Synth Service Manuals

This post consists of links to synth service manuals. I'm putting this up primarily for my own reference but putting it here it can benefit others as well. I will continually update this post.

http://manuals.fdiskc.com/tree/

http://www.synthfool.com/docs/

http://www.synfo.nl

http://www.fantasyjackpalance.com/fjp/sound/synth/index2.html -includes great quality Minimoog schematics

http://www.pro-one-synth.org/manual.php?section=5

http://www.hinzen.de/midi/juno-106/

http://www.rhodeschroma.com/?id=servicemanual

http://www.korganalogue.net/korgms/service/ms20/m20s.html

http://redchapterjubilee.tripod.com/moogsource.html

Obligatory "sorry for not posting" post

Sorry I haven't posted in the past couple of months. My excuse is as follows: Lay offs at work that I survived, super busy at work from taking on all the extra work, looking for my first house because of my new found job security, got robbed of my laptop and camera, setting up new laptop, mailing in rebate form for laptop (are they seriously still doing this shit?), being sick for the past three weeks.

I'm going to commit to a new post with sound files of the Polymoog. Looks like there won't be any pictures for a while though.

Up $#!+ Creek Without a Pedal. Polypedal Alternatives for the Polymoog (part 2)

Having gotten the expression pedal working, I then need a momentary footswitch or two. There are four back panel controls that can be connected to a footswitch:

• EXT SYN enables or disables external synth connected to S-TRIG and KYBD OUT
• TRIG MODE switches between single and multi triggering of the low-pass filter
• SUSTAIN is not a true sustain, it really switches release on and off, and release must be the same as decay
• GLIDE switches the glide (portamento) on and off. Internally to the Polymoog this would only affect the VCF when keyboard tracking is on.

I decided SUSTAIN and TRIG MODE were worth having. Unfortunately, in the days of old, Moog decided sustain pedals should use 0.206" TRS plugs rather than 1/4" TS. Their thinking was this would prevent the wrong plug going into an output. To find out more I recommend going to Fantasy Jack Palance and looking at the minimoog and Micromoog manuals (By the way, I should mention I used a minimoog scan from that site in a previous post without giving proper credit. The scans on that site are excellent). Anyway, the pedal contacts connect through the tip and ring of the TRS. The sleeve was meant for the shield. I knew I needed to get 0.206" plug so I looked at mouser. The only available option was the switchcraft plug here. Incredibly, they cost $9.10 each. Ouch. I had to buy the Cinch Jones connector too so i bit the bullet.

For the pedal I considered the Roland FS-6 because it is a dual pedal with each pedal being selectable between momentary and latching. Latching would be handy for the TRIG MODE because there is otherwise no where to select the trigger mode on the Polymoog front panel. However this pedal needs a battery, which I felt was silly. I also would prefer a piano-type pedal. I ended up choosing a dual Kurzweil pedal, KFP-2M.

My original plan was to cut off the 1/4" plugs, solder on the 0.206" plugs and be done. Once it arrived I was impressed with the 1/4" plugs and couldn't bring to cut them off.

I decided to grab some cable and make double-headed plugs instead. The switchcraft plugs actually use screw terminals instead of solder lugs, which I liked. Soldering onto the existing 1/4" plug proved to be a little difficult, but I got both working on the first attempt. Here is the final result plugged into the back of the Polymoog:

Up $#!+ Creek Without a Pedal. Polypedal Alternatives for the Polymoog (part 1)

When I got the Polymoog it didn't have the Polypedal to go with it. This is unfortunate because it can offer a lot of useful control, especially of the filter cutoff. On the other hand I really don't need a giant, dusty, probably broken, 30+ year old box that can only be used with one synth. I started doing some digging to see what my options were. First a run-down of the controls on back panel:

There's S-TRIG OUT and KB OUT. The Polymoog has a built in monophonic keyboard circuit that is used to trigger the filter, provide filter keyboard tracking, and control external synths. The Polypedal doesn't use either of these. I also didn't need either for my purposes, although the S-TRIG IN plus a screw driver to short it is very handy for troubleshooting the filter envelope.

