How can I replace a potentiometer with a fixed resistor on a buck converter?

Question

I want to use a buck converter which steps down 12 V DC to 5 V constant DV output to power my custom PCB.

I came across a buck converter in market which can do it but there is potentiometer on it which can be adjusted to vary the output voltage.

Below you can see the schematic for the buck converter and where I plan to remove the potentiometer and put fixed resistors to get a 5 V constant output.

Can someone help me in how to put fixed resistors in place of potentiometer and how to calculate the value of resistor?

MP2307 Data sheet

Answer 1

What matters is the internal reference voltage inside the circuit. The data-sheet states a reference voltage of 0.925 V. Therefore, the resistive divider made of your potentiometer and the low-side resistance \$R_2\$ must be sized so that, in regulation, the voltage at pin 5 is 0.925 V.

You can either disconnect your potentiometer once you have the correct voltage and measure its resistance or calculate the resistance value with the below expression:

This is an approximate expression neglecting the bias current contribution from the feedback pin but if the current in the divider is strong enough, it is usually not an issue. The calculation is also given in the data-sheet, page 7.

Answer 2

$$V_{FB} = V_{OUT}\ \frac {R_2}{R_1 + R_2} $$

Which means that:

$$R_1 = \left( \frac {V_{OUT}}{V_{FB}} - 1\right) \ R_2$$

So using the 8.2kΩ on the board with 5V output, you need to replace the potentiometer with 36.1kΩ (standard 2%) resistor.

Less parts to replace.