ECE4893A: Electronics for Music Synthesis
Due: Tuesday, Feb. 12 at the start of class
Semiconductor Application Note 31
(or pretty much any textbook that has op amp circuits in it)
You are free to discuss approaches to
the problems with your fellow students, and talk
over issues when looking at schematics,
but your solutions should be your own. In particular, you should never
at another student’s solutions at the moment
you are putting pen to paper on your
own solution. That’s called “copying,” and it is lame. Unpleasantness
may result from such behavior.
Moog (east coast) and Buchla (west coat) developed their ideas about
voltage controlled synthesizers independently. Moog used a pitch control
standof of 1 volt/octave, which works out to 0.08333… volts/semitone
(the pitch difference between to adjacent notes on the piano is a semitone;
there are twelve semitones per octave).
Buchla preferred to use 0.1 volts/semitone, which works out to
Hence, if to try to directly drive
a Moog oscillator from a Buchla pitch control source, or vice-versa,
everything will be horribly out of tune.
a) Design an circuit with a single op amp
that will covert pitch control voltages from
Moog standard to the Buchla standard. You may assume that your conversion
module is given an input from a voltage source with zero output impedance
and is being fed to a module with an infinite input impedance; you also do not
worry about input and output protection (assume nobody will be abusing
your module). For this part
of the exercise, assume you have perfect “zero-tolerance” resistors.
b) Off-the-shelf resistors never exactly match their listed values.
Let’s do a “worst case” analysis for the case where your circuit is given
a one volt input. If you use 5% resistors, assuming the true resistance
is uniformly distributed, what is the highest voltage you might get out?
What is the lowest voltage? How many semitones above and below the desired
value are these voltages in the Buchla pitch standard?
c) Repeat the above analysis for 1% resistors.
In this problem, we will analyze the PAiA 9710 VCA. The
is available here (and
although it is in squintovision, so
I would recommend opening it up in some sort of image viewing program
so you can zoom in a litte.
series runs off a +/- 12V power supply, i.e. V- on the schematic means
-12V, and V+ means +12V.
The 9710 actually contains two VCAs, along with a “balanced modulator” (a
full quadrant multiplier, which is good for special effects.) We will
look at the “right” VCA. You will find the core of it along the
bottom part of the diagram. J2 is the audio input, and J11 is the output.
The circuit will we look at uses one-half of a dual OTA, the LM13600; the
one of interest to us is labeled IC5:B.
R45 and R43 create a slight voltage bias to compensate for non-ideal
characteristics of the LM13600.
Real OTAs have an output bias current, which vexingly varries with the
control current. The 9710 uses a fixed resistor network to put a small
bias voltage to try to correct for a “typical” output bias current. Many other
designers use a trim pot for this. In any case, we won’t worry about such
non-ideal effects here, so you may assume the OTA is an “ideal linear
OTA” and that the + terminal of IC5:B is at ground.
The current generating circuitry for the OTA may be found in the upper
left corner of the schematic. R14, R16 (at least I think it’s R16 – it’s the
220K resistor below R14), R12, R18 (at least I think it’s R18 – it’s
a 1500 ohm resistor to round, IC2:C (an op amp), and Q2 form a current
source similar in spirit to the one from Chamberlin, p. 203 that I presented
near the beginning of lecture on 1/29/07, so if you have difficulties
analyzing it, you will want to review your notes from that day.
R44 is a small current limiting resistor
to make sure the OTA doesn’t become an ex-OTA. D2 is for protection; we
will ignore it in our analysis. Similarly, we will ignore R48, the 18M
resistor (consider it to be infinite.)
To simplify our analysis, suppose the “pan”
pot is turned all the way towards R, so the wiper is at ground.
