Limiting the output voltage of a boost converter

Question

I have a boost converter TPS61288 which can boost the input upto 18v. Adjustment of the output voltage is done via a potentiometer to the feedback of pin of the converter. 18v is dangerous for my system incase the potentiometer gets accidentally set to the wrong resistance. So i was thinking if it is possible to place a hard limit on the pot.

I want to limit the adjustment to 8V only with and input voltage of 3.7, i want to still be able to adjust between 5-7v with the pot. My plan is to place a fixed resistor before, after, or before and after the potentiometer.

^{simulate this circuit – Schematic created using CircuitLab}

According to the datasheet this is how to get the output voltage:

I can only use a 100k potentiometer. So i can maybe have a setup of skip top resistor and R2 = 10k . then my voltage window is 1.63v (pot at 0 position R1 = 0 R2 = 110k) to 6.6V(pot at 100k position R1 = 100k R2 = 10k). This is very crude as i have to brute force trial and error values till i reach 8V

can anyone help me figure out the formula for the fixed resistors to have a limit at 8V (it would be great if the formula can solve limit not just for 8v). In the mean time i guess ill brute force the calculations

Answer 1

You have to write simultaneous equations and solve them

R2min = R2fixed + 0
R2max = R2fixed + 100k
Note that total resistance in the resistor divider is R1fixed + R2fixed + 100k in whichever position of the pot

equation1: voutmax = vrefmax (R1fixed + R2fixed + 100k) / R2min

equation2: voutmin = vrefmin (R1fixed + R2fixed + 100k) / (R2max)

In the above 2 equations, substitute voutmax and voutmin that you want and find the values for R1fixed and R2fixed.