Yamaha CS80

-------------- Original message -------------- 
From: "cuari7" <diaz.jesus@...> 

> Hi, group, 
> 
> Is there a way to do pitch-bending on a CS50 (like a pedal, maybe)? 
> 
> 
> 
> 
> Yahoo! Groups Links 
> 
> 
> 

[Non-text portions of this message have been removed]

  ----- Original Message ----- 
  From: cuari7 
  To: yamahacs80@yahoogroups.com 
  Sent: Monday, November 26, 2007 9:03 PM
  Subject: [yamahacs80] Pitch bending on a CS50?


  Hi, group,

  Is there a way to do pitch-bending on a CS50 (like a pedal, maybe)?



   

[Non-text portions of this message have been removed]

> 
> 
> 
>    
> 
> [Non-text portions of this message have been removed]
>

> It could be done. But not on the cheap.
>
>   
>>   ----- Original Message ----- 
>>   From: cuari7 
>>   To: yamahacs80@yahoogroups.com 
>>   Sent: Monday, November 26, 2007 9:03 PM
>>   Subject: [yamahacs80] Pitch bending on a CS50?
>>
>>
>>   Hi, group,
>>
>>   Is there a way to do pitch-bending on a CS50 (like a pedal, maybe)?

From: david@...
Date: Thu, 29 Nov 2007 17:44:30 -0800
Subject: Re: [yamahacs80] Re: Pitch bending on a CS50?
















  


    
            This touches on a big, somewhat technical, issue of what kind of VCOs

the CS80 uses. The VCO III chip is a linear VCO, sometimes called

Hz/Volt, as opposed to the more common exponential (Volts/Octave) VCOs

(e.g. MiniMoog, Curtis & SSM chips in SCI and Oberheim polys).



Here's a pretty good explanation:

http://www.synthmuseum.com/magazine/linexpo.html



Here's a (I hope) quick one:



The most basic VCO is a sawtooth one, which can be a capacitor charged

by a current.  For non-EE types, here's my modified toilet analog (and

you though the Metasonix vacuum-tube VCO was weird) :  The capacitor is

like the water tank of a toilet.  The water filling it up is the

current.  The height of the water is like the voltage across the

capacitor.  Now, modify the float valve so that when the tank is full it

automatically flushes.  Then the cycle starts again.  If you double the

water filling rate ( = double the current), you double the frequency of

the flush cycles.



The is a basic, linear VCO (actually Water-CO).  It shows a couple of

things.  First, it's not actually voltage controlled, but current

controlled. Ignore that for now.  Also, the filling time is adjustable,

but the discharge/flushing time is fixed. This is an issue with all

sawtooth VCOs and is why many (e.g. Moog) VCOs have a

high-frequency-tracking adjustment, which helps cancel this out. Here's

the CS80 VCO: http://home.debitel.net/user/jhaible/cs80_vco.gif



Ok, so why don't all synths use linear VCOs?  As the above link

explains, human ears don't hear frequency linearly. A above middle C is

440Hz.  An octave about is 880Hz, or double the frequency. The next

octave would be 1760Hz: double that. If you graph this, it's an

exponential curve. So, the space (in Hertz) between two notes keeps

getting bigger as we get to high pitches. If you had a modular synth

with linear VCOs (like that old Paia), the top key might output 5

volts.  One octave down would be 2.5volts. The next 1.25volts, followed

by 0.625v and 0.3125v. This is a pain to generate. Also, as you get to

lower notes, smaller voltage inaccuracies start becoming bigger pitch

errors to our ears.



To avoid all this, someone (anyone know who?  Dr. Bob?  Tom Oberheim?

Don Buchla?) came up with exponential VCOs. Basically, they're just a

linear VCO with a circuit in front of them called (big surprise) an

exponential converter. This is just a circuit that takes a linear input

(1volt/octave) and outputs the doubling voltage (actually current...)

that the VCO wants. Now, everything is simple.



So, why did Yamaha go for the linear?  Two reasons, I'd guess. First,

adding the exponential converter to each VCO adds more cost to the

chips, since there's more circuitry. A bigger issue is temperature

stability. As we've been talking about lately, all circuits are affected

(i.e. knocked out of tuning) by temperature changes. The exponential

converter, for reasons I won't go into, is really sensitive to this.

People have been complaining about the tuning stability of the CS80, but

it's rock solid compared to any poly-synth with exponential VCOs (P5,

OBX, A6, etc). They all need computer-controlled auto-tuning routines to

have any chance of staying in tune.



