How capacitor blocks dc current?

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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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

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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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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LovePhys

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

I found this equation on the hyperphysics: [itex] V_{C}=V_{0}e^{-t/RC} [/itex]. 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)!?

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

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.

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!

sophiecentaur

OK
Firstly, if the load has infinite resistance (no load) the capacitor will just charge up and hold a DC value.
Secondly, a battery just needs charge (it behaves, on its own, rather like a huge value capacitor) there is no point in 'smoothing' the waveform supplied to it. It will allow current to flow in it as long as the voltage of the source is greater than the emf of the battery. Same thing for a motor: charge will flow through it for positive volts applied. There is no real advantage in smoothing the supply for it - in the end, in both these cases, the limit to how much useful charge can flow from the rectifier circuit will depend upon the total source (series) resistance (that will depend upon the rating of the transformer and even the AC supply to the primary of the transformer). If you have an inadequate transformer then no amount of smoothing can keep the volts up if your load resistance is too low. Many audio power amplifiers can suffer when they are required to supply sustained 'loud' passages, particularly with LF content - a cheap PSU 'sags' when too much current has been demanded.

carlgrace

He says it would be a straight line because R in the off-load (or "no-load") case would be infinite, and e to the 0 is 1. So it wouldn't decay. The other part of what he said means that in some applications (like battery charges) the actual value of the V doesn't need to stay stable, as it is the total charge transferred that is important. In other case, such as powering logic circuits, the stability of the voltage *is* important and in this case a smoothing capacitor would be required.

sophiecentaur

Yep. Precisely!

FOIWATER

the way I think about it, is two black boxes, which ever has a higher voltage will send current to the other, a capacitor doesn't block DC until it is charged, in which case there is no difference in potential for current to flow? But, in AC, as the capacitor is charged, the source voltage is decreasing behind it so it begins to discharge into the circuit (reactance) and later, recharge.

One can see from this explanation how reactance is frequency dependent..

sophiecentaur

And the Maths show it perfectly.

FOIWATER

Yeah, it does for sure..the rate of change in voltage show the 90 degree phase shift for a capacitor, how it will react etc.

It's pretty cool..

sophiecentaur

90 degrees for AC and an exponential change for an applied step function ('turning on the DC').

DragonPetter

I would be careful to say that there are cases that a capacitor doesn't block DC. An ideal capacitor ALWAYS blocks DC. The technicality is that when you charge a capacitor with a DC source and there is a transient charging, the signal applied is not DC but rather a step function.

psparky

There are also two kinds of DC.....flat line DC from batteries (ω=0)....and non flat lined DC produced by rectifiers (ω≠0). The cap will block true flat lined DC....but will fluctuate with rectified DC to the the dv/dt effect stated earlier in thread.

One more way of saying what everyone else has been saying.....

The impedance of a capacitor is 1/(Jωc)

When ω=0 like in a true flat lined battery...you can clearly see that you have infinite resistance......or the cap will not let current flow.

When ω equals anything but zero.....you will have current flow.

Anyone disagree with this? I've been wrong many times before....and many times to come!