The Sallen-Key filters

The Sallen-Key filters are second-order active filters (low-pass, high-pass, and band-pass) that can be easily implemented using the configuration below:

We represent all voltages in phasor form. Due to the virtual ground assumption, $V_b\approx V_{out}$ at non-inverting input is virtually the same as that at the inverting input, which is connected to the output $V_{out}$. Applying KCL to nodes a and b to get:

$$\frac{V_a-V_{in}}{Z_1}+\frac{V_a-V_{out}}{Z_3}+\frac{V_a-V_{... ...{Z_1}+(V_a-V_{out})\left(\frac{1}{Z_3}+\frac{1}{Z_2}\right)=0$$

and

$$\frac{V_{out}-V_a}{Z_2}+\frac{V_{out}}{Z_4}=0,$$

The second equation can also be written as

$$V_a-V_{out}=V_{out}\frac{Z_2}{Z_4}, \;\;\;\;\;\;\;V_a=V_{out}\frac{Z_2+Z_4}{Z_4}$$

Substituting these into the first equation we get

$$V_{out}\;\left(\frac{Z_2+Z_4}{Z_1Z_4}+\frac{Z_2}{Z_3Z_4}+\frac{Z_2}{Z_2Z_4} \right) =\frac{V_{in}}{Z_1}$$

Now the frequency response of the Sallen-Key filter can be found as the ratio of $V_{out}$ and $V_{in}$:

$$H = \frac{V_{out}}{V_{in}}=\frac{Z_3Z_4}{Z_1Z_2+Z_1Z_3+Z_2Z_3+Z_3Z_4}$$

• Second order low-pass filter

  If we let $Z_1=R_1$, $Z_2=R_2$, $Z_3=1/j\omega C_1$, $Z_4=1/j\omega C_2$, then the FRF becomes:
  

  $$H(j\omega) = \frac{1/(j\omega)^2C_1C_2} {R_1R_2+(R_1+R_2)/j\omega C_1+1/(j\omega)^2C_1C_2}$$

  $$= \frac{1/R_1R_2C_1C_2}{(j\omega)^2+j\omega(R_1+R_2)/R_1R_2C_1+1/R_1R_2C_1C_2}$$

  $$= \frac{1/R_1R_2C_1C_2}{(j\omega)^2+j\omega/R_pC_1+1/R_1R_2C_1C_2}$$

  $$= \frac{\omega_n^2}{(j\omega)^2+2\zeta\omega_n\;j\omega+\omega_n^2}... ...a+\omega_n^2} =\frac{\omega_n^2}{(j\omega)^2+\Delta\omega\;j\omega+\omega_n^2}$$

  
  
  This is a 2nd-order low-pass filter with
  
  $$\omega_n=\frac{1}{\sqrt{R_1R_2C_1C_2}},\;\;\;\;\;\;\; \Del... ..._pC_1},\;\;\;\; R_p=R_1\vert\vert R_2=\frac{R_1R_2}{R_1+R_2}$$
  
  
  and
  
  $$Q=\frac{\omega_n}{\Delta\omega}=\frac{\sqrt{R_1R_2C_1C_2}}{... ... \zeta=\frac{1}{2Q}=\frac{(R_1+R_2)C_2}{2\sqrt{R_1R_2C_1C_2}}$$
  
  
  As there are only two parameters $\omega_n$ and $\zeta$ or $Q=1/2\zeta$ to satisfy, we can arbitrarily set any two of the four variables $R_1$, $R_2$, $C_1$, and $C_2$, and then solve for the other two. For example, for convenience, if we let $R_1=R_2=1$, we get
  
  $$\omega_n=\frac{1}{\sqrt{C_1C_2}},\;\;\;\;\;\Delta\omega=\frac{1}{C_1}$$
  
  

• Second order high-pass filter

  If we let $Z_1=1/j\omega C_1$, $Z_2=1/j\omega C_2$, $Z_3=R_1$, $Z_4=R_2$, then the FRF becomes becomes:
  

  $$H(j\omega) = \frac{R_1R_2} {1/(j\omega)^2C_1C_2+R_1(1/j\omega C_1+1/j\omega C_2)+R_1R_2}$$

  $$= \frac{(j\omega)^2}{1/R_1R_2C_1C_2+j\omega\;(C_1+C_2)/C_1C_2R_2+(j\omega)^2}$$

  $$= \frac{(j\omega)^2}{(j\omega)^2+j\omega\;/C_sR_2+1/R_1R_2C_1C_2}$$

  $$= \frac{(j\omega)^2}{(j\omega)^2+2\zeta\omega_n\;j\omega+\omega_n^2... ...\omega_n^2} =\frac{(j\omega)^2}{(j\omega)^2+\Delta\omega\;j\omega+\omega_n^2}$$

  
  
  This is a 2nd-order high-pass filter with
  
  $$\omega_n=\frac{1}{\sqrt{R_1R_2C_1C_2}},\;\;\;\;\;\;\; \Delta\omega=\frac{1}{C_sR_2},\;\;\;\;\; C_s=\frac{C_1C_2}{C_1+C_2}$$
  
  
  and
  
  $$Q=\frac{\sqrt{R_1R_2C_1C_2}}{(C_1+C_2)R_1},\;\;\;\;\;\; \zeta=\frac{1}{2Q}=\frac{(C_1+C_2)R_1}{2\sqrt{R_1R_2C_1C_2}}$$
  
