Well, for anyone still paying attention, here is my take on how this Negative Slew circuit works.
First I want to say just how in awe I am of anyone who has such an intuitive feel for electronics that they can see and design such a circuit by apparently 'seeing the whole' so easily (or as the joke goes, Serge has probably _forgotten_ more electronics knowledge than I will ever know!). I can only dream of arriving at such a point.
So, first the easier stuff, and some notation. We have established that transistor 9/10/11 is actually used as a 7V zener, henceforth I shall simply refer to it as 'the zener'. The other transistors I shall call Qx, where 'x' is the pin of their base, so e.g. transistor 3/4/5 is Q4. Amplifier 10/11/12 is a simple inverting set-up, sending (approx) 0 to 4.2V at its input (the pin 9/330k), to 6V down to 0 at its output (the 4.2 is from the slight gain of 470/330, i.e. 6*33/47). Amplifier 4/3/2 is a simple adder/subtractor, providing a current into pin 13 of the main integrator (and due to the fact that there are diodes at pins 3 and 13, separated by only the 18k, I suspect this current has a very 'tanh' shape as the rate pot is swept, and the initial upswing of this curve is fairly close to being exponential): for the purposes of this description, we assume the current into pin 13 is kept constant. Q12 sets a quiescent current of about (6-0.7)/4k7 = 1.1mA being pulled from differential pair Q2/Q4. Thus the quiescent (0V at both Q2, Q4 bases) voltage at Q4 collector is 12-10k*1.1m/2 = 6.5V - in the quiescent state this is not enough to turn Q6 on.
Initially assume that both input and output ('3401 pin 10) are at 0V: because Q2 and Q4 bases are both at 0V, they share the 1.1mA equally, and so Q6 is not on; thus current into pin 13 keeps the 330n fully charged at 6V, and this is inverted to 0V at the output. Now apply a positive-going step-change at the input: this applies a differential voltage across the 18k between Q2 and Q4, turning Q2 on more, and Q4 more off; Q4 more off allows its collector to rise a little, which in turn means the zener can conduct, and this allows Q6 to turn on. Q6 on means the 330n can start to discharge, and since this is then inverted by the output amp, the output *rises*. However the output rising tends to turn Q4 on more, which in turn tends to make Q6 turn off - i.e. we have negative feedback, the end result of which is that the output will *follow* the input as it rises. If the input stays quite high, then the 330n is held at quite a low voltage.
Now apply a steep, negative-going step at the input that we expect to get slew-limited. The input drops to 0V quickly, but the output stays high: this puts a reverse differential voltage across the 18k between Q2 and Q4, which means Q4 tends to turn hard on (Q2 more off); this in turn will mean the zener is not conducting, and so Q6 is definitely off; thus the best the 330n can do is a slow charge according to the current input at pin 13 (and mirrored at pin 8), so the pin 9 output climbs linearly, and this is then inverted to give a linearly falling voltage at the output, pin 10. This is the negative slew action. If nothing changes, eventually pin 9 will max-out just below the 6V rail, and pin 10 will be just above ground.
If the negative step is not so steep as to cause the output to be slew-limited, we get the negative feedback action again (to me this seems the hardest case to explain...): at the start the voltage on the 330n is low, and the current into pin 13 (mirrored at pin 8) wants to make pin 9 climb, and hence pin 10 fall; however if pin 10 falls too much, this puts a differential across the 18k between Q2 and Q4 which tends to turn Q2 on and Q4 off; Q4 turning off allows its collector to rise, and lets the zener conduct, so Q6 is able to turn on; this effectively 'bleeds off' excess current through the 330n cap, and hence slows the rise of pin 9, in turn slowing the fall of pin 10, and again, the output (pin 10) will be forced to follow the input!
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Some months ago I came across the VCS, and since then I have fixed a Bananalogue one myself, and helped others fix theirs online - I have seen several comments to the effect that 'it is a simple enough looking circuit: slap your components in the PCB, tell us what is/isn't happening, and we'll sort the problems out, easy!'. The VCS also has a very contorted feedback loop (in fact it is probably harder to understand), which allows both the voltage to be followed, or become slew-limited when circumstances dictate. It may be that I am a bit slow on the uptake, but in my view, these circuits are emphatically NOT 'simple' circuits, but stand amongst some of the more complex synth circuits I've yet come across (I've barely looked at the positive slew, which looks equally complex!). As I intimated at the beginning - awesome!
