of Microprocessors

190 MUSICAL ApPLICATIONS OF MICROPROCESSORS
EXPONENTIAL
CURRENT 0-----1
INPUT
CI
LOW-SIAS
HIGH-SPEED
OP-AMP
C2
I
_15~Jl
NARROW
':>---+---() PULSE
OUTPUT
COMPARATOR
Fig. &-8. Sawtooth oscillator
The comparator compares the integrator output with a positive reference
voltage, Vrif. As long as the integrator output is less than Vrif, the
comparator outpur is negative, which keeps the FET switch across the integrating
capacitor off, allowing it to charge. As soon as the integrator voltage
reaches Veef, the comparator output starts to go positive. As it does, the
positive comparator input is forced even more positive through C2 giving
positive feedback and causing the comparator to snap on. The comparator
output is constrained to rise no further than ground, but this is enough to
fully turn on the high-threshold-voltage FET switch, which discharges Cl
and brings the integrator output back to ground potential in preparation for
another cycle. The comparator is prevented from responding instantly to the
drop in integrator voltage by the charge accumulated on C2 when the comparator
switched high. In effect, a one-shot is formed with a time constant of
RC2, which is arranged to be long enough for the FET to completely
discharge C 1. Even though every reasonable effort is made to speed the
discharge, the control current still loses control for a finite time each cycle.
This will cause the higher oscillator frequencies to be flat, that is, lower than
expected from the control current magnitude. This error can be compensated
for as will be shown later.
The value of C 1 is chosen to provide the required range of output
frequencies given the 0.25 MA to 0.25 rnA range of current input. The
expression for frequency is F =I/CVrej and the expression for capacitance
is C=I/FVrif where Vrif is the reference voltage in volts, F is frequency
in hertz, C is capacitance in farads, and I
is the exponential current in
amperes. The easiest way to calculate C is to first determine the highest
frequency of interest and then solve the second equation for I = 0.25 rnA.
The lowest frequency for really accurate response (and zero control voltage if
the highest frequency corresponds to a + lO-V control voltage) is 1,024
times lower. Much lower frequencies are possible with good components and
moisture-proofed circuit boards, as much as 1,000 times lower yet for a total
range of over a million to one. For a nominal audio range of20 Hz to 20 kHz
and Vrif of +5 V, C 1 would be about 2,500 pF. For optimum performance
over a 32-Hz to 8-kHz range in an 8-V system with Vref of 4.096 V, Cl