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I took the parameters from the MT-206 table of Analog Devices. I did not find in any book the chebyshev response of 0.1dB. https://www.analog.com/media/en/training-seminars/tutorials/MT-206.pdf enter image description here

enter image description here

I performed the following steps to find the value of the elements enter image description here

First: I'm pretty sure this is wrong.

Second: I have to take into account the Q of chebyshev's answer or I can only take into account the gain I want out of 2.

Third: The gain over there is almost around 5. Is there a way to "decrease the gain"?

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  • \$\begingroup\$ First: You are correct. The gain formula is wrong, it should be 1 +R3/R4 which affects Q’s from positive feedback. A Chebychev ripple of 0dB becomes a Butterworth filter, \$\endgroup\$ Apr 28, 2021 at 13:18
  • \$\begingroup\$ I don't quite recognize that topology, which one is it? Sallen-Key? Multiple feedback? ...? \$\endgroup\$ Apr 28, 2021 at 13:24
  • \$\begingroup\$ S&K... your component tolerances might exceed 0.1 dB ripple. Use TI Filter Design and choose tolerance of parts \$\endgroup\$ Apr 28, 2021 at 13:25
  • \$\begingroup\$ @TonyStewartEE75 That doesn't look like any Sallen-Key I know of. Or multiple feedback. Or any other. \$\endgroup\$ Apr 28, 2021 at 13:27
  • \$\begingroup\$ It's Sallen-Key \$\endgroup\$
    – Dragnovith
    Apr 28, 2021 at 13:28

6 Answers 6

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This is not what your Prof expects but it is how I would approach it. I assume he wants you to compute it. This way takes a few minutes.

500 Hz HPF 2nd order Chebychev response S&K filter Av=2 in passband = 6.02 dB +/- 0.1dB

enter image description here

Your link shows the analytic approach only for a unity gain 0.1 dB normalized frequency response on page 9

Table 1.2 on p.2 shows the Ripple BW f to -3dB f for 0.1 dB ripple filter and only the 2nd order has the value 1.93432= f_ripple/f-3dB which supports my comment that for this filter when Ripple = 0 equals the Butterworth Filter at 1/2 the cutoff frequency.

This is important not to confuse the two bandwidth definitions as some may assume BW is always -3dB rather than the "Ripple BW" where the ratio varies depending on the order of the filter. In the other filter designs, you can choose the attenuation level where the phase has maximum linearity like -6 dB or -10 dB. but in higher order filters, usually you defined the Stop-Band attenuation and frequency with the Pass-Band gain or attenuation with frequency.

  • Here I used Falstad's Analog filter site and chose the -3dB BW = 500 Hz /1.93432 = 259 Hz and they only offer unity gain filter tools. Again you may scale up impedance to the RC values with the same product just for low current on CMOS type Op Amps.

enter image description here

The table of parameters for unity gain from your Analog Devices link and for giggles and kicks a Quad Op Amp's worth of an 8th order filter.

enter image description here

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  • \$\begingroup\$ Okay, I'll take a closer look, thanks !! \$\endgroup\$
    – Dragnovith
    Apr 28, 2021 at 17:53
  • \$\begingroup\$ Okay any comments? \$\endgroup\$ Apr 28, 2021 at 19:55
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Here's your diagram with some annotations:

enter image description here

Although I've never seen this particular arrangement before, I'll argue that the above can be reduced to:

enter image description here

Where \$R_2=\frac{R_a\,R_b}{R_a+R_b}\$ and \$A=\frac{1}{1+\frac{R_b}{R_a}}\$.

