calculate the Q factor of LC bandpass

The Q relates to the resistance relative to the reactive components. Do you know the values for your resistances? It will depend also on whether you model the resistances as parallel or serial to the other components. Typically the L will have an ESR (equivalent series resistance). But if you are working at high frequencies such as power supply distribution, for example, the ESR of the capacitor might be the significant factor.

I just solved the equations for resonance of this circuit. But I left out the R values to simplify the math. With typical Q values the R does not change the resonance significantly.

What formula do you use for f0?

"power boy"

.... Phil

The source and load are left out, but otherwise, that's your basic

Cheers

Phil Hobbs

Should have added:

To the OP: A multi-section BPF like this doesn't have a single unique value of Q. You can talk about the Q of each section individually, but in general there isn't a single number for the whole filter.

Cheers

Phil Hobbs

Some people consider the CF/BW ratio to be the Q of a bandpass filter. It's not literally Q in the energy storage sense, but it's handy at times.

not

I wonder if the OP's professor had that in mind? ;)

Cheers

Phil Hobbs

In order to calculate the Q for the individual section, you still would have to know the parasitic components, usually the series resistance of the inductor. From there calculate separately the Q for the series LC as well as separately for the parallel LC.

I omitted the series source resistor Rs and load resistor RL in the schematic. As I said, there are many types of LC bandpass. When you do a design:

1. you can convert LC bandpass from LC low pass.
2. you can get a book, they tell you a chart, you calculate the attenuation, and ladder # to get a bandpass type, sometimes you have to convert based on LpCp=LsCs the fundamental equation. (chebychev, butterworth, elliptical, linear phase, bessel, gaussian, constant-K, etc), then convert from basic topology to final circuit.

But if all values a given, you simulate and find this is a bandpass, you want to calculate the Q or Bw. How can you do? I couldn't find a formula to do this. I can find RLC series bandpass formula and RLC parallel(resonance) formula.

I omitted the series source resistor Rs and load resistor RL in the schematic. As I said, there are many types of LC bandpass. When you do a design:

1. you can convert LC bandpass from LC lowpass.
2. you can get a book, they tell you a chart, you calculate the attenuation, inband ripple, and ladder # to get a bandpass type, sometimes you have to convert based on LpCp=LsCs the fundamental equation. (then find type like chebychev, butterworth, elliptical, linear phase, bessel, gaussian, constant-K, etc), then convert from basic topology to final circuit.

But if all values are given, you simulate and find this is a bandpass, you want to calculate the Q or Bw. How can you do? I couldn't find a formula to do this. I can find RLC series bandpass formula and RLC parallel(resonance) formula.

I'm no expert on filters, but my undestanding is that each of the two sections, series and parallel, are resonant at some frequency with bandwidths defined by the resistances in the circuits. When you connect the two with the input across the two terminals on the left of your diagram and the output across the parallel cap and inductor on the right of the diagram, you get two peaks of resonance defined by all four components. The Q of each peak is again defined by the resistance. To the best of my knowledge this is *not* a band pass circuit. I wrote the equation for this circuit in by treating it as a voltage divider using the impedances of the parallel and series circuit. When you solve the equations you get two roots, so two resonances. I only did this with the reactive components and ignored the resistance. The resistance makes the math a lot more complex. Feel like solving some very messy equations?

Vout/Vin = Zp/(Zp+Zs) I'm sure you can find the equations for Zp and Zs. If not I can provide them. Radiotron Designer's Handbook, very old, but science doesn't change much. Download this and go to page number 193 in the PDF (labeled 152 in book), section E. Z1 and Z2 are Zs and Zp. Construct the formula above and find the roots... simple right? Well, actually that gives you the resonances, but not the Q. I don't actually know how to figure the Q in this circuit. That's another section of this book I haven't read completely...

Neither of these sets of components are low pass filters, so I don't know why you refer to converting from a low pass to a band pass. I guess if you drop Cs and Lp you get a low pass...

As someone else said, are you sure you have the schematic right? Move Cp to the point between Ls and Cs and you get a band pass... or a band reject... But then they wouldn't be named with s and p for series and parallel, would they?

If this is a band pass with Q not defined in terms of the resistance, I don't get it.

BTW, try simulating this. I'm pretty sure you will get two peaks.

It's a bandpass, all right. The series inductor and shunt capacitor ensure that its response rolls off as 1/f**2 at high frequencies, and the series capacitor and shunt inductor make the response go as f**2 at low frequencies.

The detailed shape in the middle depends on the component values and the source and load impedances, and can range from two widely spaced peaks to some nice smooth bandpass with good group delay properties.

And the way to find that out is by doing the math or reading a book that does the math, not by poking SPICE.

Cheers

Phil Hobbs

Mere mortals can fiddle a 1-pole RC filter, and usually a 2-pole RLC. Beyond that, it's really easy to get "lost in space."

A narrowband top-coupled bandpass filter has pretty much orthogonal cascaded sections, so is potentially fiddleable (sp?).

Active filters are easier to fiddle, because the sections don't (or shouldn't) interact.

Sure is. The usual way to fail IME is to tweak it up to be nice looking but a bit too narrow. Getting from there to the desired response sounds easy, but is actually very difficult--you might as well start over, get the right number of bumps spread over the right bandwidth, and then tweak for prettiness without giving up BW.

That's basically Dishal's method--mistuned sections look like opens or shorts.

I don't use a lot of active filters, but I imagine that's right. Of course since you can get 1- or 2-% tolerance components for cheap, they ought not to need a lot of tuning if you get the math right. ;)

I expect that if you were to parametrize the LC sections using omega-nought and |Z(omega-nought)| instead of L and C, turd polishing ^H^H^H^H^H^H^H^H^H^H^H^ Spice tweaking would be a lot easier.

Cheers

Phil Hobbs

You can calculate this structure by hand....

-------Ls---Cs-----------o------o | | Lp Cp | | -------------------------o------o

Trivial in Laplace (Heaviside) notation. The active-circuit equivalent is all over the place in my disco boombox of the late '70's (pre-Spice)...

and Color Organ...

Fiddling happens when the opamps have limited GBW or slew rates for the application. Sometimes swapping the sections around can help, theory to the contrary.

Sometimes I don't have an objective goal for filter response, so I simulate until I like the way it looks. Lots of things are like that.

I *did* the math. This circuit has two peaks in response. This morning I went back to the equations to see if/when the two peaks could overlap to form one peak. The discriminant can't be zero with positive values of inductance and capacitance, so the circuit will *always* have two peaks and is never a simple band pass.

The width of each of the two peaks is defined by the resistances in the circuit.

The amazing thing is that all this algebra is giving me flashbacks to high school math class. I keep hearing, "What's a zero and who cares about finding them?" I wonder who was saying that...

Well, F. E. Terman would be very surprised to hear that, and so would all the rest of us who have ever built a bandpass filter. Imagine all the spectrum analyzers and radios that just quit working, and it's all your fault. ;)

Cheers

Phil Hobbs

They've stagger tuned IF's since they were first invented.

Of course, but, for some reason, Hobbs is dodging the "rickman" two peak analysis. The Larkin phenomenon I guess :-( ...Jim Thompson