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R1, R2, and R3 are as shown on the Schematic
This program solves two simultaneous equations as shown below to select values for R1 and R2 so the Max 1674/75 output voltage will vary from Vmin to Vmax when the dimmer pot moves through it's full range from R3 to 0.
If R3 is entered in kohms, then R1 and R2 will be in kohms.
For example: Vmin=2.4, Vmax=3.2, R3=9.5, gives R1=19.09, R2=13.06
From the Max 1674/75 data sheets:
V = k(R1+R2)/(R2) and k=1.3
let R2 = R2 + R3
V = k(R1+R2+R3)/(R2+R3)
Write two simultaneous equations and solve for R1 and R2. This calculation uses R3=10k ohms.
Vmin = k(R1+R2+10)/(R2+10)
Vmax = k(R1+R2+0)/(R2+0)
Vmin/k = (R1+R2+10)/(R2+10)
Vmax/k = (R1+R2+0)/(R2+0)
(Vmin/k)(R2+10) = (R1+R2+10)
(Vmax/k)(R2+0) = (R1+R2+0)
(Vmin/k)(R2)+(Vmin/k)(10)-R2-10 = R1
(Vmax/k)(R2)-R2 = R1
(Vmin/k)(R2)-R2+(Vmin/k)(10)-10 = R1
(Vmax/k)(R2)-R2 = R1
((Vmin/k)-1)(R2)+((Vmin/k)-1)(10) = R1
((Vmax/k)-1)(R2) = R1
( (Vmin/k)-1 )(R2)+( (Vmin/k)-1 )(10) =
( (Vmax/k)-1 )(R2)
( (Vmin/k)-1 )(10) =
((
Vmax/k)-1 )(R2) - ( (Vmin/k)-1 )(R2)
( (Vmin/k)-1 )(10) = (R2)( ((Vmax/k)-1 ) - ((Vmin/k)-1) )
R2 = ((Vmin/k)-1)(10) /( ((Vmax/k)-1) - ((Vmin/k)-1) )
then from above
R1 = ((Vmax/k)-1)(R2)