[sdiy] 1v /oct with ADC question

[Vinicius Brazil]

Hi Mike,

At 1volt / oct the step of each note is 1/12 = 0.08333volts.

However, in order for the ADC deviation not to be perceived as detuning,
this deviation has to be less than 1%.

(1/2) / 100 = 0.0008333volts

5 Volts / 0.000833 Volts = 6000 steps, so you need a 13-bit converter.

12 bits is already satisfactory.

regards,
Vinicius Brazil

On Wed, Nov 30, 2016 at 6:41 PM, Gordonjcp <gordonjcp at gjcp.net> wrote:

> On Wed, Nov 30, 2016 at 02:10:30PM -0500, Mike HEQX wrote:
> > Hi folks,
> >
> >  I want to use an ADC to create discreet note values so I am looking
> > at an 8bit ADC.
> >
> > I only need 6 bits to get to 64 values. ( I really only need 61 of
> > those values to make 5 octaves plus 1 )
> >
> > I want to scale the voltage across my pot so that I get 1v / oct
> > output for cv usage at full scale, and I also need to produce 61
> > discreet values from the adc at the same full scale. So it looks
> > like I have to do something mathematical, but I know not what do do.
>
> You need to run your 6-bit DAC off 5.333V because that way you will
> automatically get 1V/octave.
>
> 5.333V / 64 = 0.08333V/semitone
> 0.08333 * 12 = 1V/octave
>
> Build it just like in the DAC section of the TB303.
>
> --
> Gordonjcp MM0YEQ
>
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