Author Topic: Colorsound Octivider questions  (Read 443 times)

11-90-an

Colorsound Octivider questions
« on: August 20, 2020, 09:20:15 AM »
For the moment I’m obsessing on the colorsound octivider, which is a variant/clone of the shin ei ob-28 octave box... (when I first heard the name “shin-ei octave box” i wasn’t interested, but when I heard “colorsound octivider”, i was immediately interested, but i digress)

Here is a schem by @Rob Strand:



I get that the first 3 trannie stages, C7, R16, C8, R17, C9, R18, Q5 and other associated components are there to filter out the high frequencies (thanks @Rob for the explaination), but I’m still at a loss for the stuff after that...



How exactly does the red part work? I don’t really understand why Q11 should be there... it seems to be feeding some un-octaved signal in... Is that the case?

I’ll guess that the diode is for “one-way isolation”...

So I redrew the schem with a adjustable “clean signal” mod... would it work or potentially short any dc to ground and magic smoke galore? :icon_eek:




flip flop flip flop flip

duck_arse

Re: Colorsound Octivider questions
« Reply #1 on: August 20, 2020, 11:45:34 AM »
the flipping flopping transistors Q8 and Q9 divide whatever filtered audio is fed them, by 2. so the diode D1 is turned on every second half wave, and shunts the [buffered?] signal from Q11 away from Q10, providing the bass signal.

I tried isolated parts of this circuit on the breader, but didn't get usable results w/ the dioding bit, probably because I'd left out too much of the filterings and amplifierings section.

I don't understand the Q11 wiring, it looks short a resistor to me.
Now battery powered. Remove plug when not in use, please.

antonis

Re: Colorsound Octivider questions
« Reply #2 on: August 20, 2020, 11:53:17 AM »
C5 is effectively set in parallel with R20+D1, so the higher the signal frequency the more the signal leaking to GND instead of going into flip-flop configuration..

Or maybe didn't get you, Stephen..
"I'm getting older while being taught all the time" Solon the Athenian..
"I don't mind  being taught all the time but I do mind a lot getting old" Antonis the Thessalonian..

duck_arse

Re: Colorsound Octivider questions
« Reply #3 on: August 20, 2020, 12:15:48 PM »
mmm. more likely I didn't get the circuit.
Now battery powered. Remove plug when not in use, please.

11-90-an

Re: Colorsound Octivider questions
« Reply #4 on: August 20, 2020, 12:56:37 PM »
C5 is effectively set in parallel with R20+D1, so the higher the signal frequency the more the signal leaking to GND instead of going into flip-flop configuration..

Or maybe didn't get you, Stephen..

So the volume is essntially lower when playing high notes?  :icon_neutral:

the flipping flopping transistors Q8 and Q9 divide whatever filtered audio is fed them, by 2. so the diode D1 is turned on every second half wave, and shunts the [buffered?] signal from Q11 away from Q10, providing the bass signal.

I don't understand the Q11 wiring, it looks short a resistor to me.

Yea I get the flip flop... so what you’re saying is that it essentially “mutes” the signal every zero-crossing? (Or something like that)...

mmm. more likely I didn't get the circuit.

I”m not alone, then... :icon_biggrin: :icon_lol: :icon_mrgreen:
flip flop flip flop flip

PRR

Re: Colorsound Octivider questions
« Reply #5 on: August 20, 2020, 08:04:17 PM »
I”m not alone, then... :icon_biggrin: :icon_lol: :icon_mrgreen:

Group hug:






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Rob Strand

Re: Colorsound Octivider questions
« Reply #6 on: August 20, 2020, 08:15:13 PM »
Quote
How exactly does the red part work? I don’t really understand why Q11 should be there... it seems to be feeding some un-octaved signal in... Is that the case?
The intent of the Q11 circuit is that it sets up a voltage on C5 which represents the signal level or presence.   When the flip-flop output is high the diode D1 has no effect and C5 voltage is  passed to the Q10 stage via R20.     When the flip-flop is low the diode D1 pulls the voltage on R20 low and produces a low input voltage to the Q10 stage.   So the output of Q10 is a "square-wave" at the octave below the fundamental but the level of the square-wave is the voltage on C5.

What's not so clear by eye is if the voltage on C5 represents signal level or signal presence.    When the diode pulls the signal low C5 is discharged via R20.   The time constant is about 10ms so the time is in the same order as the signal period.   When the signal is present that kind of resets the C5 voltage level nearly on cycle by cycle basis.     You would need to simulate the circuit to see what happens in the tail of the signal.   At some point the signal will be too low for the Schmitt trigger to produce a trigger and the octave will stop due to the signal level.    The simulation will show the C5 cap voltage at the point the trigger is lost.
« Last Edit: August 20, 2020, 08:21:33 PM by Rob Strand »
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