6655_control_handout.pdf

1
VEE
-18V
VCC
18V
22kOhm
R6
R1
RTD
22kOhm
R4
VPROCESS
VARIABLE
220 Ohm
R2
22kOhm
R5
50%
8.5V
Key = A
10kOhm
R3
TEMP.
ADJUST
U1A
LF444
22kOhm
R7
U1C
U1B
LF444
LF444
(R9 ADJUSTED TO 710 ohms)
DEADBAND
ADJUST
22kOhm
R8
Key = A
2.2kOhm 50%
R9
R10
1kOhm
D1
1N4001GP
-
SOLID
+ STATE
RELAY
AC
240 VAC
2000 W
HEATER
Dr. Julio R. García Villarreal
San José State University
San José, California - USA
240 VAC
2
This material is based upon work supported by the National Science
Foundation under Grant No. 0411330.
Any opinions, findings and conclusions or recommendations expressed in
this material are those of the author and do not necessarily reflect the views
of the National Science Foundation (NSF).
3
TABLE OF CONTENTS
Introduction to Process Control ............................................................ 4
Analog Signal Conditioning ................................................................. 16
Digital Signal Conditioning ................................................................... 41
Thermal Sensors .................................................................................... 61
Optical Sensors ..................................................................................... 69
Final Control .......................................................................................... 79
Controller Principles ............................................................................. 94
Closed-Loop Systems ......................................................................... 115
4
SAN JOSE STATE UNIVERSITY
Department of Aviation & Technology
Tech 167: Control Systems
Dr. Julio R. Garcia
Introduction to Process Control
1.1 Explain how the basic strategy of control is employed in a room air-conditioning system.
What is the controlled variable? What is the manipulated variable? Is the system
self-regulating?
Solution:
The basic strategy of the room air-conditioning system can be described as follows:
1.
Measure the temperature in a room by using a “thermostat”, which is nothing more than a
sensor of temperature. Thus temperature is the controlled variable.
2.
The measured temperature is compared to a set point in the thermostat. Often this is simply a
bimetal strip which closes a contact when the temperature exceeds some limit.
3.
If the temperature is too low then the compressor and distribution fan of the air-conditioner
are turned on. This causes room air to be circulated through the unit and thereby cooled and
exhausted back into the room. Therefore you can see that the manipulated variable is the
temperature of the recirculated air.
The system is self-regulating because even without operation of the air-conditioner, the room will
adopt some temperature in equilibrium with the outside air, open windows/doors, cooking, etc.
1.2 Can you think of any other situation in which a control is employed? What is the controlled
variable? What is the manipulated variable? Is the system self-regulating?
5
1.3
Is driving an automobile best described as a servomechanism or a process-control system?
Why?
Solution:
Driving a car is a servomechanism because the purpose is to control the motion of the vehicle
rather than to regulate a specific value. Therefore the objective is to cause the vehicle to follow a
prescribed path. Of course keeping the speed constant during a trip could be considered process
control since the speed is being regulated.
1.4 A process-control loop has a setpoint
of 175°C and an allowable deviation
of ± 15°C. A transient causes the
response shown in Figure 1.
(a) Specify the maximum error and
(b) settling time.
Temperature (ºC)
195
Figure 1
185
175
165
2
4
6
8
10
Time
12
14
155
Solution:
(a) Maximum error = peak error - setpoint = 197 °C - 175 °C = 22 °C
(b) Settling time = time of first excursion beyond 175 ± 5 °C to the time that range is reacquired.
= 9.8s-1.4s = 8.4s
1.5 A process-control loop has a setpoint
of 195°C and an allowable deviation
of ± 20°C. A transient causes the
response shown in Figure 2.
(a) Specify the maximum error and
(b) settling time.
Temperature (ºC)
Figure 2
215
205
195
185
175
2
4
6
8
10
Time
12
14
6
1.6 The second cyclic transient error peak of a response test measures 6.4%. For the
quarter-amplitude criteria, what error should be the third peak value?
Solution:
Since each peak must be a quarter of the previous one, the next peak must be:
a3 = (1/4) a2
a3 = (1/4) (6.4%) = 1.6%
1.7 The second cyclic transient error peak of a response test measures 5.6%. For the
quarter-amplitude criteria, what error should be the third peak value?
1.8 Does the response of Figure 1 satisfy the quarter-amplitude criterion?
Solution:
A close observation of Figure 1 shows:
st
The 1 peak error = 197.5 – 175 = 22.5
One quarter of this is (1/4) (22.5) = 5.6 but the actual peak is 7.
The third peak should be about (1/4) (7) = 1.75 but is 2. Thus, we conclude that the tuning does not
match the quarter-amplitude exactly since each peak is higher than the predicted by the criteria.
3
1.9 An analog sensor converts flow linearly so that flow from 0 to 400 m /h becomes a current
3
from 0 to 80 mA. Calculate the current for a flow of 250 m /h.
Solution:
3
Since it is linear we can calculate the current for each m /hr of fl of flow rate.
3
3
(80 mA)/(400 m /h) = 0.2 mA/m /h. Now, we can calculate current:
3
3
I = (250 m /h)( 0.2 mA/m /h) = 50 mA
7
1.10 An analog sensor converts flow linearly so that flow from 0 to 600 m3/h becomes a current
from 0 to 100 mA. Calculate the current for a flow of 450 m3/h.
1.11 Suppose each bit change in a 4-bit ADC represents a level of 0.15 m.
(a). What would the 4 bits be for a level of 1.7 m?
(b). Suppose the 4 bits were 10002. What is the range of possible levels?
Solution:
We can make a table of changes for the 16 states of the 4-bit ADC
Binary
Level
Binary
Level
Binary
Level
0000
0
0110
0.90
1100
1.80
0001
0.15
0111
1.05
1101
1.95
0010
0.30
1000
1.20
1110
2.10
0011
0.45
1001
1.35
1111
2.25
0100
0.60
1010
1.50
0101
0.75
1011
1.65
a. We can see that a level of 1.7 m would result in an output of 10112, since the level is
greater than 1.65 but not yet 1.8 for the next bit change.
b. If the bits were 10002 then the MOST that can be said is that the level is between 1.20 m
and 1.35 m. Thus there is an uncertainty of 0.15 m.
1.12 Suppose each bit change in a 4-bit ADC represents a level of 0.25 m. (a). What would the 4
bits be for a level of 2.9 m? (b). Suppose the 4 bits were 10102. What is the range of possible
levels?
8
2
1.13 Atmospheric pressure is about 15.6 lb/in (psi). What is this pressure in pascals?
Solution:
-4
Pat = (15.6 psi)/(1.45 x 10 psi/Pa) = 107,600
2
1.14 Atmospheric pressure is about 26.8 lb/in (psi). What is this pressure in Pascals?
1.15 An accelerometer is used to measure the constant acceleration of a race car that covers a
quarter mile in 8.4 s
2
a. Using x = at /2 to relate distance, x, acceleration, a, and time, t, find the acceleration in
ft/s'.
2
b. Express this acceleration in m/s .
2
c. Find the car speed, v, in m/s at the end of the quarter mile using the relation v = 2ax.
d. Find the cm energy in joules at the end of the quarter mile if it weighs 2500 lb, where the
2
energy W = mv /2.
Solution:
1 mile = 5280 ft and 1 ft = 0.3048 m
(a) for the acceleration we find,
2
2
2
a = 2x/t = (2)(0.25 mile)(5280 ft/mile)/(8.4 ) = 37.42 ft/s
2
2
(b) In m/s we have a = (37.42 ft/s2)(0.3048 m/ft) = 11.4 m/s
2
(c) We have velocity, v = 2ax, so
2
2
v = (2)(11.4 m/s )(0.25 mile)(5280 ft/mile)(0.3048 m/ft)
2
3
2 2
v = 9.173 x 10 m /s
v = 95.8 m/s
9
(d) The weight must be converted to mass in kg
m = (2500 lb)(0.454 kg/lb) = 1135 kg
2
W = (1135 kg)(95.8 m/s) /2
6
2 2
6
W = 5.21 x 10 kg-m /s = 5.21 x 10 J
1.16 An accelerometer is used to measure the constant acceleration of a race car that covers a half
mile in 12.6 s
2
a. Using x = at /2 to relate distance, x, acceleration, a, and time, t, find the acceleration in
ft/s'.
2
b. Express this acceleration in m/s .
2
c. Find the car speed, v, in m/s at the end of the quarter mile using the relation v = 2ax.
d. Find the cm energy in joules at the end of the quarter mile if it weighs 2500 lb, where the
2
energy W = mv /2.
10
1.17 A controller output is a 4-mA to 20-mA signal that drives a valve to control flow. The
1/2
relation between current and flow is Q = 45[I - 2 mA] gal/min. What is the flow for 15 mA?
What current produces a flow of 185 gal/min?
Solution:
For a current of 12 mA we have a flow given by:
1/2
Q = 45[I - 2 mA]
1/2
gal/min = 45[15 mA - 2 mA]
gal/min = 162.3 gal/min
To find current we can derive an equation,
1/2
Q = 45[I - 2 mA]
1/2
Q/45 = [I - 2 mA]
2
(Q/45) = I – 2 mA
where
then
2
I = (Q/45) + 2 mA
2
I = (185/45) + 2 mA = 18.90 mA
1.18 A controller output is a 5-mA to 22-mA signal that drives a valve to control flow. The
1/2
relation between current and flow is Q = 50[I - 2 mA] gal/min. What is the flow for 13 mA?
What current produces a flow of 150 gal/min?
11
1.19 An instrument has an accuracy of ± 0.4% FS and measures resistance from 0 m 1200 Ω.
What is the uncertainty in an indicated measurement of 485 Ω?
Solution:
± 0.4% FS for 0 to 1200 Ω means (± 0.004)(1200) = ± 4.8 Ω.
Thus a measurement of 485 Ω actually means 485 ± 4.8 or from 480.2 to 489.8 Ω
1.20 An instrument has an accuracy of ± 0.3% FS and measures resistance from 0 m 1000 Ω.
What is the uncertainty in an indicated measurement of 345 Ω?
1.21 A sensor has a transfer function of 0.6 mV/°C and an accuracy of ± 1%. If the temperature is
known to be 50 ºC, what can be said with absolute certainty about the output voltage?
Solution:
A 0.6 mV/°C with a ± 1% accuracy means the transfer function could be 0.6 ± 0.006 mV/°C or
0.594 to 0.606 mV/°C.
If the temperature were 50 °C the output would be in the range,
(0.594 mV/°C)(50°C) = 29.7 mV to (0.606 mV/°C)(50°C) = 30.3 mV
or 30 ± 0.3 mV. This is, of course, ± 1%.
1.22 A sensor has a transfer function of 0.45 mV/°C and an accuracy of ± 1.5%. If the temperature
is known to be 70°C, what can be said with absolute certainty about the output voltage?
12
1.23 A temperature sensor transfer function is 52.5 mV/°C. The output voltage is measured at
9.28 V on a 3-digit voltmeter. What can you say about the value of the temperature?
Solution:
This is a linear transducer so it is represented by the equation of a straight line with a zero intercept,
V = KT with K = 52.5 mV/°C, or V= 0.0525T
If V= 9.28 volts then, T= V/K = 9.28/0.0525 = 176.762 °C
but we have only three significant figures, so the temperature is reported as,
T = 177°C
1.24 A temperature sensor transfer function is 45.5 mV/°C. The output voltage is measured at
8.36 V on a 3-digit voltmeter. What can you say about the value of the temperature?
13
1.25
A temperature sensor has a static transfer function of 0.15mV/°C and a time constant of
2.8 s. If a step change of 26°C to 60°C is applied at t = 0, find the output voltages at 0.5 s,
2.0 s, 3.3 s, and 9 s. What is the indicated temperature at these times?
Solution:
We do not have to use the transfer function at all since the relation between voltage and
temperature is linear. Using the equation for first-order time response,
T = Ti + (Tf – Ti )(1-e
-t/τ
)
T = 26 °C + (60 °C - 26 °C)(1-e
T = 26 °C + 34(1-e
1.26
-t/2.8s
-t/2.8s
)
) °C
T = 0.5 s Æ T = 26 °C + 34(1-e
-0.5/2.8s
T = 2.0 s Æ T = 26 °C + 34(1-e
-2/2.8s
T = 2.8 s Æ T = 26 °C + 34(1-e
-2.8/2.8s
T = 9.0 s Æ T = 26 °C + 34(1-e
-9/2.8s
) °C = 31.6°C
) °C = 44.5°C
) °C = 47.5°C
) °C = 58.6°C
A temperature sensor has a static transfer function of 0.22mV/°C and a time constant of
3.5 s. If a step change of 20°C to 50°C is applied at t = 0, find the output voltages at 0.5 s,
1.5 s, 4.8 s, and 8.5 s. What is the indicated temperature at these times?
14
1.27 A pressure sensor measures 38 psi just before a sudden change to 80 psi. The sensor
measures 46 psi at a time 3.5 s after the change. What is the sensor time constant?
Solution:
Using the equation for first-order time response,
P = Pi + (Pf – Pi)(1-e
-t/τ
46 = 38 + (80 – 38)(1-e
46 - 38 = (42)(1- e
8 = (42)(1- e
8/42 = 1 - e
-3.5/τ
-3.5/τ
) and substituting values
-3.5/τ
-3.5/τ
)
)
)
or
e
-3.5/τ
= 1 - 8/42 = 0.81
Taking natural logarithms of both sides,
-3.5/τ = ln(0.81) = - 0.211
Thus, τ = - 3.5/(- 0.211) = 16.6 s
1.28 A pressure sensor measures 45 psi just before a sudden change to 78 psi. The sensor
measures 39 psi at a time 4.25 s after the change. What is the sensor time constant?
15
1.29 Plow rate was monitored for a week, and the following values were recorded as gal/nun:
10.6, 11.2, 10.7, 8.4, 13.4, 11.5, 11.2, 12.5, 8.9, 13.5, 12.3, 10.3, 8.7, 10.9, 11.0, and 12.3.
Find the mean and the standard deviation for these data.
Solution:
There are 16 values, if xi represents the values of flow rate then the mean is found from:
X=
∑ xi = 10.6 + 11.2 + 10.7 + 8.4 + 13.4 + 11.5 + 11.2 + 12.5 + 8.9 + 13.5 + 12.3 + 10.3 + 8.7 + 10.9 + 11.0 + 12.3
N
16
X = 11.1 gal/min
The standard deviation is found from:
σ=
2
∑ (xi - x )
= 1.53 gal / min
N −1
1.30 Plow rate was monitored for a week, and the following values were recorded as gal/nun:
11.4, 12.5, 11.9, 9.9, 13.8, 12.8, 11.9, 13.9, 10.5, 14.9, 12.9, 13.2, 9.9, 12.5, 12.0, and 13.2.
Find the mean and the standard deviation for these data.
16
SAN JOSE STATE UNIVERSITY
Department of Aviation & Technology
Tech 167: Control Systems
Dr. Julio R. García
Analog Signal Conditioning
2.1
The unloaded output of a sensor is a sinusoid at 200 Hz and 5-V amplitude. Its output impedance is
2000 + 600j. If a 0.22-uF (220 nF) capacitor is placed across the output as a load, what is the sensor
output voltage amplitude?
Solution:
The circuit is as shown below:
Zo = 2000 + 600 j
Z=A+jB
Vs
5V
200Hz
The output is a voltage divider voltage: Vo =
Where
Xc =
And
Zo =
220nF
C
Vo
Vs Xc
Zo + Xc
1
1
=
= 3617 Ω
2πfC 2π(200)(0.22 x 10-6 )
(2000) 2 + (600) 2 = 2088 Ω
Then the output voltage is Vo =
5(3617)
= 3.17 V
2088 + 3617
2.2 In the circuit shown below (a) Calculate the sensor output voltage amplitude. (b) If C1 opens
determine Vo. (c) If C1 shorts find Vo.