The SWELL, FILTER, MOD AMT, and PITCH are all 1/4" TS 0-5V controls that are intended to be provided by the two expression pedals of the Polypedal. The EXT SYN, SUSTAIN, TRIG MODE, and GLIDE are all 0.206" 3-conductor jacks that are triggered by a contact closure between the tip and ring of a TRS plug. On the far right are the two accessory power outputs. In part 1 I'll be talking about the SWELL, FILTER, MOD AMT, and PITCH connections. In part 2 I'll talk about EXT SYN, SUSTAIN, TRIG MODE, and GLIDE.

The first thing I decided was to use a modern expression pedal, specifically the Moog EP-2. While some of the reviews I read weren't great (along the lines of "not nearly as bad as the EP-1"), I liked that it had an attenuator knob. Also it's from the same company, so at least it matches the Polymoog logo-wise. The EP-2 manual can be found here. It uses a 1/4" TRS plug with Sleeve = ground, Ring = 5V, and tip = output voltage. So my problem was how to get 5V to the EP-2. My options were the following:

• Add external power to the EP-2, e.g. add a 9V battery.
• Replace the Polymoog CV jacks with 3-conductor jacks and rewire to include 5V.
• Use the 5V from the accessory output like the Polypedal.

I decided on option 3. It was the least work and it would keep the Polymoog and EP-2 unmodified. I figured I needed three things-a 1/4" male TS plug, a 1/4 female TRS jack, and a 6-pin male Cinch Jones connector for the power. Then I had a revelation. Rather than get each of those and try to wire them together, I could just get a Y connector with the 1/4" female TRS jack and two 1/4" TS plug. I could cut of the appropriate TS plug and replace it with the Cinch Jones connector. The first thing I found was a product from Hosa, the YPP-136. It was inexpensive and even available at Guitar Center. However, once I saw it I realized a problem, it wouldn't be long enough to reach the accessories jack. I could extend the cable but that defeated the purpose of finding a pre-made connector. After a little more research (amazon similar items) I found a Monster Cable adaptor that could reach. It unfortunately cost $20 but I got it anyway. Please do not take this as an endorsement of Monster Cable. Monster Cable rants are welcome in the comments. After I got the cable, Cinch Jones connector, and EP-2 pedal I got to work.

The Cinch Jones connector needed to replace the 1/4 plug on the left. Verify with a multimeter. Monster doesn't even label the ends L and R.

Cinch Jones connector

Monster Cable stripped. The red wire needs to connect to 5V. Ground and shield are left unconnected since they will both be connected to ground at the Polymoog control input jack. Make sure ground and shield are cut off and aren't contacting any pins on the Cinch Jones. Heat-shrink would have been a good addition in hindsight.

Accessory connector view from inside the Polymoog. 5V is yellow. Note that the connector is "polarized" and the red and green pins are a little bit farther away then the others. Make sure you solder to the right pin on the Cinch Jones connector.

Final result with everything connected. It's definitely a stretch.

Here's what the result sounds like. In the first part I'm lightly controlling the filter cutoff on a patch based on "String." Second I'm attempting to be funky on a patch based on "Clav." Third I'm using the pedal hooked up to SWELL on a patch based on "String." Notice the high level of noise between examples. That's because I don't have the shield board under the filter/output board...and because the Polymoog is a noisy MF to begin with.

Using the Moog EP-2 pedal with the Polymoog by Bross

OTA Voltage Controls in the Polymoog - Pictures

Front panel

This is an inside view showing the front panel on top and the right control board on the bottom.

This is an up close look at the right hand control board. The 4007 chips are all socketed. Behind them are the preset resistors, also in sockets. In the center you see the resistors for the mod amt control (57k and 33k from left to right). Only two of the presets have LFO filter modulation. That's a pretty old school way to handle presets.

