# Second order passive low pass filter design

#### pinkyponky

Joined Nov 28, 2019
351
Hi,

Im planning to design the filter circuit of second order passive low pass filter (RLC circuit).

I have read about the Damping and I understand that most commonly under damping systems will be used, because which is most faster in the response and it will settle down fast.

And also I have read about the Quality factor, it says that, "Under damped systems with a low quality factor may oscillate only once or a few times before dying out. As the quality factor increases, the relative amount of damping decreases". Here, I'm not understanding that, firstly, it says that with low quality factor it will oscillate less times . Secondly, it says that, with high quality factor it will decrease the damping. My understanding is that, when low quality factor directly proportional to the oscillations, on other hand, the high quality factor should be also directly proportional to the damping. Right?. But, it is not like that.

And also trying to understand practically by simulating the parameters as follow below. From below, If Q increases and the damping of the waveform will also increasing, this is opposite to above text. Please try to simulate your self as well. Please can you observe the Q and waveform by tuning the value of R.

Please can you teach me which one is right?. How do I need to design the good Second order passive low pass filter?.

#### Ian0

Joined Aug 7, 2020
9,810
As you probably realise "Quality factor" is nothing to do with how good it is. It is simply the ratio of the energy stored in the "tank" (the inductor and capacitor) to the energy lost per cycle. It is probably best referred to simply as "Q" to avoid confusion, but it's nothing to do with Q-anon either!
High damping factor = low Q
Low damping factor = high Q
The higher the Q the more it oscillates.

If you want to design a "good" 2nd order passive filter, first, you have to decide what you mean by "good".
1) Do you want the best attenuation of frequencies just above the cutoff frequency?
2) Do you want the flattest frequency response of frequencies below the cutoff frequency?
3) Do you need constant group delay? (That means that all frequencies are delayed by the same amount, and a complex waveform tends to stay the same shape after if has been filtered.
You can't have all three!

If you want best attenuation, you will get a peak in the response around the cutoff frequency, and oscillation if you hit it with a step function or a pulse.
if you want constant group delay, you will not get best attenuation.

If you prefer 1), you need a Chebyshev filter: enter Q>1/√2
if you prefer 2), it's a Butterworth: enter Q=1/√2
and if you prefer 3) it's a Bessel: enter Q = 1/√3
(there are plenty more types, each with different features).

Once you've designed it, don't forget that it only works with the values of load- and source-impedance that you have specified. If you change either, you will change the Q.

If you're serious about passive filter design, get a copy of A. Zverev "Handbook of Filter Systhesis" ISBN 13: 9780471986805.
It's not a thrill-a-minute page-turner!

#### LvW

Joined Jun 13, 2013
1,756
My understanding is that, when low quality factor directly proportional to the oscillations, on other hand, the high quality factor should be also directly proportional to the damping. Right?. But, it is not like that.
No - your understanding is not correct. "Low quality factor" are all quality factors Q<1/SRT(2) which gives you a Butterworth (max. flat) magnitude response. For rising Q values, the magnitude shows an increasingly larger peak in the pole frequency region.
The corresponding step response shows an increasingly larger overshoot, which then transitions into several damped (decaying) oscillations.

#### pinkyponky

Joined Nov 28, 2019
351
As you probably realise "Quality factor" is nothing to do with how good it is. It is simply the ratio of the energy stored in the "tank" (the inductor and capacitor) to the energy lost per cycle. It is probably best referred to simply as "Q" to avoid confusion, but it's nothing to do with Q-anon either!
High damping factor = low Q
Low damping factor = high Q
The higher the Q the more it oscillates.

If you want to design a "good" 2nd order passive filter, first, you have to decide what you mean by "good".
1) Do you want the best attenuation of frequencies just above the cutoff frequency?
2) Do you want the flattest frequency response of frequencies below the cutoff frequency?
3) Do you need constant group delay? (That means that all frequencies are delayed by the same amount, and a complex waveform tends to stay the same shape after if has been filtered.
You can't have all three!

