<html><head></head><body><div>Vinicius,<br>Thanks. I'm not using the adc for anything other than addressing. The binary is almost irrelevant except that i need at least 61 steps to be generated by the voltage that will be present from one side of the pot to the other. <br><br></div>
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<div class="gmail_quote" >On Nov 30, 2016, at 4:15PM, Vinicius Brazil <<a href="mailto:brazil.v@gmail.com" target="_blank">brazil.v@gmail.com</a>> wrote:<blockquote class="gmail_quote" style="margin: 0pt 0pt 0pt 0.8ex; border-left: 1px solid rgb(204, 204, 204); padding-left: 1ex;">
<div dir="ltr">Hi Mike,<div><br></div><div><div>At 1volt / oct the step of each note is 1/12 = 0.08333volts.</div><div><br></div><div>However, in order for the ADC deviation not to be perceived as detuning, this deviation has to be less than 1%.</div><div><br></div><div>(1/2) / 100 = 0.0008333volts</div><div><br></div><div>5 Volts / 0.000833 Volts = 6000 steps, so you need a 13-bit converter.</div><div><br></div><div>12 bits is already satisfactory.</div></div><div><br></div><div>regards,</div><div>Vinicius Brazil</div></div><div class="gmail_extra"><br><div class="gmail_quote">On Wed, Nov 30, 2016 at 6:41 PM, Gordonjcp <span dir="ltr"><<a href="mailto:gordonjcp@gjcp.net" target="_blank">gordonjcp@gjcp.net</a>></span> wrote:<br><blockquote class="gmail_quote" style="margin:0 0 0 .8ex;border-left:1px #ccc solid;padding-left:1ex"><span class="">On Wed, Nov 30, 2016 at 02:10:30PM -0500, Mike HEQX wrote:<br>
> Hi folks,<br>
><br>
> I want to use an ADC to create discreet note values so I am looking<br>
> at an 8bit ADC.<br>
><br>
> I only need 6 bits to get to 64 values. ( I really only need 61 of<br>
> those values to make 5 octaves plus 1 )<br>
><br>
> I want to scale the voltage across my pot so that I get 1v / oct<br>
> output for cv usage at full scale, and I also need to produce 61<br>
> discreet values from the adc at the same full scale. So it looks<br>
> like I have to do something mathematical, but I know not what do do.<br>
<br>
</span>You need to run your 6-bit DAC off 5.333V because that way you will automatically get 1V/octave.<br>
<br>
5.333V / 64 = 0.08333V/semitone<br>
0.08333 * 12 = 1V/octave<br>
<br>
Build it just like in the DAC section of the TB303.<br>
<span class="HOEnZb"><font color="#888888"><br>
--<br>
Gordonjcp MM0YEQ<br>
</font></span><div class="HOEnZb"><div class="h5"><br>
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