Back view of the sliders. All are hand wired. The green wire is ground and the red wire is V₊, which are daisy chained across several controls.

This is the top right board, which handles the VCF, keyboard CV, and other functions. The three 3080s in a row are from left to right mod amt, S+H, and contour. The coresponding transistors Q6, Q5, and Q10 are below each 3080.

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.

Ben Folds Remixed Justice-Style

Take a listen to my Ben Folds remix. I did it in the style of Justice. Not a mash-up, more like when Weird Al does a "style parody." No analog content, in part because I figured it would be more authentic doing it in the box...and I didn't want to hook stuff up. In fact everything is tracks from the original song plus Reason with a lot of editing. Next time I'll invest in a plug-in to do my glitches automatically.

The winner of the remix competition is based on plays so tell your friends.

Polyphonic Spree

Now that I've covered monophonic keyboards in detail, it's time to move on to scanning keyboards as used in the early polyphonics and monophonics with last-note priority/built-in sequencers. This type of keyboard is used only on microprocessor based synths because it requires a little software to decode the notes . This is not the method used on the Polymoog and other paraphonics whose keyboards are akin to organs.

To start off, let's ask a question. How many wires do you need between a CPU and keyboard to read keys being played? If you guessed at least one for every key, hello from the future. If you guessed one, to transmit serial data, slow down. The keyboard only has simple switches and components. We need to read the keyboard before we can transmit note data by MIDI or other means. Luckily smart people have already found the answer. We can use a switch matrix with inputs and outputs instead of individual inputs for all the keys. A switch matrix is used in all sorts of devices with keypads, buttons, or switches to scan . Using this method we need at least 2x the square root of the number of keys. For a 61 key synth that would be 2 x sqrt(61) = 15.6 or 16 data lines. Below is a schematic from the Jupiter-6 Service Manual.I've labeled the connection KC-1 in the upper left "Outputs from CPU" and KC-2 in the upper right "Inputs to CPU." Here's how it works. The CPU cycles through different outputs for KC-1. It will output 10000000, 01000000, 00100000, 00010000, 00001000, 00000100, 00000010, 00000001 in some predetermined order. For each of those outputs it will scan for a particular section of keys. For example, if the output is 10000000, pin 61 will get 5V and 62-68 will be 0V. That 5V is connected to the key contacts for notes C0, G#0, E1, C2, G#2, E3, C4, G#4. Those key contacts connect to separate buss bars. Each of those eight buss bars is connected to a separate input back into the CPU. If the CPU reads 00101100, we know E1, G2#, E3 are being played. Then the CPU changes it's output and checks for a different set of seven or eight keys.

The only components in this keyboard other than wires and switches are diodes for every key. Why do they need to be there? Let's imagine they aren't there. You play F1 and G#1 simultaneously. The CPU is outputting 01000000. The 5V from the F1 will go to the buss bar 3rd from the left and be read by the CPU. There's a problem though. Without the diodes, the 5V at the buss bar also goes back through the G#1 and back to that note's CPU output, which is 0V. A 5V and 0V output get shorted together which may or may not make sparks, let out smoke, and ruin the CPU.

ARP Axxe Sliders

I had to whip an ARP Axxe into shape for a client recently. I changed out two sliders with ones sold at synthrestore.co.uk. I was very happy with the feel. It would have been nice to do all new sliders they were all pretty nasty but it's hard to justify the cost for an Axxe that cost $250 (I know because I saw the craigslist post). I used the Mountain Switch slider caps from mouser.com to replace missing caps. As prescribed, I did need to heat them up. They didn't have the exact look of the originals. That maybe a function of 30+ years of aging. I see synthrestore is going to sell their reproductions which include the elusive pink. Those will probably be worth investigating.

Speaking of slider caps, I had to do some google image searches and piece together the correct color scheme from multiple pictures. I took pictures of the final result so it will be here for reference. The correct answer (unless I'm wrong) is: black, pink, pink, yellow, red, blue, pink, red, pink, white, green, blue, black, black, yellow, pink, red, black, red, red, red, red, red.


