voltages with respect to that midpoint are 180 degrees apart. This can be inferred from the physics of
> capacitors and the o-scope waveforms, but you can not directly measure it in your circuit.

If my word as a newbie here is not good enough please take it upto DrAI and take his challange yourself.

> In this case with the capacitor and voltage, you are absolutely right. The two voltages are 180 degrees out of phase. But can you figure out how to
> come to this conclusion? The answer may not be very obvious, but it's there. I thought i would give you a little time to think about it again now that
> you have the confirmation that it really is out of phase and it must be out of phase.

I will have to declare this thread as [SOLVED] if nobody is going to sue me for it.

20. ### DGElder Member

Apr 3, 2016
347
87
ozsavran,

There is no contradiction, for the reason crutschow tried to explain. At the risk of superfluously expanding on what he said: in electronics, just as in physics, definitions are crucial. Look up the physics or electronics definition of voltage - they are the same. It is the electric potential energy difference between two points. There is no such thing as an absolute voltage at a point - only relative to some reference point. So calling the midpoint between the two plates of a capacitor zero volts has no meaning - to be meaningful it has to be explicitly stated or implicitly understood to be relative to some point. In electronics if we don't explicitly state the reference point when identifying a voltage it is generally understood to be with reference to the circuit common return or ground.

If you say a battery is 9 volts it is understood to be 9V between the two terminals. We don't say a resistor in a circuit is at 5V, we say there is 5V across the resistor, i.e. between the two leads. Either lead could be at any voltage with respect to an arbitrary reference, but the difference between the leads is 5V. As crutschow said, that is why voltmeters have two test leads, not one.

Last edited: Aug 1, 2016
atferrari and AnalogKid like this.