Suggested references: National 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 be looking 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 1.2 volts/octave.
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 the 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 need 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 schematic is available here (and mirrored here), 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. The 9700 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 at ground).
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 (I_con) as 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 frequencies).
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 we will 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 Serge Lin/Log VCA. 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 source.