So the scheme is:
Pink (x5) - LFO
Red (x8)- ADSR
Yellow (x2)- S/H
Blue (x2) - Square/Pulse
Green (x1)- Sawtooth
White (x1)- Noise
Black (x4)- Other

Synthedit

So I recently got Synthedit. Apparently I'm getting into it pretty late. It seems that all the links and posts on the internet are from 2007. For my first project I'm making a Moog Rogue clone. I've also gathered from the internet that a Moog clone is the stereotypical first VST everyone makes in Synthedit. Oh well.

So far it's been hard to balance making a true clone of the rogue with all it's limitations or make a double ADSR, three oscillator Rogue from hell. I'm staying true but overcoming the Rogue's obvious limitations. If you own one you know them by heart:

• Shared Waveform switch for both oscillators
• Shared Octave switch for both oscillators
• Pitch wheel tunes +/- 6th or something

It's also making me question user interfaces. Why on earth should any soft synth have a knob? I ironically came to the oposite conclusion about hardware. Why should any hardware synth have a slider? Knobs feel so much better. All sliders do is get dirty and you're lucky if you can find a replacement.

Sample and Hold: Not just a random LFO

Amongst some keyboard players, sample and hold has become synonymous with a random LFO. Check out this video from Automatic Gainsay that shows how the ARP 2600 sample and hold works.



So it's only a random LFO when you feed it noise. Also, if you think your synth has one sample and hold, you might be wrong. In analog monosynths the keyboard CV is controlled by a sample and hold circuit fed by the keyboard. Remember that the keyboard itself isn't doing anything after a key is released but notes must continue (with duration controlled by Release). It's also used in conjunction with the portamento control. Polysynths like the Jupiter-6 have tons of sample and hold circuits for various CVs for each voice.

To see a simple sample and hold circuit and description, see "SAMPLING" here. The Rogue uses a variation of this.

The "In" is coming from the keyboard. It is the current key's voltage or open when no keys are played. C3 is the capacitor that gets charged and holds the keyboard voltage. It is also used with R13 to form a low pass filter for portamento a.k.a. glide. R12, R14, and R15 are there for protection since the CV in and out are connected to a jack on the back. Someone could inadvertently fry the op-amp by plugging in something they shouldn't.

2810: An ARP Odyssey part 2

Like the Moog examples, the ARP Odyssey keyboard current will go through the keyboard resistor string. However, if two notes are held down at the same time the current will bypass the resistors between the notes since they are shorted out by the buss.

Up to this point we've been concerned about the buss voltage, which is responsible for pitch CV. But what about the voltage supplied by the current source to the high end of the resistor string? Let's figure out what happens to that using the Odyssey as an example. When one note (or none) is held down the high end of the keyboard voltage is 3-I*Rall where I is the current and Rall is the resistance of all resistors in the chain. For the Odyssey it's 3-.00083*36*100 = 0 V. But if two keys are held down the resistors between them are shorted out. So the voltage is 3-I*(Rall-Rbetween). But since we know 3-I*Rall = 0, this simplifies to I*Rbetween. This is how the Odyssey gets it's duophonic capability.

This voltage I*Rbetween is the CV difference between the two notes. Remember a normal CV is still generated by the keyboard voltage buss and corresponding S+H circuit. At oscillator 2 this voltage can just be added to the normal pitch CV, giving a CV corresponding to the high note.

2810: An ARP Odyssey

Up to this point I've been talking about monophonic keyboard circuits. ARP figured out a clever way to cheat, making the synth duophonic without much extra. First let's figure out what the difference is between this circuit (from the later Odysseys) and the Moog versions.

