Tutorial 13 B - Active Filters

Passive filters are simple but rather limited.  The main problems are:

We will look at four case studies.  They are based on the op-amp.


1. The Active Bass Cut Filter

We can overcome these problems by adding an active element such as an op-amp.  The circuit shows an active bass cut filter:



You will recognise the circuit as an inverting amplifier.  Generally electronic engineers ignore the phase shifts in these filters so we will ignore the minus sign in the equation.


Question 8

What is the equation that tells us the gain of an inverting op-amp?



There is a capacitor, hence a reactance in the input as well as a resistor.  We can work out the impedance using the vector sum of the reactance of the capacitor and the resistance of the resistor.

                                    Z2 = Xc2 + R2


So gain can be worked out:

  Gain =  (-) Rf/Z   


Normally we ignore the minus sign.

Worked Example

Referring to the diagram above what is the gain of the operational amplifier at 1000 Hz?

First we need to work out the reactance:


XC = __1__ = __________1___________         = 3400 W

       2pfC        2 p 1000 Hz 47 10-9 F


Now we need the impedance:

Z2 = XC2 + R2 = (3400 W)2 + (3300 W)2 = 22450000 W2 Z = 4700 W


Now we can work out the gain:

  Gain = Rf = 33000 W = 7.02 (No units for gain)

             Z       4700 W


If we reduce the frequency, the impedance increases, so the gain reduces.  We calculate the break frequency in exactly the same way as we did with a passive filter.  The break frequency is the frequency at which the resistance = reactance.  We can work this out:


What is the break frequency of the active filter above?

f0 = __1__ = __________1___________ =  1000 Hz

      2pRC     2 p 3300 W 47 10-9 F


We can show this as a graph:

A couple of points to note:

We can make a bass cut filter using a non-inverting op-amp:

It works like this:


Question 9

What is the break frequency for this circuit?



2.  Active Bass Boost Filter

The bass boost circuit is more complex:


Question 10

What is the break frequency of this circuit?



Above 150 Hz the reactance starts to get significantly less and the resistance remains constant.  Therefore the gain decreases significantly.


Question 11

Why is this circuit an active bass boost circuit?



3. Active Treble Cut Filter

The treble cut filter is a little simpler to understand:

At low frequencies:


Question 12

What is the gain of the amplifier at very low frequencies?  Assume that the reactance of the capacitor is very high indeed, almost infinite. 


Question 13 

What is the break frequency?



Above the break frequency, the reactance of the capacitor decreases, and the feedback factor increases.

This reduces the gain down to a small value, since gain = (-) Rf / Ra.  Strictly speaking we should say gain = (-) Z/Ra.  Whatever we say, Z is going to be small, so the gain will be less than 1.

  We can show this on a graph:

We can explain why the circuit acts as a treble cut filter:

Whatever the type of filter, the break frequency remains the same:


4.The Treble Boost Filter

The treble boost filter is shown in the diagram:

It looks more complex, but its not too hard if you follow the explanation.  The circuit works like this: 



Question 14 

What is the break frequency of this circuit? 


Question 15

What is the reactance of the capacitor at the break frequency?


Question 16 (Harder)

At the break frequency the input impedance is 70 kilohms, so the gain is 14.  Can you show that this statement is true?



The graph looks like this:



For ALL filter circuits:

XC = __1__


XC = Vrms



Z2 = R2 + XC2


Break frequency is for ALL circuits


Passive filters:

  • They can only attenuate (reduce) the signal.

  • The impedance of the input can alter the performance.


Active filters:

  • Can boost or attenuate signals.

  • Are based on amplifiers.

  • In these notes are based on op-amps.

  • Usually op-amps.

  • Boost or cut is governed by the gain.





Self Test