# How capacitor blocks dc current?

1. Jan 15, 2010

### samieee

hello

Capacitors are widely used in electronic circuits to block the flow of direct current while allowing alternating current to pass,how it does the job?

samieee

2. Jan 15, 2010

### Staff: Mentor

http://en.wikipedia.org/wiki/Capacitor

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3. Jan 15, 2010

### sophiecentaur

There is an insulating layer between one part and another of a circuit. Direct current cannot pass through an open circuit, can it?

4. Jan 15, 2010

### samieee

in wiki everything is given but actually nothing is given clearly

5. Jan 15, 2010

### samieee

well that good but suppose you connect a dc battery,a bulb and a capacitor in a circuit..this time dc current is flowing through the circuit isnt it?(when capacitor become charged if we disconnect the battery then capacitor itself will lighten the bulb)

6. Jan 15, 2010

Imagine it with water. A capacitor has capacity. It has two big electron containers.

DC current is just that like a flow of water in one direction. In AC the current it switches direction 50 times a second. So when you put a capacitor in the circuit the electron containers fill and empty periodically. When the electrons flow in one direction for a long time the container gets full and nothing can flow anymore, until the current reverses again.

7. Jan 15, 2010

### Averagesupernova

As far as water analogies go, the best one I've heard for a capacitor is a large vessel with a pipe coming out of each end. This vessel is split in the middle with a rubber diaphragm. Make sense now? It isn't a foolproof analogy since you can disconnect the pipes and leak all the water out. But, what analogy is foolproof?

8. Jan 15, 2010

### Staff: Mentor

Then I'll try this. Just as V=IR is the fundamental equation relating voltage, current and resistance for a resistor circuit, the following equation relates voltace, current and capacitance for a capacitor:

$$I(t) = C \frac{dV(t)}{dt}$$

Or, if you are not familiar with that calculus term with the derivative, you can think of it as:

I(t) = C * (change of voltage per time)

So when you have DC, there is no change of voltage with respect to time, so there is zero current. When you have an AC voltage signal that varies across the capacitor with time, that equation lets you calculate the current that results through the capacitor.

9. Jan 15, 2010

### sophiecentaur

And how does a paralle plate capacitor actually 'store charge'. The first thing you need to remember is that there an awful lot of electrons available for conduction in a conductor. Only a tiny percentage of the electrons, moving a tiny bit, corresponds to a huge charge. As one plate charges up relative to the other you get a build up of electrons on and near the surface of the negatively charged plate and a corresponding reduction in the number of electrons on or near the surface of the positive plate. The displacement will continue until the field due to the supply voltage is balanced out by the local fields of attraction and repulsion due to the imbalance in the number of protons and electrons. on the plates. The closer you put the plates, the higher the field between them (volts per metre) so the bigger volume of electrons that are displaced (the greater the charge and the greater the Capacitance). If a dielectric (an insulator) is put in between, the molecules of the insulator will become polarised by the field and this will lead to more electrons being displaced in the metal plates - just as if the spacing were much less but without the danger of a current flowing.
The higher the PD, the more charge is displaced and it is a linear relationship - like Ohm's Law.

Q = CV is the basic relationship so you can say
dQ/dt = C dV/dt .
But I = dQ/dt
Which shows why I= C dV/dt (as in the above post)
Hence, the Current is proportional the rate of change of the PD.

10. Jan 15, 2010

### Staff: Mentor

DC is stopped, AC passes through.

Last edited: Jan 16, 2010
11. Jan 15, 2010

### Staff: Mentor

:rofl: :tongue2: :rofl:

Kind like that old joke -- "Find x" "Here it is!"

12. Jan 16, 2010

### samieee

thanks every buddy

13. Dec 16, 2011

### sophiecentaur

Just a point. The current is Unidirectional, maybe but it is not DC because its value is varying in time (exponentially).

14. Dec 16, 2011

### gnurf

You're fired.

15. May 8, 2012

### LovePhys

Hello everyone, I apologize if I revived an old thread.

I read all above posts and I understand why capacitors do not allow DC current. However, I suspect if there is any difference if the capacitors are connected in parallel. Do they allow DC current and block AC current in this case?

Thank you.

16. May 8, 2012

### davenn

makes no difference, there is still a hole (gap) in the circuit that DC cannot cross
and AC will still flow in a circuit with multiple parallel capacitors

Dave

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17. May 8, 2012

### davenn

If AC didnt flow in a circuit with multiple capacitors in parallel then we would have a huge problem trying to smooth DC lines of power supplies either in a power supply itself or on a power rail in something like a radio transmitter, where we use different value capacitors in parallel to deal with AC signals of varying frequencies. Here is an example from a synthesiser I have ....

note the very different values of C9, 10 and C23 off the VCC1 rail

cheers
Dave

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18. May 8, 2012

### LovePhys

Yes, thank you very much Dave! You cleared the point. My teacher said that the capacitors blocked the DC current if they were connect in series, but if they were in parallel, they allowed the DC current. But I do not agree with him. (By the way, I do not really understand your circuit. I am in year 11, so it seems to be beyond my level).

Also, If I have an input voltage like this:

I think this is a DC current, right? Since it only has positive values of V.
Normally, the capacitor will block the DC current. However, in this case, I think because the DC signal varies with time, so when the voltage is 0, it will give time for the capacitor to discharge. So, the graph for the capacitor will look like:

Please correct me if I am wrong at this point. Thank you!

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19. May 8, 2012

### davenn

thats ok you are doing great for 11 yo :) yes its just a small part of a complex circuit. The main thing I wanted to show you was the 3 capacitors that were in parallel ( C9,10 and 23) and how their values were quite different. This allows for filtering of frequencies across a wide range.

the top image is not a steady DC voltage, as you can see its changing from 0V to 1 V
it is what is called a square wave. It is 1 second bursts of a DC voltage. You will mainly see square waves in digital circuits like computers etc. If it was a plain DC voltage then it wouldnt be varying from that 1V level with time.
Yes the lower diagram shows how the voltage across the capacitor would vary with a square wave applied

cheers
Dave

20. May 9, 2012

### LovePhys

Thank you Dave, I get what you say!

I just have one more question: Does the same thing happen to half-wave rectifiers with smoothing capacitors? I think that things will happen in this order: AC current (sine wave) => (diode) => varying DC current (positive voltages only) => Capacitor charges then discharges (this process is repeated several times). So in this case, we still have capacitors that conduct varying DC current.

And if we have a smoothing capacitor in a half-wave rectifier (in the picture below). Does the red "line" (I mean the part that is going down when the capacitor is discharging) a curve or a straight line? I was told by my teacher that it was a curve (just like the graph of a discharging capacitor), but it seems to be a straight line to me!?

Thank you!

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21. May 9, 2012

### davenn

now that pic shows a full wave rectifier. You can see how the capacitor voltage 'sags' drops between the pulses of voltage from the rectifiers, but it keeps the overall voltage at a higher level and smoother than if it wasnt connected across the rectifier output.

have a look at these pics.....

the top image is the AC Voltage
the middle image is a 1/2 wave rectified AC Voltage it only uses 1 diode. See the big gaps where the other half of the AC sine wave isnt being rectified.
the lower image is a full wave rectified AC Voltage
All of these are without a capacitor.

adding a capacitor to the output of the full wave rectifier gives the waveform shown as the red wavey line in your image. The difference in the Voltage between the lowest part of that red line and the top of it is what we call the "Ripple Voltage".
In a power supply we aim to get the ripple voltage as low (small) as possible. It would depend on the requirements of the power supply but a value of less than 100mV (milliVolts) would be good.

below are images of a half wave and a full wave rectifier.....

half wave = just one diode

Full wave = 2 diodes

and this last image is a full wave of 4 diodes commonly known as a Bridge Rectifier

cheers
Dave

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22. May 9, 2012

### LovePhys

Thank you Dave, and sorry since I posted the wrong graph!

I found this equation on the hyperphysics: $V_{C}=V_{0}e^{-t/RC}$. This equation proves that the graph of a discharging capacitor will be an exponential curve. I still don't know why the red discharging lines of the full-wave rectifier with smoothing capacitor still look like a straight line (even in my textbook, I have the same feeling that they are straight lines)!?

23. May 9, 2012

### davenn

the image showing the red line is ok. It gives you a really good indication of what is happening in the rectifier circuit with a capacitor. If the capacitor is too lower value then the voltage will sag lower before the capacitor recharges with the next peak.
If the capacitor value is increased, then the voltage drop across the capacitor wont be so much between peaks.

Dave

24. May 9, 2012

### sophiecentaur

Those lines 'look' straight because they are part of what would be a very long RC discharge curve. If the AC were turned off then you would see the discharge happen over a large number of AC cycles and the line would no longer look straight.
The rate of decay depends upon the resistance of the load and the value of Capacitor, in practice, is chosen to give an acceptable 'ripple'. On the charging half of each cycle, the source series resistance is low enough (normally) for the rising voltage curve to follow the emf with little reduction but the 'off-load' wavefrom (i.e. infinite load R) would be a straight horizontal line with a value equal to the source emf.
It may be worth pointing out that for some applications (like battery charging or DC motor drive, for instance) there is no point in having a smoothing capacitor because the load in 'only interested' in the total charge it gets over the cycle.

25. May 10, 2012

### LovePhys

Thank you sophiecentaur, but I don't understand this part of your post.

Can you please explain a little bit?
Thank you! I have learnt so many things!