Both analog potentiometers and digipots have output or wiper terminals that drive load resistances. How does potentiometer loading affect output linearity? How light must the loading be to achieve a desired linearity? This article addresses these questions.

The circuit of either an analog potentiometer (or *pot*, for short) or a digipot is basically the same, as shown below (left figure) using an analog pot symbol. The analog pot voltage-divider attenuation, *T*, is varied mechanically while the digipot receives a digital input value that selects where in the pot resistance the output terminal - the digital wiper - is to be attached, using many analog switches for selection.

The wiper of the pot - the arrow in the symbol - moves across a constant resistance between the top and bottom terminals, the pot resistance value of *R*. Moving the wiper causes the divider to change *T* by apportioning *R* among *R*_{T} and *R*_{B} (center figure) with

The pot terminals between *R* are shown connected to voltage sources at top, *V*_{T}, and bottom, *V*_{B}. (If the top and bottom terminals are connected to voltages in series with resistances, add the respective resistances to *R*_{T} and *R*_{B}.) The pot output terminal is loaded by resistance *R*_{L}. The open-circuit (*R*_{L} →
∞
) divider transfer function is

where the voltage across the pot is

The pot resistance at the wiper port is the parallel combination of the two resistances;

with maximum *R*_{W} = *R*/4. The wiper voltage without *R*_{L} by superposition is thus

As *T* varies, *V*_{W} varies with it linearly, where *T* is the rate of change (slope) of *V*_{W} and *V*_{B} is the offset.

Now consider the case where the pot is loaded. Then the circuit can be Thevenized so that the pot wiper port is made equivalent to an open-circuit voltage source, *V*_{W}, in series with the equivalent wiper resistance, *R*_{W}, as shown in the figure above on the right. Solving this simple divider circuit for output voltage, *V*_{O}, as a ratio of *V*_{W},

To unclutter the design formulas, let *R*_{L} be some fraction or multiple, *α*, the *loading factor* of *R*;

Then substituting for *R*_{L} ,

This voltage-divider transfer function is that of the loaded voltage normalized to the open-circuit pot divider voltage. Ideally, with no loading, *V*_{O} = *V*_{W} . As loading increases, the loaded-divider ratio decreases. As a function of *T*, the wiper position, it shows the extent to which *V*_{O} departs from the ideal linear case of *V*_{W}. Some curves of *V*_{O}/*V*_{W} as a function of *T* and loading parameter *α* are shown in the graph below, where

Two observations can be made of this graph. First, the curves compress to a lower than ideal value, with maximum departure at midscale, *T* = 1/2. Second, as loading increases, the compression flattens, with steeper sides at the extremes of the pot range, and closer to constant in the midrange. *Compression error* is

For *α* = 1 (*R*_{L} = *R*_{W}),
*ε*
is about 20 %. For *α* = 2.5, it is about 9 % and for *α* = 25, is close to 1 %. The formula evaluated at maximum error, or *T* = 1/2, is

Then maximum compression error is

Thus, to achieve a given maximum error as expressed by compression, and in bits, the following table gives some values, where the conversion of ε to number of bits, *n*, of accuracy is

Then for *n* bits of maximum error,

The maximum error at *T* = 1/2 can be verified by taking the derivative of *V*_{O}/*V*_{W} of *T*, setting it equal to zero, and solving for *T* at which the minimum value occurs:

In conclusion, digipots with 250 positions (≈ 8 bits of resolution) require that *R*_{L} be greater than 64 times the pot *R* value for less than one bit of error. For 10-turn analog pots with linear readouts having about 3 decades of resolution, loading should not exceed 256 x *R* for one bit of error. For single-turn analog pots, a loading of 25 x *R* will keep the error within a percent. Therefore, loading of a 10 kΩ pot must be no more than 250 kΩ for < 1 % compression error, 25 kΩ for < 10 %, and 10 kΩ for < 20 % error.