J5 is a “normalled” jack, meaning that if nothing is plugged into it, the
“signal” part of the jack (the top pin in the schematic, which appears
to be labled with a test point “H”)
will default to the signal going to the middle pin (which appears on
the schematic to be labeled with a test point “B”). We will assume that the
user has inserted a control signal into J5, so point “H” (the left side
of R14) will be some user-created voltage we will call “V_con”. (In
text, I will often use an underscore to indicate subscripts.)
a) Find the voltage at the output of IC3:B. You can do this quickly if
you realize that this op amp is acting in a standard “inverting mixer”
configuration. (Remember we are assuming the wiper of the “pan” pot is
b) Given all the notes above and the assumption that the current through
the base of Q2 is negligible, so we may approximate its collector and
emitter currents as being equal, find the current input to the OTA
a function of V_con.
c) Now let’s look at the main part of the VCA, with input at J2 and
output at J11. Find the gain of the VCA as a function of I_con. For now, you
may ignore C12 (i.e. open the cap, which is a reasonable assuption for low
d) Suppose V_con = 10V (from what I understand, the envelope generators
in the PAiA 9700 series can generate up to 10V, so that’s a reasonable
voltage to try). What is I_con in this case?
e) Take a look at the “Absolute Maximum Ratings” section of the
LM 13600 datasheet. What is the maximum rating for the
control current (which the data sheet will call I_ABC)? Is the value you
found in part (D) above or below this?
f) What is the gain of the VCA for V_con = 10V?
g) Without going through extensive calculations – i.e., by just reasoning
your way through the circuit – what happens to the VCA gain as the wiper of
the “pan” pot is turned away from ground and
h) Previously, we ignored C12. If we now consider it, we see that C12, R52,
and IC6:B act as a current-to-voltage single-pole lowpass filter with a
cutoff frequency (half-power point) of 1/(2*pi*RC)). What is the cutoff
frequency of this filter? Is this cutoff frequency above or below the
limit of typical human hearing?
i) What is the input impedance of the VCA?
j) What is the output impedance of the VCA? (Assume the op amps are all ideal,
i.e., they have zero output impedance).
In this problem, we’ll keep looking at the same PAiA 9710 schematic as
in Problem 2. Here, we will focus on the pan pot, and the notion of
“shaping” the curve swept by the pot.
Suppose the leg of the pot between the wiper and ground has resistance R,
where R may range from 0 to 10K;
then, the other leg (between the wiper and V+) has resistance (10K – R).
The voltage at point “C” may be found via a standard voltage divider, where
R and the 120K resistor may be simplified as a parallel combination.
On a single graph, plot (a) the voltage at “C” as a function of R,
for 0 < R < 10K, and (b) the same thing as (a), except suppose that the
120K resistor is actually a 10K resistor. I recommend using MATLAB to
make the graph, but you may use whatever computer program you prefer.
Which graph, (a) or (b), looks more linear?
In this problem
explore Dirk Lindhof’s Exponential VCA. Notice that instead of converting
the output current of the OTA by using an op amp in an inverting configuration,
this design drops the current to ground through a resistor, and then buffers
the resulting voltage with an op amp.
The OTA, a CA3080, is drawn differently than we are used to seeing it; the
control current (let’s call it I_con) is shown as being input to pin 5.
Ignore the caps at the inputs; they just block DC.
This module has
two inputs; it looks like IN2 gives
you the option of bypassing the
DC blocking cap on that channel.
Anyway, just do the analysis for one input.
Ignore the trimming circuitry, i.e. just ground the + input on the 3080 and
assume the OTA is ideal.
a) Find the gain of the VCA as a function of I_con.
b) What is the input impedance of the VCA?
c) What is the output impedance of the VCA? (Assume the op amps are all ideal,
i.e., they have zero output impedance).
Let’s look at the “AUDIO AC IN” of the
The input capacitor (470 nF) and the two resistors (68K and 330 ohm)
form a single-pole highpass filter whose “output” is measured by the
positive input terminal of the CA3080.
Find its cutoff frequency (the half-power
point). Note that the
resistor in series with the cap gives this configuration a non-unity gain.
As usual, assume that the input is being driven by an ideal voltage