What is the output to the capacitor if input is DC with a ripple?

[DEV123]

In the above circuit the input (DC with a ripple superimposed) is given between node A and GND. My doubt is whether the output is a filtered one (ie without ripples) or non-filtered one (ie input itself) ? Most of my colleagues say that output is a filtered one because of capacitor. But according to KVL the voltage across the node A and GND must be same from both directions, is not it? If a voltage is applied across a node and gnd it has to be the same at output also in this case , so what is the output?

 

[Ratch]

DEV123,

You are correct. Ignore the DC. All that does is offset the voltage. The cap will energize and de-energize according to the AC voltage across it. Provided the cap does not load the voltage source beyond its ability to supply the current, the output will look like the cap was absent. There is no diode to block the cap from de-energizing back into the source instead of the load.

Ratch

 

[ark5230]

Output is filtered. Higher the capacity, lower is the ripple at the output.
The role of the capacitor is to bypass the ac component (ripple) to the ground.

 

[Ratch]

ark5230,

Output is filtered. Higher the capacity, lower is the ripple at the output.
The role of the capacitor is to bypass the ac component (ripple) to the ground.

Did you read what I wrote in the previous post? Do you believe it? If not, do you have a cogent argument against it?

You can make the cap as large as you want. If the AC voltage source is able to supply the current, all that a larger cap will do is increase the current existing between the source and the cap. You will still see whatever voltage the souce puts out across the cap. The cap cannot do any bypassing because it is in series, and there is no other path. Remember, current does not pass through a capacitor. So there is no "bypass" path and no "filtering" to ground.
Read again what I said in the previous post about a diode, too>

Ratch

 

[RetroTechie]

Schematic-wise, output = input = DC voltage with a ripple. Simply because it's connected, that is: output & input is 1 electrical point, and thus has the same voltage at all times.

In a real-world circuit, capacitor would have a filtering effect because voltage sources are never ideal.

So answer really depends on whether this is about an existing/practical circuit, or whether this is about the theoretical circuit (with ideal components) as represented by the schematic. In the latter case, Ratch gave the answer.

 

[enjunear]

Following from RetroTechie's line of thought, think back to when you learned Thevenin and Norton equivalent circuits. All sources can be defined as a Norton or Thevenin equivalent circuit (which are interchangeable w/ some algebra). Since you can define any REAL voltage source as having some amount of equivalent series resistance, that internal R is what allows the capacitor to act like a filter.

The ripple voltage's amplitude on a real capacitor will be much less than it is at the source because the ripple has to "work across" the resistance. If the cap has a big value, then it can quickly absorb the relatively slow-moving ripple current, making the voltage across it appear very smooth.

 

[ark5230]

All the discussion assumes that there is some isolation (resistance or inductance) between input and output.
If there is no such reaistance and inductance then the the debabe piles up. every conductor has some intrinsic inductance associated with it (which may be small in general particularly at low frequencies of ripple). If the ripple is rich in high frequency components the discussion further prolongs.