Hello all, this is kind of a side project and could use some help on a passive RC bandpass filter I am constructing. The parameters I need are a center frequency at 3800 Hz and 10% tolerance on upper and lower cutoff so upper is 4180Hz and lower is 3420 Hz. I have used this link (http://aries.ucsd.edu/najmabadi/CLASS/ECE53B-LAB/05-W/LAB/filter.pdf) to solve for my resistors and capacitors and simulated through Multisim, but I am still not getting a correct graph where my cutoffs are at -3dB. I am using the bode plot instrument on Multisim and I keep getting a graph where my center frequency is correct but at the wrong gain.
My calculations thus far:

I am told to make the high pass filter resistor lets say R1 much larger than R2 on the low pass filter. So..
(High pass)
ωc=ω(lower)=2∏*3420=1/(R1C1)
R1 is chosen to be 100kΩ
C1=.4653*10^-9

(Low pass)
ωc=ω(upper)=2∏*4180=1/(R2C2)
R2 is chosen to be 10k
R1>>R2
C2=3.8*10^-9

here is the bode plot
https://www.dropbox.com/s/fqfqtmuz9wy4xk2/BodePlotofRCBandpass.pdf
Here is the center frequency
https://www.dropbox.com/s/v824ccflx9ky2ur/CenterFreqBodePlot.pdf
The gain is around -5 dB, not what I want.
Any other thing I need to add let me know please!

 

Have you accounted for insertion loss?
What are the input and output impedances?
Do you know what loaded Q is? You can't just slap two sections together and expect them to play nice with each other.

Why is your frequency F=1GHz and I=1 MHz ?? Isn't that kinda far away from 3800 Hz.?

At this frequency range you'll get way better results with a Sallen-Key bandpass filter

http://www.changpuak.ch/electronics/Sallen_Key_Bandpass_light.php

 

I am looking in my textbook. "passive RC bandpass filter" In my textbook, passive bandpass filter has three parts: resistor, inductor and capacitor. Therefore I don't know WTF you are doing.

 

I am looking in my textbook. "passive RC bandpass filter" In my textbook, passive bandpass filter has three parts: resistor, inductor and capacitor. Therefore I don't know WTF you are doing.

What he is doing is cascading an RC lowpass and an RC highpass. The corner frequency of the highpass is below the corner frequency of the lowpass. It is just about the worst possible bandpass filter you can make at that frequency.

 

What he is doing is cascading an RC lowpass and an RC highpass. The corner frequency of the highpass is below the corner frequency of the lowpass. It is just about the worst possible bandpass filter you can make at that frequency.

Yes, I saw the cascade, but my textbook uses it for active bandpass filter. Anyway, thank you for explaining it to me.

 

Yes, I saw the cascade, but my textbook uses it for active bandpass filter. Anyway, thank you for explaining it to me.

No problem. Anything you can do with a passive filter you can do with an active filter subject to the availability of suitable parts like inductors that work at audio and powerline frequencies and opamps with a 10 TeraHertz GBW product.

 

but I am still not getting a correct graph where my cutoffs are at -3dB.
The gain is around -5 dB, not what I want.
Any other thing I need to add let me know please!

What do you want as a gain value?
With two pasive RC sections in series it is not possible to achieve any gain.
At the center frequency you always have a gain lower than 0 dB.For positive gain values you need an active filter circuit.

Regarding the -3dB value:
The cutoffs always are 3dB less than the midband gain.
That means: If the gain at the center frequency is -5dB the value at the cut-off frequencies should be -8 dB.

EDIT: I should add that the last line above applies only if both calculated cut-offs are not too close to each other (they do not interfere). However, in your case, the calculation is a bit more involved.

 

What do you want as a gain value?
With two pasive RC sections in series it is not possible to achieve any gain.
At the center frequency you always have a gain lower than 0 dB.For positive gain values you need an active filter circuit.

Regarding the -3dB value:
The cutoffs always are 3dB less than the midband gain.
That means: If the gain at the center frequency is -5dB the value at the cut-off frequencies should be -8 dB.

EDIT: I should add that the last line above applies only if both calculated cut-offs are not too close to each other (they do not interfere). However, in your case, the calculation is a bit more involved.

An option in regard to gain.

Like LvW said, your passive filter circuit can not have any Gain because there are no components to provide the Gain. In an active circuit you have external power supplies, they provide the additional power/energy to the circuit. What is even worse is that in a passive circuit the components consume some of the power/energy of the input signal, and since there is not external sources of power/energy, the output signal is always slightly smaller than the input signal.

