Question about a simple RC low pass filter

[siongboon]

Hi,

A simple low pass filter consist of a series resistor,
and followed by a capacitor connected to ground.
Why can't a simple low pass filter be only a capacitor, without the resistor?

Thank you for your advise.
Siong Boon

 

[bigdogguru]

Why can't a simple low pass filter be only a capacitor, without the resistor?

Actually, single capacitors are often used as low pass filters.

Case in point, bypass or decoupling capacitors used in conjunction with digital devices.

Of course, there are still resistances present, wire, etc.

BigDog

 

[LvW]

Actually, single capacitors are often used as low pass filters.

Case in point, bypass or decoupling capacitors used in conjunction with digital devices.

Of course, there are still resistances present, wire, etc.

BigDog

I don't think that single capacitors can act as a lowpass filter.
Explanation: A lowpass must have a pass band region and a second region with increasing attenuation (stop band). Both regions are separated by the corner frequency (in most cases: 3dB frequiency).
Otherwise, it is not a lowpass.

Case 1: voltage-in and voltage-out.
A first order RC lowpasss has a 3dB frequency of wo=1/RC.
If the resistor R is decreased towards zero the frequeny wo approaches infinity - there is no frequency dependence at all.

Case 2:: Current-in and voltage-out
If in the parallel combination R||C the resistor R approaches infinity, we have wo=0 and the circuit resembles a current integrator (without any passband).
_____________
That means: In both cases we have no lowpass filtering.

- - - Updated - - -

Remark: Of course, in some cases the necessary resistance is not a compact part but a wire or something else that has resistive properties to be exploited.

 

[siongboon]

I used to think like bigdogguru, and started to realise that a resistor may be required from other forum. A capacitor alone is not correct.

LvW is trying to explain the effect of the resistor in terms of the formula.
I can understand that, but still find it difficult to understand it.

How the power supply's capacitor managed to do low pass filtering without the resistor intentionally place in front/series of capacitor?

Can the RC filter be explained in a easy way without using the standard formula?
1/RC formula must representing one of the design.
I would like to understand it from the circuit and component's point of view.
Why must the resistor be there?

I suppose it has something to do with the impedance of the load/source.
I need more guidance on this, Thank you.

I don't think that single capacitors can act as a lowpass filter.
Explanation: A lowpass must have a pass band region and a second region with increasing attenuation (stop band). Both regions are separated by the corner frequency (in most cases: 3dB frequiency).
Otherwise, it is not a lowpass.

Case 1: voltage-in and voltage-out.
A first order RC lowpasss has a 3dB frequency of wo=1/RC.
If the resistor R is decreased towards zero the frequeny wo approaches infinity - there is no frequency dependence at all.

Case 2:: Current-in and voltage-out
If in the parallel combination R||C the resistor R approaches infinity, we have wo=0 and the circuit resembles a current integrator (without any passband).
_____________
That means: In both cases we have no lowpass filtering.

- - - Updated - - -

Remark: Of course, in some cases the necessary resistance is not a compact part but a wire or something else that has resistive properties to be exploited.

 

[bigdogguru]

From a purely theoretical standpoint, I fully agree with LvW, for the reasons he cited.


However, from a practical standpoint, as both of us noted:

Of course, there are still resistances present, wire, etc.

Remark: Of course, in some cases the necessary resistance is not a compact part but a wire or something else that has resistive properties to be exploited.

It is the presence of this "resistance" which makes the implementation of a low pass filter with the application of a single capacitor possible and often utilized in the sense of a practical application.

As you did not specify the analysis of your query to be in theoretical or practical terms, you have receive both.


BigDog

 

[m.kelley]

It seems as though the equivalent resistance of the device is going to be the "R" in the lowpass filter equation. By putting a single capacitor in shunt with the device a filter is created. A single capacitor model just adds a phase shift of +/- 90 degrees to your voltage/current.

 

[LvW]

LvW is trying to explain the effect of the resistor in terms of the formula.
I can understand that, but still find it difficult to understand it.

How the power supply's capacitor managed to do low pass filtering without the resistor intentionally place in front/series of capacitor?
I would like to understand it from the circuit and component's point of view.
Why must the resistor be there?

