Chapter 1: Introduction to Mechatronics
1. What is Mechatronics?
(Lecture#1 )
Mechatronics = Mecha(nical) + (Elec)tronics
The term “Mechatronics” came from Japan in late 1960s, spread through Europe, and is growing in USA.
The whole picture of the class:
A Mechatronic system can do things that we want it to do (e.g., a copy machine, a refrigerator, an airconditioner, etc.)
(4) Plant
+
Ideal
Performance
Controller
-
(6,7)
Digital World (Digital Circuit) (3, 6)
Write this first
(1)
(2)
(3)
(4)
(5)
(6)
(7)
(8)
(9)
D/A
Actuator
(8)
(10)
A/D
Amplifier
(8)
(5)
Others
Actual
Performance
Actual
Output
Sensor
(9)
Analog/Real World (Analog Circuit) (2)
Figure 1 The Outline of the class
Chapter 2: Electric Circuits and Components
Chapter 3: Semiconductor Electronics
Chapter 4: System Response
Chapter 5: Analog Signal Processing Using Operational Amplifiers
Chapter 6: Digital Circuits
Chapter 7: Microcontroller Programming and Interfacing
Chapter 8: Data Acquisition
Chapter 9: Sensors
Chapter 10: Actuators
Dr. Winncy Du, E310F, MAE Dept., SJSU, Tel.:408-924-3866,E-mail: wdu@sjsu.edu
Course Web: www.engr.sjsu.edu/wdu/ME106Spring2010
Chapter 2 Electric Circuits and Components (Lecture # 2)
I.
Basic Electrical Elements: Resistors, Capacitors, and Inductors
DC (Direct Current) Circuit - - V (t ) or I (t ) = constant
Analog Circuit 
AC (Alternating Current) Circuit - - V (t ) or I (t ) = sinusoidal
Both DC and AC circuits may contain the three basic passive electrical elements 1: resistor,
capacitor, and inductor
1) Resistor: a dissipative element that converts electrical energy into heat (unit: Ω - Ohm).
Related law -- Ohm’s Law:
For an ideal resistor: V = IR
Resistors in series: Req =
Resistors in parallel:
1
Req
∑R
1
=∑
R
i
i
Application: in a circuit, a resistor is used as a
Voltage Divider:
Vout
R2
R1
R2
=
Vin
R1 + R2
Vout
Vin
e.g. if you need 6V for your load, but only 12V voltage source is available, then you need two
resistors in series to get the 6V.
Current Divider:
I out
R1
=
I in
R1 + R2
Iin
R1
R2
Iout
e.g. if you need 15 mA for your load, but only 0.1A current source is available, then you need
two resistors in parallel to get the 15 mA.
* Try to be familiar with the voltage and current ranges that we normally use in the lab.
Types of Resistors:
Potentiometer: Variable resistors, screw, knob, or linear slide –
1
An active component/element means that the component has to have its own power supply for operation. Since a
resistor, a capacitor, and an inductor don’t need their own power supply for their function, they are all called
“passive electrical components”.
Dr. Winncy Du, E310F, MAE Dept., SJSU, Tel.:408-924-3866,E-mail: wdu@sjsu.edu
Course Web: www.engr.sjsu.edu/wdu/ME106Spring2010
A potentiometer that is included in a circuit to adjust (trim) the resistance in the circuit is
called a trim pot.
Wire-lead Resistor: we will use this type of resistors a lot in this class. Please read Page 11 –
12 in the textbook to see how to find a resistor’s value and precision based on its color bands.
Surface Mount Resistor: this type of resistors is used a lot in the Mother/Main board of a
computer, or other commercial boards.
Others: single in-line package, dual in-line package (see figure 2.6 on page 11).
2) Capacitor: a passive element that stores energy in the form of an electric field (Unit. F, farads).
Note: F is a very big unit, the capacitance range we usually use is 1 pF ~ 1000 µF.
The voltage-current relationship of a capacitor is:
1
dV (t )
( or I (t ) = C
)
I (t )dt
∫
C
dt
1
1
Capacitors in series:
=∑
Ceq
Ci
V (t ) =
Capacitors in parallel: Ceq =
∑C
i
Application: a capacitor is used to
(1) store electrical energy
(2) protect your circuit or important components
For a capacitor, the voltage is the integration of the current with time, that means that it
will take TIME for the voltage to reach to certain level. If by accident you put too much
power in the circuit, the circuit or the components will not be burned immediately due to
the capacitor. It is a good exercise to put a capacitor in parallel to the component that you
don’t want to damage.
(3) make memory chip, filters, integrators, sensors, etc.
Types of Capacitors:
Film Capacitors: A relatively large family of capacitors, they differ pretty much just
in their dielectric properties. Available capacitance ranges from 10pF - 15uF.
Members include polyester, polystyrene, polypropylene, polycarbonate, metallized
paper, etc.
