Potentiometer Loading

For the unloaded potentiometer, the output voltage is determined by the voltage divider rule, with $R_T$ in the figure representing the total resistance of the potentiometer as shown in [Fig. 1]. Too often it is assumed that the voltage across a load connected to the wiper arm is determined solely by the potentiometer and the effect of the load can be ignored. When a load is applied as shown in [Fig. 2], the output voltage $V_L$ is now a function of the magnitude of the load applied since $R_1$ is not as shown in [Fig. 1] but is instead the parallel combination of $R_1$ and $R_L$. The output voltage is now If you want to have good control of the output voltage $V_L$ through the controlling dial, knob, screw, or whatever, you must choose a load or potentiometer that satisfies the following relationship: In general, For example, let's disregard Eq. (1) and choose a $1 MΩ$ potentiometer with a $100 Ω$ load and set the wiper arm to $1/10$ the total resistance, as shown in [Fig. 3]. Then and which is extremely small compared to the expected level of 1 V.
In fact, if we move the wiper arm to the midpoint, and which is negligible compared to the expected level of $5 V$. Even at $R_1 = 900 kΩ$, $V_L$ is only $0.01 V$, or $1/1000$ of the available voltage. Using the reverse situation of $R_T = 100 Ω$ and $R_L = 1 MΩ$ and the wiper arm at the $1/10$ position, as in [Fig. 4], we find and as desired. In general, therefore, try to establish a situation for potentiometer control in which Eq. (1) is satisfied to the highest degree possible. Someone might suggest that we make $R_T$ as small as possible to bring the percent result as close to the ideal as possible. Keep in mind, however, that the potentiometer has a power rating, and for networks such as [Fig. 4], If $R_T$ is reduced to $10 Ω$, which would require a much larger unit.