How to find Rs and Cs for 2nd order RC circuit

[LvW]

Hello,

Up to now, this thread was visited by 175 forum members – and I assume some (if not most of them) are not yet very well experienced in analog electronics. Therefore, I think it could be helpful to comment and to correct some statements and formulas as contained in the document <http://www.johnhearfield.com/RC/RC4.htm< referenced by BradtheRad.

General comment: The author always speaks about „break frequencies“ without any definition. Obviously, for a first order RC circuit this frequency is identical to the 3dB frequency (pole frequency). But by mistake he also applies this wording to a second order RC circuit. In fact, both frequencies (called w1 and w2, resp.) are complex frequencies which determine the pole location in the s-plane (in this case: two poles on the neg.real axis).
In the chapter following the BODE diagram both „break frequencies“ are calculated. Although for the selected parts values the results may be approximately correct, the calculation assumes two decoupled RC sections, which obviously is not the case.

*Quotation: And since I can clearly see two separate and predictable break frequencies, why won't the expression for G factorise into two separate break-frequency terms? What's going on?

Of course, the expression for G can be factorized in two terms – because (as mentioned above) in reality the „break frequencies“ are the „pole frequencies“ wp. And each second order denominator can be split into two expressions like (1+jw/wp).

*Quotation: ...increase the value of capacitor D from 1nF to 50nF - so that the two break frequencies become (supposedly) the same.

I don’t know how the author came to this (false) conclusion. Of course, these frequencies (whatever their name may be) are NOT the same. For the selected values we have w1=-585.8 1/s and w2=-3414.2 17s.

*Quotation: The break frequencies ω1 and ω2 can be written . . . . so that ω1ω2 = ω0

Here the author, indirectly, confirms my interpretation that the „break frequencies“ are identical to the pole locations. At this point it is to be mentioned that the given formula G=f(Q,wo) is derived for a 2nd order circuit with a complex pole pair. Here, wo is the pole frequency (magnitudes of both pole vectors); for Q=0.5 both conjugate-complex poles combine to a double pole on the neg. real axis and split again into two real poles for Q<0.5 (as in our case).

*Quotation: In fact, it's not hard to prove that the maximum value Q can have in this circuit is 0.353, which happens when D=50nF. Higher values of Q can be achieved by active filters - that is, by circuits which include op-amps.

From system theory it is known that the maximum possible value for a passive second-order RC circuit is Qmax=0.4999 (Qmax=0.5 for decoupled sections). Q=0.353 applies to the selected values only, but it is not the possible maximum.
__________________________________-
Sorry for the lengthy comment. Perhaps it helps to better understand passive 2nd order low pass filters.

LvW

 

[FvM]

To add a shorter and less fundamental comment.

I think the discussion suffers from an incomplete filter specification, stating only a cut-off frequency and second filter order. Consequently arbitrary suggestions are fed to the discussion: use buffers, implement specific filter characteristics, e.g. Butterworth. But there are no criteria to decide about it.

As previously said, 20 dB/decade is a synonym for second order, but it's an asymptotical slope that doesn't tell about stop band attenuation near the cut-off frequency. If you have specific requirements for the filter characteristic, different filter prototypes come into play. Most can't be implemented as a passive RC circuit.

 

[Kral]

If you can use a buffer between stages, or if you can use a resistor value for the second stage that is very high compared to the resistor value of the first stage, then all you need to do is design two 1st order filters with an attenuation of 1.5 dB at the desired corner frequency, and cascade the two sections. The net attenuation at the desired cutoff frequency will be 3 dB. The usual problem with the 2nd approach is that the output impedance of the filter will be relatively high.
.
If you can use an op amp, then you are better off designing an active filter such as a Butterworth or Chebyshev, rather than a cascaded lag filter, since the latter has very little to recommend it other than the fact that it is conceptually simple.

 

[BradtheRad]

Up to now, this thread was visited by 175 forum members – and I assume some (if not most of them) are not yet very well experienced in analog electronics. Therefore, I think it could be helpful to comment and to correct some statements and formulas as contained in the document <http://www.johnhearfield.com/RC/RC4.htm< referenced by BradtheRad.

