Pros & cons of Sallen-Key vs. double RC cascaded stages low pass filter topologies

Question

Here are two possible topologies to implement an active non inverting second order low pass filter, namely Sallen-Key and two cascaded RC stages.

Both circuits have the same transfer function:

$$H(j\omega) = \frac{V_\text{out}}{V_\text{in}} = \frac{H0}{1 - \left(\omega / \omega_n\right)^2 - j 2\zeta\left(\omega / \omega_n\right)}$$

Theoretically they are equivalent, but I am sure that there are also practical considerations that may render one topology more convenient than the other depending on the application.

What are the pros/cons of each topology?

I have envisaged that in the Sallen-Key topology, the feedback introduced by \$C_1\$ may reduce the phase margin so it may tend to oscillate specially when DC gain \$1+R_4/R_3\$ increases. In the two cascaded \$RC\$ stage topology, conversely, \$C_1\$ does not affect op amp tendency to oscillate.

Answer 1

In short:

• Both circuits provide a second-order lowpass function with gain

• First circuit (Sallen-Key):

• Advantage: Depending on the closed-loop gain of the opamp you can realize both - low-Q functions (2 real poles) or high-Q functions (a complex pole-pair).

• Disadvantage: High-frequency attenuation will be limited because for rising frequencies an increasing portion of the input signal will be coupled directly via C1 to the output (non-zero opamp output impedance, rising with frequency).

EDIT: An additional disadvantage of the Sallen-Key topology results from the positive feedback (which allows the mentioned Q-enhancement): Positive feedback will always increase the sensitivity of the transfer function to component tolerances. In particular, the sensitivity to the gain determining resistors R3 and R4 is critical. For this reason, the unity-gain configuration (short between opamps output and the invering input) is often preferred.

• Second (passive) circuit:

• Advantage: Better attenuation for very high frequencies (see Sallen-Key disadvantage).

EDIT: It was shown elsewhere that such a passive ladder topolgy has the smallest sensitivity to componenet tolerances -if compared with all other active filter circuits.

• Disadvantage: Only real poles are possible (bad transition from passband to stop-band). No higher-Q transfer functions are possible (Bessel, Butterworth, Chebyshev,..)

Answer 2

Both circuits have the same transfer function:

$$H(j\omega) = \frac{V_\text{out}}{V_\text{in}} = \frac{H0}{1 - \left(\omega / \omega_n\right)^2 - j\zeta\left(\omega / \omega_n\right)}$$

That is untrue. Well, maybe it is better to say it can lead to misunderstandings.

In the Sallen-Key design \$\zeta\$ is a variable whereas in the simple cascaded RC design it does not have this feature i.e. \$\zeta\$ is always unity when the two RC time constants are the same. If the two RC time constants are not the same then the equivalent value of \$\zeta\$ rises above unity i.e. it becomes a "sloppier" filter.

This is fundamentally different to the Sallen-Key design. It can readily produce values of \$\zeta\$ much smaller than unity and produce (for example), a Butterworth response where the pass-band spectrum is maximally flat. This cannot be achieved with cascaded RC sections (even if they are buffered to prevent unwanted loading of the 2nd stage on the first stage).

So, in a Sallen-Key design, \$\zeta\$ can extensively modify the shape of the filter in the resonance area as shown in these diagrams from my basic website: -

If you derive the transfer function of two unbuffered cascaded RC stages you get this: -

$$\dfrac{\omega_n^2}{s^2 + s\left(\frac{C_1R_1+C_2R_2}{C_1R_1C_2R_2}\right)+\omega_n^2}$$

Where \$\omega_n^2\$ equals \$\frac{1}{C_1R_1C_2R_2}\$

This is in the standard form for a 2nd order low-pass filter where the "s" coefficient equals: -

$$\frac{C_1R_1+C_2R_2}{C_1R_1C_2R_2} = 2\zeta\omega_n$$

If you drill down you will find that \$2\zeta\$ equals this: -

$$2\zeta = \sqrt{\frac{C_1R_1}{C_2R_2}} + \sqrt{\frac{C_2R_2}{C_1R_1}}$$

Then, if you analysed the above formula. you will find that the lowest value \$\zeta\$ can become is when both CR time constants are identical in value and, the result is that \$2\zeta = 2\$ or, \$\zeta = 1\$.

Answer 3

Your statement that the two have the same transfer function is false. This can be seen by inspection of their simplified equivalents:

^{simulate this circuit – Schematic created using CircuitLab}

Here \$K=1+\frac{R_4}{R_3}\$ is gain of the op-amp section, so that:

$$ V_X = K \times V_P $$ $$ V_Y = K \times V_Q $$

The Sallen-Key design (bottom) applies potential \$K\times V_Q\$ to the bottom end of C1, while the other applies 0V, a difference which would be very evident in their respective transfer functions:

$$ X(s) = \frac{K}{1+s(C_2R_2+C_2R_1+C_1R_1)+s^2(C_1R_1\cdot C_2R_2)} $$

$$ Y(s) = \frac{K}{1 + s \left( C_2 R_2 + C_2R_1 + C_1 R_1 (1 - K) \right) + s^2(C_1 R_1 \cdot C_2 R_2) } $$

They do not have the same transfer functions, but the difference is subtle; only the coefficient of \$s\$ in the denominator changes, with its \$C_1R_1\$ term being scaled by \$1-K\$.

For the two functions to be equal, \$Y(s)=X(s)\$, we require coefficients of \$s\$ and \$s^2\$ in each to be equal:

$$ \begin{aligned} C_{X1}R_{X1} &= C_{Y1}R_{Y1} \\ \\ C_{X2}R_{X2}+C_{X2}R_{X1}+C_{X1}R_{X1} &= C_{Y2}R_{Y2}+C_{Y2}R_{Y1}+C_{Y1}R_{Y1}(1-K) \\ \\ \end{aligned} $$

You have some flexibility with the cascaded RC design, in that the factor \$K\$ is arbitrary, not changing the position of poles in any way. You can set its \$K\$ to whatever you want in your endeavour to make \$X(s)=Y(s)\$. However, \$K\$ for the Sallen-Key design is constrained to \$K\ge 1\$, and changing it does alter poles.

I haven't examined in detail the feasibility of obtaining \$X(s)=Y(s)\$, but I think there are severe constraints on candidate values of resistance, capacitance and gain \$K\$ that can achieve such equality. There are transfer functions realisable with cascaded RC filters that Sallen-Key cannot achieve, and Sallen-Key can implement transfer functions that cascaded RC filters cannot. Consequently, the question of pros and cons seems somewhat moot, superseded by a consideration what one can do that the other cannot and vice-versa.

Having said that, there are differences not related to the transfer function that deserve mentioning. The first is that the cascaded RC design doesn't (technically) need an amplifier. As long as the following stage doesn't load node P, it can have whatever features you desire; it could be a power stage, or an audio compressor, or another low-pass filter for a third pole. The Sallen-Key design mandates an op-amp, and precludes you from shoe-horning in additional roles for it.

The second difference that comes to mind is that the output of the op-amp in the Sallen-Key design is loaded by its own feedback. This feedback path is a sink/source of current that the op-amp output must provide, while also "driving" whatever comes next. While this isn't usually considered in academic material describing Sallen-Key, in practical designs you must ensure that the op-amp can sink and source both feedback and load current simultaneously. The amplifier following a cascaded RC filter has no such feedback load to worry about.