#### Application:

Many systems have unwanted signals that can corrupt your signal of interest: digital switching noise, 60 Hz AC power, switching power supply noise, motor drivers, clock oscillators, random noise from resistors and active devices, RF pickup, etc. For example, imagine you have a 5 kHz signal to be digitized. Unfortunately, a noise source at 50 kHz has coupled onto your signal. How can you stop the noise at high frequencies and pass the signal at low frequencies? A simple RC low-pass filter with an op amp buffer might do the trick.

But, why the op amp? It's the R and C in an RC filter that determine the frequency response. The problem arises when you connect the filter output to the next stage. The input resistance of the next stage will fall directly across the C, changing the total R and the response of the filter. To avoid this problem, you need to isolate the RC components from the next stage. How? Enter the op amp. The high input resistance of an op amp (typically > Gohms) has no significant effect on the R of your filter.

1. Choose a cutoff frequency fo (Hz).

As an example, select fo=10 kHz to reduce a noise signal at 50 kHz and pass your desired signals below 5 kHz.

2. Pick a convenient cap value C1 between 100pF and 0.1 uF.

Suppose you’ve got plenty of 1000pF caps in stock, select this value for C1.

4. Calculate R1 = 1 / (2 · π · fo · C1)

R1 = 1 / (2 · π · 10kHz · 1000pF) = 15.9 k ohms

But, why the op amp? It's the R and C in an RC filter that determine the frequency response. The problem arises when you connect the filter output to the next stage. The input resistance of the next stage will fall directly across the C, changing the total R and the response of the filter. To avoid this problem, you need to isolate the RC components from the next stage. How? Enter the op amp. The high input resistance of an op amp (typically > Gohms) has no significant effect on the R of your filter.

#### Schematic Diagram:

#### LOW-PASS DESIGN:

The key specification of a low-pass filter is the cuttoff frequency fc. Well below fc, the filter passes along without attenuation. Well above fc, the filter attenuates the signal. How much? As the frequency increases by a factor of 10x, the signal decreases by a factor of 10x. Let's design a filter to suppress (as best as this filter can) noise and pass our desired signal.1. Choose a cutoff frequency fo (Hz).

As an example, select fo=10 kHz to reduce a noise signal at 50 kHz and pass your desired signals below 5 kHz.

2. Pick a convenient cap value C1 between 100pF and 0.1 uF.

Suppose you’ve got plenty of 1000pF caps in stock, select this value for C1.

4. Calculate R1 = 1 / (2 · π · fo · C1)

R1 = 1 / (2 · π · 10kHz · 1000pF) = 15.9 k ohms

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