Calculation of Potentiometer Linearity and Power Dissipation
August 1967 Electronics World
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stay the same.
August 1967 Electronics World
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Calculation of Potentiometer Linearity
and Power DissipationBy David L. Heiserman
The linearity of a potentiometer
can be completely changed by the position of the wiper arm and the resistance
of the load.
Most potentiometers used in communications and industrial electronic
equipment are specified according to three characteristics: total resistance
of the resistive material, maximum power dissipation, and the linearity
of resistance as a function of shaft position. Both engineering technicians
who must modify existing circuits and experimenters who are designing
their own circuits face the problem of choosing the right potentiometer
for the job at hand. As will be shown, this choice is not as simple
as just selecting a likely looking pot from a catalogue.
the appropriate pot is somewhat more complex than many people might
be led to believe. The discussion that follows points out the problems
involved in selecting potentiometers for loaded voltage-divider circuits
and describes how to solve the problems using a few equations and the
The circuit in Fig. 1 shows the
conventional method of controlling the voltage across a load impedance
RL. With this particular voltage-divider arrangement, a clockwise
rotation of the shaft decreases the voltage applied to the load. If
a linear voltage response is desired, the natural tendency is to choose
a pot that has a resistive element specified as linear. The fact that
the winding is linear, however, is no guarantee that it will produce
a linear response under load. The curves in Fig. 1 show how the linearity
of the output voltage changes with the ratio of load impedance to specified
When the load resistance is infinite
(no load), the response of a linear pot is truly linear. As the load
impedance decreases, however, the response becomes more non-linear.
Fig. 1. These curves show how the pot linearity varies with the load.
In theory, it is impossible to obtain a linear response from
a linear taper pot that is loaded with any impedance. In practice, though,
an RL/R ratio of 10 or more gives a response that is fairly
Likewise, a log taper pot will produce a truly log response
only if the load impedance is infinite. As the RL/R ratio
becomes smaller, the deviation from the specified log response becomes
greater. When loaded, "voltage-divider" pots with certain non-linear
characteristics will compensate for this undesirable loading effect
and produce a nearly linear output. A discussion of non-linear pots,
however, is beyond the scope of this article.
Because of the
unwanted effects of pot loading, the potentiometer resistance should
be kept as low as possible with respect to the load impedance.
However, a good linear response is bought at a high price - the
smaller the specified pot resistance, the greater the current through
its contacts and resistive elements.
The potentiometer power dissipation specified by the manufacturer
is actually a reflection of the maximum current that can pass safely
through any of the pot's three connectors or any portion of its resistive
element. The following equation enables the user to calculate this maximum
Imax = √(P/R) where Imax
is the maximum amount of current that can pass safely through any part
of the pot, P is the specified power rating of the pot, and R is the
specified resistance of the pot.
For example, a 10,000-ohm,
1-watt potentiometer can safely pass √[1/(1 x 104)] amperes,
or 10 milliamperes.
The current through a voltage-divider circuit
such as the one in Fig. 1 is at a maximum when the wiper arm is in the
position that makes the circuit strictly parallel (α
The maximum current through any part of a loaded pot may
then be determined by using the equation Imax = E [( R +
RL) /RRL] where Imax is the maximum
current through any part of the pot (at
= 0), E is the d.c. or r.m.s. value of applied voltage, and RL
is the load impedance. Substituting Imax from this latter
equation for I in the first equation, we find the relationship Preq
= E2 [(R + RL) /RRL] where Preq
is the maximum power dissipation of the pot.
Suppose the d.c. or r.m.s, value of applied voltage (E) is 10
volts and the load impedance (RL) is 10,000 ohms. If a good
linear response is desired, what are the necessary potentiometer specifications?
Consider the linearity problem. The resistance of the pot should
be no greater than 0.1 times RL, so we can select a 1000-ohm
potentiometer. The maximum power dissipation of this pot in this circuit
can be found by applying the second equation. In this case,
= 102 [( 10 x 103 + 1 x 103) / (10
x 103) (1 x 103)] or 0.1 watt.
for this particular pot should be 1000 ohms, 1 watt, and a linear taper.