General comment: The author always speaks about „break frequencies“ without any definition. Obviously, for a first order RC circuit this frequency is identical to the 3dB frequency (pole frequency). But by mistake he also applies this wording to a second order RC circuit. In fact, both frequencies (called w1 and w2, resp.) are complex frequencies which determine the pole location in the s-plane (in this case: two poles on the neg.real axis).
In the chapter following the BODE diagram both „break frequencies“ are calculated. Although for the selected parts values the results may be approximately correct, the calculation assumes two decoupled RC sections, which obviously is not the case.

*Quotation: And since I can clearly see two separate and predictable break frequencies, why won't the expression for G factorise into two separate break-frequency terms? What's going on?

Of course, the expression for G can be factorized in two terms – because (as mentioned above) in reality the „break frequencies“ are the „pole frequencies“ wp. And each second order denominator can be split into two expressions like (1+jw/wp).

*Quotation: ...increase the value of capacitor D from 1nF to 50nF - so that the two break frequencies become (supposedly) the same.

I don’t know how the author came to this (false) conclusion. Of course, these frequencies (whatever their name may be) are NOT the same. For the selected values we have w1=-585.8 1/s and w2=-3414.2 17s.

*Quotation: The break frequencies ω1 and ω2 can be written . . . . so that ω1ω2 = ω0

Here the author, indirectly, confirms my interpretation that the „break frequencies“ are identical to the pole locations. At this point it is to be mentioned that the given formula G=f(Q,wo) is derived for a 2nd order circuit with a complex pole pair. Here, wo is the pole frequency (magnitudes of both pole vectors); for Q=0.5 both conjugate-complex poles combine to a double pole on the neg. real axis and split again into two real poles for Q<0.5 (as in our case).

*Quotation: In fact, it's not hard to prove that the maximum value Q can have in this circuit is 0.353, which happens when D=50nF. Higher values of Q can be achieved by active filters - that is, by circuits which include op-amps.

From system theory it is known that the maximum possible value for a passive second-order RC circuit is Qmax=0.4999 (Qmax=0.5 for decoupled sections). Q=0.353 applies to the selected values only, but it is not the possible maximum.
__________________________________-
Sorry for the lengthy comment. Perhaps it helps to better understand passive 2nd order low pass filters.

LvW

Hearfield's article was among the first few hits of a Google search (along with the other two articles).

I did not go as far as checking all of equations. So I have no reason to doubt your critique.

I did notice that Hearfield goes into depth, in that he tries to make sure whether what is commonly thought to be going on is what's really going on. He approaches it by saying 'let's try a capacitor value that is way different, and see what frequency the graph changes direction.'

In this sense he uses a thought process which appears as likely to arrive at a correct method as anyone else's view.

If he knew his method was at variance with the 'standard' methods, then it would have helped his case if he had clarified why there's a discrepancy and where it originates.

The other two articles appear to be authoritative. My knowledge is not extensive enough to say they aren't.

The Electronics-tutorials article suggests using values for the 2nd stage which are 10 times those in the previous stage. (godfreyl recommended this in post #5.)

Speaking of going into depth...

I have not seen it mentioned anywhere how the capacitors affect each other. Meaning C2 is influenced by both the incoming volt level AND C1's charge level. In fact it is influenced more by C1 than by anything else.

And then C1 is influenced in turn by C2, as well as by the incoming signal.

Additionally, the capacitors do not act in sync. They are out of phase in the way they charge and discharge.

Maybe this goes without saying? ...considering that I am not so well versed insofar as what equations apply in which cases, and I have had to spend some time reading those articles just to get partly acquainted.

It's plain there is complex behavior going on in something as simple as a second order filter.

Sort of like finding it's much more complicated to discover how two bodies in space interact according to gravitational forces, as compared to the simplicity of one body alone.

 

[LvW]

If you can use a buffer between stages, or if you can use a resistor value for the second stage that is very high compared to the resistor value of the first stage, then all you need to do is design two 1st order filters with an attenuation of 1.5 dB at the desired corner frequency, and cascade the two sections. The net attenuation at the desired cutoff frequency will be 3 dB. The usual problem with the 2nd approach is that the output impedance of the filter will be relatively high.
If you can use an op amp, then you are better off designing an active filter such as a Butterworth or Chebyshev, rather than a cascaded lag filter, since the latter has very little to recommend it other than the fact that it is conceptually simple.

see post#6.