Zo = 1600 + 400 j
Z=A+jB
Vs
8V
300Hz
0.05uF
Vo
C1
17
Solution:
2.3 A sensor resistance varies from 560 to 2500 Ω. This
is used for R1 in the divider of the Figure shown
below, along with R2 = 390 Ω and V = 15.0 V. Find
(a) the range of the divider voltage, Vd, and (b) the
range of power dissipated by the sensor.
15 V
V
R1
560 to 2500 ohms
390 Ohm
R2
The divider output voltage is found from:
Vd =
(a)
VR2
(15)(390)
=
R1 + R2
R1 + 390
For R1 = 560 Ω, then Vd =
For R1 = 2500 Ω, then
(15)(390)
= 6.16 V
560 + 390
Vd =
(15)(390)
= 2.02 V
2500 + 390
Vd
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(b)
The range of Vd is from 2.02 V to 6.16 V.
(VR1 ) 2
Sensor dissipation is given by PR1 =
where VR1 = V - Vd
R1
For R1 = 560 Ω, VR1 = 15 – 6.16 V = 8.84 V and PR1 =
(8.84) 2
= 140 mW
560
For R1 = 2500 Ω, VR1 = 15 – 2.02 V = 12.98 V and PR1 =
(12.98) 2
= 77 mW
2500
The power dissipated by the sensor is from 77 mW to 140 mW.
2.4 A sensor resistance varies from 330 to 1800 Ω. This is used for R1 in the divider of the Figure
shown below, along with R2 = 270 Ω and V = 10.0 V. Find (a) the range of the divider voltage,
Vd, (b) the range of power dissipated by the sensor, (c) If R2 opens calculate Vd, (d) If R1 shorts
calculate Vd, (e) If R1 opens calculate Vd, and (f) If R2 shorts calculate Vd.
10 V
V
R1
330 to 1800 ohms
270 Ohm
R2
Vd
19
2.5 Prepare graphs of the divider voltage versus transducer resistance for Problem 2..3. (a) Does the
voltage (Vd) vary linearly with resistance? (b) Does the voltage (Vd) increase or decrease with
resistance?
Solution:
We can use the following equation: Vd =
VR2
15(390)
=
R1 + R2 R1 + 390
A plot of this function for R1 varying between 560 Ω and 2500 Ω is shown below:
6
5
4
3
400
600
800
1000
1200
1400
1600
1800
R1 in ohms
This is nonlinear and the output voltage (Vd) decreases with increasing resistance.
2.7 Prepare graphs of the divider voltage versus transducer resistance for Problem 2.4. (a) Does the
voltage (Vd) vary linearly with resistance? (b) Does the voltage (Vd) increase or decrease with
resistance? (c) At approximately what R1 value Vd is about 3.5 V?
20
2.8 A Wheatstone bridge, as shown below, nulls with
R1 = 319 Ω, R2 = 524 Ω, and R3 = 1265 Ω. Find R4.
a
15 V
V1
Solution
d
R1
R2
b
M
R3
R4
c
The null condition is obtained when the multiplication of
the value of two non-adjacent branches are equal to the
multiplication of the value of two other non-adjacent
branches.
In this case: R1 R4 = R2 R3, therefore, R4 = R2 R3/R1 = (524)(1265)/(319) = 2078 Ω
2.9 A Wheatstone bridge, as shown above, nulls with R1 = 456 Ω, R2 = 856 Ω, and R3 = 1543 Ω.
Find R4.
Solution
2.10
A sensor with a nominal resistance of 60 Ω is
used in a bridge with R1 = R2 = 120 Ω,
V = 12.0 V, and R3 = 150-Ω potentiometer. It
is necessary to resolve 0.1-Ω changes of the
sensor resistance.
a At what value of R3 will the bridge null?
b. What voltage resolution must the null
detector possess?
R1
15 V
V1
a
d
R2
M
b
c
R4
Sensor
R3
Solution
(a) For a null condition: R1 R4 = R2 R3, therefore,
R3 = R1 R4/R2 = (120)(60)/(120) = 60 Ω
(b) The detector resolution needed to resolve a resistance change of 0.1 Ω is found from the following
equation when R4 has changed to 60.1 Ω (or 59.9 Ω).
ΔV =
VR 3
VR 4
−
R1 + R 3 R 2 + R 4
21
For R4 = 60.1 Ω
ΔV =
(12V)(60) (12V)(60.1)
−
= -4.44 mV
60 + 120
60.1 + 120
ΔV =
(12V)(60) (12V)(59.9)
= -4.44 mV
−
60 + 120
59.9 + 120
Or,
For R4 = 59.9 Ω
2.11
A sensor with a nominal resistance of 48 Ω is used
in a bridge with R1 = R2 = 150 Ω, V = 15.0 V, and
R4 = 200-Ω potentiometer. It is necessary to
resolve 0.15-Ω changes of the sensor resistance.
R1
a
15 V
V1
Sensor R3
a At what value of R4 will the bridge null?
b. What voltage resolution must the null detector
possess?
d
M
R2
b
R4
c
2.12
A bridge circuit is used with a sensor located 120 m away. The bridge is not lead compensated,
and the cable to the sensor has a resistance of 0.36 Ω/ft. The bridge nulls with R1 = 3150 Ω,
R2 = 3835 Ω, and R3 = 1250 Ω. What is the sensor resistance?
Solution
The diagram will help understand this problem.
3150 Ohm
R1
V
a
d
M
1250 Ohm
R3
c
3835kOhm
R2
120 m
b
R4
22
If we use the null equation to find R4, it will give the resistance from b to c in the schematic, which
includes the two 120 m lead resistance. Thus, these must be subtracted to find the actual sensor
resistance.
R4 = Rbc – Rlead
Rbc = R2 R3/R1 = (1250)(3835)/(3150) = 1522 Ω
Rlead = 2(120 m)(0.3048 m/ft)(0.36 Ω/ft) = 26.34 Ω
The actual sensor resistance is then: R4 = Rbc – Rlead = 1522 Ω - 26.34 Ω = 1495.8Ω
2.13
A bridge circuit is used with a sensor located 150 m away. The bridge is not lead compensated,
and the cable to the sensor has a resistance of 0.45 Ω/ft. The bridge nulls with R1 = 2250 Ω,
R2 = 3255 Ω, and R3 = 1510 Ω. What is the sensor resistance?
2.14
A potential measurement bridge, such as the one
shown below, has:
V = 15.0 V, R1 = R2 = R3 = 11 kΩ. Find the
unknown potential if the bridge nulls with
R4 = 10.93 kΩ.
Solution
R1
V
a
d
M
R3
R2
b
R4
c
Vab = Va – Vb
Va =
V R3
12(11 kΩ)
=
= 6V
R1 + R 3 11 kΩ + 11 kΩ
Vb =
V R4
12(10.93 kΩ)
=
= 5.981 V
R 2 + R 4 11 kΩ + 10.93 kΩ
Vab = 6 – 5.981 = 19 mV
2.15
A potential measurement bridge, such as the one shown above, has:
V = 12.0 V, R1 = R2 = R3 = 12.8 kΩ. Find the unknown potential if the bridge nulls with
R4 = 15.53 kΩ.
23
2.16 A low-pass RC filter has R = 110 Ω and C = 0.5 μF. (a) Determine fc, (b) Find the
attenuation of a 900-Hz signal and (c) Determine the attenuation of a 5000-Hz signal.
Solution
(a) To determine the cutoff frequency or fc, we use the equation:
fc =
1
1
=
= 2894 Hz
2πRC 2π(110)(0.5μF)
(b) The attenuation is found from the following equation:
Vout
1
1
=
=
= 0.955
2
1
/
2
Vin
[1 + ( f / fc) ]
[1 + (900 / 2894) 2 ]1 / 2
Thus the attenuation is 1 – 0.955 = 0.045 or 4.5%.
(c) The attenuation of a 5000-Hz signal is:
Vout
1
1
=
=
= 0.501
2
1
/
2
Vin
[1 + ( f / fc) ]
[1 + (5000 / 2894) 2 ]1 / 2
Thus the attenuation is 1 – 0.501 = 0.499 or 49.9%.
2.17
In the low-pass RC filter shown below, (a) calculate the attenuation of an 1100-Hz signal
and (b) Determine the attenuation of a 10000-Hz signal..
180 Ohm
R1
0.27uF
C1
24
2.18
A high-pass RC filter must drive 60 Hz noise down to 0.8%. (a) Specify the filter critical
frequency, (b) values of R and C and (c) the attenuation of a 20-kHz signal.
Solution
(a)
We find the critical frequency for which a 60 Hz signal has an output to input voltage ratio
of 0.008 (0.8% as stated in the problem statement);
Vout
(f/fc)
=
Vin
[1 + (f/fc)2 ]1/2
From this equation we derive fc.:
Then fc =
Vout
)
(602 - 602 ( 0.008 )
Vin
=
= 44.6 kHz
Vout
0.008
Vin
(f 2 − f 2 (
fc 2 =
44600 = 668 Hz
(b) If we select C = 0.002 μF then R =
1
1
=
= 1191 Ω
2π ( fc)C 2π (668 Hz)(0.002uF )
The standard value is R = 1.2 kΩ.
(c) The attenuation of a 20 kHz signal is calculated from this equation:
Vout
( f / fc )
( 20000 / 668)
=
=
= 0.999
2 1/ 2
2 1/ 2
Vin
1 + ( f / fc )
1 + 20000 / 668)
[
]
[
]
So the attenuation is 1.00 - 0.999 = 0.001 or 0.1%.
2.19
A high-pass filter is found to attenuate a 2-kHz signal by 30 dB. What is the critical
frequency?
25
2.20
A high-pass filter is found to attenuate a 1-kHz signal by 25 dB. Find the critical
frequency.
Solution
In this case we solve for fc. Down 30 dB means that:
-30 dB= 20 log10(Vout/Vin)
-30/20 = log10(Vout/Vin)
- 1.5 = log10(Vout/Vin)
Vout/Vin = 10
-1.5
= 0.032
Vout
(f/fc)
=
Vin
[1 + (f/fc)2 ]1/2
From this equation we derive fc.:
Vout
)
(2kHz − (2 kHz)2 ( 0.032 )
Vin
=
= 121 MHz
Vout
0.032
Vin
(f 2 − f 2 (
fc 2 =
And fc =
2.21
121 MHz = 11 kHz
A high-pass filter is found to attenuate a 1-kHz signal by 25 dB. Find the critical
frequency.
26
2.22
Show how op amps can be used to provide an amplifier with a gain of +120 and an input
impedance of 1 kΩ. Show how this can be done using both (a) inverting and (b) non-inverting
configurations.
Solution
(a) In the case of the inverting amplifier we need two so that the overall gain will be +120. Thus, the
following circuit will satisfy this need. The first has a gain of -120 and an input impedance of 1 kΩ
and the second a gain of - 1.
120kOhm
Vin
120kOhm
1kOhm
1kOhm
Vout
(b) A non-inverting amplifier can be constructed with only one op amp as:
Vin
Vout
Since a non-inverting amplifier has a very high
input impedance, the 1 kΩ resistor placed in
parallel with the input terminal ensures that the
input impedance be 1 kΩ.
1.19kOhm
1kOhm
1kOhm
2.23 What change(s) would you do to provide an amplifier with a gain of +180 and an input
impedance of 5.6 kΩ in both the (a) inverting and (b) non-inverting configurations.
27
2.24 Specify the components of a differential amplifier with a gain of 18.
Solution
A differential amplifier with a gain of 18 can be built as follows:
18kOhm
V1
V2
1kOhm
Vout
1kOhm
18kOhm
2.25 What change(s) would you do to provide a differential amplifier with a gain of 22.
28
2.26
Using an integrator with RC = 5 s and any other required amplifiers, develop a voltage ramp
generator with 0.6 V/s.
Solution
Since the output is constant the output equation is:
Vo =
- (Vin)t
1
=
(Vin) t
RC
RC
However, since RC = 5 s, then Vo =
-1
(Vin)t = - 0.2(vin)t
5
Since the output should be 0.6 V/s
0.6 V/s = - 0.2 (Vin) t Æ Vin =
0.6
=- 3 V
0.2
The following circuit will provide the required output. We need to adjust the potentiometer until
the voltage at point A is – 3 V.
To get RC = 5 s, we select R = 1 MΩ and C = 5 μF. RC = 1 MΩ (5 μF) = 5 s.
VEE
-12V
C1
150 Ohm
5uF
1MOhm
Key = A
50%
100 Ohm
2.27
Vout
What change(s) would you do to provide to the integrator shown above if we need RC = 8 s to
develop a voltage ramp generator with 0.5 V/s.
29
2.28
The analysis of a signal-conditioning circuit has produced the following equation:
Vout = 2.85 Vin – 1.52
Design circuits to implement this equation using (a) a summing amplifier and (b) a differential amplifier.
Solution
(a) The respective circuit is shown below. A gain of - 2.85 is given by R5/R1; however, since the gain
is + 2.85, a gain of – 1 is given by R8/R6. The constant of - 1.52 is given by R2, R3, R4, D1 and the
inverting amplifier with unity gain (R8/R9). Notice that due to the inverting amplifier we need to
obtain + 1.52 V first. Voltage followers U3 and U4 prevent loading.
VCC
12V
150 Ohm Key = A
R3
150 Ohm
R2
10kOhm
1.52 V R7
10kOhm
R9
50%
D1
3.9 V
U4
100 Ohm
R4
28.5kOhm
R5
Vin
U3
10kOhm
R1
U1
10kOhm
R8
10kOhm
R6
(b) In this case, the equation needs to be rearranged as:
Vout = 2.85 Vin - 1.52 = 2.85 (Vin -
1.52
) = 2.85 (Vin – 0.53)
2.85
U2
Vout
30
The respective circuit is shown below. The gain of the differential amplifier is determined by R8/R9 or
R5/R6. The constant of - 0.53 is given by R2, R3, R4 and D1. Notice that since this constant voltage is
connected to the inverting input of the differential amplifier, we need to obtain + 0.53 V first. Voltage
followers U3 and U4 prevent loading.
VCC
12V
150 Ohm Key = A
R3
150 Ohm
R2
10kOhm
0.53 V R7
D1
3.9 V
Vin
50%
10kOhm
R9
U4
100 Ohm
R4
28.5kOhm
R8
U2
U3
10kOhm
R6
Vout
28.5kOhm
R5
2.29 What change(s) would you to the above circuits, the summing amplifier and the differential
amplifier to implement the following equation:
Vout = 4.25 Vin + 2.45
31
2.30
A differential amplifier has R2 = 560 KΩ and R1 = 3.9 KΩ. When Va = Vb = 2.8 V the output
is 69 mV. Find the CMR and CMRR.
Solution
The amplifier gain is determined by Ad = R2/R1 = 560/3.9 = 144
The common-mode gain is determined by Acm = Vo/Vin = 69 mV/2.8 = 0.025
CMRR is the ratio between the differential gain (Ad) and the common-mode gain (Acm); thus,
CMRR = Ad/Acm = 144/0.025 = 5760
CMRR is given in dB. The equation is:
20 log 10 (CMRR) = 20 log 10 (5760) = 75.2 dB
2.31
A differential amplifier has R2 = 680 KΩ and R1 = 4.7 KΩ. When Va = Vb = 3.2 V the output
is 98 mV. Find the CMR and CMRR.
32
2.32
A control system needs the average of temperature from three locations. Sensors make the
temperature information available as voltages, V1, V2, and V3. Develop an op amp circuit that
outputs the average of these voltages.