I should first mention there is still a keyboard voltage buss not shown where the contacts make, uhhh contact just like the Moog examples. In this schematic the + power supply (arrow pointing up) is 15 V. R13 and R17 form a voltage divider between 15 V and ground. This gives 15*R13/(R13+R17) = 3 V at the + input of the op-amp. The voltage of the - input of the op-amp will be the same. So the "keyboard low" voltage is 3 V. How about the current? It is just the current flowing through R11 and R12. The current flowing into J2-6 (up through R11 and R12) is (15-3)/(R11//R12) where R11//R12 is the parallel resistance of R11 and R12. This gives 0.83 mA. Once again this is 1/12 V per note. There is a major difference though. The current is flowing into the "keyboard low" so the keys are decreasing in voltage as you move up the keyboard. The highest key (3 octaves up) is at 0 V.

It is pretty simple to invert this voltage later to get an increasing V/oct scale. Two questions come up first, is this high or low note priority? Since the "keyboard low" is held at a specific voltage, the buss voltage will only depend on the current supplied and the resistance up to the lowest note held. So this is still low note priority like the Moog examples. The second question is why make the keyboard voltage backwards in the first place? If you've been paying attention you've probably guessed it has something to do with being duophonic. That explanation will be the topic of the next post.

Rogue vs. Prodigy

The Moog Rogue is similar to the Prodigy but came out a couple of years later. The keyboard current source circuit is almost except for a few differences.

• The op-amp is a 353 instead of 4558. The 353 has JFET input stages. This means much smaller input and bias currents (more than 1000x smaller). However the input offset voltage is 2-3x higher.
  
• There is no longer a resistor connected to the + input. I guessed that that resistor used the bias current to compensate for the input offset voltage. The smaller bias current means that won't work.
• The resistor between the trimpot and - input is 130k instead of 301k 1%. This seems like a counter-intuitive change. The current can now be adjusted between 12/14.3-12/130 = 0.747 mA and 12/14.3+12/130 = 0.931 mA. This makes it harder to adjust the trimpot for the 0.833 mA. My theory on why this change was made is compatibility. According to various sources, the Micromoog originally had a pitch CV of 0.9V/oct that could not be adjusted. The Rogue allows for a range of 0.896-1.11 V/oct.
  

Minimoogs do it discretely

How do Minimoogs do it? With discrete transistors. I've highlighted the keyboard resistor string and current source. Here are the differences between the Minimoog from the Prodigy:

• The current source uses two BJTs instead of an op-amp.
• There are 43 resistors instead of 31 because there are 44 keys instead of 32.
• The resistors are 10 ohm instead of 100 ohm. This is because we are using discrete BJTs for the current source, which will give a smaller, less ideal, current source output resistance. This is overcome by making the load smaller.
• The current needs to be 10x that of the Prodigy because of the smaller resistors.

The current shown on the schematic is 8.33 mA, which gives the 1/12 V for each key. Let's solve for this current using some quick and dirty approximations. We will assume the transistor base current is negligible and for both transistors Vbe = 0.7 V.

1. R20 and R22 form a voltage divider between 10 V and ground. Vb of Q11 is 5 V.
2. If Q11 Vbe is 0.7 V, then Q9 Vb is 4.3 V.
3. If Q9 Vbe is 0.7 V, then Q9 Ve is 5 V.
4. Q9 Ie is (10-5)/590 = 8.5 mA.
5. The current source comes from Q9 Ic, which will be almost equal to Ie = 8.5 mA.

This is pretty close to 8.33 mA. Where did I go wrong? I assumed the Vbe of the transistors was the same. First of all, they aren't match and one is NPN, the other PNP. Second of all, The current through them is not equal, so the Vbe values will be different. The current through Q11 is about 4.3/1k = 4.3 mA. We found the current through Q9 is almost twice this. The Q9 Vbe will be about 18 mV greater than Q11 Vbe based on the Ebers-Moll BJT model (26 mV*ln(2)). Q9 Ve will be about 5.018 V. Q9 Ie will be (10-5.018)/590 = 8.44 mA. Now assume Q9 has a Beta of 160. Our current source will be 8.45*159/160 = 8.39 mA. Now we're less than 1% off from the expected 8.33 mA. That's close enough considering we have 1% resistors.