Ceramic Capacitors: The common form is the multi-layer or stacked ceramic
(monolithic); single layer also exists (ceramic disk). Physically, the multi-layer looks
like the film and foil above, a dielectric stuffed between metal plates. The multi-layer
is marginally more expensive than the single layer.
Silver-mica: Another stacked capacitor. Mica is really the general family name (mica
is the dielectric); silvered mica is just the most popular form. They are popular for
Dr. Winncy Du, E310F, MAE Dept., SJSU, Tel.:408-924-3866,E-mail: wdu@sjsu.edu
Course Web: www.engr.sjsu.edu/wdu/ME106Spring2010
their high frequency characteristics (up to 500MHz). Typical values range from 2pF to
1500pF.
Electrolytics: A "breakthrough" in capacitor technology in the early 1900's. Instead of
placing a solid wedge of something (which can be quite thin), an electrolyte solution is
used. The electrolyte serves as the 2nd electrode. The electrolyte is not the dielectric.
The dielectric is a very thin layer of oxide which is grown electro-chemically in
production. The thickness of this oxide layer is on the order of .01µm, much smaller
than any piece of plastic or ceramic that could be used as a separator.
Others: Surface Mount Capacitors, etc.
Important Parameters to consider and Choose a Capacitor:
Capacitance
ESR (power dissipation in the capacitor and useful frequency range)
Tolerance (ability to plug it into frequency sensitive circuits)
Temperature/Aging Drift (capacitance changes in sensitive circuits)
ESL (useful frequency range)
3) Inductor: an energy storage element that stores energy in the form of a magnetic field (Unit:
Henry H, typical 1uH ~ 0.1H).
The voltage-current relationship of an inductor is:
V (t ) = L
Inductors in series: Leq =
Inductors in parallel:
1
Leq
dI (t )
dt
∑L
1
=∑
L
i
i
Application: Usage of Inductor
(1) Store magnetic energy
(2) Electric motors – large inductance
(3) Transformers
(4) Sensors
Types of Inductors:
i.
Radial inductor
ii.
Chip Inductor
iii.
Power Inductor
Dr. Winncy Du, E310F, MAE Dept., SJSU, Tel.:408-924-3866,E-mail: wdu@sjsu.edu
Course Web: www.engr.sjsu.edu/wdu/ME106Spring2010
II.
Basic Laws and Methods in Analog Circuit Analyses
Basic Laws: There are three laws govern the analog circuit analyses:
OHM’S LAW
The voltage-current characteristic of an ideal resistor is: V = IR .
KIRCHHOFF’S VOLTAGE LAW (KVL):
The algebraic sum of voltage drops around any closed path or loop within a circuit is equal to the
sum of the applied voltages:
∑V = ∑ I R
i
j
j
.
KIRCHHOFF’S CURRENT LAW (KCL):
The current that flows into a node is equal to the current that flows out of the node:
∑I
in
= ∑ I out .
Methods:
Loop-Current Method
The loop-current method (also known as the mesh current and Maxwell loop-current methods) is a direct
extension of KVL (Kirchhoff’s Voltage Law) and is particularly valuable in determining unknown currents.
It requires writing n-1 simultaneous equations for an n-loop system.
Step 1: Select n-1 loops (i.e., one less than the total number of loops).
Step 2: Assume current directions for the chosen loops, and show the direction with an arrow. (Note:
any current whose direction is chosen incorrectly will end up with a negative current in step 4).
Step 3: Write KVL for each of the n-1 chosen loops.
Step 4: Solve the n-1 equations for the unknown current.
Example: Find the current in the 0.5Ω resistor.
(Solution) This is a three-loop network – ABCF, FCDE, and
ABCDEF. Two simultaneous equations are required.
Step 1: Select two loops: ABCF and FCDE.
Step 2: Assumed currents i1 and i2 have the directions as
shown in the figure.
Step 3: Kirchhoff’s voltage law for loop ABCF is:
∑V = ∑ ( i R)
i
⇒
20-19 V =0.25 i1 + 0.4 (i1 - i2)
For loop FCDE is:
∑V = ∑ ( i R)
i
⇒
19 V = 0.4 (i2- i1) + 0.5 i1
The battery polarities determine the signs on the left-hand side of the equation. The polarity of the
19 V battery is opposite to the assumed current flow direction and, thus, is negative.
Step 4: Solving these two loop equations results in i1 = 20 A and i2 = 30 A.
Dr. Winncy Du, E310F, MAE Dept., SJSU, Tel.:408-924-3866,E-mail: wdu@sjsu.edu
Course Web: www.engr.sjsu.edu/wdu/ME106Spring2010
Node-Voltage Method
The node-voltage method is an extension of KCL (Kirchhoff’s Current Law). While the currents can be
determined with it, its primary use is in finding voltage potentials at various points (nodes) in the circuits.
A node is a point where three or more wires connect.
Step 1: Choose one node as the voltage reference (i.e., 0 V) node. Usually, this will be the circuit
ground – a node to which at least one negative battery terminal is connected.