Solution
A solution is to use a summing amplifier with a gain of 1/3 on all inputs because there are three input
signals, followed by a unity gain inverter to get the right polarity. Notice that R1, R2 and R3 are of the
same value (30 kΩ) and R5 is 10 kΩ (1/3 of 30 kΩ). Since U1 is an inverter we need an inverting
amplifier with unity gain. This is given by U2, R6 and R8.
V1
V2
V3
2.33
30kOhm
R3
30kOhm
R2
30kOhm
R1
10kOhm
R5
10kOhm
R6
10kOhm
R8
U1
U2
Vout
A control system needs the average of temperature from five locations. Sensors make the
temperature information available as voltages, V1, V2, V3, V4, and V5. Develop an op amp
circuit that outputs the average of these voltages.
33
2.34
Use the appropriate circuits to implement an output voltage given by
Vout = 12Vin + 5∫ Vin dt
Solution
This equation consists of the sum of a gain term and an integrator term. Thus, the circuit shown
below meets the requirement. Resistors R3/R2 determine the gain of 12. The integrator gain is
determined by 1/(R1 C1) = 1/(10 kΩ * 2 uF) = 1/(0.2) = 5.
120kOhm
R3
Vin
10kOhm
R1
U1
C1
10kOhm
R4
U3
2uF
10kOhm
R2
U2
10kOhm
R6
Vout
10kOhm
R5
2.35 What change(s) would you do to the above circuit to implement an output voltage given by
Vout = 10Vin + 4∫ Vin dt
34
2.36
Develop signal conditioning for Problem 2.3 so the output voltage varies from 0 to 6 V as
the resistance varies from 560 to 2500 Ω, where 0 V corresponds to 2500 Ω.
Solution
The input conditions and the requirements of this problem can be summarized as follows:
R
Vd = Vin
Vout
560 Ω
6.16 V
6V
2500 Ω
2.02 V
0V
The input voltage (6.16 V to + 2.02 V) is the independent variable while the output voltage (0 to
6V) is the dependent variable.
By plotting the independent variable (x axis) and the dependent variable (y axis), we have:
y
6
final point
m
0
Initial point
x
2.02
6.16
Connecting the intersection points, we find that the graph is a straight line,
y = mx + b
(1) equation of the straight line.
where:
m=
y1 - y0
x1 - x 0
(2)
slope of the straight line.
Replacing values in the slope of the straight line, we have:
m=
6-0
= 1.45
6.16 - 2.02
35
To find the value of the constant (b), in the equation of the straight line we replace the variables
(x, y) for the coordinates of the initial point (4.8, 0) or the coordinates of the final point (24, 3.5),
through this linear equation:
y = mx + b
equation of the straight line
Use coordinates of the initial point Æ (2.02, 0) Î 0 = 1.45(2.02) + b Î b = - 2.93
Use coordinates of the final point Æ (6.16, 6) Î 6 = 1.45(6.16) + b Î b = - 2.93
The result is the same; however, it is simpler to use the initial point as seen above.
Replacing the variables and the constant of the linear equation for equivalent electronic terms
we’ll have:
y = Vout
m = Av = 1.45
x = Vin
b = Voffset = - 2.93
Vout
6
Av
0
2.02
6.16
Vin
The constant of the equation can be called any name; in this case we’ll call it Voffset.
Then:
Vout = Av (Vin) + Voffset
(4)
Equation (4) indicates that we should use a circuit whose block diagram is:
Voffset = - 2.93
Vin
Av = 1.45
Replacing values in equation (4), we have:
Vout = Av(Vin) + Voffset
(4)
Vout = 1.45 (Vin) – 2.93
Vout = 1.45 (Vin – 2.02)
(5)
Summing
Amplifier
To the
Load
36
The respective
circuit using a
differential
amplifier is
shown below:
VCC
12V
150 Ohm Key = A
R3
150 Ohm
R2
10kOhm
2.02 V R7
D1
3.9 V
Vin
2.37
50%
10kOhm
R9
U4
100 Ohm
R4
14.5kOhm
R8
U2
U3
10kOhm
R6
Vout
14.5kOhm
R5
What change(s) would you to the above circuit if the output voltage varies from 0 to 8 V as
the resistance varies from 470 to 2200 Ω, where 0 V corresponds to 2200 Ω.
37
2.38
Sensor resistance varies from 22 kΩ to 1.2 kΩ as a variable changes from Cmin to Cmax.
Design a signal-conditioning system that provides an output voltage varying from -3 to + 3
V as the variable changes from min to max. Power dissipation in the sensor must be kept
below 2.0 mW.
Solution
To begin with, we establish an equation relating Vout and R from the equation for a straight line,
Vout = mR + Voffset
(1)
Using the given values,
- 3 = 22000m + Voffset
(2)
+ 3 = 1200m + Voffset
(3)
Solving, we have
0 = 23200m + 2 Voffset Æ Voffset = - 23200m/2 = - 11600m
(4)
Substituting Voffset (equation 4) into equation 3
+ 3 = 1200m – 11600m
-4
+ 3 = - 10400m Æ m = 3/(- 10400) = -2.89 x 10
(5)
So we can find Voffset (equation 5 into equation 4),
-4
Voffset = - 11600m = - 11600(- 2.89 x 10 ) = 3.352 V
The equation is then:
-4
Vout = - 2.89 x 10 R + 3.352
(6)
Equation (6) can be implemented by using an inverting amplifier with R (sensor) in the op-amp
feedback and a summing amplifier to provide the offset voltage. The current in the sensor resistor,
R, must be kept below a limit so that the dissipation does not exceed 2.0 mW. This can be obtained
by making the input resistance and fixed input voltage within certain limits since the current
through the feedback resistor is the same as the current through the input resistor.
38
2
Pmax = 2.0mW = I Rmax
2
-8
I = 2.0 mW/22000 = 9.09 x 10 Æ I = 0.3 mA
Let’s use an input current of 0.2 mA to be in the safe side. In the circuit below, the input divider
voltage is - 1 V and the input resistor is 5 kΩ. Adjust R2 until Voffset = - 3.352 V and adjust R1
until Vin = - 1 V.
VEE
-15V
150 Ohm
R3
D1
5.6 V
Key = B
150 Ohm
R2
50%
50%
Key = A
2kOhm
-1 V R1
Vin
150 Ohm
R4
2.39
10kOhm
R7
Voffset
-3.352 V
100 Ohm
R4
Sensor
1.2kOhm to 22kKOhm
10kOhm
R8
R
5kOhm
R10
U3
10kOhm
R6
U2
Vout
What change(s) would you to the above circuit if the sensor resistance varies from 18 kΩ to
0.8 kΩ as a variable changes from Cmin to Cmax. Design a signal-conditioning system that
provides an output voltage varying from -2 to + 2 V as the variable changes from min to
max. Power dissipation in the sensor must be kept below 3.0 mW.
39
2.40
A pressure sensor outputs a voltage varying as 120 mV/psi and has a 2.0- KΩ output
impedance. Develop signal conditioning to provide 0 to 3.5 V as the pressure varies from
40 to 200 psi.
Solution
As the pressure varies from 40 to 200 psi the sensor voltage will vary from (40 psi)(120 mV/psi) =
4.8 V to (200 psi)(120 mV/psi) = 24 V. The signal conditioning must convert this into 0 to 3.5 V.
The input voltage (4.8 V to + 24 V) is the independent variable while the output voltage
(0 to 3.5 V) is the dependent variable.
By plotting the independent variable (x axis) and the dependent variable (y axis), we have:
y
3.5
final point
m
0
initial point
x
4.8
24
Connect the intersection points, we find that the graph is a straight line, then:
y = mx + b
(1) equation of the straight line.
where:
m=
y1 - y0
x1 - x 0
(2)
slope of the straight line.
Replace values in the slope of the straight line, we have:
m=
24 - 4.8
= 5.49
3.5 - 0
To find the value of the constant (b), in the equation of the straight line we replace the variables (x,
y) for the coordinates of the initial point (4.8, 0) or the coordinates of the final point (24, 3.5),
through this linear equation:
y = mx + b
equation of the straight line
Use coordinates of the initial point (4.8, 0) Î 0 = 5.49(4.8) + b Î b = 26.35
Replacing the variables and the constant of the linear equation for equivalent electronic terms
we’ll have:
y = Vout
m = Av = 5.49
Vout
3.5
40
x = Vin
b = Voffset = 26.35
Av
0
4.8
24
Vin
The constant of the equation can be called any name; in this case we’ll call it Voffset.
Then:
Vout = Av (Vin) + Voffset
(4)
Replacing values in equation (4), we have:
Vout = Av(Vin) + Voffset
(4)
Vout = 5.49 (Vin) + 26.25
Vout = 5.49 (Vin + 4.78)
(5)
The respective
circuit using a
differential
amplifier is
shown below:
VEE
-15V
Key = A
150 Ohm
R2
150 Ohm
R3
10kOhm
-4.78 V R7
D1
9V
Vin
2.41
50%
10kOhm
R9
U4
100 Ohm
R4
54.9kOhm
R8
U2
U3
10kOhm
R6
Vout
54.9kOhm
R5
A pressure sensor outputs a voltage varying as 80 mV/psi and has a 1.0- KΩ output
impedance. Design a signal conditioning circuit that provides 0 to 5 V as the pressure
varies from 20 to 180 psi.
41
SAN JOSE STATE UNIVERSITY
Department of Aviation & Technology
Tech 167: Control Systems
Dr. Julio R. Garcia
Digital Signal Conditioning
3.1
Convert the following binary numbers into decimal, octal, and hex:
a. 10112
b. 1101012
c. 0101012
Solution
The basic relations for conversions of binary are defined for a binary, bnbn-1 … b1b0 where
the b’s are either a 1 or 0, then,
n
n-1
N10 = bn2 + bn-1 2
1
0
+ … + b12 + b02
For octal we just arrange the binary number in three-bit groups starting from the decimal
point and use the relations
111 = 78, 110 = 68, ………., 001 = 18 and 000 = 08
For hex we use groupings of four,
1000 = 8H
1001 = 9H
1010 = AH
1011 = BH
1100 = CH
1101 = DH
1110 = EH and
1111 = FH
so,
(a)
3
1
0
10112 = 2 + 2 + 2 = 8 + 2 + 1 = 1110
10112 = 001 011 = 138
10112 = BH
42
(b)
5
4
2
0
1101012 = 2 +2 +2 + 2 = 32 + 16 + 4 +1 = 5310
1101012 = 110 101 = 658
1101012 = 0011 0101 = 35H
(c)
4
2
0
0101012 = 2 + 2 + 2 = 16 + 4 + 1 = 2110
0101012 = 010 101 = 258
0101012 = 0001 0101 = 15H
3.2
Convert the following binary numbers into decimal, octal, and hex:
a. 110112
b. 1010012
c. 0100112
43
3.3
Convert the following binary numbers into decimal, octal, and hex:
a. 10110102
b. 0.10012
c. 1101.01102
Solution
(a)
6
4
3
1
10110102 = 2 + 2 + 2 + 2 = 64 + 16 + 8 + 2 = 9010
10110102= 001 011 010 = 1328
10110102= 0101 1010 = 5AH
(b)
For this we must use the base 10 fractional to binary fractional relationship,
Given a binary fraction, 0.b1b2 … bn then
-1
-2
-n
0.N10 = b12 + b22 + … + bn2
Octal and Hex fractional are found by the regular 3 and 4 groupings of bits. So,
-1
-4
0. 10012 ≈ 2 + 2 = 0.5 + 0.0625 = 0.562510
0. 10012 = 0.100 100 = 0.448
0. 10012 = 0.9H
(c)
1101.01102 treating the whole and fractional parts separately,
44
3
2
0
1101 = 2 + 2 + 2 = 8 + 4 + 1 = 1310
and
-2
-3
0.0110 = 2 + 2 = 0.25 + 0.125 = 0.37510
thus, 1101.0110 = 13.37510
1101.01102 = 001 101.011 000 = 15.38
1101.01102 = D.6H
3.4
Convert the following binary numbers into decimal, octal, and hex:
a. 10101012
b. 0.11012
c. 1011.01012
3.5
Convert the following decimal numbers into binary, octal, and hex:
a. 29 10
b. 530 10
c. 627 10
Solution
If we find binary first then the octal and hex can be found easily using the
groupings of 3 and 4 bits
(a)
29/2 = 14 + Remainder = 1
so b 0 = 1
14/2 = 7 + Remainder = 0
so b 1 = 0
45
7/2 = 3 + Remainder = 1
so b 2 = 1
3/2 = 1 + Remainder = 1
so b 3 = 1
1/2 = 0 + Remainder = 1
so b 4 = 1
Therefore, 29 10 = 111012
and,
111012 = 011 101 = 358
111012 = 0001 1101 = 1DH
(b) Lets do successive division by 8 instead of finding the octal first,
530/8 = 66 + Remainder = 2
so d0 = 2
66/8 = 8 + Remainder = 2
so d1 = 2
8/8 = 1 + Remainder = 0
so d2 = 0
1/8 = 0 + Remainder = 1
so d3 = 1
53010 = 10228
using binary groupings we see that,
10228 = 001 000 010 010 = 100000100102 and,
100000100102 = 0100 0001 0010 = 412H
(c) On this one let’s successively divide by 16 to get the hex first,
627/16 = 39 + Remainder = 3
so a 1 = 3 10 = 3H
39/16 = 2 + Remainder = 7
so a 2 = 7 10 = 7H
2/16 = 0 + Remainder = 2
so a 3 = 2 10
627 10 = 273H
Using groupings we get the binary and then the octal,
273H = 001001110011 = 1001110011 2
and 1001110011 2 = 001 001 110 011 = 11638
3.6
Convert the following decimal numbers into binary, octal, and hex:
a. 56 10
b. 850 10
46
c. 1210 10
3.7
Find the 2s complement of
a. 1101 2
b. 10111101 2
Solution
(a) Complement of
1101 = 0010
+
2’s complement:
1
0011
(b) Complement of 10111101 = 01000010
+
2’s complement:
3.8
Find the 2s complement of
a. 10111 2
b. 11001001 2
1
01000011
47
3.9
Simplify the Boolean equation AB + A( AB ).
Solution
Let’s assume that Y represents the output:
Y = AB + A( AB )
Y = AB + A( A + B )
Y = AB + A A + A B
but
AA = 0
Y = A( B + B ) but B + B = 1
Y=A
3.10
Simplify the Boolean equation AC + BC + C ( AC )
3.11
A process involves moving speed, load weight, and rate of loading in a conveyor system.
The variables are provided as high (1) and low (0) levels for digital control. An alarm
should be activated whenever any of the following conditions occur:
a. Speed is low; both weight and loading rate are high.
b. Speed is high; loading rate is low.
Find a Boolean equation describing the required alarm output. Let the variables be S for
speed, W for weight, and R for loading rate.
Solution
48
(a) We have an alarm when speed (S) is low, weight (W) is high and loading rate (R) is high,
so:
S WR
(b) Speed is high and loading rate is low, S • R
The combination is OR’ed to give the Boolean equation,
Y = S WR + SR
3.12
A process involves moving speed, load weight, and rate of loading in a conveyor system.
The variables are provided as high (1) and low (0) levels for digital control. An alarm
should be activated whenever any of the following conditions occur:
a. Speed is high; both weight and loading rate are low.
b. Speed is low; loading rate is high.
Find a Boolean equation describing the required alarm output. Let the variables be S for
speed, W for weight, and R for loading rate.
3.13
Implement Problem 3.11 with
a. AND/OR logic and
b. NAND/NOR logic
Solution
The equation Y = S WR + SR is implemented as follows,
(a) AND/OR logic
S
AY
W
R
49
(b) in NAND/NOR logic we have,
S
W
R
3.14
Implement Problem 3.12 with
a. AND/OR logic and
b. NAND/NOR logic
AY
50
3.15
A tank shown in the Figure below (Figure 3.1) has the following Boolean variables: flow
rates, QA, QB, and QC; pressure, P; and level, L. All are high if the variable is high and
low otherwise. Devise Boolean equations for two alarm conditions as follows:
a. OV = overfill alarm
1. If either input flow rate is high while the output flow rate is low, the pressure is low
and the level is high.
2. If both input flow rates are high while the output flow rate is low and the pressure
is low.
L
QA
P
QB
QC
Figure 1
b. EP = empty alarm
1. If both input flow rates are low, the level is low and the output flow rate is high.
2. If either input flow rate is low, the output flow rate is high and the pressure is high.
Solution
We simply translate the statements directly into Boolean expressions
a. OV: 1. (QA + QB) QC L P
2. QA QB QC P
OV = (QA + QB) QC L P + QA QB QC P
b. EP: 1. QA QB L QC
2. (QA + QB) QC P
EP = QA QB L QC + (QA + QB) QC P
51
3.16
A tank shown in the Figure above (Figure 1) has the following Boolean variables: flow
rates, QA, QB, and QC; pressure, P; and level, L. All are low if the variable is high and
high otherwise. Devise Boolean equations for two alarm conditions as follows:
a. OV = overfill alarm
1. If either input flow rate is low while the output flow rate is high, the pressure is high
and the level is high.
2. If both input flow rates are low while the output flow rate is high and the pressure
is low.
b. EP = empty alarm
1. If both input flow rates are high, the level is high and the output flow rate is low.
2. If either input flow rate is high, the output flow rate is low and the pressure is low.
52
3.17 Devise logic circuits using NAND/NOR logic that will provide the two alarms of problem
3.15.
The following logic circuits will provide the needed alarms.
QA
QB
P
L
QC
OVERFILL
ALARM
QA
QB
L
P
EMPTY
ALARM
QC
3.18 Devise logic circuits using NAND/NOR logic that will provide the two alarms of problem
3.16.
53
3.19 A sensor provides temperature data as 430 μV/°C. Develop a comparator circuit that goes
high when the temperature reaches 620°C.
Solution
If a transfer function is 430 uV/ºC then a temperature of 620 ºC will result in an output
voltage of,
-6
V = (430 x 10 V/ºC)(620 ºC)
V = 0.2666 volts or 0.267 to three significant figures.
We can construct a divider from a + 5 volt supply to obtain this required alarm voltage for
the comparator. One possible circuit then is shown below. Adjust R1 until Va = 0.267 V.
VCC
5V
Va = 0.267V
Key = A
3kOhm
R1
50%
100 Ohm
COMPARATOR
430 uV/ºC
3.20 What change(s) would you do to the circuit shown above if the sensor provides temperature
data as 560 μV/°C and the comparator circuit should go high when the temperature reaches
965°C.
54
3.21 An 8-bit DAC has an input of 101001012 and uses an 8.0-V reference.
a. Find the output voltage produced.
b. Specify the conversion resolution.
Solution
For the 8-bit DAC with a 101001012 input and an 8.0 V reference,
(a) The output is given by,
-1
-2
-3
-8
Vout = Vref (b0 2 + b0 2 + b0 2 +……… b0 2 +)
-1
-3
-6
-8
Vout = 8(2 + 2 + 2 + 2 )
= 5.156V
-n
(b) The resolution is ΔV = Vref2 so
-8
ΔV = (8)(2 )
= 0.031 V
3.22 A 10-bit DAC has an input of 10111001012 and uses a 6.0-V reference.
a. Find the output voltage produced.
b. Specify the conversion resolution.
55
3.23 A 6-bit DAC must have a 10.00-V output when all inputs are high. Find the required
reference.
Solution
We have
-1
-2
-3
-4
-5
-6
Vmax = Vref (2 + 2 + 2 + 2 + 2 + 2 )
10 = Vref (0.5 + 0.25 + 0.125 + 0.1625 + 0.031 + 0.016)
10 = Vref (1.085)
Thus, Vref = 10/1.085 = 9.217 V
3.24 An 8-bit DAC must have a 12.00-V output when all inputs are high. Find the required
reference.
3.25 A 10-bit ADC with a 12.0-V reference has an input of 4.869 V. (a) Find the digital output
word. (b) What range of input voltages would produce this same output? (c) Suppose the
output of the ADC is 11101101112. What is the input voltage?
Solution
(a) The ratio of input to reference is, (4.869/12) = 0.406
This fraction of the total counting states will provide the output as,
10
(0.406)(2 ) = 415.744
but since the output is the integer part only it will be just 415, so
415 ≈ 19FH or 01100111112
(b) This same output would be produced by input voltages which range from,
12(415/1024) = 4.863 V to 12(416/1024) = 4.875 V
(c) An output of 11101101112 = 95110 so the input is at least 12(951/1024) = 11.145 volts
but could be as high as 12(952/1024) = 11.156 volts.
56
3.26 An 8-bit ADC with a 10.0-V reference has an input of 3.453 V. (a) Find the digital output
word. (b) What range of input voltages would produce this same output? (c) Suppose the
output of the ADC is 110110012. What is the input voltage?
3.27 An ADC that will encode pressure data is required. The input signal is 548.4 mV/psi.
a. If a resolution of 0.4 psi is required, find the number of bits necessary for the ADC. The
reference is 8.0 V.
b. Find the maximum measurable pressure.
Solution
The pressure transducer converts pressure to voltage at 548.4 mV/psi or 0.5484 V/psi.
(a) We need a resolution of 0.4 psi with an 8.0 volt reference. This means a voltage
resolution of (0.4 psi)(548.4 mV/psi) = 219.4 mV. So,
ΔV = .2194 V = Vref2
.2194/8 = 2
-n
-n
0.027= 2
-n
Taking logarithms,
log(0.027) = - n log(2)
-1.569 = - 0.30303n
n = 5.178
So, we must use a 6-bit ADC.
= (8)2
-n
57
(b) The maximum measurable pressure occurs when the output is 1111112, so we can use
Vmax = Vref (2
n-1 n
/2 ),
Vmax = 8(63/64) = 7.875 volts for a pressure of,
Pmax = (7.875V)/(0.5484 V/psi) = 14.36 psi
3.28 An ADC that will encode pressure data is required. The input signal is 635.8 mV/psi.
a. If a resolution of 0.5 psi is required, find the number of bits necessary for the ADC. The
reference is 10.0 V.
b. Find the maximum measurable pressure.
58
3.29 A sample-and-hold circuit like the one shown below has C = 0.56 μF, and the ON
resistance of the FET is 50 Ω. (a) For what signal frequency is the sampling capacitor
voltage down 3 dB from the signal voltage? (b) How does this limit the application of the
sample hold?
U1
Vin
Q1
Vout
C
S/H Voltage
Solution
(a) A model of the “ON” FET and capacitor shows that the system acts like a low-pass
filter of R = 50 Ω and C = 0.56 μF. The voltage appearing across the capacitor is down
1
3 dB at the critical frequency of the filter, fc =
, so,
2πRC
fc =
1
2π (50)(0.56 x 10-6 )
= 5684 Hz
(b) The limitation is the fact that the system cannot be used to sample signals with a
frequency greater than about 5.684 kHz due of attenuation.
3.30 A sample-and-hold circuit like the one shown above has C = 0.068 μF, and the ON
resistance of the FET is 60 Ω. For what signal frequency is the sampling capacitor voltage
down 3 dB from the signal voltage?
59
3.31 A S/H and ADC combination has a throughput expressed as 50,000 samples per second.
Explain the consequences of using this system to take samples every 5 ms.
Solution
A throughput of 50,000 samples per second means that there must be at least (1/50,000) =
20 μs between samples. If samples are taken every 5 ms = 5000 μs then the time available
for signal processing between samples is 5000 μs - 20 μs = 4980 μs.
3.32 A S/H and ADC combination has a throughput expressed as 40,000 samples per second.
Explain the consequences of using this system to take samples every 2 ms.
3.33 A data-acquisition system has ten input channels to be sampled continuously and
sequentially. The multiplexer can select and settle on a channel in 4.2 μs, the ADC converts
in 29 μs, and the c omp ut e r processes a single channel of data in 325 μs. What is the
minimum time between samples for a particular channel?
Solution
The total time for selecting, inputting and processing one channel is,
t = (4.2 + 29 + 325) μs = 358.2 μs
Therefore the total time for all 10 channels is,
T = 10t = 10(358.2 μs) = 3582μs
This is the minimum time between samples of a particular channel.
3.34 A data-acquisition system has twelve input channels to be sampled continuously and
sequentially. The multiplexer can select and settle on a channel in 3.1 μs, the ADC converts
in 18 μs, and the c omp ut e r processes a single channel of data in 265 μs. What is the
minimum time between samples for a particular channel?
60
3.35 A 10-bit ADC has a 12.0-V reference.
a. Find the output for inputs of 4.3 V and 8.2 V.
b. What range of inputs could have caused the output to become A5H?
Solution
Given a 10-bit ADC with a 12.0 volt reference.
(a) For an input of 4.3 volts we find the output as,
N10 =
4.3 10
Vin n
(2 ) =
(2 ) = 366.933 ≈ 366 ≈ 16EH
Vref
12
For an input of 8.2 volts the output is;
N10 =
8.2 10
Vin n
(2 ) =
(2 ) = 699.733 ≈ 699 ≈ 2BBH
12
Vref
(b) For A5H we first find that A5H = 16510 Then,
Vin =
Vin =
Vref (Input in decimal)
(12)(165)
210
2n
=
(12)(165)
= 1.934 V
1024
But the output will stay A5H until the input changes by the voltage of one LSB,
AV =
Vref
2
n
=
12
210
= 0.012V , so the range is 1.934 V to (1.934 + 0.012) = 1.946 V.
3.36 A 12-bit ADC has a 10.0-V reference.
a. Find the output for inputs of 3.8 V and 6.8 V.
b. What range of inputs could have caused the output to become B8H?
61
SAN JOSE STATE UNIVERSITY
Department of Aviation & Technology
TECH 167: Control Systems
Dr. Julio R. Garcia
Thermal Sensors
1. (a) What is a sensor?
A sensor is a transducer that converts a physical variable such as pressure, temperature, flow,
etc., into an analog quantity (voltage or current) or in resistance.
(b) Provide an example of a temperature sensor:
2. Describe an RTD and a thermistor.
RTD (Resistance-temperature detector): variation of metal resistance with temperature.
Thermistor: variation of semiconductor resistance with temperature. Two types:
NTC (Negative Coefficient Temperature): Resistance decreases when temperature increases
PTC (Positive Coefficient Temperature): Resistance increases when temperature increases.
3. Provide two types of NTC thermistors and type of PTC thermistor.
4. See the following Figure and answer the following questions:
3
Nickel
R(T)
R(25 º C)
Platinum
2
1
-100
0
100
200
300
400 500
Temperature (ºC)
600
a) Are the two curves nearly linear? →Yes.
b) Which metal has a better linear response with temperature? → Platinum.
700
800
62
5. An RTD has α (25 °C) = 0.006/°C. If R= 108 Ω at 25°C, find the resistance at 35°C.
Solution
The equation to be used is: R(T) = R(T0)[1 + α0(T-T0)]
Where
R(T) is the approximation of resistance at temperature T
R(T0) is the resistance at temperature T0
α0 is the fractional change in resistance per degree of temperature at T0
Therefore: R(35°C) = 108[1 + 0.006(35 - 25)] = 114.48 Ω
6. If α (25 °C) = 0.005/°C and R = 112 Ω at 25°C, Find the resistance at 20 °C:
and at 30 °:
7. The RTD of question 5 is used in the bridge circuit shown below. Calculate the voltage the
detector must be able to resolve in order to resolve a 1.0°C change in temperature.
R1
120 Ohm
R2
120 Ohm
V1
12 V
V
R3
120 Ohm
R4
RTD
108 Ohm
Solution
Note that the bridge is not nulled at 25°C since the RTD is 108 Ω at that temperature, not 120 Ω.
We find the off-null voltage at 25 °C and then the voltage at 26 °C. The difference will be the
required detector resolution for a 1°C change.
We use the equation:
R4 ⎞
⎛ R3
−
ΔV = V⎜
⎟
⎝ R1 + R 3 R 2 + R 4 ⎠
12
108 ⎞
⎛
ΔV(25) = V⎜
−
⎟ = 0.316 V
⎝ 120 + 120 120 + 108 ⎠
63
The RTD resistance at 26 °C is,
R(26 °C) = 108[1 + 0.006(26 - 25)] = 108.648 Ω
So the off-null voltage is,
ΔV = 12[120/240 - 108.648/228.648] = 0.298 V.
Thus the difference, which is the required resolution, is
VRes = 0.316 V - 0.298 V = - 0.018 V or - 18 mV
8.
If in problem 7, we need to resolve a 1.5°C change in temperature calculate the voltage the
detector must be able to resolve.
9.
Use the values of RTD resistance versus temperature shown in the table to find the equations
for the linear approximation of resistance between 95°C and 125°C.
T(ºC) R(Ohm)
Assume T0 = 110 °C.
90.0 553.45
Solution
95.0
569.63
100.0
574.70
Where:
105.0
579.82
R2 is the resistance at temperature T2.
110.0
585.31
R1 is the resistance at temperature T1.
115.0
590.16
here, T0 = 110°C, T1= 95 °C, T2 = 125 °C and the corresponding resistances
(from the table) are,
120.0
595.42
125.0
603.21
R0 = 585.31 Ω, R1 = 569.63 Ω and R2 = 603.21 Ω so,
130.0
606.81
We use the following equation:
α0 = [1/ R(T0)] [(R2-R1)/( T2-T1)]
α0 = (1/585.31)[( 603.21 - 569.63)/(125 - 95)] = .001912
α0 = 0.0019 /°C
Thus, the equation for the linear approximation of resistance between 95°C and 125°C is:
R(T) = 585.31 [1 +0.0019(T - 110)]
64
10. Find the equations for the linear approximation of resistance between 90°C and 130°C.
Assume T0 = 110 °C.
11. Suppose the RTD of Problem 5 has a dissipation constant (PD) of 30 mW/°C and is used in a
circuit that puts 10 mA through the sensor. If the RTD is placed in a bath at 120°C, (a) what
resistance will the RTD have? (b) What then is the indicated temperature?
Solution
(a) In a bath at a temperature of 120°C the resistance of the RTD should be,
R(120°C) = 108[1 + 0.006(120 - 25)] = 169.56 Ω
However, if there are 10 mA through the sensor then the self heating will cause a
temperature rise from the power dissipation. The power dissipated is,
P = I2R = (0.01)2(169.56) = 0.017 W = 17 mW
Thus the temperature rise will be,
ΔT = P/PD = 17 mW/30 mW/°C = 0.567 °C
So the resistance will be,
R(120.567 °C) = 108[1 + 0.006(120.567 - 25)] = 169.927 Ω
(b) If you didn't know about the self-heating temperature rise you would think the temperature was
120.0°C. The temperature is actually 120°C + 0.567 °C = 120.567 °C.
65
12. If in problem 11 the dissipation constant (PD) is 20 mW/°C, the current through the sensor is
15 mA and the RTD is placed in a bath at 130°C, (a) what resistance will the RTD have?
(b)What then is the indicated temperature?
13. Using an RTD with α = 0.0045/°C and R = 110 Ω at 20 °C, design a bridge and op amp
system to provide a 0.0- to 12.0-V output for a 20 °C to 130 °C temperature variation.
Solution
First we find the resistance of the RTD at the two temperature extremes,
R(20 °C) = 110 Ω (given)
R(130 °C) = 110[1 + .0045(130 - 20)] = 164.45 Ω
If we use this in a bridge with all arms at 110 Ω then it will null at 20 °C, which is good.
The equation to be used is:
ΔV =
VR 3
VR 4
−
R1 + R 3 R 2 + R 4
Assuming a 15 volt bridge excitation voltage, we find the off-null voltage at 130 °C as,
164.45
⎛ 110
ΔV = 15 ⎜
−
⎝ 110 + 110 110 + 164.45
⎞
⎟ = - 1.49 V
⎠
So, to get the required output of 12 volts we need a gain of,
Av = 12/1.49 = 8.05
66
The following circuit will provide this result.
R1
110 Ohm
R2
110 Ohm
V1
12 V
10kOhm
R7
U1
10kOhm
R5
R3
110 Ohm
R4
RTD
U3
Vout
U2
10kOhm
R6
80.5kOhm
R8
14. If α = 0.0035/°C and R = 120 Ω at 25 °C, design a bridge and op amp system to provide a 0.0to 10.0-V output for a 25 °C to 125 °C temperature variation.
67
15. A calibrated RTD with α = 0.0056/°C, R = 295.8 Ω at 20°C, and PD = 25 mW/°C will be used
to measure a critical reaction temperature. Temperature must be measured between 40 and
120°C with a resolution of 0.1 °C. Devise a signal conditioning system that will provide an
appropriate digital output to a computer.
Solution
From the conditions of the problem, 40 to 120 °C is a span of 80 °C and a resolution of 0.1°C
means 80/0.1 = 800 increments. A 9-bit computer provides only 512 but a 10-bit provides 1024, so
we must use a 10-bit ADC, unipolar and with a 5.000 V reference. A 10-bit ADC is common so we
are in good shape.
The expected resistance variation will be,
R40 = 295.8 [1 + 0.0056 (40 - 20)] = 328.9 Ω
R120 = 295.8 [1 + 0.0056 (120 - 20)] = 461.5 Ω
Let's use a bridge for the RTD (although an op amp circuit could be used). We must keep the
self-heating below 0.01°C to maintain the 0.1°C resolution. Thus, P/PD = 0.01°C, and
P = PD(0.01) = (25 mW)(0.01) = 0.25 mW.
At the maximum temperature, 120 °C, R = 461.5 Ω; thus,
P = I2 R → I =
P / R = 0.00025 / 461.5 = 0.74 mA
Thus the bridge voltage across the RTD should be about, Vamax = (461.5 Ω)(0.74 mA) ≈ 0.34
volts. We design so that Va = 0.34 volts at 40°C, which means Vb will be 0.34 volts also so that the
output is AV = 0 volts. This is shown in the schematic below for the bridge.
At 40 °C, RTD = 328.9 Ω.
Va =
RTD
328.9
V = 0.34V →
5V = 0.34V → R1 = 4.502 kΩ.
RTD + R1
328.9 + R1
Assuming R4 = 1 kΩ,
Vb =
R4
1k
V = 0.34V →
5V = 0.34V → R2 = 13.8 kΩ.
R2 + R4
R2 + 1k
Now, at 120 °C we will have a bridge offset voltage of,
ΔV = Va – Vb = 5(461.5)/(461.5 + 4502) - 0.34 = 0.125 V
68
Since the input to the ADC needs to be, 5.000 - 5.000/210 ≈ 5.000 V, thus an amplifier with a gain
of = 5.000/0.125 = 40 is required. The whole equation is,
Vout = Av (Va – Vb)
Va =
RTD
5RTD
and Vb = 0.34 V, so
V=
RTD + R1
RTD + 4502
⎡ 5RTD
⎤
− 0.34⎥
Vout = 40 ⎢
⎣ RTD + 4502
⎦
And the circuit is given below.
R1
4502 Ohm
R2
13.8 Ohm
V1
5V
40kOhm
R7
U1
1kOhm
R5
R3
RTD
R4
1kOhm
U3
Vout
U2
1kOhm
R6
40kOhm
R8
16. If in Problem 15 the temperature must be measured between 50 and 150°C with a resolution of
0.1 °C; what change(s) would you do?
69
SAN JOSE STATE UNIVERSITY
Department of Aviation & Technology
TECH 167: Control Systems
Dr. Julio R. Garcia
Optical Sensors
•
•
•
•
•
•
•
•
•
Sensors should have negligible effect on the measured environment (the process).
Example: Heat developed by an RTD can alter the environmental temperature.
Electromagnetic (EM) radiation allows that transducers do not affect the process-variable
measurements. No physical contact is made.
In process control, EM radiation in either the visible or infrared band is frequently used in
measurement applications.
The techniques of such applications are called optical because such radiation is close to visible
light.
Fundamentals of EM Radiation
EM radiation is a form of energy that is always in motion, that is, it propagates through space.
An object that releases or emits such radiation loses energy.
An object that absorbs radiation gains energy
Frequency and Wavelength
•
•
•
Frequency: oscillations per second.
Wavelength: spatial distance between two successive maxima or minima of the wave in the
direction of propagation.
Speed of Propagation: EM radiation propagates through a vacuum at a constant speed
independent of both the wavelength and frequency.
c = λf
c = 3 x 108 m/s (speed of EM radiation in a vacuum)
λ = wavelength in meters
f = frequency in Hz or cycles/sec (S-1)
1. Determine the wavelength for an EM radiation frequency of 500 kHz.
Solution
c = λf ⇒ λ =
c 3 (108 m/s)
= 600 m.
=
f
5 (105 s -1 )
70
2. If the wavelength is 250 m determine the radiation frequency.
•
When such radiation moves through a nonvacuum environment, the propagation velocity is
reduced to a value less than c.
•
The new velocity is indicated by the index of refraction (n)
n = c/v
v = velocity of EM radiation in the material (m/s)
3. A certain material has an index of refraction of n = 1.26. Find the velocity of EM radiation in
this material.
Solution
n=
3 (108 m/s)
c
c
⇒v= =
= 2.38 x 108 m/s
v
n
1.26
4. A certain material has a velocity of EM radiation of 1.46 x 107 m/s. Find the index of refraction
in this material.
5. A certain source of light has a frequency of 4.0 x 1012 Hz. What is its wavelength in nm, um,
and Å?
Solution
c 3 (108 m/s)
c = λf ⇒ λ = =
= 75 μm = 75,000 nm = 750,000Å
f 4 (1012 s -1 )
1 Å = 10-10 m
Wavelength Units
Angstrom (Å) = 10-10 m or 10-10 m/Å
71
6. A certain source of light has a frequency of 2.8 x 1012 Hz. What is its wavelength in Å, um, and
nm?
Characteristics of Light
•
Photon: EM radiation at a particular frequency can propagate only in discrete quantities of
energy.
•
These discrete units or quanta are called photons.
Wp = hf = hc/λ
Wp = photon energy (J)
h = 6.63 x 10-34 J-s
(Planck’s constant)
The energy of one photon is very small compared to electron energy.
7. A microwave source emits a pulse of radiation at 1.3 GHz with a total energy of 0.8 J.
Determine:
a) The energy per photon.
Solution
Wp = hf = (6.63 x 10-34 J-s)(1.3 x 109 s-1) = 8.62 x 10-25 J
b) The number of photons in the pulse.
Solution
N=
W
0.8J
= 9.28 x 1023 photons
=
Wp (8.62)(10− 25 J/photon)
8. A microwave source emits a pulse of radiation at 3.1 GHz with a total energy of 1.2 J.
Determine (a) the energy per photon and (b) the number of photons in the pulse.
72
Intensity
I = P/A
I = Intensity in W/m2
P = Power in W
A = Beam cross-sectional area in m2.
Intensity is better expressed in mW/cm2
Divergence
•
Radiation travels in straight lines.
•
Intensity of the light may change even though the power remains constant.
9. Calculate the intensity of a 12-watt source whose radius is 0.04 m at (a) the source in W/m2 and
mW/cm2 and (b) 1.3 meters away if the divergence is 1.8º.
Solution
a) At the source in W/m2 and mW/cm2.
I=
P 12W
12W
=
=
= 2388.5 W/m2
2
2
A1 π r1
π (0.04m)
2388.5
m2
W 1,000mW
= 238.85 mW/cm2
2
2
W
m
10,000cm
b) 1.3 meters away if the divergence is 1.8º.
r2 = r1 + L tan(θ)
θ
r1
L
R2 = 0.04 m + (1.3 m)(tan 1.8°) = 0.0808 m.
I=
P
A
2
=
12W
π r2
2
=
12W
π (0.0808m)
2
= 585.071 W/m2
73
10. Calculate the intensity of a 20-watt source whose radius is 0.05 m (a) at the source in W/m2 and
mW/cm2 and (b) 2.1 meters away if the divergence is 2.4º.
Photodetectors
•
In most process-control-related applications, the radiation lies in the range from IR through
visible and sometimes UV bands.
•
The measurement sensors are called photodetectors
•
Four types of photodetectors:
–Photoconductive
–Photovoltaic
–Photoemissive
–Photodiode
Photoconductive Detectors
•
Also called photo resistive cells
•
Resistance changes with light intensity
•
As intensity increases, the semiconductor resistance decreases, making the resistance an
inverse function of radiation intensity.
Photovoltaic Detectors
•
They generate a voltage that is proportional to incident EM radiation intensity
•
They convert the EM energy into electrical energy
74
Equiv Circuit for a photovoltaic cell
Irad
Rc
+
Vc
=
•
Photovoltaic cells have a range of spectral response within which a voltage will be produced
•
Vc varies with light intensity in an approximately logarithmic fashion
•
The internal resistance of the cell also varies with light intensity. This complicates the design
of systems to derive maximum power from the cell, since RL optimum = Rc.
Signal Conditioning
Error!
Rc
Irad
+
Isc
Vc
R
U1
Vout
•
ISC (short-circuit current) can be obtained by connecting the cell directly to an op-amp.
•
Since the current is linearly proportional to light intensity, so is the output voltage
11. A CdS cell has a dark resistance of 120 kΩ and a resistance in a light beam of 25 kΩ. The cell
time constant is 60 ms. Design a system to trigger a 2-volt comparator within 6 ms of the beam
interruption.
Solution
The equation to be used is:
-t/τ
R(t) = Ri + (Rf – Ri)[1 – e
]
-6/60
R (6 ms) = 25 KΩ + (120 – 25) KΩ [1 – e
] = 34.04 KΩ
75
This means that at 6 ms, the CdS’s resistance is 34.04 KΩ. The circuit that can be used is:
1kOhm
R2
V1
1kOhm
R1
Photocell
U1
Vo1
U2
Vout
VCC
2V
Vo1 = 2 V =
− R2
V1
R1
+2V=
− 34.04 K
V1
R1
Assume V1 and determine R1.
Example, If V1 = -1 V, then R1 =
(- 34.04 K)(- 1 V)
= 17.02 KΩ
2V
12. A CdS cell has a dark resistance of 150 kΩ and a resistance in a light beam of 20 kΩ. The cell
time constant is 50 ms. Design a system to trigger a 1.2-volt comparator within 4 ms of the
beam interruption.
76
13. A photovoltaic cell is to be used with radiation of intensity from 4 to 15 mW/cm2.
Measurements show that its unloaded output voltage ranges from 0.15 to 0.45 volts over this
intensity while it delivers current from 0.4 to 1.9 mA into an 80 Ω load. (a) Determine the
range of short-circuit current. (b) Develop signal conditioning to provide a linear voltage from
0.3 to 1.5 V as the intensity varies from 4 to 15 mW/cm2.
Solution
a) Determine the range of short-circuit current.
2
Intensity = 4 to 15 mW/cm
Vc (no load) = 0.15 to 0.45 V
I = 0.4 to 1.9 mA
RC
RC
Isc
IL =
80 Ohm
RL
Vc
Rc + 80
Rc IL + 80 IL = Vc
Rc =
Vc - 80 I L
IL
At 4 mW/cm2, Vc = 0.15 V & IL = 0.4 mA
Rc =
0.15 − 80(0.4mA)
= 295 Ω
0.4mA
At 15 mW/cm2, Vc = 0.45 V & IL = 1.9 mA
Rc =
0.45 − 80(1.9mA)
= 157 Ω
1.9mA
The short-circuit current (Isc) is given by Isc =
Vc
Rc
At 4 mW/cm2, Vc = 0.15 V & Rc = 295 Ω
Isc =
Vc 0.15 V
=
= 0.51 mA
Rc 295 Ω
At 15 mW/cm2, Vc = 0.45 V & Rc = 157 Ω
Isc =
Vc 0.45 V
=
= 2.87 mA
Rc 157 Ω
The range of short-circuit current is from 0.51 mA to 2.87 mA.
80 Ohm
RL
Vc = Isc Rc
77
b) Develop signal conditioning to provide a linear voltage from 0.3 to 1.5 V as the intensity
varies from 4 to 15 mW/cm2.
The circuit that we will use is shown below:
VCC
5V
330 Ohm
R5
U1
R6/100
R3
R4
50%
U3
80 Ohm
U2
V2
R1
R2
Vc/Cell
At 4 mW/cm2,
Isc = 0.51 mA
We need to obtain Vout = 0.3 V (condition of the problem)
V2 = (0.51 mA) (80) = 40.8 mV
At 15 mW/cm2,
Isc = 2.87 mA
We need to obtain Vout = 1.5 V (condition of the problem)
V2 = (2.87 mA) (80) = 229.6 mV
1.5
Vo (V)
m (Av)
0.3
40.8
229.6
V2 (mV)
Vout
78
Vout = Av V2 + Voffset (1)
Av =
1.5 − 0.3
= 6.356
( 229.6 − 40.8) mV
0.3 = 6.356 (40.8mV) + Voffset
Voffset = 40.68 mV
From Eq. (1): Vout = 6.356 V2 + 40.68 mV = 6.356 (V2 + 6.4 mV)
•
Adjust R6 until Voffset = - 6.4 mV
•
Make R1 = R3 = 1 kΩ
•
Make R2 = R4 = 6.356 kΩ
Another solution:
V2
10kOhm
63.56kOhm
10kOhm
+6.4 mV
10kOhm
U1
V210kOhm
U2
Vout
14. A photovoltaic cell is to be used with radiation of intensity from 6 to 20 mW/cm2.
Measurements show that its unloaded output voltage ranges from 0.24 to 0.56 volts over this
intensity while it delivers current from 0.3 to 2.5 mA into a 60 Ω load. (a) Determine the range
of short-circuit current and (b) Develop signal conditioning to provide a linear voltage from 0.2
to 2.0 V as the intensity varies from 6 to 20 mW/cm2.
79
SAN JOSE STATE UNIVERSITY
Department of Aviation & Technology
Tech 167: Control Systems
Dr. Julio R. Garcia
Final Control
1. A 4 - 20-mA control signal is loaded by a 100 Ω resistor and must produce a 20 - 40 V motor
drive signal. Find an equation relating the input current to the output voltage.
Solution
The 100 Ω resistor provides Va = 100 I so as I varies from 4 mA to 20 mA, this voltage will vary
from 0.4 V (Va = 100 Ω x 4 mA = 400 mV = 0.4 V) to 2.0 volts (Va = 100 Ω x 20 mA = 2 V).
There must be a linear circuit that converts this voltage variation into 20 to 40 volts. So,
Vout = mVa + Vo
(Eq. 1)
Using the given conditions provide the equations,
20 = 0.4m + Vo
40 = 2.0m + Vo
subtracting,
20 = 1.6m or m = 20/1.6 = 12.5 then,
20 = (0.4)(12.5) + Vo
Vo = 20 - 5 = 15
Therefore, from Eq 1:
Vout = 12.5 Va + 15
(Eq. 2)
Since, Va = 100 I
then,
Vout = 1250 I + 15
(I in amperes)
2. Implement the equation of Problem 1 if a power amplifier is available that can output 0-100 V
and has a gain of 10.
Solution
Since the power amplifier has a gain of ten, the equation above can be reduced by a factor of ten.
Using equation 2,
Vout = 1.25 Va + 1.5
Or
Vout =1.25 (Va + 1.2)
This can be provided by a differential amplifier as follows,
80
VCC
12V
V1
R2
50%
125kOhm
R5
200 Ohm
R3
100kOhm
R9
1.2V
Adjust R2 until V1 = 1.2 V
100kOhm
Key = A
R6
100 Ohm
I
100 Ohm
R1
U1
10
Vout
125kOhm
R4
3. Implement the equation of Problem 1 if a power amplifier is available that can output 0-80 V
and has a gain of 12.
81
4. The power in the load must be 2 kW, determine the triggering angle, α.
D4
D1
220 VAC
D3
D2
10 Ohm
RL
50%
SCR
DIAC
Solution
Since the SCR will work as a full-wave device due to the bridge rectifier,
SCR is ON
SCR is OFF
0 α
π
VL = Vav =
1 π
Vmax sinwt dwt
π ∫α
VL = Vav =
− V max
(cos π − cos α)
π
(Eq. 3)
The power across the load is calculated as:
PL =
VL =
(VL )2 , where
RL
PL R L = (2000)(10) = 141.4V
From equation 3:
− 220 2
( −1 − cos α)
π
141.4 π V= 220 2 + 220 2 cos α
141.4 V =
⎛ 141.4 π - 220 2 ⎞
⎟ = 64.5°
The triggering angle, α is: α = cos-1 ⎜⎜
⎟
220 2
⎝
⎠
82
5. The power in the load must be 4.2 kW. Determine the triggering angle, α.
D4
D1
330 VAC
D3
D2
6 Ohm
RL
50%
SCR
DIAC
83
6. Calculate the power dissipated in the load. α = triggering angle = 50°
18 Ohm
RL
50%
220 VAC
TRIAC
DIAC
Solution
A TRIAC is a full-wave device. Thus, the equation for calculating the PL or Vav is the same as
shown in Eq. 3:
VL = Vav =
− V max
(cos π − cos α)
π
VL = Vav =
− 220 2
(cos π − cos 50°) = 162.70 V
π
The power across the load is:
PL =
(VL )2 = (162.70)2 = 1,470.6 W
RL
18Ω
7. Calculate the power dissipated in the load. α = 63°
22 Ohm
RL
50%
660 VAC
TRIAC
DIAC
84
8.
In the following circuit, α = triggering angle = 35°. (a) Calculate PL (b) If D1 and D3 open
find PL (c) If D2 opens determine PL. (d) If it is required that PL = 1200 W, calculate α.
D4
D1
330 VAC
D3
D2
12 Ohm
RL
50%
TRIAC
DIAC
Solution
a)
Calculate PL
From Eq. 1:
VL = Vav =
PL =
b)
− 330 2
(cos π − cos 35°) = 270.2 V
π
(VL )2 = (270.2)2 = 6,084 W
RL
12Ω
If D1 and D3 open find PL.
If D1 and D3 open, then the TRIAC will function as a half-wave device. Therefore,
VL = Vav =
− V max
(cos π − cos α)
2π
VL = Vav =
− 330 2
(cos π − cos 35°) = 135.1 V, and
2π
PL =
c)
(VL )2 = (135.1)2 = 1,521 W
RL
12Ω
If D2 opens determine PL.
If D2 opens, the solution is the same as part (b)
85
If it is required that PL = 1200 W, calculate α.
d)
PL =
(VL )2 ⇒ (V
L)
RL
2
VL =
= PL RL
PL R L = (1200)(12) = 120V
From Eq 1:
120 V =
− 330 2
− 330 2
(cos π − cos α) =
( −1 − cos α)
π
π
⎛ 120 π - 330 2 ⎞
⎟ = 101°
α = cos-1 ⎜⎜
⎟
330
2
⎝
⎠
9.
In the following circuit: (a) Calculate PL, (b) If D3 opens determine PL. α = triggering
angle = 48°
D4
D1
500 VAC
D3
D2
8 Ohm
RL
50%
TRIAC
DIAC
86
10. In the circuit shown below: (α = triggering angle = 35°). (a) Determine PL. (b) If the diode
shorts, calculate PL. (c) If D1 diode opens, calculate PL.
4 Ohm
RL
50%
D1
440 VAC
SCR
DIAC
Solution
a. Determine PL.
The SCR is a half-wave device.
VL = Vav =
1 π
Vmax sinwt dwt
2π ∫α
VL = Vav =
− V max
(cos π − cos α)
2π
VL = Vav =
− 440 2
(cos π − cos 35°) = 180.16 V, and
2π
PL =
(VL )2 = (180.16)2 = 8,114.4 W
RL
4Ω
b. If the diode shorts, calculate PL.
If the diode shorts, the SCR’s gate will receive both positive and negative voltages. The SCR
cannot withstand large negative voltages at its gate, therefore, the SCR will blow up. Thus, PL = 0
c. If the diode opens, calculate PL.
If the diode opens, the SCR won’t receive any excitation at its gate. Thus, the SCR won’t conduct
and no current will flow through the load. Therefore, PL = 0.
87
11. In the circuit shown below: (α = triggering angle = 35°)
12 Ohm
RL
660 VAC
50%
D1
SCR
DIAC
a. Determine PL.
b. If D1 diode opens, find PL.
c. If D1 diode is reversed, find PL.
d. If DIAC opens, find PL.
e. If SCR shorts between A and K, find PL.
f. If the SCR is replaced by a TRIAC, find PL.
g. If the load shorts, find PL.
88
12. For each circuit identify the formula that you would use to find Vav and PR1 (load).
1kOhm
R1
R2
50%
D1
330 VAC
1uF
C1
1kOhm
R1
R2
50%
330 VAC
D2
1uF
C1
1kOhm
R1
R2
50%
D1
330 VAC
D2
1uF
C1
89
1kOhm
R1
R2
50%
D1
330 VAC
D2
1uF
C1
1kOhm
R1
R2
50%
D1
330 VAC
D2
1uF
C1
D1
D4
330 VAC
D2
D3
1kOhm
R1
R2
50%
D5
D6
1uF
C1
90
D1
D4
330 VAC
D2
D3
1kOhm
R1
R2
50%
D5
D6
1uF
C1
D1
D4
330 VAC
D2
D3
1kOhm
R1
R2
50%
D5
D6
1uF
C1
91
13. (a) If diode D4 opens and θ = 42° then the SCR is =
(b) If diode D4 shorts and θ = 55° then the SCR is =
(c) If R1 opens and θ = 60° then the SCR is =
Select:
(a) working as a λ/2
(b) working as a λ
(c) open
(d) shorted.
560 VAC
100 Ohm
R1
R2
50%
D4
D2
1uF
C1
14. (a) TRIAC is
(b) If D4 is reversed then TRIAC is
(c) If D4 is shorted then TRIAC is
Select:
(a) working as a λ/2
(b) working as a λ
(c) open
(d) shorted.
12 Ohm
R1
R2
50%
D4
800 VAC
D2
1uF
C1
92
15. (a) PR1 is
(b) If D4 opens then PR1 is
(c) If D4 shorts then PR1 is
(d) If R1 opens then TRIAC
500 VAC
R2
50%
Select:
(a) zero
(b) greater than zero
(c) less than zero
(d) blows up
(e) remains intact
5 Ohm
R1
D4
D2
1uF
C1
16. In the following circuit (α = 48°)
(a) PRL1 is
(b) PRL2 is
(c) If D5 shorts then PRL1 is
(d) If D5 and D6 open then PRL2 is
500 VAC
5 Ohm
RL1
D5
D6
Select:
(a) zero
(b) greater than zero
(c) less than zero
10 Ohm
RL2
93
17. In the following circuit (α = 98°)
(a) PL is
(b) SCR is
(c) If diode shorts then SCR is
(d) If diode opens then PL is
Select:
(a) zero
(b) greater than zero
(c) less than zero
(d) blows up
(e) remains intact
(f) working as a λ/2
(g) working as a λ
50%
440 VAC
94
SAN JOSE STATE UNIVERSITY
Department of Aviation & Technology
TECH 167: Control Systems
Dr. Julio R. Garcia
Controller Principles
1. What is a control system?
A control system is a group of properly arranged devices and components that maintain a
certain process at a desired level.
2. Provide two examples of a control system.
3. Why is the control in industrial processes very critical?
The control in industrial processes is very critical because some areas are very hazardous or
impossible for human operators to work in such as high-temperature environments and
high-voltage surroundings.
4. What is the classification of Control Systems?
Process
Being
Controlled
Temperature control systems → Temperature
Flow control systems → Fluid flow
Level control systems → Height of material in holding bins or reservoirs
Nature of
Controlling
Components
Analog
Digital
Open-loop
Feedback
Closed-loop
95
5. Closed-loop control system. Figure 1 below shows a typical closed-loop control system
Error
amplifie
Forward path
Error
signal
Controller
Setpoint
Vref
Process variable signal
Figure 1
Output signal
conditioning
Disturbance
Output
actuator
Input signal
conditioning
Process
Input
sensors
Feedback path
a. Is this a closed-loop or an open-loop control system? Why?
b. What is Vref and with what other names is also called?
Vref is the desired operating point for the process. Other names are Set-point, Command,
or Reference.
c. What does vf represent and with what other names is also called?
vf is the signal that represents current process status. Other names are process variable,
measured value or controlled variable.
d. What does the Error Amp represent? How can we implement the Error Amp?
The Error Amp is a circuit that represents whether the process is under control. The
Error Amp can be implemented through an Error detector, Comparator or Summing
amplifier.
e. What is vε, what other names is also called and what is the equation?
vε is the Error Amplifier output. Other names are Error signal or System deviation signal.
The equation is:
vε = Av (vref – vf); If Av = 1 ⇒ vε = vref - vf
96
f.
Briefly describe the Controller, Output signal conditioning, Output actuator, Input
sensor and Input signal conditioning:
The Controller provides a corrective signal. Output will depend on vε.
The Output signal conditioning or Signal conditioner is the interface between the
controller output (a signal) to the output actuator.
The Output actuator or Final correcting device directly affects a process change: motor,
heater, solenoid, etc.
The Input sensor detects any changes in the process respect to the set-point.
The Input signal conditioning converts the output from the input sensor to a process
variable signal.
6. What is the Controller?
The Controller is the heart of any electronic control system and possesses the following
characteristics:
a. It maintains the process variable within acceptable limits of the set point.
b. The smaller the variations of the process from the set point the better the controller.
c. The faster the controller responds when the process variable deviates from the set point the
better the controller.
7. What are the types of Controllers?
•
•
•
•
•
•
•
ON-OFF
Proportional
Integral
Derivative
Proportional-Integral
Proportional-Integral Derivative (PID)
Digital Proportional-Integral
Selection of type of controller depends on speed of response, allowable system error and process
dynamics.
ON/OFF Controllers. (Two-position controllers)
a. Output is fully ON or fully OFF. It is inexpensive but limited.
Process variable > Setpoint
Process variable < Setpoint
Direct acting controller
Inverse acting controller
(Process variable and controller
output move in the same
direction)
(Process variable and controller
output move in the opposite
direction)
ON
OFF
OFF
ON
97
b. An ON/OFF controller must have some degree of hysteresis. Otherwise, the output will
oscillate. This may destroy the system.
Controller output ON 100%
Figure 2
Controller output OFF
- Error
0
Error
Deadband
+ Error
The controller output will remain OFF until the error signal decreases to the level of -Error. When
this level is reached the controller output is ON. The output will remain ON until the error signal
reaches the level of +Error. At this point, the controller output is OFF.
Deadband = ErrorON - ErrorOFF
The deadband is the difference between the error signal that turns the controller output fully ON
and the error signal that turns the controller output fully OFF.
Inherent part
of the system
Deadband
Implemented
electronically
- Mechanical hysteresis
- Thermal lag
98
Analyze Figure 3.
Error!
10kOhm
R3
+V
Figure 3
10kOhm
R2
Process
variable
Controller
output
U1
U2
R1
Setpoint
adjust
10kOhm
R4
50%
R8
R6
10kOhm
R5
D
R7
-V
U1 and associated components
Error amplifier with unity gain.
vo1 = Error signal = Av1 (Vset-point - Vprocess variable) = Av1 (VREF – vf)
Since Av1 = 1 ⇒ vo1 = Vset-point - Vprocess variable = VREF – vf
U2 and associated components
Comparator
VUTP =
R7
(+ Vsat)
R6 + R7
VLTP =
R7
(- Vsat)
R6 + R7
When Error signal > VUTP ⇒ vo2 = - Vsat
When Error signal < VLTP ⇒ vo2 = + Vsat
⎛ R7 ⎞
Deadband = VUTP – VLTP = 2 ⏐Vsat⏐ ⎜
⎟
⎝ R6 + R7 ⎠
Controller output is between – 0.7 V and + Vz.
In actual applications, the controller output must be limited between 0 and V+ or 0 and V-. R8 and
D serve this purpose.
99
Problem 1. See Figure 3. If ± V = ± 18 V, R6 = 120 kΩ, R7 = 27 kΩ and D = 5.6 V.
a. Find the circuit Deadband.
b. Draw the circuit Transfer Curve.
Solution
⎛ 27 ⎞
⎛ R7 ⎞
a.) Deadband = 2 ⏐Vsat⏐ ⎜
⎟ = 2 ⏐16 V⏐ ⎜
⎟ = 6.8 V
⎝ 120 + 27 ⎠
⎝ R6 + R7 ⎠
b)
Controller output
Maximum
Controller output
+5.6V
Minimum
Controller output
-0.7V
-3.4V
- Error
Figure 4
0
+ 3.4V
+ Error
E
Deadband = 6.8V
Deadband Endpoints
Problem 2. See Figure 3. If ± V = ± 15 V, R6 = 150 kΩ, R7 = 33 kΩ and D = 6.8 V.
a. Find the circuit Deadband.
b. Draw the circuit Transfer Curve.
100
Proportional Controllers
In many applications, the ON/OFF controller output is not acceptable. The output of a
proportional controller varies between fully ON and fully OFF depending on the magnitude of the
error signal. A proportional controller usually has a linear response.
Analyze Figure 5.
Controller output
ON 100%
Figure 5
Offset
Controller output
OFF
Min. Error
0
Max.Error
Proportional band
Proportional band =
Vout,max −Vout,min
, Av = controller gain
Av
3 points of interest
Max Error: Magnitude of error signal that makes output = full-ON.
Min Error: Magnitude of error signal that makes output = full-OFF.
Offset: Point where the curve crosses the Y-axis or controller output when Error = 0. Offset does
not affect the magnitude of the proportional band.
Error, max =
Error, min =
Vout,max
- Voffset
Av
Vout,min
- Voffset
Av
101
Analyze Figure 6.
+V
Process
variable
10kOhm
R2
Figure 6
10kOhm
R3
R9
R11
U1
R8
Error!
R1
Setpoint
adjust
R10
10kOhm
R4
50%
10kOhm
R5
Controller
output
U2
U3
R12
R1
Offset
adjust
50%
D1
R7
U1 and associated components
Similar to Figure 3.
U2 and associated components
Inverting summing amplifier
vo2 = - (Verror + Voffset) ≠ 0 (always)
The offset facilitates correction of the process variable, but it will not allow the controller to
maintain an error of zero.
