RLC Bandpass filter help.

Factual Questions
1

I am about to pull my hair out. I am trying to design a very basic RLC bandpass filter. I am using a series setup. I want the passband to be 100Hz - 40kHz. I have arbitrarily chosen R=1K. So, knowing that R/L should equal my bandwidth of 39900Hz and 1/(sqrt(LC) should equal my center frequency of sqrt(10040000)=200Hz I have calculated L=25mH and C=1uF .

When run an AC sweep on this thing in spice I am getting 3dB points at 156hz and 6kHz!

What the heck am I doing wrong? I need this to pass 100hZ thru 40kHz and it seems to be have a passband of 5kHz or so.

I know I have configured the circuit correctly and set up the freq sweep correctly in spice.

Can somone tell me if I have calculated the correct values for RLC for this bandpass filter? This is killing me here.

2

n=? T? pi? What?

3

Huh?

4

What is the configuration?

5

Its a series RLC bandpass circuit. I am not sure what else you want to know. I chose R=1K out of thin air.

6

Or is it just that you’ve miscalculated “sqrt(100*40000)=200Hz” instead of 2000 Hertz?

7

I am sorry, thats a typo. I did calculate it as 2Khz not 200Hz.
Sorry.

8

A 100 Hz to 40 KHz RLC band-pass filter? Wow, that’s a pretty wide pass-band for a resonate circuit. Have you thought about using an active LP and HP in series?

If I have time I’ll scratch out the equations to see what’s going on…

9

IIRC, when you use those equations, you need to convert your frequency values from Hz to rad/s. So your center frequency will be 2pi2000 rad/s and your bandwidth will be 2pi39900 rad/s. I came up with L=3.99 mH and C=1.59 uF, but you may want to double-check my math since I haven’t had coffee yet. :slight_smile:

10

What do you mean by “a series RLC bandpass circuit”. I assume you mean one R, one L, one C.

Is it all series, or is it T, pi or L? If T, pi or L, which element is the shunt?

It may seem obvious to you what you’re talking about, but it isn’t to me.

11

I did it using radians as well as regular freq just to see if that was the problem and it is still hosed up.

I understand what you are asking now D, It is all in series L C R with the output between the C and R.

On a side note wouldnt a T or Pi setup be a parelell setup?

12

IMNAEE

Desmostylus, R is the shunt IMHO.

Assuming all the math is correct (converting n to /omega and all), try a smaller resistor - say around 100-200 ohms for a better absolute tolerance.

Electrolytic Capacitors don’t have a “good frequency response” at higher frequencies.

13

FTR after poking around in PSPICE for a while I was able to make a functional BPF with the correct passband using a series config with L=20mH C=300nF and R=5K.

But my question still stands because these numbers don’t work in my equations and I really don’t like guessing like this.

What is the advantage of using active filters instead of a resonate circuit? Would using active filters give it a steeper rolloff? It rolls off pretty slow with the circuit as is.

14

Go with what headshok said.

15

Center frequency = 1/(2.pi.root(LC))

16

It would have helped a lot if you’d presented the circuit configuration and the equations in the first place. You also have omitted any kind of design criteria. Either provide the proper information, or work it out yourself.

17

Something is just not jiving here. My center frequency should be sqrt(wlwh) or sqrt(10040k2pi) which is coming out to about 5k radians/sec . If I take the L and C of the working BPF I modeled in SPICE I get a center freqency of 26Mhz in Radians. This isn’t even close, but the filter IS displaying the correct bandwidth in the simulation of SPICE.

Im sorry to be frustrating you Desmostylus but there are no design requirements other than I need to have a passband of 100hz-40khz. That is the only thing I need. I chose to use a basic RLC filter because it seems to be the simplest way to accomplish this.

18

Your center freq is sqrt(4pi[sup]2[/sup]10040k) or 2pisqrt(10040k), which will give you 12566 rad/s. The equations should work from there.

19

Update:

Like a true schmuck my calculator was the problem. I had accidentally set x= some number at some time. Since I was using X as my variable in my equations and I didn’t know it had been set already it was throwing out bogues values. I was doing ti right the whole time.

Thanks for the help though.

Although, I would still like to hear why active filtering would be better than passive in thie case. Does it affect the rolloff?

20

If you want simplicity, but better cutoff performance, all you need to do is add more stages, resulting in a cascaded T or pi network. Look up any or all of the following: “Classical filter design”, “Chebychev filter design”, “Butterworth filter design”.

The advantage of using active filters is that you don’t need inductors. This can be a big advantage at low frequencies, because sometimes the inductances required are large.

To improve cutoff performance with an active filter, you cascade stages, just as with a passive filter.