Yes the drawing is wrong. I think you understand the pot is from FEED BACK to ground.
The supply can not output more than about 18 volts because the input is 19V. Even when the pot=0 ohms the output can not get above 18.x volts. Under heavy load it probably can't make it to 18 any way. If you want to stop at 18V then in software limit the pot to some minimum value or add a series resistor.
R27 and (R18+POT) makes a voltage divider for 3V
R27 and R18 makes a voltage divider for 18V.
View attachment 84030
Hi Ron,
Yes that might work, i found out the 10k pot version has a max current of 5ma too at the top end only, but that might be good enough.
Two equations that describe what we need are:
3=(1.25*R27)/(R18+Rp)+1.25 [eq1]
18=(1.25*R27)/R18+1.25 [eq2]
Solving these two simultaneously for R18 and R27 in terms of the pot total resistance Rp we get:
R27=(469*Rp)/300
R18=(7*Rp)/60
Now it is just a matter of selecting an available pot value and testing to see if it meets the specs.
Using a pot value of 1k we get values for R27 and R18 that lead to a maximum current through the pot wiper to B (or A) of around 10ma, which is too high. However, using a pot value of 10k we get:
R27=46900/3
R18=3500/3
or in decimal:
R27=15633.3333
R18=1166.6666
Checking the max current it comes out to around 1.07ma when the output is 18v, so i think this will work without any additional active elements.
LATER:
Usually we can not get the exact component values we want, and often the IC spec's are not exact either. So to insure we can actually get the range 3 to 18v we might want to solve for something like 2.9v to 18.3v and that should give us at least 3.0 to 18.0v in the real life circuit.
If dv1 is the voltage decrease from the 3v spec and dv2 is the increase in voltage of the 18v spec then we have solutions:
R27=((7-4*dv1)*(4*dv2+67)*Rp)/(20*(dv2+dv1+15))
R18=((7-4*dv1)*Rp)/(4*(dv2+dv1+15))
Setting dv1=0.1 and dv2=0.3 we get resistor solutions:
R27=(1023*Rp)/700
R18=(3*Rp)/28
or in decimal with again Rp=10000:
R27=14614.2857
R18=1071.42857
The max current is then about 1.167ma which is still ok.
Finally, given any two voltage set points VL is the lower (like 3v) and VH is the higher (like 18v) the solutions for the two resistor values are:
R27=(Rp*(5-4*VH)*(4*VL-5))/(20*(VL-VH))
R18=(Rp*(5-4*VL))/(4*(VL-VH))
This allows finding resistor values for any two set point voltages. Note we get the same resistor values as previously if we set VL=2.9 and VH=18.3.
There may be other restrictions so you'll have to test after finding the two resistor values for any given circuit.