# A basic question about capacitors in AC circuits

Discussion in 'General Electronics Chat' started by boogie, Jan 10, 2011.

1. ### boogie Thread Starter New Member

Jan 10, 2010
25
0
Hello friends.

I have a very basic question about capacitors which might sound very idiotic but I really must understand it.
A capacitor basically opposes changes in voltage. But what happens when we have a sinusoidal waveform between 90° and 180°
Consider the following circuit:

Between 0° and 90° the current flows clockwise and the capacitor charges and at the same time opposes the change with the maximum opposition occurring at 90°. Between 90° and 180° the current still flows clockwise but the voltage starts to descend in amplitude. This means that the capacitor should oppose the descending voltage and does so by discharging counter clockwise and herein lies my problem - by discharging counter clockwise it seems to actually exasperates the descend in amplitude instead of resisting it!!!

What am I missing here?
Thanks

2. ### hgmjr Moderator

Jan 28, 2005
9,030
214
I think you would benefit from reading the introductory material on capacitance contained in our AAC ebook at www.allaboutcircuits.com.

hgmjr

Feb 19, 2009
6,357
718
Frequency.

4. ### boogie Thread Starter New Member

Jan 10, 2010
25
0
I did - more than once. Still don't understand

5. ### boogie Thread Starter New Member

Jan 10, 2010
25
0
What do you mean?

6. ### thatoneguy AAC Fanatic!

Feb 19, 2009
6,357
718
The reactance of the capacitor is different for different frequencies.

It seems you are analyzing the circuit as if it were a Clamper

Though I'll admit to not quite understanding your description of what are interpreting as happening. Capacitors pass AC and block DC. Looking at caps from the impedance view makes more sense than looking at them from the mini-battery viewpoint.

Apr 5, 2008
15,647
2,346
8. ### boogie Thread Starter New Member

Jan 10, 2010
25
0
That's exactly what causes the confusion for me.
In the first quarter of the cycle between 0-90°, the voltage is rising and the current is running clockwise.
after 90° it passes the crest of the waveform and the voltage starts descending. The capacitor, should in theory counteract the descend by discharging, the problem is that it's discharging counterclockwise which should (the way I see it) causes the voltage to descend even faster instead of resisting the descend.
Had the capacitor discharged clockwise it would have compensated for the descend thereby resisting the change but it's discharging counterclockwise!!

I'm totally baffled by this.

9. ### thatoneguy AAC Fanatic!

Feb 19, 2009
6,357
718
Don't think of the capacitor as "holding a charge" in a series circuit as you have shown above. When the voltage on one plate changes, the voltage on the other plate changes as well. If this is a continual cycle (sine wave), no charge actually "builds up" on the capacitor, as the continually changing voltage is passed on to the next plate, rather than stored.

Caps will store energy when in parallel with a source as well, but only if that voltage has an overall bias, such as in the clamp linked above, compare the first cycle shown to the rest. That is what your description sounds most like to me, anyway.

10. ### t_n_k AAC Fanatic!

Mar 6, 2009
5,448
782
This may help - trying a different approach. Hope the image is clear enough.

File size:
68.6 KB
Views:
66
11. ### #12 Expert

Nov 30, 2010
16,261
6,770
I think that one guy nailed the exact point of misconception. In this case, the capacitor does not "charge up". Thinking in those terms will not solve this circuit.

12. ### boogie Thread Starter New Member

Jan 10, 2010
25
0
Thank you everyone.
I found the answer in module 2 of NEETS (http://www.tpub.com/neets/book2/4b.htm) . It all boils down to the phase shift.
Actually there IS charge building on the capacitor.
My mistake was looking at the capacitor as a voltage source, where in fact it should be seen more as a current source.

13. ### Jony130 AAC Fanatic!

Feb 17, 2009
3,957
1,097
No, capacitor act very similarly to a voltage source, not a current source.
The key equation is this
I = C*dV/dt
Current in the capacitor is proportional to the rate of change in voltage.
So, the faster the voltage change, the larger the current flow through the capacitor.
When capacitor is full charged then he act like open circuit ( no current flow) But empty capacitor act like short circuit, voltage across capacitor is equal 0V but max current is flow through the capacitor.

When the voltage on the capacitor rise, capacitor is charging ("absorbing" the current).
When the voltage on capacitor "drops", capacitor is discharging (supply the current).

So when we connect empty capacitor to AC voltage from 0° to 90°
Capacitor is charging and at 90° reach Vin an no current is flow.
( rate of change of a sine wave is very small at peak voltage so the current that is flow through the capacitor must also be small)
And now from 90° to 180° Vin star to drop.
So capacitor start to discharge (the current will be flow in opposite direction). And when Vin reach 0V at 180° capacitor is empty (Vc= 0V).
But now when Vin reach 0V and start to rise in opposite direction,the rate of change for sin wave reach the max value.
So we have empty capacitor (act like short circuit) so we have max current flow through the capacitor.
And from 180° to 270° capacitor is again start charging but in opposite direction. And again form 270° to 360° capacitor start to discharge ( current change the direction again).
And this discharge process ends at 360° when voltage across capacitor reach 0V.

• ###### 121.PNG
File size:
19.2 KB
Views:
35
Last edited: Jan 10, 2011
14. ### boogie Thread Starter New Member

Jan 10, 2010
25
0
10X Jony

But if the the capacitor acts like a voltage source, how is it resisting the change between 90 and 180?
it seems that it's pushing against Vin which will cause the voltage to drop even faster instead of slower.

15. ### Jony130 AAC Fanatic!

Feb 17, 2009
3,957
1,097
Simply the capacitor at 90 start to discharge it self with the rate of change of a input voltages. Vin is connect parallel to capacitor, so voltage on capacitor must be equal input voltages ( second Kirchhoff's law ).
And when input voltages reach 0V, capasitor voltage also reach 0V (discharge ends).

I = C *dU/dt
and for Vin=10V ; F=50Hz and C=1uF
The rate of change for 10V and 50Hz is equal :
S = dU/dt = 2 * pi * F * Uin= 3.141mV/μs
And this rate of change reach his maximum ( 3.141mV/μs ) when input voltage cross 0V.
So the max current in capacitor will be equal:

I = 1uF * 3.141mV/1us = 3.141mA

The same value we get by calculating the reactance of a capacitor

Xc = 1/ ( 2 * pi * F * C ) = 3.183K

I = 10V/3.183K = 3.141mA

And capacitor try to keep voltage constant by supplying the extra current (discharging) or by absorbing the current (charging). Nothing more the capacitor is able to do.

Last edited: Jan 11, 2011
16. ### theyikes New Member

Aug 27, 2012
1
0
not sure if this is the right thread to post in but i didn't want to start a new thread as that seems a bit extreme. Anyway I have a samsung flat screen tv with two bulging surface mount electrolytic capacitors. The text on the capcitors says 77nk 1000 10v. I've googled this and found nothing except an article in a french forum. does anyone know where i might get replacements or a substitute. not even ebay has anything.

17. ### matty204359 Member

Apr 6, 2011
105
3
a new thread would be less extreme than reviving a post that is this old.