If you want best attenuation, you will get a peak in the response around the cutoff frequency, and oscillation if you hit it with a step function or a pulse.
if you want constant group delay, you will not get best attenuation.

If you prefer 1), you need a Chebyshev filter: enter Q>1/√2
if you prefer 2), it's a Butterworth: enter Q=1/√2
and if you prefer 3) it's a Bessel: enter Q = 1/√3
(there are plenty more types, each with different features).

Once you've designed it, don't forget that it only works with the values of load- and source-impedance that you have specified. If you change either, you will change the Q.

If you're serious about passive filter design, get a copy of A. Zverev "Handbook of Filter Systhesis" ISBN 13: 9780471986805.
It's not a thrill-a-minute page-turner!
Thank you for nice explanation. Still I have few more questions.

1) What is difference between the Damping and Oscillating?. I thought that both are same. What do you say?.
2) Is that good idea to design a circuit with high Q, because I'm using this filter to filtering out the noise, but when I use high Q then the system will oscillate more, right?. Is that fine if we have more oscillation when filtering the signal?.

#### Ian0

Joined Aug 7, 2020
9,810
1) What is difference between the Damping and Oscillating?. I thought that both are same. What do you say?.
Damping stops oscillation: - they are opposites.
2) Is that good idea to design a circuit with high Q, because I'm using this filter to filtering out the noise, but when I use high Q then the system will oscillate more, right?. Is that fine if we have more oscillation when filtering the signal?.
The question is: Is there any energy in the system at or around the cutoff frequency?
If your cutoff frequency is, say, 1kHz, but there is nothing going on in your signal around 1kHz, then there's no energy there to excite the oscillations. The Q could be as high as you like, and it won't cause a problem.
When I say "no energy at that frequency", don't forget that step functions and pulses have energy at all frequencies.

#### pinkyponky

Joined Nov 28, 2019
351
Damping stops oscillation: - they are opposites.

The question is: Is there any energy in the system at or around the cutoff frequency?
If your cutoff frequency is, say, 1kHz, but there is nothing going on in your signal around 1kHz, then there's no energy there to excite the oscillations. The Q could be as high as you like, and it won't cause a problem.
When I say "no energy at that frequency", don't forget that step functions and pulses have energy at all frequencies.
As I see that, the RLC second order passive low pass filter and Butter worth filter, the both filters are different. So, which filter is good to use to filter the signal of MOSFET and passing to the ADC. In this application, which filter is better to use?.

#### Ian0

Joined Aug 7, 2020
9,810
There's no generic answer to that question. Please draw out your circuit. It depends a lot on what the MOSFET is doing, and what signal you wish to remove, and what you wish to do with the signal once you have sampled it with the ADC.

#### pinkyponky

Joined Nov 28, 2019
351
Damping stops oscillation: - they are opposites.

The question is: Is there any energy in the system at or around the cutoff frequency?
If your cutoff frequency is, say, 1kHz, but there is nothing going on in your signal around 1kHz, then there's no energy there to excite the oscillations. The Q could be as high as you like, and it won't cause a problem.
When I say "no energy at that frequency", don't forget that step functions and pulses have energy at all frequencies.
If the Q value is High, then the signal oscillations are more. So, Don't you think that it won't be a problem?. Since I'm filtering the DC voltage?.

#### LvW

Joined Jun 13, 2013
1,756
Damping stops oscillation: - they are opposites.

The question is: Is there any energy in the system at or around the cutoff frequency?
If your cutoff frequency is, say, 1kHz, but there is nothing going on in your signal around 1kHz, then there's no energy there to excite the oscillations. The Q could be as high as you like, and it won't cause a problem.
When I say "no energy at that frequency", don't forget that step functions and pulses have energy at all frequencies.
Just for clarification:
* When we speak about oscillations of filters we do NOT refer to any kind od self-oscillations. Instead, we only mean oscillations as a form of step response.

