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I know how to calculate the resistance values for a voltage divider without a load on Vout.

I would like to know how I would like to calculate the resistance values of a pot (Example 500kΩ pot) used as voltage divider with a load on it.

With No Load

$$ Vo = V_i*\frac{R_2}{R_1+R_2} $$ Is this the correct equation to use? With a load across Vout and 0V $$ Vo = V_i*\frac{\frac{R_2*R_L}{R_2+R_L}}{R_1+\frac{R_2*R_L}{R_2+R_L}} $$

If $$ V_o=10V $$ $$ V_i=12.2V $$ $$ R_L=150,000Ω $$ $$ R_1+\frac{R_2*R_L}{R_2+R_L}=500,000Ω $$

Then $$ 10 = 12.2*\frac{\frac{R_2*150,000}{R_2+150,000}}{R_1+\frac{R_2*150,000}{R_2+150,000}} $$

I input all of this in to Wolfram Alpha here and I get $$R_1=90163.9$$$$ R_2= -236593$$

How can $$R_2$$ be negative? Does that mean that it is not possible to get my desired out put with a 500kΩ pot?

ETA: $$ 10 = 12.2*\frac{\frac{R_2*150,000}{R_2+150,000}}{500000} $$ $$ R_2 = -236593.059936909 $$

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ For a next time: Altough the circuit is simple, a schematic would be nice to help understanding your question, using appropriate units even more. \$\endgroup\$ – bunto1 Sep 8 '17 at 13:13
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    \$\begingroup\$ @bunto1: Added that information to OP \$\endgroup\$ – TheColonel26 Sep 8 '17 at 13:51
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This is your combined potentiometer+resistor schematic:

schematic

simulate this circuit – Schematic created using CircuitLab

Here above, we know that \$R_{1_X}+R_{1_Y} = R_1\$. Let's define a new variable, \$0 \le x\le 1\$, that determines the percent of rotation of the potentiometer, such that \$R_{1_X} = x R_1\$ and \$R_{1_Y} = \left(1-x\right) R_1\$.

Suppose there were three resistors meeting in the middle, each with a different voltage source at the other end, then the equation for the voltage in the central node would be:

$$\begin{align*} V &= \frac{V_1 R_2 R_3+V_2 R_1 R_3+V_3 R_1 R_2}{R_1 R_2+R_1 R_3+R_2 R_3} \end{align*}$$

From the above, plugging in the voltage nodes from the schematic above, the combination of these three resistors yield:

$$\begin{align*} V_C &= \frac{V_B R_{1_X} R_2+V_AR_{1_Y}R_2+V_BR_{1_X}R_{1_Y}}{R_{1_X} R_2+R_{1_Y}R_2+R_{1_X}R_{1_Y}}\\\\ &= \frac{V_B x R_1 R_2+V_A \left(1-x\right) R_1 R_2+V_B x R_1 \left(1-x\right) R_1}{x R_1 R_2+\left(1-x\right) R_1 R_2+x R_1 \left(1-x\right) R_1}\\\\ &= \frac{R_1 V_B\left(x-x^2\right)+R_2\left(\left[1-x\right]V_A+x V_B\right)}{R_1\left(x-x^2\right)+R_2}\\\\ R_{A-B} &= x R_1 + \frac{\left[1-x\right]R_1 R_2}{\left[1-x\right]R_1 + R_2} \end{align*}$$

Above, \$R_{A-B}\$ is the resistance from terminal \$A\$ to terminal \$B\$, with terminal \$C\$ open (unconnected.) The value for \$V_C\$ above is also with terminal \$C\$ open, of course.

There are other useful simplifications. For example, if you know that \$V_B=0\:\textrm{V}\$, then:

$$\begin{align*} V_C &= V_A\frac{1-x}{1+\frac{R_1}{R_2}x\left[1-x\right]} \end{align*}$$

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$$R_1+\frac{R_2*R_L}{R_2+R_L}=500,000$$

This is a step which is incorrect. With a 500k pot, R1+R2 always equals 500k (assuming R1 is the 'top' bit and R2 is in parallel with the load, shown in the diagram below).

schematic

simulate this circuit – Schematic created using CircuitLab

You get the right ratio with R1 approximately 25k, and R2 approximately 475k.

As the load is so much less than the end to end value of the pot, the control is more by R1 acting as a variable resistor, than R1 and R2 potting the voltage down.

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  • \$\begingroup\$ I am confused. When I substitute 500,000 for R1+R2 I still get -236593 =R2. Can you please show more steps? Also see my ETA in my Question. \$\endgroup\$ – TheColonel26 Sep 8 '17 at 14:02
  • \$\begingroup\$ @TheColonel26 that's because we're using different meanings for R1 and R2. You seem to have switched from calling the load RL to calling it R2 in your schematic. The equation you have for output voltage, with your R2//RL calculations, is why I guessed that you were referring to R1 as a the top part, and R2 as the bottom part, I made this guess explicit in my answer. I didn't bother to solve simultaneous equations, I just whacked the formula into a spreadsheet and varied R1 until I got the required output voltage, so there are no further steps. Try R1 and R2, to my diagram, in your formula. \$\endgroup\$ – Neil_UK Sep 8 '17 at 14:40

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