Same Polarity Charges Accumulating on Both sides of a Capacitor Simulataneously?

[ozsavran]

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:
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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.

Thanks in advance for taking the time and effort to answer. 

[Simon]

Firstly when you are measuring your DC voltage you are not measuring the plates relative to ground by usually measuring either side of the capacitor so one plate relative to the other. When you apply an AC voltage you are effectively discharging the capacitor every cycle and recharging it in the opposite polarity. If this happens fast enough your perception could be that it is both negatively and positively charged at the same time but assuming you are looking at this with an oscilloscope you should see that voltage on the capacitor is going to vary with the alternating voltage of your supply.

[Ysjoelfir]

Firstly when you are measuring your DC voltage you are not measuring the plates relative to ground by usually measuring either side of the capacitor so one plate relative to the other. When you apply an AC voltage you are effectively discharging the capacitor every cycle and recharging it in the opposite polarity. If this happens fast enough your perception could be that it is both negatively and positively charged at the same time but assuming you are looking at this with an oscilloscope you should see that voltage on the capacitor is going to vary with the alternating voltage of your supply.

That would be the way I also would expect it. But ozsavran wrote, both the scope and a simulation shows the wrong signal, even if chosen very low frequencys or very high capacitances. So I assume there is some mistake in the way he measures / places the measuring points in the simulator - but what exactly could be done wrong here to get that result?

[tatus1969]

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.

Just to make sure: you are using a bridged symmetrical lab supply that can provide + and - voltage with respect to ground, right? Then your statement is correct. 

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.

If you have a simulation, please provide the circuit so we know that we are talking about the same thing.

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? 

When you connect a capacitor to a voltage source (with low impedance, as I assume from the above), no matter AC or DC, then you will just measure the voltage of your source across the capacitor. The voltage source "dictates" the voltage.

[Brumby]

The DC case is very simple - and your understanding will do for the moment.

The AC case is not that much different .... if you take things s l o w l y.....

Consider a 4 step process, where one connection to the capacitor is taken as our reference point and the other is the one we are monitoring...

1. Charge up a capacitor with a DC source of positive polarity that takes 10 seconds for the capacitor to reach the DC voltage
2. Discharge the capacitor with a load that takes 10 seconds for it to effectively have zero volts.
3. Charge up a capacitor with a DC source of negative polarity that takes 10 seconds for the capacitor to reach the DC voltage
4. Discharge the capacitor with a load that takes 10 seconds for it to effectively have zero volts.

This is 4 steps involving DC - but put them together and you have AC .... and if the AC frequency, supply and capacitance were to produce the same time constants, then the observed characteristics would be comparable.  (Yes, the 'signal' would be an unusual sort of square wave.)

But now lets look a little bit more closely....

In the very first instant of step 1, a DC supply is connected to a capacitor with no voltage across it.  A capacitor CANNOT change it's voltage instantaneously - (and here is the important bit:) it needs to build up that charge over time.  To do this, current needs to flow.  At the start, this current flows quickly limited by the resistance in the system, then as the charge builds up across the capacitor, the voltage increases - and increases more and more slowly until it eventually reaches equilibrium with the supply.  At that point, current flow stops.

Over the next 3 steps, the exact same process occurs - just with different voltage levels.

Now, at 10 seconds per 'stage' this exercise is somewhat straightforward and a bit boring ... but let's speed things up a bit - say 1 second per step, but using the exact same capacitor and supply.

At step 1, we start to charge up the capacitor.  A big current flows because of the voltage difference, but there is no time for it to climb up to the supply voltage before we cut it off and start step 2.
At step 2, we discharge.
At step 3, we repeat the charging in the opposite direction - with the same cutoff.
At step 4, we discharge.


Now let's peed things up even more - say to 1/100th of a second per stage (25Hz)

At stage 1, the current is still high when it gets switched off and the voltage hasn't had time to increase much at all.
Stages 2, 3 and 4 follow the same pattern as before.

Now let's speed things up even more - say 1kHz

The current is essentially at it's peak value for all of the short time it's on, only changing from positive to negative and back.  The voltage across the capacitor does not have any time at all to change by any measurable amount.  Notice that the higher the frequency, the greater this phenomenon becomes.  (This is important - and the relationship between frequency, capacitance and reactance is one you will see a lot more of...)

Now we have a fair idea about what is happening inside the capacitor, let's see how it fits into a couple of situations...

If you were to have this capacitor across the power supply, then it would - in effect - short it out.  The AC part of it, that is.  The DC is left behind.

However, if you were to have this capacitor in series with a signal, then the voltage difference between both sides will be essentially unchanged, since there won't be time for charge to build up in one direction before that direction is reversed.  Once you've got a grasp on this, we can go one step further and look at a DC offset.


Edit:  Change some wording for better accuracy.