2014 Edition © by D. G. Meyer
Microcontroller-Based
Digital System Design
Module 4
Embedded System Design Issues
1
Learning Outcomes
A student who successfully fulfills the course
requirements will have demonstrated:
1. an ability to program a microcontroller to
perform various tasks
2. an ability to interface a microcontroller to
various devices
3. an ability to effectively utilize microcontroller
peripherals
4. an ability to design and implement a
microcontroller-based embedded system
2
Mini-Project




Embedded system design based on Freescale
9S12C32 Microcontroller Kit
Basic requirement is to design a stand-alone
(“turn-key”) product that makes good use of the
processor’s computational / interfacing resources
Should make meaningful / effective use of at least
four of the microcontroller’s on-chip peripherals
(ATD, SCI, SPI, TIM, PWM)
Done in teams of 2-4 students (self-selected)
3
Learning Objectives






form a project team consisting of 2-4 students
choose a project idea and write a proposal
design the hardware and software necessary to
realize the project idea
package the finished system
demonstrate the project’s functionality to the
course staff
write a formal report documenting the embedded
system design process
4
5
6
ECE 362 Embedded Microcontroller Mini-Project  Fall 2013
Lucky 13 / Aqualizer
Filter Design
Polycarbonate Case Design
Transfer Functions of Filters
1
0.9
0.8
|H(jω)|
0.7
0.6
0.5
Treble
Bass
0.4
Mid
0.3
0.2
0.1
0
0
10
10
1
10
20Hz
2
10
250Hz
3
2,000Hz
10
4
20,000Hz
10
Aqualizer is in action.
5
Band-Pass
Low-Pass
Frequency (Hz)
Software Design
High-Pass
Filter Type
Electronics
Schematic
TIM interrupt
(every 1 ms)
Low-Pass
Vin
RTI interrupt
(every 2.048 ms)
Processor
Boot Up
Vout
6.2k
Increment
tens
0.1u
counter
0
No
Is tenths
Was UV
pushbutton pressed?
flag set?
Band-Pass
Vin
Vout
20mH
2.2u
270
Does tens
counter ==
10?
0
Yes
High-Pass
Vin
Is UV flag set?
Update LED PWM
0
Set push button flag
No
conversion
Set tenths flag
820
Yes
Yes
Perform ATD
Vout
0.1u
Digijock(ette)-Strength
Digital System DesignTM
No
No
Yes
Toggle UV LED
Update Pump PWM
lights
https://www.youtube.com/watch?v=ovuchyM2fDg&feature=youtu.be
ECE 362 Embedded Microcontroller Mini-Project − Spring 2014
Everett Berry, Allen Chien, Jacob Stevens, Kyle Woodworth / Flappy Bits
Block Diagram
Overview
Accelerometer
Data
The Flascii Bird is a real life simulation
of the game flappy bird. The player is
asked to wear two specially designed
gloves that have accelerometers
embedded. The game is entirely
controlled by movements of the player
and the data collected from the
accelerometers is sent wirelessly to the
HCS12 microcontroller. The Papilio One
FPGA is responsible for outputting VGA
graphical interface to the monitor.
Xbee Transmitter
Xbee Receiver
Left to Right: Allen, Everett, Jacob, and Kyle
Component List
•
•
•
•
•
PCB Design
ADC
HCS12
Microcontroll
er Unit
HCS12 Microcontroller
Papilio One FPGA
Xbee 1mW Trace Antenna Series 1
ADXL 335 Accelerometer
Printed Circuit Boards
PWM
Buzzer
Papilio One – 500K
The Papilio One is an open-source
development platform. It utilities the
power of FPGA through implementation
VGA driver with SPI communication.
Features:
•4Mbit SPI Flash Memory
•Two-Channel USB Connection for
JTAG and Serial Communications
•48 I/O lines
Digijock(ette)-Strength
Digital System DesignTM
Low Pass Filter
http://youtu.be/LIovdlLAlTs
Game Logic
TIM
SPI
Papilio One
FPGA
VGA
Monitor Output
ECE 362 Embedded Microcontroller Mini-Project  Spring 2014
Dabonda / Surveillance Video Camera
Microprocessor used: 9S12C32
Peripherals used:
RTI: The two buttons
SPI: LCD screen
ATD: Potentiometers
TIM: Reset
PWM: Motors
The purpose of this project is to build a surveillance video camera, whereby the
direction of the camera can be controlled by a user using two potentiometers. The
camera covers a range from 60 degrees left to 60 degrees right and 60 degrees up
to 60 degrees down. The push button then toggles the machine between run mode
and hold position mode. The Liquid Crystal Display (LCD) panel will display the
angle of the camera and the corresponding the Light-Emitting Diode (LED) rows
will also light up as an additional indicator. If there is no user input after a whole
minute, the machine will return to its initialization stage and the welcome screen
will display on the LCD panel.
Digijock(ette)-Strength
Digital System DesignTM
https://www.youtube.com/watch?v=cHe7elEZAPk
ECE 362 Embedded Microcontroller Mini-Project  Spring 2014
Raulmaha Digital Synthesizer
Andrew Pawling, Aimad Md Uslim, John Sterrett, Chris Pierson
Project Design Goals
The creation of a
digital synthesizer that integrates
the following features:
• Mono-phonic, single octave
keyboard range synthesizer
• Analog volume and distortion
control
• Variety of audio waveforms
(sine, triangle, saw tooth, and
square)
• Accelerometer-based pitch
bend ability
Digijock(ette)-Strength
Digital System DesignTM
https://www.youtube.com/watch?v=148smSUeaGM
Module 4