So, what issues/problems/advantages does the CS80 having linear VCOs create?



Good things:

1) modulation - linear vibrato sounds a bit different than v/oct

vibrato, probably closer to acoustic vibrato (e.g. violin).  Also, as

the modulation speed increases, you start getting into F.M. land, which

requires linear modulation (you don't want to know the math!). This is

why some modular VCOs have linear FM inputs in addition to the normal

v/oct controls.



2) sweep to D.C. - my favorite. If you start a pitch bend at the right

end of the ribbon and slide all the way to the left, the pitch of the

VCOs all go down to 0Hz / D.C. / flat-line. This is because the input to

the VCOs goes to 0 volts and the frequency equals the voltage times a

constant.  With a exponential VCO this is impossible.  Going 1 volt less

on the control input goes down one octave.  Mathematically, you can't

get to zero Hz.  You'd need to input -infinity volts!  Also, many other

limitations in the circuit block the VCO from even getting close.  Big

win for linear VCOs!



Bad things:

1) Keyboard voltages - as I wrote above, the keyboard has to generate

exponential voltages. This is a big pain. In a digitally-controlled

analog (like the CS80, P5, etc), the keyboard voltage comes from a DAC

(digital-analog-converter). 99.99% of DACs are linear. The CS50/60/80

(and others in the family) have bizarre, custom exponential DACs. This

makes interfacing the CS80 to other synths and/or MIDI-CV converters a pain.



2) CV mixing.  Finally, we get to the original question of adding a

pitch-bend input to the CS80.  In the volts/octave world, everything is

easy: you just add voltages together. Adding voltages is simple to do -

just an op-amp and a few resistors.  Let's say you had the following

voltages come out of a v/oct keyboard: 1v, 2v, 4v.  This could represent

a low C (c1), C one octave up (c2), and C two octave above that (c4). 

To make it simple, let's say we have a pitch wheel or pedal add 1 volt

to this (2v, 3v, 5v).  This would be c2, c3, c5, so we've just

transposed the sequence up an octave.



Ok, what happens if we try this with a linear voltage.  For the same c1,

c2, c4 notes, we might have 1volt, 2volt, 8volt. Adding one volt gives

2volt, 3volt, 9volt.  The first note is correctly up an octave, but the

next is only up about a 5th and the third note is only transposed up

about a semitone. This, obviously, doesn't work.  What we need to do,

instead, is multiply the voltages. To transpose up an octave, double the

voltages. To transpose down an octave, halve them. This is easy for a

fixed transpose, but if you want a variable, like a pitch-bend pedal

input, you need to multiply voltages.  Just like it's much, much easier

for people to add and subtract than multiply and divide, so it is for

analog (and digital) circuitry.



If you follow the schematics or block diagram of the CS80, you can see

that the voltage to the VCOs comes through a long chain of

multiplications. The ribbon is actually the initial voltage source for

the whole instrument.  If the ribbon isn't pressed it outputs some fixed

voltage (not sure the actual value - call it 2 volts).  If the ribbon is

slid up, all the way, from the left to the right, it would output double

this voltage, which corresponds to one octave up.  If the ribbon is slid

the other way, it outputs zero volts, as mentioned above.  Next, the

voltage is sent through the concentric pitch knobs. Any normal

potentiometer is a voltage multiplier, which can multiply the input by

anything from zero to one.



This voltage then becomes the reference input to the exponential DAC on

the KAS board, which multiplies it by it's exponential resistor network

to create the CVs for each of the either voices. These voltages go to

the VCO chips on the M-Boards. Are we done - nope - one more CS80

weirdness.  In a v/oct synth, the octave/foot switches would just

generate a voltage that would be added to the keyboard CV (e.g.

MiniMoog). The CS80 VCO, instead, has a special footage input that needs

an exponential current for each feet setting.  Because this is difficult

to do accurately over a wide range, we end up with the wonderful VR4,

VR5, and VR6 trimmers to get the feet switching calibrated separately

for each of the 16 VCOs. Yuch!



Getting back to the original question (remember Alice?  There's a song

about Alice...), a pitch bend input would need to control a voltage

multiplier.  This could be an added circuit, after the ribbon circuit,

or could probably be merged with the ribbon voltage. I haven't figured

out the details, but it's not rocket science.  However, it is a lot more

work than it would be on something like a Prophet 5.



Ok, I guess that wasn't quick, but at least I didn't have an graphs or

get into transistor curves or Bessell functions.



David



kent_spong wrote:

> It could be done. But not on the cheap.

