Under controlled conditions, a MOSFET can function as a variable power resistor.
Bogdan Raducanu, Bucharest, Romania; Edited by Martin Rowe and Fran Granville -- EDN, July 28, 2011
Testing power supplies or discharging
batteries usually requires
a constant-current load. Sometimes,
however, you must study the behavior
when the load is a resistor. Using a high-power
potentiometer is an expensive
approach that might not be worth the
cost. The circuit in Figure 1, which performs
like a high-power resistor that connects
between P1 and P2, provides an
alternative approach.
To understand how the circuit works,
assume that the op amp is ideal and that
the total resistance of R2 and R3 exceeds
that of a high-power resistor (not
shown). R2 and R3 form a divider that
produces an output voltage, according
to the following equation:
The operational amplifier maintains a
voltage, such that R1’s voltage equals the
reference voltage, that causes the current
through R1 to be:
Substituting the first equation in the second
equation yields:
If you neglect the current through R2 and
R3, then R1’s current equals the input current,
as the following equation describes:
This equation shows a linear relationship
between the input current and the input
voltage. Thus, the circuit between P1 and
P2 acts as a resistor. The equation then
becomes:
where k=(R2+R3)/R2 is a factor greater
than 1, which multiplies R1. Making
either R2 or R3 variable lets the circuit
function as a variable resistor. The cost
of a suitable transistor and R1, along
with the rest of the components, is
smaller than that of a variable potentiometer
that can dissipate the same
amount of power.
The circuit has some limitations, however.
First, it can accept input voltages of only one polarity, which might limit
its use in some applications. Second, the
minimum resistor value is the value of R1
plus the transistor’s minimum on-resistance.
Other factors such as op-amp offset,
the values of R2 and R3, and input voltage
influence the circuit’s linearity, but the
circuit still achieves high performance
with low-cost components. Depending
on the op amp’s input range, the circuit
requires an external dual power supply.
Figure 2 shows a prototype of the tested
and built circuit using a potentiometer for
changing the equivalent resistance and
no heat sink on the power transistor.
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Testing power supplies or discharging
batteries usually requires
a constant-current load. Sometimes,
however, you must study the behavior
when the load is a resistor. Using a high-power
potentiometer is an expensive
approach that might not be worth the
cost. The circuit in Figure 1, which performs
like a high-power resistor that connects
between P1 and P2, provides an
alternative approach.