Desired Outcome: “An ability to design and
implement a (‘turn-key’) microcontroller-based
embedded system”
– Part A: PCB Design and Fabrication
– Part B: External Microcontroller Interfaces
– Part C: Basic Regulated Power Supply Design
– Part D: Embedded System Hardware and
Software Development Tips
11
2013 Edition © by D. G. Meyer
Microcontroller-Based
Digital System Design
Module 4-A
PCB Fabrication and Layout
12
Outline
Overview
 PCB fabrication process
 Typical PCB fabrication tolerances
 Basic layout guidelines
 EMI considerations (reference: AN1259)
 Layout tips

13
Overview

Objective: To understand what printed
circuit boards are, how they are created, and
basic guidelines on PCB design
14
Anatomy of a PCB

Quick overview:
– Copper foil
– Substrate
– Lamination
– Etch
– Drill
– Plate
– Solder mask
– Silkscreen
15
Anatomy of a PCB

Copper foil
– electro-plated onto a large drum then
scraped off (typical)
– one side very smooth, one rough
16
Anatomy of a PCB

Substrate
– FR4 (fiberglass reinforced multi-functional
epoxy)
– others: FR2

Lamination
– copper foil applied to both sides of
laminate and then bonded using heat and
pressure
17
Anatomy of a PCB (3/8)

Etching
– Circuit first realized here
– Etch-resist applied, pattern is exposed
18
Anatomy of a PCB

Etching
– Uncured resist is washed off and then
the pattern is etched
– common etchants: FeCl, Ammonia
– solvent/abrasive wash to remove etchresist
– PCB then washed to remove residues
from solvents and abrasive process
19
Anatomy of a PCB

Drill / Plate
– This is how the connections are made
between layers
– Holes drilled through where connections
are desired
20
Anatomy of a PCB

Drill / Plate
– PCB then immersed in a plating solution
where a thin layer of copper forms inside
the barrel of the hole
– Once enough copper is deposited this
way, then on to electro-plating, where
~1 mil of copper is plated on
– If a gold-plate finish is required, typically
applied at this time
21
Anatomy of a PCB

Solder mask
– Protects metal from corrosion, short
circuits
– Also prevents solder from sticking
– applied as a liquid, then cured with UV
– Many colors available, green most common
22
Anatomy of a PCB

Silkscreen (legend)
– labels everything: components, notes,
warnings, logos
– similar process to making T-shirts
– applied as a liquid, then cured
– several colors, white most common
23
PCB Terminology





Pin – A plated through hole used to
connect the terminal of a part
Pad – A flat conductive surface for
connecting the terminal of a part
Via – A plated through hole used for signal
routing
Trace – A wire or 1 dimensional electrical
connection
Signal Plane – A 2 dimensional electrical
connection (commonly used for
ground)
24
Typical PCB fabrication tolerances

Notation: 1 mil = 1 “milli-inch” = 0.001 in.