* Quote: "The Q could be as high as you like, and it won't cause a problem."
To me, this sounds a bit misleading.
As I have mentioned in my former answer - any Q-value larger than 1/SQRT(2)=0.7071 (max. flat Butterworth resonse) will exhibit a magnitude peak around the pole frequency.
So - I think you cannot and you should not choose a Q-value "as high as you like".
Example: A value Q=1.305 will cause a gain peaking of 3 dB.

#### BobTPH

Joined Jun 5, 2013
8,957
There's no generic answer to that question. Please draw out your circuit
Good luck with that request. This poster has consistently asked fir specific answers ti general questions and refuses to give more details, mostly because he has no idea of what he is trying to accomplish.

Bob

#### Papabravo

Joined Feb 24, 2006
21,225
One more time. Your approach to designing a filter is putting the cart before the horse. You have decided on a filter type before articulating what you want the filter to do. Those in the know write filter specifications and let the desired performance dictate the filter type. Time to get your head out of the dark place that it is in.

#### Ian0

Joined Aug 7, 2020
9,810
So - I think you cannot and you should not choose a Q-value "as high as you like".
Example: A value Q=1.305 will cause a gain peaking of 3 dB.
For example, if the cutoff frequency is 1kHz, but the signal has no 1kHz content, the gain at 1kHz doesn't matter because there won't be any 1kHz output.

Take as an example a buck regulator. The output is filtered by 100uH/220uF which is a second-order low-pass filter with a cutoff frequency about 1.1kHz. The Q of that filter is set by the load resistance. With no load, the Q becomes extremely high. There is a huge peak in the response about 1kHz, but the switching part of the buck regulator is switching at 100kHz. Unless something goes horribly wrong with the feedback loop there is no output at 1.1kHz, the size of the response peak at 1.1kHz is immaterial.

#### crutschow

Joined Mar 14, 2008
34,427
I see that, the RLC second order passive low pass filter and Butter worth filter, the both filters are different.
You are confusing two different things.
They are not different filters, they are different characteristics of a filter.

RLC second order passive low-pass filter describes what components are used to make the filter, and the response type (high-pass or low-pass) but not the filter's behavior otherwise.

Butterworth describes the behavior of the filter in the passband and the nature of the rolloff in the stop-band.

#### Audioguru again

Joined Oct 21, 2019
6,691
We are guessing that the frequency is 1kHz and the load is a high resistance.
What is the cutoff frequency and load for the passive lowpass filter?
Do you want it to have a Butterworth flat response below its cutoff frequency?

#### Ian0

Joined Aug 7, 2020
9,810
If the load is high resistance, then the inevitable consequence is that the inductance will be unfeasibly large, and would make a Sallen & Key a much better proposition.
If this is something to do with current sensing in a power supply, then the 180° phase shift from a 2nd order filter will really mess things up.

#### dcbingaman

Joined Jun 30, 2021
1,065
If the load is high resistance, then the inevitable consequence is that the inductance will be unfeasibly large, and would make a Sallen & Key a much better proposition.
If this is something to do with current sensing in a power supply, then the 180° phase shift from a 2nd order filter will really mess things up.
Ian0 is correct best choice here is Sallen Key. It is a second order filter and technically to achieve 40db per decade roll-off you only need capacitors and resistors. The values determine the characteristics of such filter. Look up and read aboud Sallen Key filter. This is only true if you don't care about phase shift.

#### BobTPH

Joined Jun 5, 2013
8,957
If the load is high resistance, then the inevitable consequence is that the inductance will be unfeasibly large, and would make a Sallen & Key a much better proposition.
If this is something to do with current sensing in a power supply, then the 180° phase shift from a 2nd order filter will really mess things up.
Except that what he is filtering is a DC power supply. See his other thread.

Bob

#### Ian0

Joined Aug 7, 2020
9,810
Except that what he is filtering is a DC power supply. See his other thread.

Bob
Silly of me to think that because the other thread was about filtering power supplies, this thread would be about filtering something different!

#### Ian0

Joined Aug 7, 2020
9,810
You can use a Sallen and Key filter to smooth a power supply. It suffers from the usual Sallen & Key problem of a frequency response that starts to rise again at higher frequencies, but you can get a very useful amount of attenuation for a few cheap components.