Filter circuits working- Lets start with RC filter

Now, it's time to explore the workings of the filter circuits. I have been thinking about where to start for too long. But we have to start it anyway; that's what matters.

So the filter circuit is basically a voltage divider circuit. In case of resistance, based on the resistance value, the voltage gets divided, and we get the voltage across the resistance.

In the above example, a 12V 10Hz AC signal drops to a 6V AC signal across the resistor R2, which obeys Ohm's law. Simple.

In the case of a capacitor, the capacitive reactance depends on the frequency of the signals. Means that capacitive reactance increases with a decrease in frequency.

The equation for the RC low-pass filter is given in the image below. 

Let's try to calculate the output voltage.

For the above circuit, we are getting an output voltage of around 6.22V across the capacitor for the 2 kHz signal. Let's try to simulate the same, and let's see the result.

Let's play with different frequency values.

For a frequency of 5 kHz, the output voltage across the capacitor is around 3V. If we do simulate the same simulation in LTSpice, we get the result as below.

From the above example, it is clear that as frequency increases, the capacitive reactance decreases, which in turn decreases the voltage across the capacitor. 

By using this simple principle, we can use this circuit to eliminate the high-frequency noises in the signal.

Now, let's see the critical frequency (fc) of the RC lowpass filter circuit. I have no idea why they named it critical frequency, but the simple definition is that the frequency at which the output voltage is 0.707 times the input voltage is called Critical frequency.

Actually, 0.707 is nothing but the square root of 2. Also, we can say that at the critical frequency, the gain of the filter circuit is -3db. Wait a minute..? You can ask where this concept of Gain is coming from, actually, the function of the filter can be expressed in terms of Gain(dB) also. We will discuss that also. Derivation will take one page long, let's skip directly to the result.

After trying for a long time, I finally drew the Bode plot for the RC low-pass filter.

A Bode plot shows the relation between Frequency and gain.

Usually, Gain is represented in a logarithmic scale, such as decibels.

Here, -20 dB means 10 times, and -3 dB means 0.707 in the case of voltage.

Here, -20 dB decade means that when the frequency increases 10 times, the voltage decreases by 10 times.

In the above example, if we pass a 20 kHz signal instead of a 2 kHz signal, then the output voltage will be 10 times lower. Let's check...

See, the output voltage almost decreases 10 times. That's what this Bode plot is all about.

If we want to further decrease the gain, then we have to use higher-order filters. It's nothing, just multiplying the resistors and capacitors connected like below.

Here, one rule is that in order to avoid the loading of resistors, R2 should be greater than 10 times R1 (R2>>10R1).

The equation for higher-order filters can be given.

That's it, I covered almost most of the things related to the RC low-pass filter. Finally, finished the article that I wanted to write for a very long time.

Please provide your valuable feedback if any😉