Prodigy Keyboard Current Source

Let's look at the keyboard current source used in the Prodigy. The lowest key is connected to the "0.00V" inverting (-) op amp terminal. The highest key is connected to the "2.66V" op amp output. In case you don't remember, here are the rules for ideal op amps with negative feedback: There is no current going into the + or - inputs. The voltage difference between the + and - is so small we can assume the two voltages are equal. The output does whatever it can to make this happen.

Let's figure out the current.

1. First off, no current into the + input, therefore no voltage drop across R84.
2. The voltage of the + input is 0.
3. The voltage of the - input must also be 0.
4. No current into - input so the current going through the keyboard resistor string is equal to the current going through R83 and R179 by Kirchoff's Current Law.

There's a trimpot so I will solve for current at both extremes of the trimpot position. First when the pot wiper is at +12V, I = 12/14.3k-12/301k = 0.799 mA. At -12V I = 12/14.3k+12/301k = 0.879 mA. Between those positions, the trimpot can be adjusted to our magic 0.833 mA to give 1/12 V across each 100 ohm resistor.

A question I have looking at this: Why is the high note keyboard voltage 2.66V? We have 32 keys, but 31 resistors between them. V = 31*1/12 = 2.58 V. Someone made a math error and figured 32*1/12 = 2.66 V.

Another question I have looking at this: What is R84 for? Why not connect the + input directly to ground? In our ideal op amp world it works but notice the op amp is a 4558. In the not-so-ideal world this op amp uses bipolar transitors which have a bias current coming out of the + input. So the + input is a little bit above ground in our example. From the datasheet the input bias current is 50 nA. The + input voltage is 50 nA*18k = 0.9 mV. Also from the datasheet the input offset voltage is 1 mV. That means the - input is held very close to ground (roughly 0.9-1 = 0.1 mV) because of R84.

R83 and R179 are 1% to ensure the proper current range. C22 is presumably there for stability while still allowing the output voltage to change quickly enough when new notes are played.

There are plenty more where that came from. Check back for the Rogue, Minimoog, and more.

The Prodigy Keyboard Circuit part 2

So we determined a voltage source to the keyboard circuit won't work. Here is a current source used instead. The lowest key is still connected to ground. What should be the current I? Each resistor is 100 ohm and should have 1/12 V across it. By Ohms Law I = V/R = 1/(12*100) = .833 mA. What happens when two keys are held down at the same time? Like the previous example with the voltage source, the current will bypass the resistors between the two keys. To solve for Vout, we see the point on the far left is grounded. Vout will just be I*(Resistance from ground to the lowest key held down). Pressing higher keys won't do anything to Vout. This is the reason for low note priority.

It would be just as easy to make a keyboard with high note priority. The highest note could be held at 2.58 V. The current source would be connected the opposite way to the lowest note to sink current. Making a last-note-priority keyboard would be much more difficult to make. Figuring out the last note played really requires digital logic, so it wasn't until the introduction of microprocessors in synths that last note priority became common. The SH-101 and Pro-One come to mind. The SH-101 has a last-note/low-note priority switch so the synth shredders of the day could still use their Moog soloing techniques.

Now that we've established why we need a current source, we can ask why 100 ohm 1% resistors are used. (It looks like .1% on this scan of the schematic. Other scans definitely have 1% and I'm fairly certain .1% aren't manufactured). The 1% is easy. Our ears are very sensitive to pitch. If a note is 5% (5 cents) out of tune, it will be noticeable. Why 100 ohm? Consider this. If the resistors were large, say 100k, we could save power. The power used by the keyboard when no notes are played is V^2/R. We can't change V because we are using 1V/oct. But if we switched from 100 ohm to 100k ohm, it would require 1/1000 the power. What's the problem?

We are using a current source. In a perfect world it wouldn't matter, but real current sources have an output resistance. It is a large resistance in parallel with the current source. If we are driving a load with small resistance, this output resistance won't matter. But if the load is comparable to the output resistance, the current source will start misbehaving.