Since \$A\$ is in series with \$R_2\$, we can swap positions when performing the nodal analysis as seen below:

Using sympy and assuming \$C=C_1=C_2\$:

eq1 = Eq( va/ra + va/rb + va/(1/s/c) + va/(1/s/c), vp/(1/s/c) + vi/(1/s/c) + (A*vo)/r2 + 0/ra)
eq2 = Eq( vp/(1/s/c) + vp/r1, va/(1/s/c) + 0/r1 )
eq3 = Eq( vm/r3 + vm/r4, vo/r3 + 0/r4 )
eq4 = Eq( vo/r2 + vo/r3, vm/r3 + (va/A)/r2 + io )
eq5 = Eq( vp, vm )
ans = solve( [eq1, eq2, eq3, eq4, eq5], [io, va, vp, vm, vo] )
tf = simplify( ans[vo]/vi )
n0 = factor( -expand(fraction(simplify(ans[vo]/vi))[0]), s )
d0 = factor( -expand(fraction(simplify(ans[vo]/vi))[1]), s )
np = Poly( n0, s )
dp = Poly( d0, s )
w0 = sqrt( simplify(dp.coeffs()[2] / dp.coeffs()[0]) )
alpha = simplify( 1/2 * dp.coeffs()[1] / dp.coeffs()[0] )
zeta = simplify( alpha / w0 )
K = expand( simplify( np.coeffs()[0] / dp.coeffs()[0] ) )

So now I find:

$$\begin{align*} R_2&= R_a\mid\mid R_b = \frac{R_a\,R_b}{R_a+R_b} \\\\ A&=\frac1{1+\frac{R_b}{R_a}} \\\\ K&=1+\frac{R_3}{R_4} \\\\ \omega_{_0}&=\frac1{C\sqrt{R_1\,R_2}} \\\\ \zeta&=\sqrt{\frac{R_2}{R_1}}-\frac12\sqrt{\frac{R_1}{R_2}}\bigg[A\,K-1\bigg] \\\\ \mathcal{H}\left(s\right)&=K\frac{s^2}{s^2+2\zeta\,\omega_{_0}s+\omega_{_0}^{\:\!2}} \end{align*}$$

The Chebyshev function you are looking to achieve (if I understand it) is:

eta0 = solve( Eq( 10**(-0.1/10), 1/(1+x**2) ), x )[1]
alp0 = arcsinh( 1 / eta1 ) / 2
chr0 = Poly( simplify((s-(-sinh(alp1)*sin(pi/4)+i*cosh(alp1)*cos(pi/4)))*(s-(-sinh(alp1)*sin(3*pi/4)+i*cosh(alp1)*cos(3*pi/4)))).n(), s )
w0chr = sqrt( chr0.coeffs()[2] / chr0.coeffs()[0] )
d0chr = chr0.coeffs()[1] / sqrt( chr0.coeffs()[0] * chr0.coeffs()[2] )
d0chr, w0chr

    (1.30317045127507, 1.82044969090790)

Let's look at a table from R. P. Sallen and E. L. Key's "TR-50: A Practical Method of Designing RC Active Filters," 6 May 1954:

enter image description here

Similar values there.

As the frequency shape only depends on the value of \$d=2\,\zeta\$, we can set \$\zeta\approx 0.6516\$ for this case and then solve the above equations for the filter.

(Do keep in mind that I believe the \$0.1\:\text{dB}\$ down point will be \$\approx 1.82\times\$ your designed \$\omega_{_0}\$, based upon the above calculations I just performed. This could mean it's located at about \$f_c\approx 910 \:\text{Hz}\$. So this may not be expected or desired. If not, you'll have to take this factor into account, setting \$f_c= 500 \:\text{Hz}\$ and computing a now-reduced \$\omega_{_0}\$, before designing the filter. But I don't know what's expected. So I'm leaving it here.)

I think that spells out enough to get going. And I believe it is enough to say that your odd arrangement with \$R_a\$ and \$R_b\$ (which I've not seen) can be used. I'm not sufficiently familiar with it, yet, to know if there's a great purpose in arranging it with the extra resistor. Perhaps that will settle in, later. But at least I think it can be used to get where you need to go.

