2-pole active filter question

[Fovakis]

Hello! I am desiging in PSpice a 2-pole active filter. the values are taken from a wikipedia example to be sure about the fcut.

[URL=https://s1284.photobucket.com/user/fovos1/media/activefilter2polesallenkeytopology_zps30b85696.png.html]**broken link removed**[/URL]

this is the circuit **broken link removed**

and the fcut=16KHz. but when i am doing it in pspice the fcut is at 10khz.

i tried with other values and again problem. i read about FREQUENCY SCALING FACTOR accoring to Butterworth filter but i don't understand
why for example wiki has wrong fcut. what is this value?

at pages 9 to 10 texas instruments has some tables with FSF depending of what filter design you like.BUT ONLY ABOUT LOW PASS FILTERS....
https://www.electro-tech-online.com/custompdfs/2013/06/sloa049b.pdf

please someone to explain.thanks a loT

 

[alec_t]

From your graph the -6dB point seems to be ~16kHz, not 10kHz. I don't see a problem, either in your sim or in wiki. (But your equation should include a square root sign)).

The frequency scaling factor means that if, for example, you doubled each of the resistor (or capacitor) values the fcut frequency would be halved.

 

[Fovakis]

From your graph the -6dB point seems to be ~16kHz, not 10kHz. I don't see a problem, either in your sim or in wiki. (But your equation should include a square root sign)).

The frequency scaling factor means that if, for example, you doubled each of the resistor (or capacitor) values the fcut frequency would be halved.

i forgot to put the square at the form sorry about it. if you make the calculation with these values R1R2C1C2 you make fcut=16Khz. fcut is the frequency where the Vout=0.7Vin right SO -3dB at BODE DIAGRAM?

But at -3dB at my diagram i have 10khZ not 16

 

[ericgibbs]

hi,
This is what TI filter pro shows.

E

 

[Fovakis]

hi,
This is what TI filter pro shows.

E

hi,thanks !!

So the wiki R and C values are wrong? why when i run it is come 10khz the cutoff. i think something with FSF is the solution

 

[ericgibbs]

hi,thanks !!

So the wiki R and C values are wrong? why when i run it is come 10khz the cutoff. i think something with FSF is the solution

hi,
This what I see with LTSpice.

E

 

[Fovakis]

hi,
This what I see with LTSpice.

E

thanks for the reply. so what is this? i don't understand

 

[ericgibbs]

thanks for the reply. so what is this? i don't understand

hi,
Sorry, which bit is confusing.?

This image is using the TI component values

E

EDIT:
If you help with PBW and BW look at this link.
**broken link removed**

 

[alec_t]

Odd. The post #6 pic is not what I see with LTspice (using the default opamp). Which opamp model did you use, Eric? Why is the ref voltage at -2dB?

 

[ericgibbs]

Odd. The post #6 pic is not what I see with LTspice (using the default opamp). Which opamp model did you use, Eric? Why is the ref voltage at -2dB?

hi,
Using a CA3140 OPA, the circuit has an overall attenuation of around 2dB.

Post your sims , I will compare.

E.

 

[Fovakis]

hi friend i am still not understand why this confusion with fcut when i use butterworth,bessel etc......

i thought it is a formula fcut = 1/2π SQRT(R1R2C1C2)

BUT depends on what filter design there is a correction the FSF

 

[ericgibbs]

Odd. The post #6 pic is not what I see with LTspice (using the default opamp). Which opamp model did you use, Eric? Why is the ref voltage at -2dB?

hi alec,

Woops! I used a single supply configuration!

E

 

[alec_t]

Here's the result using the TI values:

 

[ericgibbs]

hi friend i am still not understand why this confusion with fcut when i use butterworth,bessel etc......

i thought it is a formula fcut = 1/2π SQRT(R1R2C1C2)

BUT depends on what filter design there is a correction the FSF

hi,

Your original values give 15915.49431Hz

This attached Filter Design PDF should answer your questions.

