OTA Voltage Controls in the Polymoog

Here's a use for OTAs (courtesy of the Polymoog) besides the VCA we already looked at. Technically this could be called a VCA as well. It's taking a voltage from a front panel slider to control the amplitude of filter modulation signals. We will look at the Modulation Amt. control. It affects the amount of LFO going to the filter cutoff. The top half of the schematic shows the front panel slider and other components from the right hand control panel board. The slider itself is a 10k audio taper pot with its wiper terminal in series with a 27k resistor. The 4007 is a CMOS switch. The ZZ control line determines if the output should be in PRE (preset resistors, upper right) or VAR (slider) mode. In VAR mode pin 12 is connected to pin 9 via an n-channel FET. The output of the 4007 is named PVMAN.

The lower half of the schematic is from the top right filter board of the Polymoog. Notice the PVMAN output connects to transistor Q6 which is in turn connected to the control current pin of OTA A9. This is essentially the same pnp transistor configuration we saw with the MG-1 VCA. The input of the OTA is a voltage divider, not a differential input like before. Note that the LFO is ±2.5V. The voltage divider reduces this to 100/(22000+100)*±2.5 = ±11mV. This is within the ±25mV input limit of the 3080. What about the control current I_ABC? With the slider all the way down we can see no current will flow and the OTA will be off. With the slider all the way up to V₊ = 15V the current will be roughly (15 - 0.7)/27k = 0.53 mA. This is also a reasonable result considering I_ABC for the 3080 can be 0.1µA to 1mA.

The output of the OTA goes to pnp emitter follower Q11 which adds several VCF control signals, S+H, LFO, contour, and keyboard amt, which all use 3080s and cutoff (PVCON), which is directly connected from the 4007 without a 3080. The emitter follower in turn drives an npn transistor (not shown) that sinks current dirrectly from the ladder filter. Next time I'm going to take some pictures while my Polymoog is still openned up so I can show what this stuff actually looks like.

Theory of the MG-1 VCA

I will be referencing the schematic in the previous post.

Q: Why is Q28 needed?
A: We want an envelope to be based at 0V, go to some positive voltage, then back to 0V. This is the control signal we expect for a filter, oscillator, and VCA as well. The problem is the 3080's control current pin (5) is internally connected to a diode which is connected to the negative supply voltage. If we wanted to connect a control voltage (with series resistor) directly to this pin, our CV would have to based at the negative supply instead of ground. This single PNP transistor circuit is the simplest solution. Here's how it works. When the envelope is 0V, the transistor doesn't conduct and no current is passed to the OTA. Note the -15V on Q28's collector at the start and finish of the envelope. During the peak of the envelope the transistor conducts, so the base-emitter voltage will be roughly 0.6V. The voltage across R104 is (4.5-0.6), giving a current of (4.5-0.6)/15k = 0.26mA. The collector voltage is then roughly 0.26m*10k+(V_ABC). V_ABC is a diode drop above the negative supply, so the collector voltage is 2.6+(-15+0.6)=-11.8. This corresponds to the -12 shown at the envelope peak on the MG-1 schematic. Actually, if we neglected the diode drop voltages of the transistor and OTA instead of using 0.6V, we get -12V exactly.

Q: What are R96 and R97 for?
A: According to an article on OTAs from Nuts and Volts magazine posted here (follow the link at the bottom to UsingOTAs2.pdf), the input resistors "help equalize the source impedances of the two signals and thus maintain the DC balance of the OTA."

Q: Why are they 100 ohm?
A: If they were larger, they would do a better job of equalizing the source impedance. However, there are two concerns. The OTA has a finite input resistance. The data sheet gives a typical value of 27k. If the resistors were larger, the differential input voltage would be reduced. The output of the ladder filter is already quite low. The second concern is noise. The larger the resistor, the noisier. These resistors are at the input of an amplifier, so any noise will be amplified as well.

Q: What is the gain of the OTA?
A: We found I_ABC is 0.26 mA at the peak of the envelope. We also know the transconductance (g_m) of the OTA is 19.2*I_ABC. This gives a g_m of 5mS. We can see our input voltages are two reverse polarity 2mV_pp signals, so a total differential input of 4mV_pp. The output current swing is then 4mV*5mS=0.02mA.

Q: What is the output voltage?
A: The OTA output is fed to the master volume pot in series with a 1k resistor. This is a clever arrangement that allows a standard pot to be used while the resistance, dominated by the smaller resistor, can be designed. The max output voltage peak-to-peak is 0.02mA*1k//10k = .02*10//11=18mV_pp. This corresponds to the 17mV_pp shown on the schematic.

The MG-1 VCA

From this list of venerable synths, I have decided to focus on the one sold at Radio Shack, the Realistic MG-1. Why? Because my scan of the schematics is the most legible.

The VCA and master volume circuit is shown. I've photoshopped (actually Paint.NET'd) a little so all the labels are visible. The 3080 is in the center. It gets it's differential input from the output of the ladder filter (not shown). The control current I_ABC is coming from the lower right. The contour generator puts out a 0-4.5V envelope. Q28 and corresponding resistors convert the CG voltage to the control current I_ABC. The output current of the OTA passes through R102//R103 to give the output voltage. R103 is set up as a standard pot volume control. The addition of R102 in parallel allows the overall voltage gain to be controlled while using a standard pot size for R103.

In the next post I'll show the math behind the labeled values.

VCA the OTA way

To recap, an OTA has a voltage input, control current input, and output current. An obvious synth application would be as a VCA. The challenge is a VCA is voltage-controlled, not current-controlled. Also the output current must turn into an output voltage, which isn't a huge deal since that really just requires a resistor. When I talk about VCAs, I am referring to the amplifier that is controlled by the envelope to shape the sound, creating the illusion of a note playing. Final level controls especially in polysynths can also be VCAs and use OTAs, but those are boring.

The following is a list of synths using the 3080 or 3280 OTAs for the VCA:

CA3080: Moog Taurus (I & II), Prodigy, Rogue, Opus 3, (Realistic) MG-1, ARP Odyssey, Octave CAT
CA3280: Sequential Circuits Pro-One

Not all synths use OTA chips for the VCA. Roland mostly used the BA662 which is a VCA rather than OTA. Also many later synths (especially polysynths) used SSM or CEM chips for the VCA. Korg synths (at least the MS-10 & MS-20) use a different solution. In conclusion, if you don't like OTAs, you don't like America.

OTA Synopsis

Here are the important points about OTAs that set them apart from opamps:

1. The output is a current, not a voltage.
2. The output level is determined by a control current I_ABC, not fixed by feedback resistors.
3. The input is a small differential voltage, but cannot be assumed to be 0V.

The equation for the output current is the following:

I_{out} = V_{in}\frac{\alpha I_{ABC}}{2V_T}

V_T = kT/Q and is known as thermal voltage. It frequently shows up in semiconductor equations. For normal conditions, this equation simplfies to:

I_{out} = V_{in} I_{ABC} \cdot 19.2

Another important point from the application notes is the voltage of the control current pin. It is one diode drop above the negative supply voltage. So we know roughly what the voltage is, but it will vary with varying control currents. For this reason a current buffer is usually placed before the OTA control current input. In the next post I'll track down some real life applications of OTAs in synthesizers.

OTA, how can I explain it?

OTAs or operational transconductance amplifiers are commonly used in analog synths for a variety of purposes. Most commonly they are used as VCAs. After all, how would you make a VCA with standard op-amps? I graduated with an electrical engineering degree without ever hearing about OTAs. Then again, I could say that about a lot of things. I digress. We must teach ourselves about these triangles.

I was first shown the ways of the triangle by Prof. Aaron Lanterman's online videos of his Electronics for Music Synthesis class at Georgia Tech. See session 6 and 14. That would be a good place to start if you're like me and you long for the days of sitting in a classroom watching somebody solve a math problem. If only those problems were about synthesizers. Well now they are!

