best way to adjust an RLC bandpass filter

If you’re working with a series RLC circuit (with the R and L in series, and the cap to ground) – then you actually have a second order low-pass filter – as opposed to a band-pass. As you note – there are theoretically an infinite number of solutions to a second order filter design with a specific resonant frequency. The difference will be the Q of the filter, and the peaking at the point of resonance. A filter with too high a Q will be under-damped – and will be likely to ring as a result of transient signal changes. In practice – an engineer will typically select a ratio of component values to damp at some specific level under the point of critical damping (Q=1 – no ringing). A Q of about 1.2 is the point our own filter guru tends to shoot for. He recently demonstrated this to me by comparing two RLC filters – both with a 1 ohm resistor. One had a 10uH inductor and a 10uF cap; while the other had a 100uH inductor and a 1uF cap. Both had the same resonant frequency and attenuation slope (12 dB / octave) – but the low Q circuit (with the higher capacitance) had very little ringing – while the High Q circuit went through 8 or 9 ringing cycles before settling during a transient response test. The Q of the circuit can be calculated as 1/R(√L/C).

 

One other point of note is that unless you are winding your own inductors and building your own caps in practice the values you solved for are impossible. Building your own cap is not at all practical so you start by selecting a value that you can obtain in the range your working(generally larger caps for smaller frequencies). Next you have some flexibility in making an inductor so you should try to calculate for a standard value but its not the end of the world if you have to make one. And last of all is the resistor since they come in the broadest part value selection.

That said when I make filters I try to keep resistors out of it since they equal loss. And selecting the correct LC values is a matter of filter types. One can make many different types of filters with a handful of L&C's. For instance Butterworth, Tchebychev, bessel, Gaussian... which all have distinct pro's and con's that make them well suited certain tasks.

 

I agree with your inclination - as long as we're talking about series resistive elements, that increase insertion loss without adding value to the equation. I do however, add resistance into almost any filter design - as part of the optimization process. I will usually split the value of the X caps into two halves - and then add resistance in series with one half of the X capacitance - to set Zeta to a controlled, desirable value - limiting filter Q to a set limit. The damping factor added by the resistance is essential to getting where I need to be in terms of EMI performance.

 

I have seen resistors added to caps to bring their Q down to the same range as the coils. I just have never needed to do it yet. And there are some filters that just require resistors(zero's) like ellipticals but the group delay on those are so bad that I rarely consider them for LC implementations. But then again I think perhaps we are building filters for different applications. I primarily build small signal filters and I think you are operating more in the realm of power or large signal filtering(that is just a guess).

Also to add some value to the OP, If you are filtering small signals in the frequency range you described then I believe active filters(op amp) to be absolutely the way to go. It is just so much easier to use caps and resistors than it is to find good Q in a high value inductor plus you can add gain at the same time. Filtering high power is another story.

 

Good guess - my filters are typically in the input and output lines of power supplies or high-power sine-shaping transformer-rectifiers. We do build active electronic filters, and even program software Kalman filters, particle filters, etc. for some applications - but when I'm working with high voltage spikes, or currents up over 1000A, an active filter just won't quite cut it.