Tim
__________________________________________________________
Tim Stinchcombe
Cheltenham, Glos, UK
www.timstinchcombe.co.uk
First I want to say just how in awe I am of anyone who has such an intuitive feel for electronics that they can see and design such a circuit by apparently 'seeing the whole' so easily (or as the joke goes, Serge has probably _forgotten_ more electronics knowledge than I will ever know!). I can only dream of arriving at such a point.
So, first the easier stuff, and some notation. We have established that transistor 9/10/11 is actually used as a 7V zener, henceforth I shall simply refer to it as 'the zener'. The other transistors I shall call Qx, where 'x' is the pin of their base, so e.g. transistor 3/4/5 is Q4. Amplifier 10/11/12 is a simple inverting set-up, sending (approx) 0 to 4.2V at its input (the pin 9/330k), to 6V down to 0 at its output (the 4.2 is from the slight gain of 470/330, i.e. 6*33/47). Amplifier 4/3/2 is a simple adder/subtractor, providing a current into pin 13 of the main integrator (and due to the fact that there are diodes at pins 3 and 13, separated by only the 18k, I suspect this current has a very 'tanh' shape as the rate pot is swept, and the initial upswing of this curve is fairly close to being exponential): for the purposes of this description, we assume the current into pin 13 is kept constant. Q12 sets a quiescent current of about (6-0.7)/4k7 = 1.1mA being pulled from differential pair Q2/Q4. Thus the quiescent (0V at both Q2, Q4 bases) voltage at Q4 collector is 12-10k*1.1m/2 = 6.5V - in the quiescent state this is not enough to turn Q6 on.
Initially assume that both input and output ('3401 pin 10) are at 0V: because Q2 and Q4 bases are both at 0V, they share the 1.1mA equally, and so Q6 is not on; thus current into pin 13 keeps the 330n fully charged at 6V, and this is inverted to 0V at the output. Now apply a positive-going step-change at the input: this applies a differential voltage across the 18k between Q2 and Q4, turning Q2 on more, and Q4 more off; Q4 more off allows its collector to rise a little, which in turn means the zener can conduct, and this allows Q6 to turn on. Q6 on means the 330n can start to discharge, and since this is then inverted by the output amp, the output *rises*. However the output rising tends to turn Q4 on more, which in turn tends to make Q6 turn off - i.e. we have negative feedback, the end result of which is that the output will *follow* the input as it rises. If the input stays quite high, then the 330n is held at quite a low voltage.
Now apply a steep, negative-going step at the input that we expect to get slew-limited. The input drops to 0V quickly, but the output stays high: this puts a reverse differential voltage across the 18k between Q2 and Q4, which means Q4 tends to turn hard on (Q2 more off); this in turn will mean the zener is not conducting, and so Q6 is definitely off; thus the best the 330n can do is a slow charge according to the current input at pin 13 (and mirrored at pin 8), so the pin 9 output climbs linearly, and this is then inverted to give a linearly falling voltage at the output, pin 10. This is the negative slew action. If nothing changes, eventually pin 9 will max-out just below the 6V rail, and pin 10 will be just above ground.
If the negative step is not so steep as to cause the output to be slew-limited, we get the negative feedback action again (to me this seems the hardest case to explain...): at the start the voltage on the 330n is low, and the current into pin 13 (mirrored at pin 8) wants to make pin 9 climb, and hence pin 10 fall; however if pin 10 falls too much, this puts a differential across the 18k between Q2 and Q4 which tends to turn Q2 on and Q4 off; Q4 turning off allows its collector to rise, and lets the zener conduct, so Q6 is able to turn on; this effectively 'bleeds off' excess current through the 330n cap, and hence slows the rise of pin 9, in turn slowing the fall of pin 10, and again, the output (pin 10) will be forced to follow the input!
----------
Some months ago I came across the VCS, and since then I have fixed a Bananalogue one myself, and helped others fix theirs online - I have seen several comments to the effect that 'it is a simple enough looking circuit: slap your components in the PCB, tell us what is/isn't happening, and we'll sort the problems out, easy!'. The VCS also has a very contorted feedback loop (in fact it is probably harder to understand), which allows both the voltage to be followed, or become slew-limited when circumstances dictate. It may be that I am a bit slow on the uptake, but in my view, these circuits are emphatically NOT 'simple' circuits, but stand amongst some of the more complex synth circuits I've yet come across (I've barely looked at the positive slew, which looks equally complex!). As I intimated at the beginning - awesome!
Tim
__________________________________________________________
Tim Stinchcombe
Cheltenham, Glos, UK
www.timstinchcombe.co.uk