Voltage divider formula

Question

How to find value for both R1 and R2 for Vout=1.23 volts, where Pot=5K, Vin=4.5 - 12 volts?

^{simulate this circuit – Schematic created using CircuitLab}

In reality, I want to adjust lm2596 buck converter output to 4.5 volts to 12 volts with any potentiometer, by adjusting R1 and R2 value.

For long time, I was using the trial and error method with a calculator to solve that problem :). Hope that someone have easier method to solve it.

Answer 1

You can use the straightforward method described in the other answer.

Another approach is to observe that the total resistance of the divider is constant (ignoring the regulator bias current) and therefore the current will be proportional to the regulator output voltage, so we can say that:

\$ \frac{V_{REF}}{R_2} = \frac{12}{4.5}\cdot\frac{V_{REF}}{R_2 + R_{POT}}\$

so

\$R_2= \frac{5\text k \Omega}{(12V/4.5V) -1} \$

whereupon you can easily solve for R1

If this is a real project, I would suggest using an additional parallel resistor to deal with the (usually crummy) pot element tolerance and taking resistor and Vref tolerances into account, as well as selecting standard E96 (1%) or whatever values.

Answer 2

You have two unknowns so you need two equations.

With 12V the wiper of the pot will be at the bottom. That means the voltage divider will be R1 + Rpot above the output and R2 below.

^{simulate this circuit – Schematic created using CircuitLab}

With 4.5V the wiper will be at the top of the pot.

^{simulate this circuit}

No trial and error is required. With the pot at one extreme you'll get 12V as shown in the top schematic. If you understand how to work with voltage dividers, then you can write down an equation for that condition: $$1.23{\rm\,V} = 12{\rm\,V} \frac{R_2}{R_1+R_{pot}+R_2}$$

With the pot at the other extreme you'll get 4.5V as shown in the bottom schematic. The corresponding equation is: $$1.23{\rm\,V} = 4.5{\rm\,V} \frac{R_2+R_{pot}}{R_1+R_{pot}+R_2}$$

You know the value of Rpot. (You wrote 5k in your question, but it could be anything.) That means you have two equations with two unknowns: R1 and R2.

Now you need to solve the simultaneous equations, which I'm not going to do for you and is beyond the scope of this question.