Second order passive low pass filter design

Thread Starter

pinkyponky

Joined Nov 28, 2019
318
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!
I would like to attenuate the signal above 20khz and implement the second order passive low pass filter to filter out the signal effectively which are above 20khz.

I would like to chose one of these filters but help me which filter is good to filter out the signal above 20khz and also maintain the flattest frequency response of frequencies below the cutoff frequency.

1) RLC passive low pass filter. 2) pi passive filter. 3) T passive filter.

Actually, please can you provide how to implement these filters and description of these filter such when it will be used and so on...
 

Ian0

Joined Aug 7, 2020
6,698
Here's the relevant page from Zverev.
9ABBF367-17BD-454B-ABC1-882559BC07DC.jpegIt's a second order filter that your want, so refer to the n=2 section, and
First you need to know the ratio of your load resistance to your source resistance, Rl/Rs. Refer to that line in the n=2 section.
Then read off the two coefficients - it's a Thevenin source with the inductor first, so refer to the bottom row (the first column is L1 and the second column is C1).
Work out the inductor
\(
L_1 = \frac{k R_l}{2\pi f}
\)
where k is the coefficient you have read from the table, f is the cutoff frequency, and Rl is the load impedance.
Same for the capacitor
\(
C_1 = \frac{k}{2\pi fR_l}
\)

You can't make Pi- and T-section 2nd order filters. Pi-section and T-section filters must be an odd number of orders.
 
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Thread Starter

pinkyponky

Joined Nov 28, 2019
318
Here's the relevant page from Zverev.
View attachment 247173It's a second order filter that your want, so refer to the n=2 section, and
First you need to know the ratio of your load resistance to your source resistance, Rl/Rs. Refer to that line in the n=2 section.
Then read off the two coefficients - it's a Thevenin source with the inductor first, so refer to the bottom row (the first column is L1 and the second column is C1).
Work out the inductor
\(
L_1 = \frac{k R_l}{2\pi f}
\)
where k is the coefficient you have read from the table, f is the cutoff frequency, and Rl is the load impedance.
Same for the capacitor
\(
C_1 = \frac{k}{2\pi fR_l}
\)

You can't make Pi- and T-section 2nd order filters. Pi-section and T-section filters must be an odd number of orders.
The values are in the above image is K-values?.

Ok, Even the pi-filter and t-filter are not 2nd order, fine. But, Why can't use the pi and T filters in this case?. Please can you explain?.
 

Thread Starter

pinkyponky

Joined Nov 28, 2019
318
Yes.

You can, but you said you wanted a 2nd order filter.


Yes, I said. But, which filter is more effective. The output of this filter is connecting to the ADC.

Can I chose any K value in the column?. If so, which value should I chose?.

Why n=2 section?. Does it mean second order?.
 

Ian0

Joined Aug 7, 2020
6,698
Not clearly understand. But, my source impedance is 350uohm and load is 30Gohm.
So, use the "INFINITY" row, because the source impedance is effectively zero; but you're going to need a much lower load impedance.
A quick inspection of the equation for L should tell you that the value of L is proportional to the load impedance.
There's a reason most people use active filters or RC filters in this situation!
 

Thread Starter

pinkyponky

Joined Nov 28, 2019
318
So, use the "INFINITY" row, because the source impedance is effectively zero; but you're going to need a much lower load impedance.
A quick inspection of the equation for L should tell you that the value of L is proportional to the load impedance.
There's a reason most people use active filters or RC filters in this situation!
If I chose the infinity row, then I got below values:
L value is '661,990 H' and C value is '1.47E-15 F'. I think these values are impossible to get it in real time.
 

Papabravo

Joined Feb 24, 2006
19,576
I too am curious about the circuit you are talking about, and what problem you have that you think a filter is going to solve.
 

Papabravo

Joined Feb 24, 2006
19,576
If I chose the infinity row, then I got below values:
L value is '661,990 H' and C value is '1.47E-15 F'. I think these values are impossible to get it in real time.
Time has nothing to do with the ability to realize those values. That said, the capacitance might be easier to realize than the inductance.
Going back to post #12, we still lack the actual requirements you are aiming for the filter to fulfill.
 

Ian0

Joined Aug 7, 2020
6,698
If I chose the infinity row, then I got below values:
L value is '661,990 H' and C value is '1.47E-15 F'. I think these values are impossible to get it in real time.
That's a lot of turns of fine wire. Your thumbs would hurt by the time you had finished winding it.
If you really want to use a passive filter, add a load resistance of between 50Ω and 1k.
Alternatively. . .
If you want a second order filter, use an RCRC filter, but Q can be no higher than 0.5
or
Use an active filter
but
Do you really want a second order filter - a phase shift of 180° sometimes isn't such a good idea, especially if it ends up inside a feedback loop.
 

ericgibbs

Joined Jan 29, 2010
16,785
Current sensor as an input of this filter and output of this filter is connecting to the amplifier.
hi pp,
Please post a circuit diagram of this 'amplifier' circuit , showing the location of the Current sensor/filter network and a short explanation of why you need the filter.?

If we are to give meaningful answers, we need to know this information, so that we can stop having to guess its purpose.

E
 

Thread Starter

pinkyponky

Joined Nov 28, 2019
318
That's a lot of turns of fine wire. Your thumbs would hurt by the time you had finished winding it.
If you really want to use a passive filter, add a load resistance of between 50Ω and 1k.
Alternatively. . .
If you want a second order filter, use an RCRC filter, but Q can be no higher than 0.5
or
Use an active filter
but
Do you really want a second order filter - a phase shift of 180° sometimes isn't such a good idea, especially if it ends up inside a feedback loop.
Shall I use these values, Just I was simulated online.
 
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Ian0

Joined Aug 7, 2020
6,698
You have simulated a 1Ω source impedance, when your source impedance is almost zero, you have no load resistance, and your filter cutoff frequency is 5kHz when you said it should be 20kHz.
 
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