Fast analog sw. as VC resistor

Christopher_List at Sonymusic.Com Christopher_List at Sonymusic.Com
Thu Jan 9 16:16:19 CET 1997

>> It also occurred to me that you could get a log pot reponse (good for
>> ADSR) without too much trouble. A log pot is, afterall just two linear
>> sections. So, you attenuate your sawtooth, then run it through an
>> with some zener diodes that will amplify it by 1 when it's below the
>> threshold, and by 10 when it's above. Set the threshold right in the
>> of the wave and presto - you've got a log pot reference!

> Wow. How do you do this?
> Put the zener diodes in the feedback loop?
> I haven't heard of that. Sounds great.

OK, I haven't experimented with this, so this is pure speculation....

A log pot covers 15% of it's resistence in the first 50% of it's travel,
and 85% in the last 50%.

This means that if you've got a 0 to +10v sawtooth, you want to convert it
into a 0 to 1.5v sawtooth for the first half of the cycle and 1.5v to 10v
for the second half. You should be able to do this with a single follower -
like so;

Use a 27K/3K divider to change the 0 to 10v wave to 0 to 1.2v

Run it into a non inverting op-amp with the usual voltage divider beween
the output and the (-) pin. Make the voltage divider 10K/10K, so the output
will be double the input (Vout = Vin * (1+ Rfb/Rin)). Now, when the wave is
at 50%, the output is 1.2v, and the voltage at the middle of the divider is
.6v - the diode threshold voltage! So we throw a diode and a 1.6K resistor
in parallel with the 10K resistor on the ground side of the feedback votage
divider. This means that when the output voltage gets higher than 1.2v (Ok,
so it's not 1.5 volts!) it will start getting amplified by (1+10K/1.38K =
8.25)... so that max input voltage of 1.2v will come out at 9.9v!

Sounds good on paper - but I'm a lowly hobbiest.

- CList

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