I have mentioned the word Loading, frequently in my posts. In the post voltage divider, I bring up that the passive filters are subjected to the phenomenon called loading. Since I have not explained this term, though I am using it frequently, it needs explanation.
Loading is not to load rounds in the magazine. We do not talk about weapons here. Loading means the effect on the performance of the circuit when another circuit element is attached to its output, in this case, the filter. Let me explain this with an example. I take the RC filter as an example.
For simplicity, let it be a resistor. How does the resistor hook to the output of the filter affect the performance of the filter? Calculating the time constant of the circuit, which is denoted by tau, helps to understand this. Calculating the time constant of this type of circuit is a simple task. You take the value of Resistance and Capacitance and multiply, you get the time constant. Let R=100k \Omega and C=100nF. You gets time constant \tau=10ms .
To understand the effect of load on the filter, let us Thevenize the circuit. For Thevenizing the circuit, you have to remove all the sources. As soon as you remove the voltage source (Short circuit the voltage source), you will see that the load resistor and R are in parallel (Figure 2). Their equivalent resistance is 50k\Omega . You see, there is a significant change in the equivalent resistance of the circuit. And the value of the time constant is \tau=5ms. And as you all know that the output of this circuit is dependent on the time constant, you can see that the performance of the filter is affected.
A buffer is introduced to the circuit with the intent to make the filter active. The advantage of an active filter is that you can get a predictable response from the active filter. The buffer is nothing but an op-amp, which has a very high input impedance (in the range of megaohm). After introducing op-amp in the circuit, no matter what you attach to the output, the performance of the filter is not compromised.
The calculation for the time constant after adding buffer is left for the readers. Take impedance of buffer to be 10M\Omega and send me the answer.