I want to know what the cutoff frequency is when I change C6 and C11 to 100pF (I don't have 82p/150p)


I'm not even sure what kind of a filter it is.. it says it's an elliptic Cauer, but I can't find anything on how to design it or calculate L/C.

I need the cutoff to be anywhere between 30 and 50 MHz

  • \$\begingroup\$ If you have 3 100pF caps, you have 150pF. I don't know what else is in your junkbox but I expect you can make 82pf (75-90pf) easily too. So build it as designed ...or simulate as Neil suggests. \$\endgroup\$
    – user16324
    Commented Aug 2, 2017 at 9:59

1 Answer 1


Increasing C6 will reduce the frequency of that stopband zero from 45.4MHz to 41.1MHz.

Reducing C11 will tend to widen the passband and upset the passband shape.

I suggest you get a copy of LTSpice, a free, fully featured simulator, enter the circuit, and see whether you like the result. This will allow you to play with the results of other component swaps.

A hint. Changing the values of odd components in an existing design rarely give you a nice filter, especially when zeroes (Cauer filter) are involved. The flatness of the passband, the depth of the stopband, tend to require all of the components to be at their design ratios.


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