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    \$\begingroup\$ Jonk - interesting modification (second diagram in your answer). Here is another modification with simplified design equations (4-element version): Use of an additional buffer after the first node (decoupling of both RC sections). This modification allows that both gain stages have unity gain with reduced component spread (if compared with the single gain stage version). \$\endgroup\$
    – LvW
    Apr 29, 2021 at 7:37
  • \$\begingroup\$ @LvW That had crossed my mind. I only avoided it to focus on the main thrust I wanted to convey to aconcernedcitizen. But it's a really good idea. I may add a section later. It's actually worth the trouble. (But I have other things to do in the meantime, though, and I'd also like to see what aconcernedcitizen thinks, as well. He may find something I need to adjust. I think I got it right... and I do make mistakes. ;) \$\endgroup\$
    – jonk
    Apr 29, 2021 at 7:45
  • \$\begingroup\$ @LvW Ah. I found my earlier comment made before I wrote that answer: "To me, that active opamp in the OP's question here has its output fed to two, not one, voltage dividers. One that goes to an opamp input and the other that feeds back." So yes, it had crossed my mind. \$\endgroup\$
    – jonk
    Apr 29, 2021 at 8:01
  • \$\begingroup\$ @jonk Yes, that's what I tried, and it reduces to the S&K version. As I replied to LvW, this is just the generic SAB derived from the bridged T network, with the shunt R split. This is the start of S&K, but not S&K. But, yes, this is what I have seen, too. \$\endgroup\$ Apr 29, 2021 at 9:28
  • \$\begingroup\$ To complete the picture: Less known - but there are also negative-gain S&K structures. However, due to relatively large gain values (if compared with pos.-gain SK) these filters have larger active sensitivities (influence of real opamp gains) and the application is restricted to small quality figures. \$\endgroup\$
    – LvW
    Apr 29, 2021 at 9:54
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Dragnovith, may I give you some general information?

1.) You can use the shown S&K topology with 5 passive components - however, it is not necessary to have the first grounded resistor. The design with only 4 elements is somewhat simpler.

2.) For Chebyshev responses it is quite common to use not the -3dB criterion to define the begin of the passband (cut-off). Instead, the first crossing of the maximum gain line (in your case "2") is used. That means, it is the ripple which defines the passband. This is important to know when you are trying to use tabulated values for the pole data (or for the parts values). Therefore, check which kind of definitions is used for the data.

3.) Normally, the tabulated figures are normalized to the corresponding pole frequency wp. This is important to know because there is a difference between pole frequency and cut-off frequency wc. So - before using these values you must do "denormalization".

4.) For each 2nd-order filter structure - and for each basic filter function - there are formulas which relate the pole data (pole frequency wp and pole quality Qp) to the various passive parts values.

5.) Example: Highpass, 2nd order, Chebyshev with 0.1dB ripple: wc/wp=1.82 and Qp=0.767.

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  • \$\begingroup\$ Darn. Although I'd write more, you've hit on the key points and now I don't want to write. +1! (My first experience with Tschebyscheff was with approximation theory using nodes/nonlinear minimax/Clenshaw–Curtis quadrature -- 1973/74 for me as I was faced with writing transcendental function libraries. So it was a pleasure to revisit it towards the end of the 1970's when I first designed and built a 600 bps FSK modem with async detector and needed to 1st learn about filters generally, as I was then totally ignorant, and then design a practical 6-pole transmit and 10-pole receive filter.) \$\endgroup\$
    – jonk
    Apr 28, 2021 at 18:32
  • \$\begingroup\$ I should add that R. P. Sallen and E. L Key's "TR-50: A Practical Method of Designing RC Active Filters," 6 May 1954 (before their official public paper in 1955, which was a complete white-wash of their earlier report -- secret at the time, unfortunately) covers Tschebyscheff filters well (and uses that spelling.) \$\endgroup\$
    – jonk
    Apr 28, 2021 at 18:33
  • \$\begingroup\$ TR-50 covers at least (I'd need to recheck to be sure) 25 different filter topologies, including quite a few multi-active-amplifier ones (described as "2 active inputs".) Just FYI. This includes formulas needed to develop each of them. They focus more on Butterworth and Tschebyscheff, with regard to specific tables, though. \$\endgroup\$
    – jonk
    Apr 28, 2021 at 18:42
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    \$\begingroup\$ Note: Just to make clear LvW's point (2) above, see this picture. I annotated a picture I found at Analog Devices: DSP Guide: Chapter 20. Do note that this is not (necessarily) the -3 dB point. \$\endgroup\$
    – jonk
    Apr 28, 2021 at 19:15
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    \$\begingroup\$ Finally, the mystery solved! That's just a generic single amplifier biquad, derived from the bridged T network, where the shunt R is split into R/a and R/(1-a), one of them going in the input. S&K derived their circuit based on this one, noting that the R/(1-a) and R/a can be combined -- just like in @jonk's answer -- to form one single R (since a would only lower the Q). That's how the S&K topology was born. I felt like going nuts over this. :-) \$\endgroup\$ Apr 29, 2021 at 9:26
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If you can choose the topology then use the proper Sallen-Key that you see in the link I gave. That only uses 2xR+2xC (plus 2xR, the gain network). The link on ADI you gave is part of a book called Op Amp Applications (Walter G. Jung), and it's freely available. In there you'll find the Sallen-Key highpass (fig. 5-68) and ready-to-use formulas.