>

>   

>>   ----- Original Message ----- 

>>   From: cuari7 

>>   To: yamahacs80@yahoogroups.com 

>>   Sent: Monday, November 26, 2007 9:03 PM

>>   Subject: [yamahacs80] Pitch bending on a CS50?

>>

>>

>>   Hi, group,

>>

>>   Is there a way to do pitch-bending on a CS50 (like a pedal, maybe)?




      

    
    

















[Non-text portions of this message have been removed]

  ----- Original Message ----- 
  From: David Rogoff 
  To: yamahacs80@yahoogroups.com 
  Sent: Friday, November 30, 2007 2:44 AM
  Subject: Re: [yamahacs80] Re: Pitch bending on a CS50?


  This touches on a big, somewhat technical, issue of what kind of VCOs
  the CS80 uses. The VCO III chip is a linear VCO, sometimes called
  Hz/Volt, as opposed to the more common exponential (Volts/Octave) VCOs
  (e.g. MiniMoog, Curtis & SSM chips in SCI and Oberheim polys).

  Here's a pretty good explanation:
  http://www.synthmuseum.com/magazine/linexpo.html

  Here's a (I hope) quick one:

  The most basic VCO is a sawtooth one, which can be a capacitor charged
  by a current. For non-EE types, here's my modified toilet analog (and
  you though the Metasonix vacuum-tube VCO was weird) : The capacitor is
  like the water tank of a toilet. The water filling it up is the
  current. The height of the water is like the voltage across the
  capacitor. Now, modify the float valve so that when the tank is full it
  automatically flushes. Then the cycle starts again. If you double the
  water filling rate ( = double the current), you double the frequency of
  the flush cycles.

  The is a basic, linear VCO (actually Water-CO). It shows a couple of
  things. First, it's not actually voltage controlled, but current
  controlled. Ignore that for now. Also, the filling time is adjustable,
  but the discharge/flushing time is fixed. This is an issue with all
  sawtooth VCOs and is why many (e.g. Moog) VCOs have a
  high-frequency-tracking adjustment, which helps cancel this out. Here's
  the CS80 VCO: http://home.debitel.net/user/jhaible/cs80_vco.gif

  Ok, so why don't all synths use linear VCOs? As the above link
  explains, human ears don't hear frequency linearly. A above middle C is
  440Hz. An octave about is 880Hz, or double the frequency. The next
  octave would be 1760Hz: double that. If you graph this, it's an
  exponential curve. So, the space (in Hertz) between two notes keeps
  getting bigger as we get to high pitches. If you had a modular synth
  with linear VCOs (like that old Paia), the top key might output 5
  volts. One octave down would be 2.5volts. The next 1.25volts, followed
  by 0.625v and 0.3125v. This is a pain to generate. Also, as you get to
  lower notes, smaller voltage inaccuracies start becoming bigger pitch
  errors to our ears.

  To avoid all this, someone (anyone know who? Dr. Bob? Tom Oberheim?
  Don Buchla?) came up with exponential VCOs. Basically, they're just a
  linear VCO with a circuit in front of them called (big surprise) an
  exponential converter. This is just a circuit that takes a linear input
  (1volt/octave) and outputs the doubling voltage (actually current...)
  that the VCO wants. Now, everything is simple.

  So, why did Yamaha go for the linear? Two reasons, I'd guess. First,
  adding the exponential converter to each VCO adds more cost to the
  chips, since there's more circuitry. A bigger issue is temperature
  stability. As we've been talking about lately, all circuits are affected
  (i.e. knocked out of tuning) by temperature changes. The exponential
  converter, for reasons I won't go into, is really sensitive to this.
  People have been complaining about the tuning stability of the CS80, but
  it's rock solid compared to any poly-synth with exponential VCOs (P5,
  OBX, A6, etc). They all need computer-controlled auto-tuning routines to
  have any chance of staying in tune.

  So, what issues/problems/advantages does the CS80 having linear VCOs create?

  Good things:
  1) modulation - linear vibrato sounds a bit different than v/oct
  vibrato, probably closer to acoustic vibrato (e.g. violin). Also, as
  the modulation speed increases, you start getting into F.M. land, which
  requires linear modulation (you don't want to know the math!). This is
  why some modular VCOs have linear FM inputs in addition to the normal
  v/oct controls.

