# Design a two-stage LC low-pass filter with one fixed stage

I need to design a two-stage LC low-pass filter just as the picture below. The values of L and C in the second stage(L6 and C4) are fixed. I can only change L8 and C5. The required cutoff frequency is around 2MHz. The Q factor is not so matter.

So my question is: is there any special point I need to pay attention to when designing L8 and C5. (I know that for the RC filter, the resistance of the second stage needs to be much larger than that of the first stage. But I dont know whether it would be the case for LC filter. Or any similar rules)

Update: the source impedance is about 50ohm. And the load is about 30ohm. Of course at the end some resistors would be added into the circuit, but at this moment, I think I can just ignore them.

• LTSpice is free. – Andy aka Dec 16 '16 at 13:50
• So is ngspice :). – Al Bundy Dec 16 '16 at 14:30
• @Andyaka I need to learn some basic ideas first. I dont want to select the components randomly and then simulate them. – billyzhao Dec 16 '16 at 14:31
• He has encouraged you to think and then check your ideas using simulator not to select components values randomly. – Al Bundy Dec 16 '16 at 14:33
• What is the source and load impedance? Are we to assume 50 ohms for each? – George Herold Dec 16 '16 at 14:38

I have derived the transfer function of this 4th-order network but the trick is to factor it into two second-order polynomial forms. As there are no zeros in this network, we can use the Fast Analytical Techniques (FACTs) to determine the denominator coefficients from $$\b_1\$$ to $$\b_4\$$ or use a simple Thévenin approach and ask Mathcad to factor the $$\s\$$ terms. The only difference is that Mathcad won't arrange terms in series-parallel forms what the FACTs would let you do. Anyway, here, the transfer function came quite easily. Below is the Mathcad sheet showing the raw transfer function (using Thévenin) and the other one using a 4th-order polynomial form:

The difficulty now lies in trying to factor-in the 4th-order denominator into two 2nd-order filters. If we consider that the second resonance is above the first one, then a simplification is feasible. This is what the below sheet shows and the deviation with the full-blown formula is not too ugly. Considering your fixed values for $$\L_6\$$ and $$\C_4\$$, you should be able to extract the values for $$\L_8\$$ and $$\C_5\$$.

As usual, I encourage you to discover the FACTs which are of invaluable help for determining transfer functions for passive or active circuits. You can start with the Extra-Element Theorem from Dr. Middlebrook (https://en.wikipedia.org/wiki/Extra_element_theorem) and look at the presentation taught at APEC in 2016: http://cbasso.pagesperso-orange.fr/Downloads/PPTs/Chris%20Basso%20APEC%20seminar%202016.pdf

You also have an example of a 4th-order $$\LC\$$ network tuned for a maximally-flat response here http://cbasso.pagesperso-orange.fr/Downloads/Book/List%20of%20FACTs%20examples.pdf.

"I know that for the RC filter, the resistance of the second stage needs to be much larger than that of the first stage. "

No - this is NOT a general requirement, neither for RC nor for LC. Rather, this is a "design option" only - with the following background:

It is very easy to design a direct coupled passive two-stage filter when both stages are sufficiently decoupled. Without using an active buffer between both stages, it is often sufficient to have a second stage that does not load the first stage considerably. And the advantage is: You can design (calculate) both stages separately.

In your case, we have a lossless 4th-order lowpass, which is somewhat unrealistic - really no resistors ? This makes no sense because the poles will be on the imaginary axis (infinite Qp).

Update: OK - as one can see you have updated your question (resistors at the input and output). Therefore, I propose the following approach:

1.) Using the classical rules of voltage division you can find the general transfer function (finite Rin and Rout) for the last stage only (setting L8,C5=0).

2.) Implemeting the particular values, you can roughly plot the magnitude of the output voltage as a function of the input voltage (variable frequency).

3.) From the result, you can decide what you want to do (what the task of the first stage should be).

• Thank you. I just want to learn the general idea first. Of course I would add resistance into my circuit at the end. So can I just say that, the values in the second stage have no direct influence on the choice of the first stage(except for the cutoff frequency or the bandwidth)? – billyzhao Dec 16 '16 at 14:29