However, tables are really not my thing and, when I tried to use them, the values I got for R1 and R2 didn't seem to fit the response, so I calculated them based on the transfer function:

$$\left\{\begin{aligned} \omega^2&=\dfrac{1}{R_1R_2C^2} \\ Q^2&=\dfrac{1}{R_1R_2C^2}\cdot\dfrac{1}{\left(\dfrac{2}{R_2C}+\dfrac{1-K}{R_1C}\right)^2} \end{aligned}\right.$$

with 4 solutions, out of which two are negative and the 3rd giving the correct magnitude but the wrong phase (poles end up in the right-hand plane, unstable). Which leaves the two, correct values, given by:

$$\left\{\begin{aligned} R_1&=\dfrac{\sqrt{8Q^2+1}+1}{4\omega QC}&=3.52\;\mathrm{k}\Omega \\ R_2&=\dfrac{\sqrt{8Q^2+1}-1}{2\omega QC}&=2.88\;\mathrm{k}\Omega \end{aligned}\right.$$

test

The gain is 2, the ripple ~0.1 dB, the corner frequency is at -3 dB, 500 Hz (usually considered the end of the passband for a Chebyshev). I would have chosen a multiple feedback, it saves one resistor and handles Q better, but at 0.1 dB, it's fine as it is.

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  • \$\begingroup\$ The high pass ripple BW of 0.1dB is not 500 Hz. what is it? That’s the book I used in the 70’s \$\endgroup\$ Apr 28, 2021 at 17:34
  • \$\begingroup\$ Interestingly, I didn't know this book or that method. Thanks, I'll take a look \$\endgroup\$
    – Dragnovith
    Apr 28, 2021 at 17:45
  • \$\begingroup\$ @TonyStewartEE75 There is a comma between the -3 dB and the 500 Hz which changes the meaning of the phrase to a list: 1) the gain is two, 2) the ripple is 0.1 dB, and 3) the attenuation at the corner frequency is -3 dB @ 500 Hz. I did not say anywhere that the ripple is in Hz. \$\endgroup\$ Apr 28, 2021 at 18:19
  • \$\begingroup\$ yet interestly, the Filter Pro app. assumes 0.1dB ripple is the passband of 500 Hz and the -3dB freq. is listed as “Cutoff Freq. 274.657 Hz, which is why I indicated to clarify this distinction and not assume, as there are 2 BW definitions. But then I always doubt answers with excess sigfigs ;) yet appreciate the choice of component tolerances. \$\endgroup\$ Apr 28, 2021 at 18:37
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    \$\begingroup\$ Tony Steward, the ratio 500/274.657=1.82 can be found in the filter tables. It is the normalized frequency wc/wp (cutoff-to-pole frequency ratio). Therefore, it is important to check which passband edge is used for the tabulated figures. (see the last line in my detailed answer) \$\endgroup\$
    – LvW
    Apr 28, 2021 at 18:43
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I proposed an answer to a similar problem.

This uses a table solution. For your 0.1dB Cheby filter the coefficients are:

B = 2.372356
C = 3.314037

The corner frequency is where the ripple crosses the 0dB (in your case, the 6dB) point as opposed to the -3dB point.

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How many comments, I'm reading and learning. But I would really like to thank those who helped, thank you very much!

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