E.

 

[MrAl]

Hello guys,


An analytical solution for the angular frequency w is:

w=sqrt[(sqrt(C2^2*R2^4+(4*C2^2-4*C1*C2)*R1*R2^3+(6*C2^2-8*C1*C2+8*C1^2)*R1^2*R2^2+
(4*C2^2-4*C1*C2)*R1^3*R2+C2^2*R1^4)-C2*R2^2-2*C2*R1*R2+2*C1*R1*R2-C2*R1^2)/(2*C1^2*C2*R1^2*R2^2)]

and then:
f=w/(2*pi)

However, if we make R2=R1 and C2=C1 then the formula reduces to:
w=sqrt(sqrt(2)-1)/(R1*C1)

So for R2=R1=10000, and C2=C1=1e-9, we get:
w=64359.425291

so we get:
f=w/2pi=10243.12 Hz approximately.

So the -3db cutoff point occurs at that frequency for the original circuit with the original values.

Since the constraint R2=R1 and C2=C1 leads to a much simpler expression but limits the type of response we can obtain, we can instead just constrain C2=C1 and then we still get a much simpler formula for w:
w=sqrt(sqrt(R2^4+6*R1^2*R2^2+R1^4)-R2^2-R1^2)/(C1*R1*R2)

[LATEX]w=\frac{\sqrt{\sqrt{{R1}^{4}+6\,{R1}^{2}\,{R2}^{2}+{R2}^{4}}-{R1}^{2}-{R2}^{2}}}{\sqrt{2}\,R1\,R2\,C1}[/LATEX]

[LATEX]with \,\,C2=C1.[/LATEX]

 

[Fovakis]

Hello guys,


An analytical solution for the angular frequency w is:

w=sqrt[(sqrt(C2^2*R2^4+(4*C2^2-4*C1*C2)*R1*R2^3+(6*C2^2-8*C1*C2+8*C1^2)*R1^2*R2^2+
(4*C2^2-4*C1*C2)*R1^3*R2+C2^2*R1^4)-C2*R2^2-2*C2*R1*R2+2*C1*R1*R2-C2*R1^2)/(2*C1^2*C2*R1^2*R2^2)]

and then:
f=w/(2*pi)

However, if we make R2=R1 and C2=C1 then the formula reduces to:
w=sqrt(sqrt(2)-1)/(R1*C1)

So for R2=R1=10000, and C2=C1=1e-9, we get:
w=64359.425291

so we get:
f=w/2pi=10243.12 Hz approximately.

So the -3db cutoff point occurs at that frequency for the original circuit with the original values.

Since the constraint R2=R1 and C2=C1 leads to a much simpler expression but limits the type of response we can obtain, we can instead just constrain C2=C1 and then we still get a much simpler formula for w:
w=sqrt(sqrt(R2^4+6*R1^2*R2^2+R1^4)-R2^2-R1^2)/(C1*R1*R2)

[LATEX]w=\frac{\sqrt{\sqrt{{R1}^{4}+6\,{R1}^{2}\,{R2}^{2}+{R2}^{4}}-{R1}^{2}-{R2}^{2}}}{\sqrt{2}\,R1\,R2\,C1}[/LATEX]

[LATEX]with \,\,C2=C1.[/LATEX]

Hi mr.Al and many thanks !

So everytime i must solve the transfer function to learn what is the fcut?

The wikipedia is wrong so?

 

[ericgibbs]

hi,
I would download this free TI Filter calculator.

https://www.ti.com/tool/filterpro

E

 

[MrAl]

Hi mr.Al and many thanks !

So everytime i must solve the transfer function to learn what is the fcut?

The wikipedia is wrong so?

Hello again,


Well if you want to do an unknown circuit then you have to solve it which means evaluating the transfer function, or else use ready made software. When you evaluate it you can find the formula that explicitly solves for a given thing like fc.

