Active Filters

Load the PSPICE file activef1.sch, Figure 27.

**Figure 27:** Active filter.
$\begin{figure} \begin{center} \epsfig{file=images/clab4img3.eps}\end{center}\end{figure}$

Lecture Notes: Frequency Response : Active Filters
For the filter of Figure 27, Z₁(s) = R₁, $Z_2(s) = R_2 \parallel (1/sC) = R_2/(1+s R_2 C)$ . Then

$\begin{displaymath}H(s) = - \frac{Z_2(s)}{Z_1(s)} = -\frac{(R_2/R_1)}{1+s R_2 C} \end{displaymath}$

and so

$\begin{displaymath}H(j\omega) = - \frac{(R_2/R_1)}{1+j\omega R_2 C} \end{displaymath}$

The DC gain corresponds to $\omega = 0$ and equals in magnitude

$\begin{displaymath}\vert H(j0) \vert = R_2/R_1, \ \ \ {\rm or} \ \ \vert H(j0) \vert (dB) = 20 \log_{10}(R_2/R_1) \ {\rm dB} \end{displaymath}$

The high frequency gain corresponds to $\omega = \infty$ and equals in magnitude

$\begin{displaymath}\vert H(j\infty) \vert = 0, \ \ \ {\rm or} \ \ \vert H(j\infty) \vert(dB)=-\infty \ {\rm dB} \end{displaymath}$

The 3dB point is

$\begin{displaymath}\omega_{3dB} = \frac{1}{R_2 C} \ \ {\rm rad/sec}, \ \ {\rm or} \ \ f_{3dB} = \frac{\omega_{3dB}}{2\pi} \ \ {\rm Hz} \end{displaymath}$

Exercise:

1.: Simulate the active filter circuit and obtain the magnitude and phase Bode diagrams.
2.: What type of filter is it? Find the 3 dB point. What is the DC gain? What is the high frequency gain?
3.: By changing component value(s), increase the 3 dB point by one decade (Hz) and re-simulate to check.
One decade means a factor of 10, so work out new values of R₁, R₂ or C to increase f_3dB by 10, keeping the DC gain unchanged.
4.: By changing component value(s), change the magnitude of the pass band gain to 10 dB, and re-simulate to check.
Work out new values of R₁, R₂ or C to achieve a new DC gain of 10 dB (what is this in the absolute scale?) keeping the 3dB point unchanged.

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