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