Passive Filters

Op amps are active filters because they require power. Passive filters use passive components (Resistors, Inductors, Capacitors) to achieve a similar effect. They are constructed using a potential divider with reactive components. The diagram below shows a potential divider with two impedances, $Z_1$ and $Z_2$:

$$\frac{V_{out}}{V_{in}}=\frac{Z_2}{Z_1+Z_2}$$

Transfer Functions

The transfer function is the ratio of input to output (see ES197 - Transfer Functions for more details.). For a passive filter, this is the ratio of output voltage to input voltage, as shown above. For a filter, this will be a function of the input waveform, $H(j\omega )$. When $Z_1$ and $Z_2$ are both identical resistors $R$:

$$H(j\omega )=\frac{R}{R+R}=\frac{1}{2}$$

However, if $Z_2$ was a capacitor $C$, $Z_2=\frac{1}{j\omega C}$:

$$H(j\omega )=\frac{Z_2}{Z_1+Z_2}=\frac{1}{1+j\omega RC}$$

The gain and phase of the output are then the magnitude and argument of the transfer function, respectively: $$|H(j\omega )|=\frac{1}{\sqrt{1+(\omega RC{)}^2}}$$ $$\angle H(j\omega )=\frac{\angle 0^{\circ }}{{\tan }^{-1}(\omega RC)}=-{\tan }^{-1}(\omega RC)$$

Cutoff Frequency

Similar to active filters, passive filters also have a cutoff frequency $f_c$. This is the point at which the power output of the circuit falls by $\frac{1}{2}$, or the output gain falls by -3dB, a factor of $\frac{1}{\sqrt{2}}$. Using the above example again (a low pass RC filter):

$$|H(j\omega )|=\frac{1}{\sqrt{1+(\omega RC{)}^2}}=\frac{1}{\sqrt{2}}$$ $$2=1+(\omega RC{)}^2$$ $${\omega }^2=\frac{1}{R^2{C}^2}$$ $$\omega =\frac{1}{RC}$$ $$f_c=\frac{1}{2\pi RC}$$

This is also the point at which $H(j\omega )=1+j$

The filter bandwith is the range of frequencies that get through the filter. This bandwith is 0 to $f_c$ for low pass filters, or $f_c$ and upwards for high pass.

RC High Pass

$$H(j\omega )=\frac{j\omega RC}{1+j\omega RC}$$ $$|H(j\omega )|=\frac{\omega RC}{\sqrt{1+(\omega RC{)}^2}}$$ $$\angle H(j\omega )=\frac{\angle 90^{\circ }}{{\tan }^{-1}(\omega RC)}=90-{\tan }^{-1}(\omega RC)$$ $$f_c=\frac{1}{2\pi RC}$$

RC Low Pass

$$H(j\omega )=\frac{1}{1+j\omega RC}$$ $$|H(j\omega )|=\frac{1}{\sqrt{1+(\omega RC{)}^2}}$$ $$\angle H(j\omega )=\frac{\angle 0^{\circ }}{{\tan }^{-1}(\omega RC)}=-{\tan }^{-1}(\omega RC)$$ $$f_c=\frac{1}{2\pi RC}$$

RL High Pass

$$H(j\omega )=\frac{j\omega L}{j\omega L+R}$$ $$|H(j\omega )|=\frac{\omega L}{\sqrt{R^2+(\omega L{)}^2}}$$ $$\angle H(j\omega )=90-{\tan }^{-1}(\frac{\omega L}{R})$$ $$f_c=\frac{R}{2\pi L}$$

RL Low Pass

$$H(j\omega )=\frac{R}{j\omega L+R}$$ $$|H(j\omega )|=\frac{R}{\sqrt{R^2+(\omega L{)}^2}}$$ $$\angle H(j\omega )=-{\tan }^{-1}(\frac{\omega L}{R})$$ $$f_c=\frac{R}{2\pi L}$$

2nd Order Circuits

For circuits more complex than those above, to find the transfer function, either:

• Find a thevenin equivalent circuit, as seen from the element
• Combine multiple elements into single impedances

Note that any of the above techniques only work for simple first order circuits.

Example

Using $H(j\omega )=\frac{Z_2}{Z_1+Z_2}$, where $Z_1=R_1$, and $Z_2=R_2||j{X}_C$:

$$Z_2=\frac{\frac{R_2}{j\omega C}}{R_2+\frac{1}{j\omega C}}=\frac{R_2}{1+j\omega R_{2}C}$$ $$H(j\omega )=\frac{Z_2}{Z_2+Z_1}=\frac{\frac{R_2}{1+j\omega R_{2}C}}{R_1+\frac{R_2}{1+j\omega R_{2}C}}=\frac{R_2}{R_1+j\omega R_1{R}_{2}C+R_2}$$ $$|H(j\omega )|=\frac{R_2}{\sqrt{(R_1+R_2{)}^2+(\omega R_1{R}_{2}C{)}^2}}$$ $$\angle H(j\omega )=-{\tan }^{-1}\frac{\omega R_1{R}_{2}C}{R_1+R_2}$$