Voltage divider ratio limitation applied to potentiometer

Question

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

Answer 1

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.

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Generally you will be constrained by the available potentiometer values. Pick one that suits and calculate the other resistors afterwards.

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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.

^{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.

Answer 2

1) Choose R total for the string.

2) Choose Range difference , thus 70-50% = 20% which is your RV ( variable resistor) 20% of R total

3) Calculate the remaining fixed R's using Min and Total-Max, i.e. 50% and 30% of R total.

^{simulate this circuit – Schematic created using CircuitLab}

Answer 3

Work it backwards....

A simple trick is to figure out the percentage range you want from the pot.

In your example you want 50 to 70% of Vcc or 20% range.

Not pick a pot which is a nice multiple of 20. Perhaps 200R, but lets say 2K.

Now it is a simple matter to deduce that 50% would need 5K and 30% would need 3K

Hey presto.

^{simulate this circuit – Schematic created using CircuitLab}

But say you wanted 60 to 75%, 15% range

Pick a 1.5K pot, and the bottom resistor R2 is now 6K and R1 is 2.5K.

^{simulate this circuit}

Of course, you can not always get pots of the right value so you may need to scale it to a different total value.

Answer 4

^{simulate this circuit – Schematic created using CircuitLab}

Consider the circuit above. The pot can be adjusted from 0 to 100% so first label the node between R1 and R2 with the higher percentage voltage, 70% of 5V is 3.5V.

Now label the node between R2 and R3 with the lower percentage voltage, 50% of 5V is 2.5V.

Now since you know the value of your pot and the voltage across it you calculate the current in all 3 resistors using ohms law. Since you know the voltage across R1 and R3 together with the current we can calcu1ate the value of R1 and R3 again using ohms law.

Simple.

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My answer above assumes there is no current being taken from the wiper of the pot as do all the other answers posted here so far which may be a reasonable assumption but may not be the case if the current taken from Vout is not insignifiant.

So lets add a load resistor RL between Vout and 0V

We now have to solve two simultaneous equations assume you know RL and R2

$$\begin{align}\\ BottomPercent & = 100 \times \dfrac{R3 || RL}{R1 + R2 + R3 || RL} \\ & = 100 \times \dfrac{R3 \cdot RL}{(R1 + R2) \cdot (R3+RL) + R3 \cdot RL} \end{align} $$

$$\begin{align}\\ TopPercent & = 100 \times \dfrac{(R2+R3) || RL}{R1 + (R2 + R3) || RL} \\ & = 100 \times \dfrac{(R2 + R3) \cdot RL}{R1 \cdot (R2+R3+RL) + (R2+R3) \cdot RL} \end{align} $$

Note if the output is loaded then output will not adjust in a linear fashion.