OK, siongboon, I'll try another explanation without formulas.
At first, the main task of a power supply capacitor is not lowpass filtering in the classical sense. Rather, these capacitors shall serve as an ac short of the power line.

Regarding the RC lowpass: If you need a passive circuitry that delivers at its output a smaller voltage than at its input you need a voltage divider.
If the division ratio should be frequency-dependent you need a frequency dependent voltage divider.
Low pass case: Large division ratio for the lower part of the frequency spectrum and smaller for increasing frequencies.
Thus, knowing the properties of a capacitor it is clear that we need the classical R-C arrangement for low pass filtering.
Assuming zero source impedance (ideal voltage source) there is no voltage divider at all without this resistor R - and as the result we have input=output=const (no filtering) .
Of course, a real signal voltage source always has a source resistance Rin which can take over the task of the filter resistor - if the value of Rin is known (and constant).
Therefore, to be (nearly) independent on this unknown value Rin, the filter resistor should be selected much larger than the expected value for Rin.
Further questions?

 

[Kral]

singboon,
Regarding your comment on power supply filters performing without the aid of a series resistor: The formula for cutoff frequency depending on a series resistor is based on linear circuit theory. A power supply filter has highly non-linear circuit element(s), the diode(s). In this case, the capacitor is used strictly as an energy storage component.

 

[albbg]

In a theoretic (non realistic) situation where you have an ideal voltage generator connected to a (ideal) capacitor with no series resistor, the voltage will be imposed by the generator and there is no way the capacitor can lower this voltage. Independently from the frequency.
If we instead suppose to have the voltage drop, this means that voltages across generator and capacitor will be different. But since the resistance in between is zero (ideal condition) means the current will be V/0 --> infinite regardless from the level of V.

During the initial transient, if the capacitor is not charge, there will be an ideally infinite current pulse of duration --> 0 that will charge istantaneously the capacitor.

this means single capacitor can be used as filter, only using the parasitics L and/or R of wiring and non-ideal internal resistance of voltage sources that will dissipate the power generated by the flowing current and the voltage drop (depending from the frequency).

 

[siongboon]

Thank you for the various explanation.
I think I have got some of the insight after the explanation.

 

[BradtheRad]

A resistor inline limits current to the capacitor. The current is what 'fills up' the capacitor.

Thus a higher series resistance makes it take longer to charge to a given volt level.

The volt level on the capacitor opposes an incoming signal (concept of a low pass filter).

I have Youtube videos which portray capacitor action visually. Current through wires is portrayed. Dynamic charge level on a capacitor is portrayed.

www.youtube.com/watch?v=eIWEU4pObJw

Equivalent Series Resistance comes into play when capacitors are used to filter a power supply. A capacitor with low ESR has a short time constant, and it works well at filtering ripple V at 50-60 Hz.

A capacitor with high ESR becomes 'sluggish' in comparison. It has a longer time constant. It becomes ineffective as a high pass filter.

Link to my Youtube video demonstrating this visually, over a range of ESR values:

www.youtube.com/watch?v=vm8wf7y94rM

 

[siongboon]

Thank you BradtheRad.
I like your explanation very much.
The videos are great too.

A resistor inline limits current to the capacitor. The current is what 'fills up' the capacitor.

Thus a higher series resistance makes it take longer to charge to a given volt level.

The volt level on the capacitor opposes an incoming signal (concept of a low pass filter).

I have Youtube videos which portray capacitor action visually. Current through wires is portrayed. Dynamic charge level on a capacitor is portrayed.

www.youtube.com/watch?v=eIWEU4pObJw

Equivalent Series Resistance comes into play when capacitors are used to filter a power supply. A capacitor with low ESR has a short time constant, and it works well at filtering ripple V at 50-60 Hz.

A capacitor with high ESR becomes 'sluggish' in comparison. It has a longer time constant. It becomes ineffective as a high pass filter.

Link to my Youtube video demonstrating this visually, over a range of ESR values:

www.youtube.com/watch?v=vm8wf7y94rM