Active Filters

Passive RLC filters are simple and easy to design and use. However, can we implement them on-chip? Let us look at a simple example to give us a bit more insight regarding this question.

Example: A passive band-pass filter

Consider the filter shown in Fig. 1.

We can write the transfer function as:

${\begin{aligned}H\left(s\right)&={\frac {v_{o}}{v_{i}}}={\frac {sL\|{\frac {1}{sC}}}{R+sL\|{\frac {1}{sC}}}}={\frac {sL\cdot {\frac {1}{sC}}}{R\cdot \left(sL+{\frac {1}{sC}}\right)+sL\cdot {\frac {1}{sC}}}}\\&={\frac {sL}{s^{2}RLC+R+sL}}={\frac {s\cdot {\frac {1}{RC}}}{s^{2}+s\cdot {\frac {1}{RC}}+{\frac {1}{LC}}}}\\\end{aligned}}$

(1)

If we let $Q=\omega _{0}RC={\tfrac {R}{\omega _{0}L}}$ and $\omega _{0}={\tfrac {1}{\sqrt {LC}}}$, then we can rewrite our expression for $H\left(s\right)$ as:

$H\left(s\right)={\frac {s\cdot {\frac {\omega _{0}}{Q}}}{s^{2}+s\cdot {\frac {\omega _{0}}{Q}}+\omega _{0}^{2}}}$

(2)

Notice that the transfer function has two zeros, $s_{z}=\left(0,\infty \right)$, and two poles located at:

$s_{p}=-{\frac {\omega _{0}}{2Q}}\pm {\frac {1}{2}}\cdot {\sqrt {{\frac {\omega _{0}^{2}}{Q^{2}}}-4\omega _{0}^{2}}}$

(3)

We get complex conjugate poles if $4\omega _{0}^{2}>{\tfrac {\omega _{0}^{2}}{Q^{2}}}$ or when $Q^{2}>{\tfrac {1}{4}}$, or equivalently, when $Q>{\tfrac {1}{2}}$. If the band-pass filter has $\omega _{0}=2\pi \cdot \left(1\,\mathrm {kHz} \right)$, $Q=20$, and $R=50\,\mathrm {\Omega }$:

$L={\frac {R}{\omega _{0}Q}}={\frac {50\,\mathrm {\Omega } }{2\pi \cdot \left(1\,\mathrm {kHz} \right)\cdot 20}}=398\,\mathrm {\mu H}$

(4)

$C={\frac {Q}{\omega _{0}R}}={\frac {20}{2\pi \cdot \left(1\,\mathrm {kHz} \right)\cdot \left(50\,\mathrm {\Omega } \right)}}=64\,\mathrm {\mu F}$

(5)

Let us now consider a lossy inductor with $Q_{\text{ind}}=40={\tfrac {\omega _{0}L}{R_{s}}}$. The loss can then be modeled by the series resistance, $R_{s}$, as shown in Fig. 2, with:

$R_{s}={\frac {\omega _{0}L}{Q_{\text{ind}}}}={\frac {2\pi \cdot \left(1\,\mathrm {kHz} \right)\cdot \left(398\,\mathrm {\mu H} \right)}{40}}=0.0625\,\mathrm {\Omega }$

(6)

We can convert the series RL circuit to its parallel circuit equivalent in Fig. 3 for frequencies around $\omega _{0}$ by first writing out the admittance of the series RL circuit as:

${\begin{aligned}Y&={\frac {1}{R_{s}+j\omega _{0}L}}={\frac {1}{R_{s}+j\omega _{0}L}}\cdot {\frac {R_{s}-j\omega _{0}L}{R_{s}-j\omega _{0}L}}\\&={\frac {R_{s}-j\omega L}{R_{s}^{2}+\omega _{0}^{2}L^{2}}}={\frac {R_{s}}{R_{s}^{2}+\omega _{0}^{2}L^{2}}}-{\frac {j\omega L}{R_{s}^{2}+\omega _{0}^{2}L^{2}}}\\&={\frac {1}{R_{s}}}\cdot {\frac {1}{1+{\frac {\omega _{0}^{2}L^{2}}{R_{s}^{2}}}}}+{\frac {1}{j\omega _{0}L}}\cdot {\frac {1}{1+{\frac {R_{s}^{2}}{\omega _{0}^{2}L^{2}}}}}\\&={\frac {1}{R_{s}}}\cdot {\frac {1}{1+Q_{\text{ind}}^{2}}}+{\frac {1}{j\omega _{0}L}}\cdot {\frac {1}{1+{\frac {1}{Q_{\text{ind}}^{2}}}}}\\&={\frac {1}{R_{p}}}+{\frac {1}{j\omega _{0}L^{\prime }}}\end{aligned}}$

(7)

Thus, we get:

$R_{p}=R_{s}\cdot \left(1+Q_{\text{ind}}^{2}\right)$

(8)

(9)