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 opamp.
1.
The Active Bass Cut Filter
We can overcome these problems by adding an active element such as an opamp. 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 opamp? 
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.
Z^{2} = Xc^{2} + R^{2}
So
gain can be worked out:
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:
X_{C}
= __1__ = __________1___________ = 3400
W
2pfC
2 × p ´
1000 Hz
´
47 ´
10^{9}
F
Now
we need the impedance:
Z^{2}
= X_{C}^{2 }+ R^{2}
= (3400 W)^{2} +
(3300 W)^{2} = 22450000
W^{2}
Þ
Z = 4700 W
Now we can work out the 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? 
f_{0} = __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 10fold 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 noninverting opamp:
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:
What is the break frequency of this circuit? 
Why is this circuit an active bass boost circuit? 
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.
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 = () R_{f } / R_{a}. 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 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.
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? 
Summary For ALL filter circuits:
X_{C}
= __1__ 2pfC
X_{C}
= V_{rms} I_{rms } _{ }
Z^{2}
= R^{2} + X_{C}^{2}
Passive
filters:
Active
filters:
