College of Science and Engineering, James Cook University
Last updated: 03 February 2022
Question 1
A 1 kΩ linear potentiometer is used to deliver a voltage to a load RL, as shown in Figure Q1.
Figure Q1:
A potentiometer is used to control a voltage VL. The potentiometer has the equivalent circuit shown in the circle, where R=1kΩ is the potentiometer’s resistance and α is the fractional position of the potentiometer.
Zoom:
Suppose that the potentiometer is at its halfway point (α=0.5). When there is no external load connected (RL open circuit), the potentiometer forms a simple voltage divider, and the output voltage VL is obviously half of the supply voltage.
However, in practical situations, the potentiometer must be connected to some downstream circuit. If that circuit has an input resistance of RL, then the resistive loading of the potentiometer will affect the voltage VL.
Given α=0.5, what is the minimum value of RL such that the voltage VL is maintained within 5% of its unloaded value?
Answer
Applying the voltage divider formula, the loaded voltage is
VL=12×500+500∣∣RL500∣∣RL,
where ∣∣ means “in parallel with”.
After some algebraic simplification:
VL=2RL+50012RL.
The unloaded voltage (when RL is disconnected) is half the supply,
i.e. 6 V. The required threshold is 5%, i.e.
VL=0.95×6=5.7V.
Solving for RL we obtain
RL=4750Ω.
Question 2
The power rating of a potentiometer can affect the maximum allowed
driving voltage. The worse case for power dissipation in the potentiometer
will occur in the limit of RL→0, i.e. when the maximum current
flows in the load. If the voltage source is modelled by a Thevenin
equivalent then the following circuit is obtained:
Figure Q2:
A potentiometer with its external load shorted.
Zoom:
(a) Calculate the power dissipated in the potentiometer when the wiper
is at a fractional position α (measured from the top), i.e.
α=0 means the wiper is at the top and α=1 means the
wiper is at the bottom.
(b) Find the value of α that maximises the dissipated power.
(c) Using your result from part (b), find the worst case dissipated power
for RTH=10 Ω, R=25 Ω, and V=12 V.
A 500 Ω strain gauge with gauge factor G=200 is attached to
a steel beam that is supporting a load of 3 kN. The steel beam has
a cross-sectional area of 250 mm2 and a Young’s modulus
of 190 GPa. What is the total strain gauge resistance when the beam
is loaded?
Answer
The beam is supporting a strain of σ=3000/(250×10−6)=12×106N/m2,
which results in a stress of ϵ=σ/E=(12×106)/(190×109)=63.16με.
Consequently there is a change in resistance
ΔR=RGϵ=500×200×63.16×10−6=6.3158Ω.
The total resistance of the gauge will be
R+ΔR=506.3158Ω.
Question 4
A 120 Ω strain gauge with G=2.1 is used to measure a strain of
1.05×10−5. What is the resistance change from the unloaded
to the loaded state?
Answer
RΔRΔR=Gϵ=120×2.1×1.05×10−5=0.002646Ω.
Question 5
The resistance of a strain gauge changes by 0.5% when a strain of
25 με is applied. Calculate the gauge factor.
Answer
From the equation
RΔR=Gϵ
we have
0.005G=G×25×10−6=200.
Question 6
A nickel RTD has a transfer function
R=R0[1+α(T−T0)].
The device has a resistance of 500 Ω at 0 °C, and a temperature coefficient
of α=0.00618K−1.
(a) Calculate the sensitivity of the temperature sensor.
(b) Determine its resistance at 100 °C.
Answer
Substituting the known values into the transfer function:
R=500[1+0.00618(T−0)]=500+3.09T.
Here it is OK to use temperature in °C because T−T0 is the same
regardless of whether °C or K are used.
(a) The sensitivity is dTdR=3.09 Ω/K.
(b) The resistance at 100 °C is R(100)=809 Ω.
Question 7
A coil of wire is used to activate a relay. However, the coil’s resistance
changes with temperature, resulting in a different activation threshold
at different temperatures. It is proposed to use an NTC thermistor
to compensate for the temperature dependence in the coil, such that
the total resistance of the circuit remains constant as the temperature
varies. The circuit diagram is shown below:
Figure Q7:
A series connection of a thermistor and a relay coil.
Zoom:
The coil has a resistance of 5 kΩ at 25 °C, and a temperature coefficient
of resistance (TCR) of 0.0069 K−1. The thermistor also has a
resistance of 5 kΩ at 25 °C. Its temperature dependence is given by
RT=R0eB/T.
Find a value of B such that RT has the opposite TCR to the coil at 25 °C. In other words,
find B such that the series combination of the two devices has no temperature dependence (within
the range over which the linear TCR is a valid approximation).
Hint: TCR is defined as
TCR=RT(∂T∂RT).
Answer
The TCR for the thermistor is
TCR=T2−B=(298.15)2−B.
Solving for TCR=−0.0069, we obtain
B=613.4K.
Question 8
Building upon the previous question, suppose that a thermistor with
the required value of B is not available. The only available thermistor
has too large a temperature dependence. Hence, to reduce the temperature
sensitivity of the compensation circuit, you shunt the thermistor
with a fixed resistance R:
Figure Q8:
Modified temperature compensation circuit.
Zoom:
The goal in this question is to choose the value of R such that the TCR of the
compensation circuit is -0.0069 K−1.
Define Rp to be the parallel combination of the resistor and the thermistor:
Rp=R+RTRRT.
Assume that resistor R has no temperature dependence.
(a) Show that the partial derivative of Rp with respect to temperature is
∂T∂Rp=(R+RT)2R2∂T∂RT.
(b) Show that the TCR of Rp is
TCR=R+RTRRT1∂T∂RT.
(c) Assume that the available thermistor has resistance
RT=100e1000/T,
find a value of R such that the TCR of Rp is -0.0069 K−1
at 25 °C.
Answer
(a) Use the quotient rule to evaluate this derivative, noting that R is a constant because it is assumed to be temperature independent. Let u=RRT and v=R+RT. Then u′=RRT′ and v′=RT′. This gives:
