I'm trying to find the general formula or more like understanding of how to choose resistor values to limit the voltage ratio when using a potentiometer.
A potentiometer is equal to a voltage resistor divider where the ratio depends on the resistor value. Okay, so you can tune from 0% to 100% which means for a Vin of 5V for example, from 0 to 5V.
Now let's say I want to limit this ratio to a certain value as from 50% to 70% to obtain a Vout ranged from 2.5V to 3.5V. So then, I'm sure when I will change the potentiometer value the voltage will be limited between this voltage range.
I understand by adding a grounded series resistor the ratio will then change to constrained value since when the potentiometer will be at min/max value, there will still be this resistor to take into account:
Which will give a new ratio of 50% to 100% (so if VCC = 5V, 2.5V to 5V) due to the 2k series resistance added. (2k/4k for min potentiometer state and 4k/4k for max potentiometer state)
But I can't get my head around how to now get arbitrary ratio as 50% to 70%.
What I am not seening ?
Thank you
Best Answer
If your 2k pot is to give you a span of 20% from 50% to 70% then you need \$ \frac {2k}{20} = 100 \ \Omega \$ per %.
So the bottom, 50%, resistor will be 50 x 100 = 5k.
The top, 30%, resistor will be 30 x 100 = 3k.
Generally you will be constrained by the available potentiometer values. Pick one that suits and calculate the other resistors afterwards.
simulate this circuit – Schematic created using CircuitLab
Using min and max in percentage form:
$$ R_1 = \frac {R_2 \cdot (100 - max)}{max - min} $$
$$ R_3 = \frac {R_2 \cdot min}{max - min} $$
Testing for your example with a 2k pot and a 50% to 70% adjustment range:
\$ R_1 = \frac {R_2 \cdot (100 - max)}{max - min} = \frac {2k \cdot (100 - 70)}{70 - 50} = 3k \$
\$ R_3 = \frac {R_2 \cdot min}{max - min}= \frac {2k \cdot 50}{70 - 50} = 5k \$
Obviously, min can be set as low as 0% and max to 100% which will result in R3 or R1 being 0 Ω.