Gain of U2 determines the slope of the line.
Magnitude of the offset positions the entire curve above the zero-error point.
Analyze Figure 7. This is a typical proportional controller response curve
Figure 7
+ 6.0V
+2.0V
-0.7V
-1.35V
0
Error
+ 2.0V
Proportional band
102
Integral Controllers
The advantage of an integral controller respect to an ON/OFF controller is that an integral
controller can drive the error to zero and maintain it. An ON/OFF controller will never stabilize at
the desired set point; therefore, some error is expected.
1. The principal circuit of an integral controller is
.
2. The output equation of the circuit shown below is:
C
Verror
Vout
R
If Verror is a steady DC, then
Therefore, as t increases, Vout increases (ramp).
Analyze Figure 8.
+V
Process
variable
10kOhm
R3
10kOhm
R2
C
R8
U1
R6
R1
Setpoint
adjust
R7
10kOhm
R4
50%
Controller
output
U2
U3
R10
10kOhm
R5
R9
Figure 8
D
103
Note: Unless Verror is a simple step function, Vout may become very difficult to calculate. See
Figure 9.
Input
signal
Input
signal
Input
signal
Figure 9
Output
signal
Output
signal
A. Step Input
Output
signal
B. Ramp Input
C. Parabolic Input
Problem 1.
The error signal indicated below appears at the integral controller input. R = 20 kΩ and
C = 0.02 uF. If C has an initial voltage of + 1 V, determine the controller output.
Figure 10
+1
+0.5
ms
0
-1
0
1
2
3
104
Proportional-Integral Controller
Integral
Controller
Error = 0
Slow response to
Proportional
Controller
Error
≠
ΔVerror
0
Slow response to
ΔVerror
By combining these two controllers it is possible to obtain Error = 0 and a Fast response to
ΔVerror.
Analyze Figure 11.
Process
variable
Setpoint
Proportional
block
Inverting
Summing
amplifier
Error
amplifier
Controller
output
Integral
block
A. Parallel arrangement
Process
variable
Setpoint
Error
amplifier
Proportional
block
Figure 11
-
Output
Diferential
amplifier
Controller
output
+
Error!
Integral
block
B. Series arrangement
The series arrangement responds faster than the parallel arrangement to ΔVerror. This is because
the integral block receives an amplified error signal. Therefore, it forces the error to zero more
rapidly.
105
Analyze Figure 12. Response of Parallel Proportional-Integral Controller to a Step Input.
Process
variable
Proportional
Block
Av = 1
Error
Amplifier
Av = 1
Setpoint
Integral
Block
RC = 1s
Inverting
Summing
Amplifier
Av = 1
Figure 12
Error
signal
+1V
–1V
1s
2s
3s
Proportional +1V
Block output
–1V
Integral
Block output –1V
+2V
+1V
Controller
output
–1V
Circuit response
4s
Controller
output
106
Analyze Figure 13. The gain of the proportional block = - 2, the RC of the integral block = 1
second, and the gain of the differential amplifier is 1.
Process
variable
Setpoint
Error
amplifier
Proportional
block
+
Integral
block
Figure 13
B. Series arrangement
Error
signal
Proportional
Block output
+2V
+1V
–2V
–4V
+4V
Integral
Block output
+2V
+8V
Controller
output
+6V
+4V
+2V
Output
Diferential
amplifier
Controller
output
107
Derivative Controllers
In many cases, a process has an inherent inertia or hysteresis. This means that a disturbance will
not produce a deviation from the set point immediately. It also means that there is a lag from the
time the process deviates from the set point and the corrective action.
To overcome this sluggish response and prevent oscillations the controller must produce a large
corrective action signal initially but tapers off as time goes on. A Derivative Controllers does this
job.
1. The basic element of a Derivative Controller is a
.
2. Analyze Figure 14. Ideal Differentiator Output Responses.
Input
signal
Input
signal
Input
signal
Figure 14
Output
signal
Output
signal
A. Step Input
Output
signal
B. Ramp Input
C. Parabolic Input
Disadvantages
a.
Noise (high-frequency transients) will produce larger outputs that can saturate the amplifier.
This action can be reduced by inserting a resistor in series with the input capacitor.
b.
The Derivative controller responds only to changes in the error signal. It will not produce a
corrective signal if the system has a steady-state error.
Derivative Controllers are never used alone. Proportional-Integral- Derivative (PID) Controllers
are the industry standard.
108
Proportional-Integral- Derivative (PID) Controllers
PID
Controller
Fast response to system disturbances.
Error = 0
Overcomes lag time without sarurating amp.
Analyze Figure 15.
Proportional
Block
Process
variable
Setpoint
Error
Amplifier
Integral
Block
Derivative
Block
Summing
Amplifier
Controller
output
Figure 15
Several PID variations are possible.
In this case, a parallel configuration is shown.
Each block receives the same error signal and their outputs are added through a summing amp.
Tuning is the process of adjusting each of the 3 blocks. Tuning depends on:
a. the configuration of the controller,
b. the characteristics of the process being controlled, and
c. the desired controller performance.
Tuning is not a simple procedure. Computer simulation programs make this task easier but
accuracy of the results depends on how the system response can be modeled.
Precautions
The action of the integral or derivative block can mask the effect of the other blocks in the
controller.
a. A sudden change (step) in error will saturate the derivative block causing a saturation in the
summing amp. The result may be an overcompensation and this will make the process to
oscillate.
c. If a large error is present in the system for a large period of time, the output of the integral block
may be forced into saturation and will remain in this state even though the error becomes zero.
This output will make the process to overshoot. This condition is corrected when the resultant
negative error brings the integral block out of saturation.
109
Analyze Figure 16. The gain of the proportional block = - 1, the RC of the integral block = 1
second, the RC of the derivative block = 0.2 seconds, and the gain of the summing amplifier is - 1.
Proportional
Block
Process
variable
Setpoint
Error
Amplifier
Integral
Block
Controller
output
Summing
Amplifier
Derivative
Block
Figure 16
1s
Error
signal –1V
2s
3s
Proportional +1V
Block output
+2V
Integral
Block output
+8V
+ Vsat
Derivative
Block output
–8
– Vsat
V
+ Vsat
+8V
Controller
output
– Vsat
8
4s
110
Points to consider
Verror is applied to the three blocks simultaneously.
The integral block output is given by:
The derivative block is driven to:
a. Positive saturation in response to a negative step.
b. Negative saturation in response to a positive step.
Since RC = 0.2 S, the capacitor will charge in
. At this time, the output is 0.
1. Draw the controller output.
The initial voltage in the integrator block is - 1 V. ± Vsat = ± 16 V.
Proportional
Block
Av = -2
Process
variable
Setpoint
Error
Amplifier
Integral
Block
RC = 1s
Derivative
Block
RC = 0.5s
Summing
Amplifier
Controller
output
111
+2V
0
Error
signal -1V
-2V
1
3
6
7 Sec
112
2. Draw the controller output. (time in seconds).
Process
variable
Proportional
Block
Av = -2
Error
amplifier
Setpoint
+
Output
Diferential
Amplifier
Av = 2
Controller
output
Integral
Block
RC = 4S
1
+6
Error
Signal
+4
+1
0
-1
4
5
10
15
19
20
113
SAN JOSE STATE UNIVERSITY
Department of Aviation & Technology
Tech 167: Control Systems
Dr. Julio R. Garcia
Closed-Loop Systems
1.
In the figure 1 shown below:
VEE
-18V
VCC
18V
22kOhm
R4
VPROCESS
VARIABLE
220 Ohm
R2
Figure 1
22kOhm
R6
R1
RTD
22kOhm
R5
50%
U1A
LF444
22kOhm
R7
8.5V
Key = A
10kOhm
R3
TEMP.
ADJUST
R10
U1C
U1B
LF444
LF444
(R9 ADJUSTED TO 710 ohms)
DEADBAND
ADJUST
22kOhm
R8
1kOhm
-
SOLID
+ STATE
RELAY
D1
1N4001GP
Key = A
2.2kOhm 50%
R9
AC
240 VAC
2000 W
HEATER
240 VAC
Analyze the circuit.
The circuit is an ON/OFF temperature control. This circuit controls the temperature of a reaction
chamber shown in Figure 2.
Control valves
Glass reaction
chamber
Reactant
inputs
From RTD sensor
Figure 2
Nichrome heating
element wire
To heater
Controller
Product –of-reaction
outputs
The internal temperature must be between 300°C and 500°C throughout the reaction.
U1A: Differential amplifier, Gain = 1 because R4 = R5 = R6 = R7 = 22 kΩ.
U1B: Window comparator
U1C: Voltage follower (buffer)
114
Vprocess variable
Process-variable feedback is provided by R1 (RTD sensor) assembly suspended inside the reaction
chamber.
RTD: Resistance Temperature detector
R1-R2: voltage divider
R1: RTD sensor. R = 100 Ω at 0°C.
PTC = 0.385Ω/°C Æ resistance increases 0.385Ω/°C
Usable temperature = - 200°C to 750°C, see Figure 3.
Temperature ( ºC )
550
500
450
400
RTD Sensor
Output response
350
300
250
200
150
100
50
Figure 3
2
4
6
8
Process Variable voltage
10
12
14
16
V
Vprocess variable = [R2/(R1 + R2)] (VCC)
As T ↑ ⇒ R1 ↑ ⇒ Vprocess var ↓
R1 = 100 Ω + 0.385Ω/°C (Temp in °C)
However, due to the overshoot in ON/OFF controllers, the thresholds are established between 350
°C to 450 °C.
When the heater is shut off at 350°C it will actually fall below this value maybe closer to 300°C.
Likewise, when the heater is shut off at 450°C, the temperature will continue to rise to maybe
500°C.
From Figure 3 the Vprocess variable for 350°C is 9 V and for 450°C is 8 V.
115
Vtemp adj
Knowing these values, the Vtemp adj is adjusted to 8.5 V to allow the error voltage to vary – 0.5 V
at 350°C to + 0.5 V at 450°C.
Window Comparator (U1B)
To maintain the 350°C - 450°C temperature range, the comparator must switch when one of these
two thresholds (UTP = +0.5 V and LTP = -0.5 V) is reached.
UTP = [R9/(R8 + R9)] (+ Vsat) = [710/(22k + 710)](16 V) = + 0.5 V
LTP = [R9/(R8 + R9)] (- Vsat) = [710/(22k + 710)]( - 16 V) = - 0.5 V
The deadband is:
+16 V
0
-0.7 V
-0.5V
0
Deadband
0.5V
Buffer (U1C)
It isolates the load from U1B to prevent asymmetrical saturation voltages.
Solid-state relay (ZVS = Zero Voltage Switching)
DC input: 3 V – 32 V, I = 5 mA
Output: controls from 24 VAC to 280 VAC, Iout = 10 A.
ZVS: Output will not switch into conduction unless the load potential at the time of triggering is
about 10 VAC.
Output limiter: R10 – D1.
116
System Operation
Let’s assume that the system has been OFF for a period of time.
This means that Vo, U1C = - Vsat = - 16V
Æ Vout, comparator = - 16 V Æ Vpin 5, U1B = LTP = - 0.5 V.
a) Vprocess var = + 13 V (See Figure 3, T = 25°C)
b) Verror = Vtemp adj – Vproc var = 8.5 – 13 = - 4.5 V
c) Since the voltage at pin 6 of comparator (U1B)= - 4.5 V (Verror) and the voltage at pin 5 of
comparator = - 0.5 V, Æ Vout, comparator = + Vsat = + 16 V.
Thus,
1) The heater is turned ON, and
2) The voltage at pin 5 of comparator becomes UTP = + 0.5 V.
d) See Figure 3. As T ↑ ⇒ Vprocess var ↓ ⇒ Verror ↑ (Verror = 8.5 V – Vproc var)
e) When T > 450°C ⇒ Vprocess var < 8 V ⇒ Verror > + 0.5 V
⇒ Vout, comparator = - Vsat = - 16 V.
1) Heater is OFF. However, chamber temperature may continue to rise slightly, and
2) The voltage at pin 5 of comparator is LTP = - 0.5 V.
Problem 1: If the R9 wiper is adjusted to 50% draw the transfer curve.
Solution
50% of R9 = 50% (2.2 kΩ) = 1.1 kΩ
VUTP = [R9/(R8 + R9)] (+ Vsat) = [1.1/(22 + 1.1)] (16 V) = 0.76 V
VLTP = [R9/(R8 + R9)] (- Vsat) = [1.1/(22 + 1.1)] ( - 16 V) = - 0.76 V
+16 V
0
-0.7 V
-0 76V
0
Deadband
0.76V
117
Problem 2: If the R9 wiper is adjusted to 70% draw the transfer curve.
Problem 3: If R1 is substituted by an NTC thermistor, explain the circuit operation.
Solution
Heater will be ON at high temperatures and OFF at low temperatures. To correct this problem,
switch the positions between R1 and R2.
Problem 4: What change(s) would you do to control the temperature between 200°C and 450°C.
Solution
We want 200°C ≤ Temp ≤ 450°C
Due to overshoot we need to control temperature between 250°C and 400°C.
From Figure 14-3:
250°C → V ≈ 10 V
400°C → V ≈ 8.6 V
Setpoint (Temp Adj) = (10 + 8.6)/2 = 9.3 V
Deadband = ± (10 – 9.3) = ± 0.7 V
Adjust R9 until voltage at pin 5 of U1B = ± 0.7 V
Problem 5: What change(s) would you do to control the temperature between 250°C and 400°C.
118
Problem 6: If the wipers of all potentiometers have been adjusted to 50% respect to ground and
the temperature is 150°C, indicate the voltages and/or waveforms that appear at the following test
points: (assume that the heater has been OFF for a long time).
U1A: Pin 1, U1B: Pin 7, U1C: Pin 8
Solution
R3 at 50% produces Vtemp adj = 9 V. (1/2 of VCC, ½ of 18 V = 9 V)
At 150°C, Vprocess variable = 11 V (See Figure 3)
Vout of Error Amp = 9 – 11 = - 2 V ⇒ U1A, pin 1 = - 2 V.
Since the heater has been OFF for a long time, the voltage at pin 7 of U1B = - Vsat. Thus, voltage at
pin 5 of U1B = - 0.76 V See problem 1).
Voltage at pin 6 of U1B = - 2 V. This makes the output (pin 7) of U1B = + Vsat = + 16 V.
Since U1C is a voltage follower, U1C, pin 8 = + Vsat = + 16 V.
Problem 7: If the wipers of all potentiometers have been adjusted to 70% respect to ground and
the temperature is 200°C, indicate the voltages and/or waveforms that appear at the following test
points: (assume that the heater has been OFF for a long time).
U1A: Pin 1, U1B: Pin 7, U1C: Pin 8
Problem 8: What would happen if the gain of U1A were increased?
Solution
Unless some changes are made, the heater’s temperature span will be lower than the 350°C and
450°C.
119
Problem 9: What would happen if R8 opens?
Solution
Comparator (U1B) will switch between +Vsat to –Vsat with a few mV at pin 6. Therefore, the
circuit cannot effectively control the temperature at the reaction chamber.
Problem 10: What would happen if D1 opens?
Solution
Solid-state relay will damage due to the excessive negative voltage applied to its internal LED
when Vo of U1C (pin 8) reaches – Vsat. D1 protects the internal LED of the relay against high
negative voltage.
Problem 11: If diode D1 were reversed, what would happen with the operation of the circuit?
Solution
If diode D1 were reversed then solid-state relay will be damaged during the –Vsat swing from the
output of U1C (pin 8).
Problem 12: If diode D1 were shorted, what would happen with the operation of the circuit?