Wikipedia is not wrong, but they are not giving the information that is usually desired by a designer. Namely, they are not providing a formula for fc (or wc) they are providing a formula for f0 (or w0), which is not the same thing. So you can not use their formula to find the cutoff frequency without taking additional steps. The formulas i provided earlier calculate wc directly from which you can easily get fc.

The Texas Instruments paper does the same thing. They calculate some things explicitly and others im not even sure if they calculate at all, and these are things that they take for granted in the formulas. For example, FSF, they dont seem to provide an explicit calculation for this do they? If we are to use w0 or f0 from Wikipedia then we also need to know how to calculate FSF.
Note also that in the example for the low pass Butterworth Sallen-Key they coincidentally use an FSF=1, which to me says they are cheating on their own paper.

So if you want simple formulas for known circuits that's one thing, but if you want to know how to solve every circuit like this even with different topologies, then you should learn how to calculate the transfer function one way or another.



I went ahead and calculated the FSF for any 2nd order filter with passband gain equal to 1. The result is:
FSF=sqrt(sqrt(2)*sqrt(2*d^4-2*d^2+1)-2*d^2+1)

[LATEX]FSF=\sqrt{\sqrt{2}\,\sqrt{2\,{d}^{4}-2\,{d}^{2}+1}-2\,{d}^{2}+1}[/LATEX]

where d is the damping factor. This means if you calculate w0 from the Wikipedia site you can then calculate wc from:
wc=FSF*w0

or:
fc=FSF*f0

The damping factor is calculated from:
d=(R1+R2)/(2*w0*C1*R1*R2)

and again w0=1/sqrt(R1*R2*C1*C2)

and just to note, when R2=R1 and C2=C1 the damping factor is exactly equal to 1, but FSF is not necessarily equal to 1. In fact for the original circuit with original values 10k and 1nf, we get:
FSF=0.64359425290558

and multiplying this by 15.9kHz we get 10.2kHz which is the right value for the cutoff frequency fc.

We might also note that the only time we get FSF=1 is when the damping factor d is equal to 1/sqrt(2).

Also note that the FSF given here is the reciprocal of the FSF given in the Texas Instruments paper.

 

[audioguru]

When the resistor values are the same and the capacitor values are the same then the lowpass filter type is Bessel that has a droopy response.
The response is droopy because there are two poles, each dropping the level -3dB so the total loss at the calculated cutoff frequency is -6dB.

When the resistor values are the same and the feedback capacitor has double the value of the capacitor to ground then the lowpass filter type is Butterworth that has a sharp cutoff and a steep drop of level with frequency.
The feedback capacitor boosts the response a little at the cutoff frequency then the level is -3dB.

 

[MrAl]

When the resistor values are the same and the capacitor values are the same then the lowpass filter type is Bessel that has a droopy response.
The response is droopy because there are two poles, each dropping the level -3dB so the total loss at the calculated cutoff frequency is -6dB.

When the resistor values are the same and the feedback capacitor has double the value of the capacitor to ground then the lowpass filter type is Butterworth that has a sharp cutoff and a steep drop of level with frequency.
The feedback capacitor boosts the response a little at the cutoff frequency then the level is -3dB.

Hi audioguru,

The cutoff frequency is defined as the point where the normal passband gain drops to -3db. Thus you cant have a cutoff frequency 'fc' that is -6db.
When we solve for the cutoff frequency we solve for the response when the amplitude is 3db down from the normal passband gain, and this is for a single pole filter, two pole filter, or any number of poles filter.

What changes is the rate of decline after the cutoff frequency. As the filter order increases the 'cut' after the -3db point falls faster roughly two times faster for a two pole filter, so by the time we get to another frequency like 10 times fc we see much less gain with a two pole filter than with a single pole filter, yet at the cutoff frequency the gain is -3db for BOTH filters.

I'm pretty sure you know this already so maybe you worded your post a little incorrect.