  

We further consider a band-pass filter shown below:

By voltage divider and virtual ground, we get

$$V_2=\frac{R_a}{R_a+R_b}V_{out}=kV_{out},\;\;\;\;\;\;\;\left(k=\frac{R_a}{R_a+R_b}\right)$$

Apply KCL to node $V_2$ to get:

$$\frac{V_1-V_2}{Z_2}=\frac{V_2}{Z_4},\;\;\;\;\;\;\;\mbox{i.e... ...}+\frac{1}{Z_4}\right)Z_2 =V_2\left(1+\frac{Z_2}{Z_4}\right)$$

Apply KCL to node $V_1$ to get:

$$\frac{V_{in}-V_1}{Z_1}+\frac{V_{out}-V_1}{R_f}+\frac{V_2-V_1}{Z_2} =\frac{V_1}{Z_3}$$

Rearrange the terms, and replace $V_1$ by $V_2(1+Z_2/Z_4)$ to get

$$\frac{V_{in}}{Z_1}+\frac{V_{out}}{R_f}+\frac{V_2}{Z_2} =V_1\left(\frac{1}{Z_1}+\frac{1}{R_f}+\frac{1}{Z_2}+\frac{1}{Z_3}\right)$$

$$= V_2\left(1+\frac{Z_2}{Z_4}\right)\left(\frac{1}{Z_1} +\frac{1}{R_f}+\frac{1}{Z_2}+\frac{1}{Z_3}\right)$$

Further rearrange the terms and replace $V_2$ by $kV_{out}$ to get

$$\frac{V_{in}}{Z_1}+\frac{V_{out}}{R_f} =kV_{out}\left[\lef... ..._f} +\frac{1}{Z_2}+\frac{1}{Z_3}\right)-\frac{1}{Z_2}\right]$$

Further rearrange the terms

$$\frac{V_{in}}{Z_1} = kV_{out}\left[\left(1+\frac{Z_2}{Z_4}\right) \left(\frac{1}{Z_1} ... ..._f}+\frac{1}{Z_2}+\frac{1}{Z_3}\right)-\frac{1}{Z_2}\right]-\frac{V_{out}}{R_f}$$

$$= V_{out}\left\{k\left[\left(1+\frac{Z_2}{Z_4}\right)\left(\frac{1}... ...}+\frac{1}{Z_2}+\frac{1}{Z_3}\right)-\frac{1}{Z_2}\right]-\frac{1}{R_f}\right\}$$

Finally we get the frequency response function:

$$H=\frac{V_{out}}{V_{in}} = \frac{1}{Z_1\left\{k\left[\left(1+\frac{Z_2}{Z_4}\right)\left(\fr... ...+\frac{1}{Z_2}+\frac{1}{Z_3}\right)-\frac{1}{Z_2}\right]-\frac{1}{R_f}\right\}}$$

$$= \frac{1/k}{Z_1\left[\left(1+\frac{Z_2}{Z_4}\right)\left(\frac{1}{... ...{R_f}+\frac{1}{Z_2}+\frac{1}{Z_3}\right)-\frac{1}{Z_2}\right]-\frac{Z_1}{kR_f}}$$

$$= \frac{1/k}{1+\frac{Z_1}{R_f}+\frac{Z_1}{Z_3}+\frac{Z_2}{Z_4}+\frac{Z_1Z_2}{Z_4R_f}+\frac{Z_1}{Z_4}+\frac{Z_1Z_2}{Z_3Z_4}-\frac{Z_1}{kR_f}}$$

If we let

$$Z_1=R_1,\;\;\;\;\;Z_4=R_2,\;\;\;\;Z_2=\frac{1}{j\omega C_2}, \;\;\;\;Z_3=\frac{1}{j\omega C_1}$$

the frequency response function becomes

$$H = \frac{\left(1+\frac{R_b}{R_a}\right)\frac{1}{R_1C_1}j\omega}{(j\o... ...1}{R_1C_1}-\frac{R_b}{R_aR_fC_1}\right)j\omega+\frac{R_1+R_f}{R_fR_1R_2C_1C_2}}$$

$$= \frac{Aj\omega}{(j\omega)^2+\Delta\omega j\omega+\omega_n^2}, \;\;\;\;\;\;\;\;\; A=(1+R_b/R_a)/R_1C_1$$

This is a band-pass filter with the peak frequency equal to the natural frequency:

$$\omega_n=\sqrt{\frac{R_1+R_f}{R_fR_1R_2C_1C_2}}$$

the bandwidth

$$\Delta\omega=\frac{1}{R_2C_1}+\frac{1}{R_2C_2}+\frac{1}{R_1C_1}-\frac{R_b}{R_aR_fC_1}$$

The gain of the filter is controlled by $1+R_b/R_a$.

More Examples of op-amp filters are listed below (with detailed analysis in here):

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