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Re: vco octave switches

2004-07-01 by selfoscillate

hello florian,

many thanks for the great explanation :-)

best wishes

ingo



--- In Doepfer_a100@yahoogroups.com, Florian Anwander 
<Florian.Anwander@c...> wrote:
> Hi Ingo
> 
> > is it a simple modification? soldering is not one
> > of my biggest strengths ;-)
> Yes, quite simple.
> 
>  From the center connector of the poti (=slider) a resistor leads 
to the 
> mixing point. The less the resistor the wider is the range of the 
poti.
> 
> As far as I remember the value is 1 Megaohm for the coarse.
> You do not have to remove the existing resistor. Simply add a 
resistor 
> with lower value parallel to the existing resistor.
> 
> I added an 220kOhm resistor for the coarse tune. The value for fine 
tune 
> I do not remember at the moment.
> 
> 
> 
> 
> 
> 
> Further explanations:
> 
> 
> PP=Poti
> Ra=existing Resistor
> Rb=new Resistor
> 
> 
> Before:
> 
> ----PP
>      PP
>      PP<----RaRaRa-----
>      PP
> ----PP
> 
> 
> After:
> 
> ----PP
>      PP
>      PP<--+-RaRaRa-+---
>      PP   |        |
> ----PP   +-RbRbRb-+
> 
> 
> 
> R'a and R'b appear electrical like one Resistor R'eff.
> 
> 
> The new value can be determined by the following formula:
> 
> R'eff = R'a * R'b / (R'a + R'b)
> 
> Where is:
> R'a is the existing resistor
> R'b is the new add on resistor
> R'eff is the resulting value
> 
> 
> 
> Example:
> Old resistor=1Megohm
> New resistor=1Megohm
> 
> 1*1/(1+1) = 1/2 = 0,5
> 
> So the resulting value is 0,5Megaohm. This would double the range 
of the 
> poti.
> 
> 
> Florian

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