The Prodigy Keyboard Circuit

The previous posts shows what 1V/oct keyboard CV should look like. So how do we get it? There are two steps. 1.) We need to spit out a CV when a key is depressed. 2.) We need a sample and hold circuit to hold that voltage until the next key is depressed. This is needed because of the "Release" portion of the note's envelope. The pitch must stay constant until after the note fades out (or another note is played). First I will talk about the first step, spitting out the CV when the key is depressed. The Moog Prodigy keyboard circuit/switches are shown here.

Each key is connected to two switches which make contact when the key is depressed (double pole single throw, or DPST). I will concentrate on the bottom switches that are responsible for CV. The top switches are for gate/trigger signals. Each key CV switch is separated by a 100 ohm resistor. When the key is depressed it makes contact with the bus bar labeled "3" in the schematic. this is the output that goes to the sample and hold stage. From Ohms Law we know the resistors will act as voltage dividers. In the last post we figured each key should be 1/12 V higher than the previous key. So let's try to run the CV circuit with a voltage source. This will be the wrong solution for reasons I will explain.

Each of the 31 resistors has 1/12 V. The whole chain should have 31*1/12 = 2.58 V. So we could hook-up this voltage source like so:

Let's say a note an octave above the lowest note is played. Treating this like a voltage divider, the voltage Vout will be 2.58*12(100)/(31(100)) = 1 V. This is the correct voltage. In fact all keys will put out the correct voltage. So what's the problem?

Even though we're talking about a monophonic synth, often two keys are held down at the same time. What happens in that case? Our voltage divider gets screwed up. Let's say the note an octave above the lowest note AND the note an octave above that are held down. The current will bypass the resistors between the notes since they are shorted out.













What does this do to the voltage? In the example I just gave the voltage Vout will be
2.58*12(100)/((31-12)(100)) = 1.63 V. This is not the correct voltage of either note. In fact it's not the correct voltage of ANY note, at least not in the musical/CV scale we're using.

How do we solve this problem? Luckily the synth forefathers already did it for us. Use a current source instead. That will be explained in the next post.

Keyboard CV Demystified

There seems to be a lot of misunderstanding about keyboard a.k.a. pitch CV on analog synths. You've probably heard most Moogs and ARPs are 1 Volt/Octave. I've made a video with my Moog Rogue to show what 1V/oct means.

I have the probes of my multimeter connected to the "KYBD CV OUT" and ground. First I play the low F, which puts out close to 0 V. Then I play the F an octave up (1 V) and the F two octaves up (2 V), hence 1V/oct. Note that the CV stays the same until the next note is played.

Next I play F, G#, B, D, F. What should the voltages be? If we have a 1V/oct scale and 12 notes = 1 octave, then 1V/oct = (1/12)V/note. So we get the following:

F = 0 V
G# = 3*(1/12) = 1/4 V
B = 6*(1/12) = 1/2 V
D = 9*(1/12) = 3/4 V
F = 1 V

Next I turn up the portamento. The Rogue's portamento kicks ass and I have yet to hear a softsynth duplicate it. That will be the topic of a future post. With the portamento turned up I play the lowest note and highest note to show the CV slide between the values. The high note incidentally sounded like the slow down part in "Rockafeller Skank."

Then I turn the multimeter to mV. I play the low F and F#. We would expect the voltages to be 0 mV and 83.3 mV, respectively. They are actually 2.7 mV and 85.7 mV. Why the deviation? Analog isn't perfect and that's why it's great.


Measuring Pitch CV for my Moog Rogue from Robert Hume on Vimeo.

First!

This blog will be my investigation into analog synthesizer electronics. How do analog synths work? Why do they act and sound the way they do?

I've decided the best place to start is with monophonic keyboards. Ever wonder why your Moog has low-note priority? Why doesn't it have last-note priority like the Pro-One? Check back here to find out.