Solution
If diode D1 were shorted, op-amp U1C might overheat, solid-sate relay will be OFF all the time and
the heater will be OFF.
120
See Figure 4 below.
VEE
VCC
R1
10kOhm
R2
Key = A
3.3kOhm
U2
XR2209
-12V
12V
1.47uF
C
50%
2.2kOhm
R3
+V
Timing cap
2000W
Heater
Triangle out
+
U1D
1kOhm
R18
LF444A
Bias
Timing res
D1
-
Solid State
Relay
240VAC
1N4001GP
Gnd
R4
Key = A
100kOhm
R13
50%
50%
Key = A
3.3kOhm
Frequency
adjust
VCC
10kOhm
R8
12V
Key = A
10kOhm
R5
Temp
Adjust
+
Type B
LF444A
10kOhm
R6
50%
U1A
VCC
Gain
adjust
U1B
-
10kOhm
R12
Key = A
10kOhm
10kOhm
R3
R10
50%
10kOhm
R16
LF444A
LF444A
12V
10kOhm
R7
Thermocouple
transmitter
120 VAC
10kOhm
R11
4.7kOhm
R14
10kOhm
R15
U1C
4.7kOhm
R17
Offset
adjust
Figure 4
Analyze the circuit.
This circuit is a PWM Temperature controller. This circuit will maintain the temperature in the
reaction chamber much closer to a set-point by varying the ratio of ON time to total cycle time (ON
time + OFF time) for the solid-state relay.
U1A: Error amplifier. Gain = 1
U1B: Inverting summing amplifier. Gain = - 1.
U1C: Inverting amplifier. Gain = - 1.
Note: U1B and U1C constitute a non-inverting summing amplifier with Gain = 1.
U2: Triangle wave oscillator
U1D: Pulse-width modulator
121
Thermocouple sensor
1)
Type E thermocouple
2)
Thermocouple transmitter
a) Produces linear output over selected temperature range.
b) Provides cold-junction compensation for the thermocouple.
3)
Provides DIP switches and trim pots for temperature range and output.
a) Current loop: 4 – 20 mA, etc
b) Voltage: 0 – 5 V, 0 –10 V, etc.
In this case, the thermocouple transmitter has been configured for an output of 0 – 10 V over a
temperature range of 0°C - 600°C. See Figure 5.
Summing Amp, Inverting Amp (U1B, U1C)
Form a non-inverting summing amp. They provide an output offset adjustment to the controller.
This adjustment is critical because it prevents the controller from turning OFF the heater
completely when error is zero (0).
R13: adjust controller gain and sensitivity to temperature changes.
Temperature ( ºC )
600
550
500
450
400
TX 30E Sensor
Output response
350
300
250
200
150
100
50
Figure 5
2
4
6
8
10
12
14
Process Variable voltage
16
V
122
Triangle-wave oscillator.
U2 = XR2209 or equivalent such as XR 2206. It is actually a square/triangle oscillator.
F = 1/(R4 C). In this case, R4 is adjusted to obtain a freq of 6 Hz.
6V
Vout =
½ Vcc = 6 V
0
Pulse-width modulator
+12V
VU1C
+Vsat = +10 V
741
U1D
Vout
-Vsat = -10 V
-12V
When V
> VU1C Æ Vout = + Vsat = + 10 V
When V
< VU1C Æ Vout = - Vsat = - 10 V
Oscillator
output
U1C output
+6V
Figure 6
+10V
U1D output
Output Limiter
R18 – D.
+Vsat = +10 V
-0.7 V
(See Figure 6)
Cycle time = 167 ms
123
Solid-State Relay
Similar to the one described in Figure 2.
The duty cycle of output U1D related to load voltage is shown in Figure 7. The duty cycle is
variable between 0 and 100% in increments of 5%.
Line
Voltage
U1D
Output
10%
duty cycle
60%
duty cycle
80%
duty cycle
Load
Voltage
U1D
Output
Load
Voltage
U1D
Output
Load
Voltage
Figure 7
System Operation
1. Let’s assume the following:
a)
The heater has returned to ambient temperature (25°C)
b)
Vtemp adj = 6 V (R5)
c)
Voffset adj = 1.5 V (R10)
d)
R13 = 20 kΩ (This yields a gain of 2)
2. At ambient temperature (25°C), Vproc var = 0.5 V (see Figure 5)
3. Verror = Vproc var – Vtemp adj= 0.5 – 6 = - 5.5 V
124
4. VU1C = Av (Verror + Voffset adj) = 2 ( - 5.5 + 1.5) = - 8 V
Since V
> VU1C Æ VU1D = + Vsat Æ Heater ON
Mathematically, the output of U1C is:
VU1C = 2 [(Vproc var – Vtemp adj) + Voffset adj]
VU1C = 2 [(Vproc var – 6V) + 1.5 V]
(Eq. 1)
5. As Temp ↑ ⇒ Vproc var ↑ ⇒ VU1C ↑
6. The heater will be ON continuously until VU1C = 0 V. This will happen when:
VU1C = 2 [(Vproc var – 6V) + 1.5 V] = 0 ⇒ Vproc var = 4.5 V
From Figure 5 ⇒ Temp ≈ 270°C
7. When Temp = 300°C ⇒ VU1C = + 1.0 V
Therefore, solid-state relay will not be ON until V
>+1V
8. When Temp = 350°C ⇒ VU1C = + 2.8 V
9. When Verror = 0 (Vproc var = Vtemp adj), VU1C = + 3 V (From Equation 1).
Thus, 350°C corresponds to zero error. This is the theoretical temperature setting of the
controller.
10. It is difficult to predict whether a 50% duty cycle will cause the chamber temperature to
rise, to fall, or to stay the same.
It can be predicted, however:
a) If Temp ↑ ⇒ Duty cycle ↓
b) If Temp ↓ ⇒ Duty cycle ↑
11. The controller will vary the duty cycle proportionally over a temperature range of
approximately 250°C - 450°C. The actual chamber temperature will not vary nearly this
much. It will vary slightly around the set point.
12. If Gain ↑ (R13) ⇒ Duty cycle will change more drastically for smaller temperatures
(sensitivity ↑).
13. If Vtemp adj ↑ ⇒ chamber will reach a higher temperature before duty cycle decreases
enough to reduce power to heater.
125
Problem: If R13 is adjusted to 40 kΩ draw the transfer curve.
+10V
6V
-0.7V
-1.675V
0
+1V
Gain = 4
Since Voffset = 1.5 V, then Vo = 4 (1.5V) = 6 V (When error = 0)
Error min = Vout,min/Av – Voffset = (-0.7V/4) – 1.5 = - 1.675 V
Error max = Vout,max/Av – Voffset = (10V/4) – 1.5 = 1 V
Proportional band = (Vout,max - Vout,min)/Av = [10 – (- 0.7)]/4 = 2.675 V
Problem: If R13 is adjusted to 60 kΩ draw the transfer curve.
Problem: If the oscillator frequency is adjusted to 1 kHz, would the controller operate properly?
Explain.
Solution:
No, heater will not have continuous power.
126
Problem: If a random-trigger solid-state relay were substituted for the ZVS relay, would the
controller operate properly? Explain.
Solution
No, heater may be damaged because hen heater is OFF, resistance is about 0 ohms. If solid-state
relay is ON and VAC is at its peak, too much current will be developed.
Problem: Is it possible to eliminate U1C? Explain.
Solution
Not in this circuit because we need a noninverting output at U1C.
Assume that R13 = 25k, Temp adjust = 3.965V and Offset adjust = 1.25 V.
Problem: When the error is zero, calculate the exact chamber temperature.
10V
V
0
T
600°C
600°C/10 V = T/V
600°C/10 V = T/3.965V
T = [(600°C) (3.965V)]/10V = 237.9°C
Problem: If resistor R13 shorts, what would happen with the operation of the circuit?
Solution
If R13 shorts then the gain of the non-inverting summing amplifier is zero. Thus, the voltage at pin
13 of U1D = 0 and since the voltage at pin 12 of U1D > 0, the output of U1D = + Vsat. This makes the
heater ON fully.
127
Problem: If capacitor C were reduced by half, what would happen with the operation of the
circuit?
Solution
If capacitor C were reduced by half, then the frequency of the triangle wave doubles. It will go
from 6 Hz to 12 Hz. Period will be 1/12 = 83 ms, and a 10% duty cycle will be 8.3 ms. Since the
line frequency is 60 Hz, the period is 16.6 ms, and at 10% duty cycle only half of a sine wave will
feed the heater. In this case, some adjustments need to be made to keep the chamber temperature
within the desired temperature within a narrow range.
Problem: If capacitor C were reduced by fourth, what would happen with the operation of the
circuit?
Problem: Assume that R13 = 25k, Temp adjust = 5V and Offset adjust = 1.4 V. If the output from
the thermocouple is 3.5 V, calculate the voltages at the following points:
Pin 2 (U1A); Pin 3 (U1A); Pin 6 (U1B); Pin 5 (U1B); Pin 9 (U1C); Pin 10 (U1C)
Solution
The voltage at pin 3 of U1A is equal to VR3.
VR3 = [R3/(R3 + R7)] (Vthermocouple) = [10k/(10k + 10k)](3.5 V) = 1.75 V.
Since the voltage between pins 2 and 3 of U1A = 0 (ed = 0), voltage at pin 2 of U1A = 1.75 V.
Thus,
Pin 2 (U1A): 1.75 V
Pin 3 (U1A): 1.75 V
Pin 5 (U1B): 0
Pin 9 (U1C): 0
Pin 6 (U1B): 0
Pin 10 (U1C): 0
Problem: Assume that R13 = 25k, Temp adjust = 5V and Offset adjust = 1.4 V. If the output from
the thermocouple is 4.5 V, calculate the voltages at the following points:
Pin 2 (U1A); Pin 3 (U1A); Pin 6 (U1B); Pin 5 (U1B); Pin 9 (U1C); Pin 10 (U1C)
128
Problem: When the temperature reaches 250°C, indicate the voltages and/or waveforms that
appear at the following test points. (Assume that the heater has been OFF for a long time). U1A;
Pin 1; U1B: Pin 7; U1C: Pin 8
Solution
According to the condition of the circuit, Vtemp adj = 6 V.
At 250°C, Vprocess var = 4 V
U1A: Pin 1 = Vprocess var – Vtemp adj = 4 – 6 = - 2 V
In addition, R13 has been adjusted to 20k and Voffset adj = 1.5 V.
U1B: Pin 7 = Av U1B (U1A,out + Voffset adj) = - 2 (- 2 + 1.5) = + 1 V
U1C: Pin 8 = - 1 V (Inverting amplifier)
Problem: When the temperature reaches 350°C, indicate the voltages and/or waveforms that
appear at the following test points. (Assume that the heater has been OFF for a long time). U1A:
Pin 1; U1B: Pin 7; U1C: Pin 8
Problem: If R14 shorts then the voltage at pin 7 of U1B is:
Solution
If R14 shorts practically nothing will change.
Problem: Assume that R13 has been adjusted to 15k and we need to maintain the temperature at
around 400°C. Indicate all the change(s) that you would do.
Solution
1)
Adjust R5 until Vtemp adj ≈ 6 V (see figure 5)
2)
Adjust R10 until Voffset = 4.4 V/1.5 = 2.9 V.
This is because at 400°C the U1C output = + 4.4 V. In addition, the gain of the noninverting
summing amp is R13/R11 = 15k/10k = 1.5.
129
Problem: Assume that R13 has been adjusted to 25k and we need to maintain the temperature at
around 300°C. Indicate all the change(s) that you would do.
Problem: When the chamber temperature reaches 410°C, indicate the voltage(s) at pin 8 of U1C.
Solution
From Figure 5.
400°C – 6 V
410°C – X
X = [(410°C)(6 V)]/400°C = 6.15 V.
VU1A, pin 1 = 6.15 – 6 = 0.15 V
VU1B, pin 7 = - 1.5 (0.15 + 2.9) = - 4.58 V
VU1C, pin 8 = + 4.58 V
Problem: When the chamber temperature reaches 330°C, indicate the voltage(s) at pin 8 of U1C.
130
See Figure 9 below.
VCC
12V
LM7805
Vin
Vout
+5V
GND
1kOhm
R2
Figure 9
10kOhm
R4
+V
0 - 5V
Vout
Ultrasonic
transducer
Gnd
2.2kOhm
R6
LM339
RESET
U1A
1.1V
1kOhm
R1
2.2kOhm
R7
4.6V
U1B
Key = A
50% 10kOhm
R5
High-level
adjust
Set Reset Pump Control
L
L
---------------
L
H
H
L
L(Pump ON)
H
H
Unchanged
+ Solid State
Relay
-
SET
LM339
Key = A
50%
33kOhm
R3
Low-level
adjust
7400N
U2A
7400N
U2B
Condition
This condition will never occur
H (pump OFF) Water level has fallen below low threshold
Water level has risen above high threshold
Water level is between two thresholds
Analyze the circuit.
At least do this on your own.
Pump
120VAC
131
Problem: If R3 = 2.5 kΩ and R5 = 1 kΩ draw the transfer curve.
Solution
VR3 = [R3/(R2 + R3)] (VCC) = [2.5/(2.5 + 1)](5 V) = 3.57 V
VR5 = [R5/(R4 + R5)] (VCC) = [1/(10 + 1)](5 V) = 0.45 V
5V
0
0.45 V
0
3.57 V
Problem: If R3 = 1.5 kΩ and R5 = 2 kΩ draw the transfer curve.
Problem: If the voltage across R5 were higher than the voltage across R3, would the controller
operate properly? Explain.
Solution
With the values shown in Figure 9, the pump will be ON when the water reaches the level of 10
inches from the sensor. The pump will be OFF when the level of the water drops to 28 inches from
the sensor.
According to Figure 10, when the distance from the sensor is 10 inches, the output voltage from
the sensor is 1.1 V. Likewise, the output voltage is 4.6 V when the distance reaches 28 inches.
132
Sensor output
5V
4V
4.6V
Distance/output response
3V
2V
1.1V
1V
Figure 10
10
15
20
25
Distance from sensor (in.)
28
30
Now, assume that the voltage across R5 were 3 V (High-level adj) and the voltage across R3 were 2
V (Low-level adj). When the sensor output is 2.9 V, the voltage at pin 4 of U1A (inverting input) is
higher than the voltage at pin 5 of U1A (non-inverting input). This makes the output of U1A (pin 2)
= 0. Thus, RESET = 0. By the same token, the voltage at pin 6 of U1B (inverting input) is higher
than the voltage at pin 7 of U1B (non-inverting input). This produces a zero output at pin 1 of U1B.
Therefore, SET = 0.
We then have the condition that SET and RESET are both active. This is unacceptable in a
RS-latch. In conclusion, the circuit won’t operate properly.
Problem: What would happen if SET and RESET become LOW? Explain.
The output is unpredictable. In a RS-latch, SET and RESET cannot be active simultaneously.
The pump will be ON when the output of U2A is
output of U2B is
(HIGH/LOW).
(HIGH/LOW), and the
Problem: Resistors R6 and R7 are needed because…..
Solution
the LM 339 IC has open-collector outputs.
Problem: When the output of the ultrasonic transducer is 1.09 V, the output of U2A is
(HIGH/LOW), and the output of U2B is
(HIGH/LOW).
133
Problem: If R3 and R5 are adjusted to 50% respect to ground, calculating the values of
VLOW LEVEL and VHIGH LEVEL
Solution
VLOW = [16.5/(1 + 16.5)] (5 V) = 4.7 V
VHIGH = [5/(10 + 5)] (5 V) = 1.67 V
Problem: If R3 and R5 are adjusted to 70% respect to ground, calculating the values of
VLOW LEVEL and VHIGH LEVEL
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