  2) sweep to D.C. - my favorite. If you start a pitch bend at the right
  end of the ribbon and slide all the way to the left, the pitch of the
  VCOs all go down to 0Hz / D.C. / flat-line. This is because the input to
  the VCOs goes to 0 volts and the frequency equals the voltage times a
  constant. With a exponential VCO this is impossible. Going 1 volt less
  on the control input goes down one octave. Mathematically, you can't
  get to zero Hz. You'd need to input -infinity volts! Also, many other
  limitations in the circuit block the VCO from even getting close. Big
  win for linear VCOs!

  Bad things:
  1) Keyboard voltages - as I wrote above, the keyboard has to generate
  exponential voltages. This is a big pain. In a digitally-controlled
  analog (like the CS80, P5, etc), the keyboard voltage comes from a DAC
  (digital-analog-converter). 99.99% of DACs are linear. The CS50/60/80
  (and others in the family) have bizarre, custom exponential DACs. This
  makes interfacing the CS80 to other synths and/or MIDI-CV converters a pain.

  2) CV mixing. Finally, we get to the original question of adding a
  pitch-bend input to the CS80. In the volts/octave world, everything is
  easy: you just add voltages together. Adding voltages is simple to do -
  just an op-amp and a few resistors. Let's say you had the following
  voltages come out of a v/oct keyboard: 1v, 2v, 4v. This could represent
  a low C (c1), C one octave up (c2), and C two octave above that (c4). 
  To make it simple, let's say we have a pitch wheel or pedal add 1 volt
  to this (2v, 3v, 5v). This would be c2, c3, c5, so we've just
  transposed the sequence up an octave.

  Ok, what happens if we try this with a linear voltage. For the same c1,
  c2, c4 notes, we might have 1volt, 2volt, 8volt. Adding one volt gives
  2volt, 3volt, 9volt. The first note is correctly up an octave, but the
  next is only up about a 5th and the third note is only transposed up
  about a semitone. This, obviously, doesn't work. What we need to do,
  instead, is multiply the voltages. To transpose up an octave, double the
  voltages. To transpose down an octave, halve them. This is easy for a
  fixed transpose, but if you want a variable, like a pitch-bend pedal
  input, you need to multiply voltages. Just like it's much, much easier
  for people to add and subtract than multiply and divide, so it is for
  analog (and digital) circuitry.

  If you follow the schematics or block diagram of the CS80, you can see
  that the voltage to the VCOs comes through a long chain of
  multiplications. The ribbon is actually the initial voltage source for
  the whole instrument. If the ribbon isn't pressed it outputs some fixed
  voltage (not sure the actual value - call it 2 volts). If the ribbon is
  slid up, all the way, from the left to the right, it would output double
  this voltage, which corresponds to one octave up. If the ribbon is slid
  the other way, it outputs zero volts, as mentioned above. Next, the
  voltage is sent through the concentric pitch knobs. Any normal
  potentiometer is a voltage multiplier, which can multiply the input by
  anything from zero to one.

  This voltage then becomes the reference input to the exponential DAC on
  the KAS board, which multiplies it by it's exponential resistor network
  to create the CVs for each of the either voices. These voltages go to
  the VCO chips on the M-Boards. Are we done - nope - one more CS80
  weirdness. In a v/oct synth, the octave/foot switches would just
  generate a voltage that would be added to the keyboard CV (e.g.
  MiniMoog). The CS80 VCO, instead, has a special footage input that needs
  an exponential current for each feet setting. Because this is difficult
  to do accurately over a wide range, we end up with the wonderful VR4,
  VR5, and VR6 trimmers to get the feet switching calibrated separately
  for each of the 16 VCOs. Yuch!

  Getting back to the original question (remember Alice? There's a song
  about Alice...), a pitch bend input would need to control a voltage
  multiplier. This could be an added circuit, after the ribbon circuit,
  or could probably be merged with the ribbon voltage. I haven't figured
  out the details, but it's not rocket science. However, it is a lot more
  work than it would be on something like a Prophet 5.

  Ok, I guess that wasn't quick, but at least I didn't have an graphs or
  get into transistor curves or Bessell functions.

  David

  kent_spong wrote:
  > It could be done. But not on the cheap.
  >
  > 
  >> ----- Original Message ----- 
  >> From: cuari7 
  >> To: yamahacs80@yahoogroups.com 
  >> Sent: Monday, November 26, 2007 9:03 PM
  >> Subject: [yamahacs80] Pitch bending on a CS50?
  >>
  >>
  >> Hi, group,
  >>
  >> Is there a way to do pitch-bending on a CS50 (like a pedal, maybe)?



   

[Non-text portions of this message have been removed]