Voltage Divider Equation with Load on output

Question

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

^{simulate this circuit – Schematic created using CircuitLab}

Answer 1

This is your combined potentiometer+resistor 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:

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

Answer 2

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

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).

^{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.