[sdiy] Re: Linear detuning !

[Simon Brouwer]

Hi Karl,

At 02:31 3-8-2005, you wrote:
>Hi simon
>--- simon.oo.o at xs4all.nl skrev:
>
> > Hi Karl,
> >
> > karl dalen zei:
> > > Here we go.
> > >
> > > Now, if  we add a tiny DC voltage to Iref
> > > summing point we offset the VCO what if we
> > > use the Expo CV voltage as the offset voltage?
> > >
> > > Constant multiplication of a varying linear voltage! Ugh!
> >
> > Linear detuning means that the frequency is increased by a constant
> > amount, for example, instead of 100 Hz or 5000 Hz you want 100.1 or 5000.1
> > Hz.
>
>Yes i know, i was just one of my "fun" ideas!

It would be fun if it worked...

>Lets expand the wicked idea, we have two inputs to
>the NPN expo, B1 and B2, apply positive voltage V1
>to B2 and frequency increases apply V1 voltage
>to B1 and frequncy decreases. If we match the input
>resistors R1,R2 but let one of the divider resistors
>become a pot, we can under compensate the expo.
>
>As the V1 increases on B2 the voltage on B1 also increases
>wich counteract the V1 voltage on B2 but if we make this "same"
>voltage a fraction of V1 on B1 , a varying constant of the beat
>over octaves then beat would stay constant!
>(or close to).
>
>However impedance on B1 might be wicked but not as a
>sensitive matter as on B2.
>
>The idea are, create a "scale voltage that equals the beat you want!
>Ie, 1.000V scale on B2, and 0,002V scale on B1 automatically
>logs to constant beat over octaves if trimmed carefully!

So let me get this straight, instead of connecting the second base of the 
dual transistor pair to ground, you want to apply a fraction of the control 
voltage to it.

Your example would result in a scale of 0.998 V/octave. Relative to 
1V/octave you can have zero beating at one frequency, then one octave up 
you have a frequency difference of 0.002 octave.

If the frequency difference is zero at one frequency and nonzero at 
another, then it is obviously not constant...


Vriendelijke groet,
Simon Brouwer.

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