Get an analog voltage source (not a digial one) and monitor it with a digital voltmeter, to at least hundreths precision.
Set your Erica to quantize and feed that voltage to it, from 0v to 5v.
Observe the voltage when the oscillator jumps to the next half step.
Is it when the input is just past ~((1volt / 12.0) * note), or just before?
Or does it flip when the to-be-quantized voltage is past half way between?
Round up or round down or mid-nearest. Or maybe some internal rules table or something.
I couldn't find the quantizer's rules in the vco manual, it just says how to turn it on and set it.Anyway, now read the voltages from your digital sequencer as it runs a chromatic scale.
Are they each a 'precise' 0.0833333333333333333333 volts apart from each other?
Or 0.0830, or 0.0820, or 0.085?
Analog. Squishy.
There is an implied imprecision down in the 0.003333333s, especially when you're adding them together.
And so there is no "C", or 'in tune', just 'sounds in tune', even when it never really can be.
An analog quantizer can only approximate what your intent for 'C' is, and they all can do it differently.
It's a best guess, because your sequencer isn't sending 'A', it's generating ~0.8333333 volts. Give or take.Now, knowing the vco's quantizing behaviour, you can see why it is doing what it does and when.
I still think your oscillator is working correctly.Don't quantize already quantized CVs. You'll get stray semitone errors.
I never like it when someone asks for help, and the annoying reply is 'Why are you doing that'?
But, why are you doing that?-- noodle_hut
Hello noodle_hut.
I have just measured ( for the last time ) the voltage coming from my sequencer for the first step ( the C ) and its 0.000V
So a precise voltage. I can send you a picture if you do not believe me. Its about as perfect a voltage as you could hope for.
The GVCO plays C when its given this voltage as I would expect it to. If I turn on the intermal quantizer the GVCO produces C sharp. I don't know why you are insisting that this is anything other than wrong.
I don't think we are getting anywhere with this so probably best I stop asking for help now.
Once again, thanks for your input.