Electronic – Voltage divider ratio limitation applied to potentiometer

analogcircuit analysispotentiometervoltagevoltage divider

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.

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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:

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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.


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.

schematic

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.