Tutorial 13 B - Active Filters

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

• They can only attenuate (reduce) the signal.

• The impedance of the input can alter the performance.

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.

 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 Þ 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:

• The scales look odd because they go up in powers of 10.  They are logarithmic.

• The gain goes down by 10 for every 10-fold reduction in frequency.  This means that bass frequencies are cut.

• Above 1000 Hz the resistance becomes much greater than the reactance.  Therefore the gain is almost flat.

We can make a bass cut filter using a non-inverting op-amp: It works like this:

• The amplifier as shown is a voltage follower.

• Its gain is 1.

• Its input is the output of the voltage balance.

• If the frequency is low, the voltage across the 3k3 resistor is low, so the output voltage is low.

• If the frequency is high, then the input voltage is high.

• The behaviour is quite like the passive filter.

 What is the break frequency for this circuit? 2.  Active Bass Boost Filter

The bass boost circuit is more complex: • At very low frequencies the capacitor would have a very high reactance.

• The 2.2 MW resistor is there to stop the gain rising to open loop at zero frequency.

• The maximum gain of this circuit is 2.2 ´ 106 W ÷ 10000 W = 220.

• The gain reduces with increasing frequency.

• We can work out the break frequency with the formula f0 = 1/(2pRC).

 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. 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:

• The reactance of the capacitor is very high;

• The only path is through the resistor.

 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. 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:

• At low frequencies, the reactance of the capacitor is high

• Therefore the gain is a maximum

• At high frequency the capacitor has a low reactance

• So the effective resistance of the feedback is less

• Hence the gain is less.

 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 it’s not too hard if you follow the explanation.  The circuit works like this:

• At a low frequency the reactance of the capacitor is very high.  The input impedance is governed by the 100 kW resistor.  The gain is about 10.

• As the frequency rises above the break frequency, the reactance of the capacitor gets much less, and the gain rises.  However we don’t want the gain to be too much, as the result of this would be that the bandwidth will get less.

• So there is a 1 kW resistor in the input that restricts the maximum gain to 100.

 What is the break frequency of this circuit? 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: Summary For ALL filter circuits: XC = __1__          2pfC XC = Vrms                         Irms   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.