Step 2: Identify the unknown voltage potentials at all other nodes referred to the reference node.
Step 3: Write KCL for all unknown nodes. (This excludes the reference node.)
Step 4: Write all currents in terms of voltage drops.
Step 5: Write all voltage drops in terms of the node voltages.
Example:
Find (1) the voltage potential for node A and (2) the current i1.
(Solution)
(1) Step 1: Choose node D as the reference voltage. Thus, VD = 0.
Step 2: Node A is the only other true node.
Step 3: Kirchhoff’s current law for node A is: i1 + i2 = i3
Step 4: Write the current equation in terms of the voltage
drops and resistances. The directions chosen determine the polarity.
VBA VCA VAD
+
=
2
4
10
Steop 5: Write the voltage drops in terms of the node voltage:
VB − VA VC − VA VA − VD
+
=
2
4
10
⇒
50 − VA 20 − VA VA − 0
+
=
2
4
10
Solving, VA = 35.3 V.
(2) i1 =
VBA VB − VA 50 − 35.3
=
=
= 7.35 A
2
2
2
Dr. Winncy Du, E310F, MAE Dept., SJSU, Tel.:408-924-3866,E-mail: wdu@sjsu.edu
Course Web: www.engr.sjsu.edu/wdu/ME106Spring2010
NORTON’S THEOREM
Given a pair of terminal (A and B) across a load in a linear network, the network can be replaced
by the Norton equivalent circuit consisting of a single current source (the Norton equivalent current
IN) and a resistor (the Norton equivalent resistor RN ) in parallel.
Step 1: Find RN: (1) remove the load between the two terminals A and B; (2) set all independent
sources to zero (i.e. for a voltage source, short it; for a current source, open it); (3) find the equivalent
resistance across terminals A and B in the new circuit.
Step 2: Find IN: (1) remove the load and short A and B together; (2) calculate the current the flows
through the terminals in the new circuit.
Note that the Norton equivalent voltage, VN, can be found using VN = I N RN
A
remaining
circuit
network
or
load
R1
+
Portion of circuit to be
replaced with Norton
equivalent
R2
VS
B
A
IN
RN
B
remaining
circuit
network
or
load
Dr. Winncy Du, E310F, MAE Dept., SJSU, Tel.:408-924-3866,E-mail: wdu@sjsu.edu
Course Web: www.engr.sjsu.edu/wdu/ME106Spring2010
THEVENIN’S THEOREM
Given a pair of terminal (A and B) across a load in a linear network 2, the network may be replaced
by the Thevenin equivalent circuit consisting of a voltage source (the Thevenin equivalent voltage
VTh ) in series with a resistor (the Thevenin equivalent resistor RTh).
Step 1: Find VTh: (1) remove the load between the two terminals A and B; (2) find the voltage across
the terminals A and B in the remaining circuit network.
Step 2: Find RTh: (1) remove the load between the two terminals A and B; (2) set all independent
sources to zero (i.e. for a voltage source, short it; for a current source, open it); (3) find the equivalent
resistance in the new circuit between A and B.
A
remaining
circuit
network
or
load
R1
+
Portion of circuit to be
replaced with
Thevenin equivalent
R2
VS
B
A
RTh
+
VTh
remaining
circuit
network
or
load
The Thevenin and Norton equivalents are independent of the load. This is very
useful because it is possible to make changes in the load without reanalyzing the
Thevenin or Norton equivalent.
2
A linear network means that only resistors are involved in the circuit (recall that the Voltage-Current relationship
of a resistor is linear)
Dr. Winncy Du, E310F, MAE Dept., SJSU, Tel.:408-924-3866,E-mail: wdu@sjsu.edu
Course Web: www.engr.sjsu.edu/wdu/ME106Spring2010
HW problem3:
(1) Find the current through resistor R3 using a Norton equivalent
circuit.
(Solution)
Step 1: Find Norton equivalent resistance RN by removing the load
R3 between the two terminals A and B and set the voltage
source to zero (i.e., short it).
RN =
A
B
B
1
1
1
+
2Ω 4Ω
= 1.333 Ω
Step 2: Find Norton equivalent current IN that flows through the terminals A and B in the equivalent
circuit by removing the load and short the terminal A and B.
A
IN =
= 30 Α
B
Step 3: the current through R3 is found from the Norton current and the current division principle:
i3 = 30 A (
?
?
+ 10
) = 3.53A
(2) Solve the above problem using a Thevenin equivalent circuit.
(Solution)
Please follow the given steps and do this part yourself. The answers are:
(The thevenin equivalent voltage is VTh = i Th R Th = 30 × 1.333 = 40 V
The current i3 is
i3 =
40
V
=
= 3.53A )
1
.
333
+ 10
Req
Dr. Winncy Du, E310F, MAE Dept., SJSU, Tel.:408-924-3866,E-mail: wdu@sjsu.edu
Course Web: www.engr.sjsu.edu/wdu/ME106Spring2010