High and Low Pass Filters

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Hello!

Something on the order of three weeks ago, I said that I would start posting every Tuesday and Thursday. Seeing as this has been my only post since then, we can all see how we’ll that worked out.

At any rate, on to today’s topic – high- and low-pass RC filters.

Sometimes it’s necessary to filter an AC signal – for example to eliminate DC offset (or vertical shift). This requires the use of an RC filter – in this case, a high-pass filter.

So what is an RC filter?

In essence, an RC filter is, aptly enough, a filter made of a resistor and a capacitor. It’s formation is like that of a voltage divider, with the input on one end of the chain, the other end connected to ground, and the junction connected to the output. As a high-pass filter, the resistor is grounded; as a low-pass filter, the capacitor is grounded.

How does an RC filter function?

The filter hinges on a capacitor’s impedance as a function of voltage, or more specifically,

This causes the divider’s output to be for a high-pass filter or or for a low-pass filter.

Since the output depends on three factors – resistance, capacitance, and frequency – and two of them remain constant – resistance and capacitance – a filter can be made to divide voltage relative to frequency.

Selecting Resistor and Capacitor Values

In order to choose values for the RC filter, it’s important to know about the -3dB breakpoint. The -3dB point is the frequency at which the signal will be attenuated (un-amplified) by -3dB, or about half. Frequencies above the -3dB breakpoint are considered to pass through the filter, whereas those below the breakpoint are considered to not pass. To find the formula for the breakpoint, all we need to is set the filter’s output to .5 and solve for frequency, as is done below for the high-pass filter below:

$\begin{align*} \frac{1}{2} & = \frac{R}{R + \frac{1}{2 \pi f C}} \\ \frac{1}{2} R + \frac{1}{4 \pi f C} & = R \\ \frac{1}{4 \pi f C} & = \frac{1}{2} R \\ 1 & = 2 R \pi f C \\ f & = \frac{1}{2 \pi R C} \end{align*}$

Now we can choose values. To do this, I suggest choosing a resistor near the center of the allowable range for the application (for example, for op-amps, 1 KΩ to 100 KΩ), rounding off to the nearest calculated capacitor value, and then recalculating to get a resistor value.

For example, let’s say I’m filtering off DC offset (f < 100 Hz) for an amplifier that uses op-amps. I'll use 10 KΩ as my chosen resistor value. From this I get:

$\begin{align*} f & = \frac{1}{2 \pi R C}\\ 100 \mathrm{Hz} & = \frac{1}{2 \pi R C}\\ 628.3 \mathrm{Hz} & = \frac{1}{10 \: \mathrm{K} \Omega \times C}\\ 6283000 \: \mathrm{F}^{-1} & = \frac{1}{C}\\ C & = 15.9 \: \mathrm{nF} \end{align*}$

The nearest value to this is 150 nF. Recalculating I get:

$\begin{align*} f & = \frac{1}{2 \pi R C}\\ 100 \mathrm{Hz} & = \frac{1}{2 \pi R C}\\ 628.3 \mathrm{Hz} & = \frac{1}{15 \: \mathrm{nF} \times R}\\ .00000094245 \: \Omega^{-1} & = \frac{1}{R}\\ R & = 10.6 \: \mathrm{K} \Omega \end{align*}$

Or approximately 11 KΩ, the nearest common resistor value.

That’s it! Thanks for reading. Put any comments or questions below and subscribe to future posts if you want updates in the future!

Capacitor Impedance, Capacitors, Filter, RC Filter

This entry was posted on October 30, 2012, 9:16 PM and is filed under Electronics. You can follow any responses to this entry through RSS 2.0. Both comments and pings are currently closed.

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