If you're more hardcore, check out the application notes for the CA3080. These are filled with all kinds of good information if you can actually stand to read them. When you're finished, put down your pencils and I will continue.

Switches are an Illussion. They're Really the Matrix.

In a previous post I talked about how keyboard scanning works. The same method can be also used for switches that are not attached to keyboard keys, such as the rainbow colored switches of the Roland Jupiter-8.

Here we see the schematics of Panel Boards G and E. The switches are shown with corresponding LEDs above them and, at the top, the function name, e.g. "TUNE", "A", "SPLIT." This should look very familiar; it's just like the keyboard scanning schematic. The output from the CPU (actually Panel Board A, more on that later) is fed to AG2-1,2,3,4 and AE2-1,2,3,4,5. As in keyboard scanning, the outputs cycle. First AG2-1 is high, everything else is low, then AG2-2 will be high, everything else will be low, and so on. During each cycle AG1-3,4,5,6,7,8 and AE1-1,2,3,4,5,6 will be read to see if any switches in a subset are currently pressed. The diodes are there for protection. So the only new thing is the LEDs, which are connected to the same outputs as the switches. How do they work?

The LEDs will illuminate when the diodes are forward biased. This occurs when a positive voltage is supplied from the Panel Board A outputs and current is being sunk from AG1-9,10,11,12,13,14 or AE1-7,8,9,10,11,12 back to Panel Board A. But we know that a positive voltage is only supplied a fraction of the time from the Panel A outputs. So the LEDs must only be "on" a fraction of the time. Whatever flicker is present must be imperceptible.

So the operation of the LED matrix is just like that of the switches except slightly backwards. The outputs are cycled through, corresponding to a subset of LEDs. Then, instead of a human pressing a switch and the CPU reading it, the CPU turns on an LED by sinking current and the human "reads" it. For example, suppose the "A" button LED on Panel Board G needs to be on. When AG2-2 is high, "6", "7", "8", "MANUAL", "A", or "B" can be turned on. Now the CPU has to sink current from AG1-10 to illuminate the "A." To understand this current sink, we'll take a look at Panel Board A.


First notice the LED Matrix in the upper right hand corner. The eight horizontal lines are the cycling outputs from Panel Board A. The six vertical lines are the current sinks for the LEDs going back to the "LED DATA LATCH." Here are a few interesting things to note:

• Boards E and G LED current sink lines are connected together to the same six LED lines.
• AE2-5 and AG2-1 are also tied together. There are really only eight outputs from Panel Board A to these switches/LEDs.
  
• The two boards are physically separate but effectively act as one.

Let's take a closer look at the current sink labelled "LED DATA LATCH." IC13 is a D flip-flop. The inputs are on the bottom. Every rising edge clock (not the same as CPU clock) sets the outputs on top to the input logic states. Each flip-flop output drives a single transistor switch circuit. When the flip-flop output is high (5 V) AND the LED is biased with a positive voltage, the transistor will saturate, sinking current to put the collector near 0 V (0.05-0.2 V typically according to Art of Electronics). This will turn on the LED. R40-45 are the current limiting resistors for the LEDs. The LED current will be about (5 - V_led)/100. Assuming Vled = 1.8 V, the current will be 32 mA. R46-51 have been designed with a resistance of 10k. I'll do some calculations to show why. The transistor base current is (5-V_be)/R_b = I_C/Beta. Assuming V_be = 0.7 V, I_C = .032 A, and Beta = 200, R_b comes out to 26k ohm. Using a lower resistance of 10k ensures there is ample current to put the transistor into saturation.

Let's back up and and figure out why there is a need for an LED DATA LATCH. Why do we need a flip-flop to store values? Notice on this block diagram the data bus is shared between the switches, LEDs, Numerical LED display latches, Interface board, and more. The LED DATA LATCH is needed in order to store values so the data bus can be used for other things. The flip-flop clock only triggers when the appropriate values are on the data bus.