Therefore. Here is what you can do. You can split the whole system into two parts. The filter part. The gain part. Since you are required to use passive circuit for the filter, use the passive circuit. Then feed the output of the filter into the amplifier circuit which is an active circuit that uses an op-amp. Depending on your instructor, this might be acceptable so consult them.

So you have choices:
* passive filter circuit, no gain
* passive filter circuit with active circuit to provide gain
* complete active circuit which will have active circuit for low pass filter, for high pass filter and for amplifier, three stages altogether (if allowed)

 

As Papabravo noted your high-pass section has a corner frequency below the low-pass section which is backwards. Correct that and your filter response should look more like you would expect. But as also noted, a single passive RC filter section has poor filter characteristics even when correctly made.

 

Do you know what loaded Q is? You can't just slap two sections together and expect them to play nice with each other.

I'm not sure of loaded Q, but I thought Q was the quality factor and to for wide-band pass filters the Q has to be Q<=.35. I used the formula:
q=√(ωL/ωU)/(1+ωL/ωU) and my Q = .4974. From what it looks like, I most likely can not achieve this band pass with this circuit because of the Parameters I want. Is this correct?

 

What do you want as a gain value?
With two pasive RC sections in series it is not possible to achieve any gain.
At the center frequency you always have a gain lower than 0 dB.For positive gain values you need an active filter circuit.

I am aiming for a 0dB at the center frequency and 3dB less than the center frequency at the cutoffs. However I have had a result where my center freq. was at -5 dB and the cutoffs where -8 dB.

 

I am aiming for a 0dB at the center frequency and 3dB less than the center frequency at the cutoffs. However I have had a result where my center freq. was at -5 dB and the cutoffs where -8 dB.

It is established fact that you cannot get 0 dB insertion loss with a passive filter. The presence of pure resistance components guarantees that you cannot achieve 0 dB or even come much closer than you already have.

 

I'm not sure of loaded Q, but I thought Q was the quality factor and to for wide-band pass filters the Q has to be Q<=.35. I used the formula:
q=√(ωL/ωU)/(1+ωL/ωU) and my Q = .4974. From what it looks like, I most likely can not achieve this band pass with this circuit because of the Parameters I want. Is this correct?

That is correct. Loaded Q means that when you cascade two filter sections as you have done the second one changes the Q of the first one which changes its characteristics. Normally when you do this with passive sections you need to do an impedance match of output impedance from the first section to input impedance of the second section.

 

What configuration would be best for my parameters? the Salen-key you mentioned? Or is there another minimal component filter I could implement?

 

I'm not sure of loaded Q, but I thought Q was the quality factor and to for wide-band pass filters the Q has to be Q<=.35. I used the formula:
q=√(ωL/ωU)/(1+ωL/ωU) and my Q = .4974. From what it looks like, I most likely can not achieve this band pass with this circuit because of the Parameters I want. Is this correct?

The problem is that you don't have a wide bandpass filter.

My textbook defines broadband bandpass filter in the following way:
upper cutoff frequency divided by lower cutoff frequency is equal or greater than 2

In your case the:
lower cutoff frequency=3420 Hz
upper cutoff frequency=4180 Hz
4180/3420=1.22
1.22<2
Therefore you should not/can not use broadband bandpass filter design.

 

Another thing.

Q, the quality factor, is ratio or center frequency, w0, to bandwidth, beta.

Your bandwidth, beta, is 4180-3420=760

You center frequency, w0, is 3800 Hz.

Q=3800/760=5

The above is from RLC bandpass filter material in my textbook. I have no clue how you can use Q in RC filter circuit since Q never mentioned in the RC filter material in my textbook.

 

What configuration would be best for my parameters? the Salen-key you mentioned? Or is there another minimal component filter I could implement?

A Sallen-Key filter is an active filter that uses an operational amplifier. I'll give you a link this time, but you really need to learn how to do this on your own.

http://en.wikipedia.org/wiki/Sallen–Key_topology

A good text on filter design like Van Valkenburg wouldn't hurt either.

http://www.amazon.com/Analog-Filter-Electrical-Computer-Engineering/dp/0195107349

 

What configuration would be best for my parameters? the Salen-key you mentioned? Or is there another minimal component filter I could implement?

RLC bandpass filter has 3 parts: 1 resistor, 1 capacitor, 1 inductor.

Your earlier cascade RC bandpass filter had 4 parts.

 

RLC bandpass filter has 3 parts: 1 resistor, 1 capacitor, 1 inductor.

Your earlier cascade RC bandpass filter had 4 parts.

An alternative yes. The inductor will be a challenge to find or fabricate, and the insertion loss will still be there.