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
Message
Re: vco octave switches
2004-07-01 by selfoscillate
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