of Microprocessors

396 MUSICAL ApPLICATIONS OF MICROPROCESSORS
Building a Chebyshev Filter
The first decision to be made in building a filter for an aduio DAC is
whether it is to be passive, that is, use real inductors, or active, using only
resistors, capacitors, and amplifiers. Nowadays, an active implementation is
chosen almost automatically for any audio-frequency filter. The list of apparent
advantages is long, but most of them are the result of eliminating inductors.
Two possible pitfalls must be avoided when designing a sharp active
low-pass filter for use with a DAC, particularly 16-bit units. One is noise in
the amplifiers, especially since several will be in the signal path. The other is
distortion and possible overload. Remember that a 16-bit DAC is capable of
distortions on the order of 0.00 15%, so just about any amount of amplifier
distortion is going to be excessive. Active filter configurations that require
high-gain wide bandwidth amplifiers should therefore be avoided.
Of the three best-known configurations for a resonant low-pass section
(SaIlen and Key, multiple feedback, and state variable), the Sallen and Key
circuit shown in Fig. 12-23 has many important advantages. First, it is hard
to imagine a circuit using fewer components. Also, the filter characteristics
are relatively insensitive to component variations. Most important, however,
is the amplifier requirement, a simple unity-gain buffer with ideally infinite
input impedance and zero output impedance. Such an amplifier is very easily
constructed with an ordinary op-amp, special voltage-follower op-amp, or
discrete components. Since the gain is unity, the output noise amplitude is
nearly the same as the input referred noise level, which is what appears on the
amplifier spec sheet. Also, the frequency-independent 100% feedback
minimizes the amplifier distortion. About the only negative aspect of the
circuit is that high Q factors require a large spread in capacitor values.
OUTPUT
1
F "" 271" VR2C1C2
0~05~
If R,F, and Q are Known, then
C, =_0-
rrFR
Cz = 4g~
If C2, F. and Q are known. then
C, = 40 2 C2
1
R = 2rrF vc:fC2
Fig. 12-23. Sallen and Key resonant low-pass section