Drills: +5/-3 mils diameter, 5 mil center
smallest drill size 20 mil
Layer-to-layer alignment: +/- 3 mils
Etched feature size: +/- 1 mil
Trace size (isolate): 6 mil
 ≥ 8 mil trace/isolate recommended
Solder mask size: +/- 3 mil
Silkscreen size: +/- 10 mil




25
Basic Layout Guidelines






Minimum recommended trace/space 10-12 mil
Power and ground traces should be sized for
current being passed (width/current charts
available online)
Follow all manufacturer layout recommendations
Decoupling capacitors should be placed as
close to each IC as possible
Provide space and mechanical support for
connectors, heat sinks, and standoffs (used for
mounting board)
Incorporate headers or vias for verification and
debugging
26
27
28
Basic Mini-Project PCB Guidelines







Maximum board size is 60 square inches
Minimum trace/space size should be 10-12 mil
Power and ground traces should be as “big as
possible” (at least 40-60 mil, 100 mil better)
Decoupling capacitors should be placed as
close to each IC as possible
Use 32-pin DIP socket for 9S12C32 “stamp”
module and DIP sockets for other ICs (GALs,
optical isolators, buffers, etc.)
Generally stick with “through hole” passives
(resistors and capacitors), but surface mount OK
Provide space and mechanical support for
connectors, heat sinks, and standoffs (used to
29
mount board in packaging)
COM port (can
use standard
9-pin “D”
connector)
30
General Layout Guidelines

Separation of circuits on a PCB
– pay close attention to the potential routing of
circuits between subsystems
31
An Undersized Trace
32
General Layout Guidelines
Ground layout is the most important PCB layout design
consideration – most EMI problems can be resolved using
practical and efficient grounding methods
 Noise can be coupled into other circuits by common
impedance
 Dynamic DC offset can be created that produces a highfrequency AC component of noise that affects low-level
analog circuitry

33
General Layout Guidelines

Ground layout design tips
– separate digital logic and low-level analog circuits
– provide as many parallel pathways to ground as possible
(becomes a “ground plane” in the limit, and decreases
inductance of ground return)
– if ground plane is uneconomical, use single-point (starpoint) grounding – lowers common impednace coupling
among subsystems
– to decrease trace inductance, use as short and wide a
trace as possible
– use 45-degree turns instead of 90-degree turns to
decrease transmission reflections
– decrease the size of all ground loops as much as
possible (e.g., single-point power system)
34
General Layout Guidelines

Single-point power system for 2-layer PCB
35
General Layout Guidelines

IC decoupling capacitor placement
– should be as physically close to IC as possible (for
surface mount components, place capacitor halfway
between VDD and GND)
– use 0.1 µF (surface mount, ceramic) decoupling
capacitors for system frequencies up to 15 MHz
– above 15 MHz, use 0.01 µF decoupling capacitors
36
General Layout Guidelines

Power terminal decoupling capacitor placement
– sometimes called a “bulk” capacitor – should be placed as
close to the power input terminal (connector) as possible
– a small (0.1 µF capacitor) should also be used to decouple
high frequency noise at the power input terminal
– purpose of bulk capacitor is to help recharge the IC
decoupling capacitors
– value is not critical, but should be able to recharge 15-20 ICs
(multiple bulk capacitors may be used if there are more than
15-20 ICs)
37
General Layout Guidelines

Signal layout
– most sensitive digital signals are clock, reset,
and interrupt lines
– if analog and digital signals have to be mixed,
make sure the lines cross each other at 90degree angles (reduces cross-coupling)
– ATD performance can be severely impacted by
reference lines (route directly from power
supply and low-pass filter using combination of
1 K resistor and 1.0 µF capacitor
38
General Layout Guidelines

Typical crystal or ceramic resonator circuit layout
39
Layout Tips - 1





Keep parts that belong close together on
layout close together on schematic
Print layout in 1:1 scale and compare
footprints with your actual parts before
ordering PCBs
Position parts carefully first, route second
(“measure twice, cut once”)
Avoid trace angles ≤ 90° in routing whenever
possible (PCB layout tool enforces this)
Separate digital/analog grounds, and tie
together at a single point only
40
Layout Tips - 2




Don’t use 90° (or acute) angles when routing
traces – chamfer/bevel them whenever
possible (routing tool typically enforces this)
Take mounting into consideration (route
connectors on bottom side, and save space
for standoffs/screws)
Avoid acute angles in traces (they are “acid
traps” and can cause problems with etching)
Use larger power traces and orient your power
supplies and supply lines near where they are
used, and put them in a place where they will
have minimum interference (e.g., away from
GPS or RF modules)
41
Layout Tips - 3




Provide an ample number of test points
(suggest header that breaks out all
significant microcontroller signal pins)
Be sure to include relevant board
information in the silkscreen layer of your
board (name, date, board revision, etc.)
Be aware that connector pin-out may differ
based on gender (watch out for inadvertent
“mirroring” of pin-out)
Start even earlier than you thought you
would have to start
42
Clicker Quiz
43
1. The primary reason to “chamfer” corners (using 45°
turns, instead of using 90° turns) on PCB layout traces is to:
A.
B.
C.
D.
E.
decrease the size of ground loops
decrease noise radiation
decrease trace capacitance
decrease trace inductance
none of the above
2. The primary goal of a single-point power and ground
strategy on a PCB layout is to:
A.
B.
C.
D.
E.
decrease the size of ground loops
decrease noise radiation
decrease trace capacitance
decrease trace inductance
none of the above
3. The main reason for avoiding acute angles between
traces on a PCB layout is to:
A.
B.
C.
D.
E.
decrease the size of ground loops
decrease noise radiation
decrease trace capacitance
decrease trace inductance
none of the above
4. The purpose of a “bulk” capacitor, connected in parallel
with the input power and ground terminals of the PCB, is to:
A.
provide additional “ripple” filtering of the power
supply voltage
B. suppress noise generated by the board from
being radiated
C. reduce the power supply regulation requirement
D. recharge the decoupling capacitors
E. none of the above
2014 Edition © by D. G. Meyer
Microcontroller-Based
Digital System Design
Module 4-B
External Microcontroller Interfaces
48
Learning Objectives







design an interface that allows a microcontroller
port pin to control a D.C. load (e.g., motor)
describe how optical isolation can be used to
protect microcontroller port pins used as either
inputs or outputs
describe how a scanned keypad works
describe the operation of a rotary pulse generator
(RPG) and cite some potential applications
describe how a stepper motor works and be able
to distinguish between full- and half-step modes
describe how port expansion can be
accomplished using shift registers
design an interface that allows a microcontroller
to communication with an LCD
49
Outline
















Switching D.C. Loads
Optically Isolated Inputs
Keypads (Switch Matrices)
Rotary Pulse Generators (RPG)
Switch (SPST Contact) Debouncer
Position Control (Steppers and Servos)
LCD Interface / Serial I/O Expansion
Digitally Controlled Potentiometer
Temperature and Humidity Sensors
Digital Compass
Accelerometers
Hall Effect Sensors
Pressure/Force Sensors
Ultrasonic Range Finders
IR Remote Decoders
RF Links (Transmitters/Receivers)
50
Switching D.C. Loads

Basic BJT-based switching circuit
(“saturated mode”) V
L
LOAD
Port
Pin
“current-controlled
switch”
C
B
NPN
BJT
IC = hFE x IB
E

Choose BJT based on following parameters
 ICmax continuous
 VCE breakdown
 hFE (D.C. current gain), also called 
51
Switching D.C. Loads

Use either BJTs or power MOSFETs
– BJTs: cheap, but may require a significant amount of
base current to drive into saturation
– MOSFETs: tend to be more expensive, but require
virtually no gate current to operate

Inductive loads require arc suppression diode
VL
Energy stored in an inductive load
must be dissipated, otherwise the
“inductive kickback” can damage
the switching device
52
Switching D.C. Loads

BJT limitations
– have to use Darlington (more expensive) to get
high IC capacity with large hFE
– generally need base current on the order of (a few)
milliamps to switch transistor
– if “something bad” happens to transistor, can take
port pin (and possibly microcontroller) with it – use
optical isolator to provide protection
C
NPN Darlington
B
E
53
Switching D.C. Loads

Optically isolated BJT
VL
LOAD
NPN
BJT
54
Switching D.C. Loads

Optically isolated BJT
Vcc
VL
LOAD
NPN
BJT
active low
port pin
R1
55
Switching D.C. Loads

Optically isolated BJT
VL
Vcc
LOAD
NPN
BJT
active low
port pin
R1
Assume Vcc = 5 V, VLED = 1.5 V, and VOL @ 10 mA = 0.8V
 R1 = 2.7/0.01 = 270
56
Switching D.C. Loads

Optically isolated BJT
VL
Vcc
VL
LOAD
R2
active low
port pin
NPN
BJT
R1
Assume Vcc = 5 V, VLED = 1.5 V, and VOL @ 10 mA = 0.8V
 R1 = 2.7 / 0.01 = 270
57
Switching D.C. Loads

Optically isolated BJT
Vcc
VL
VL
LOAD
R2
NPN
BJT
active low
port pin
R1
Assume switching 1 amp load, and that hFE of transistor
is 100  need 10 mA of base current to saturate
transistor (assume VL is 12 V, VBEsat of transistor is 0.7 V,
and that VCEsat of photo transistor is 0.3 V)
 R2 = 11.0 / 0.01 = 1100
58
Switching D.C. Loads

Basic MOSFET-based switching circuit
VL
LOAD
Port
Pin

G
D
S
“voltage-controlled
switch”
N-channel
power
MOSFET
Choose MOSFET based on following parameters
 IDmax continuous
 VDS breakdown
 rDS (on) (drain-to-source “on” resistance)
59
Switching D.C. Loads

MOSFET limitations
– be careful to choose a MOSFET with a
small rDS(on) when switching high current
loads (heat, voltage drop)
– tend to be more “fragile” than BJTs
(MOSFETs can be damaged by ESD)
– if “something bad” happens to transistor,
can take port pin (and possibly
microcontroller) with it – can use same
optical isolation circuit used with BJT
60
Optically-Isolated Inputs

Off-board (external, remotely located)
switch/data inputs should be optically isolated
–
–
–
–
helps reduce noise
helps prevent ESD-induced damage
prevents “strange” voltages from entering board
eliminates ground loops
61
“Ben Franklin Experiment”
62
“Ben Franklin Experiment”
63
“Ben Franklin Experiment”
64
“Ben Franklin Experiment”
65
Optically-Isolated Inputs
Off-board (external, remotely located)
switch/data inputs should be optically isolated
–
–
–
–
helps reduce noise
helps prevent ESD-induced damage
prevents “strange” voltages from entering board
eliminates ground loops
|
+

Isolated
Power Supply
(or Battery)
Remote
Switch
66
Optically-Isolated Inputs
Off-board (external, remotely located)
switch/data inputs should be optically isolated
–
–
–
–
helps reduce noise
helps prevent ESD-induced damage
prevents “strange” voltages from entering board
eliminates ground loops
Vcc L = closed
|
+

H = open
Isolated
Power Supply
(or Battery)
Remote
Switch
67
Level Translation

Needed for interfacing CMOS families operating at
different supply voltages
Level Translation
Line Drivers and Receivers

Needed for driving (long) cables based on various
standards (e.g., RS 232, RS 422, RS 485, etc.)
Line Drivers and Receivers

Example: RS 232
Scanned Keypad
73
Scanned Keypad
row
return
lines
active low
column
scan lines
74
Scanned Keypad
H
row
return
lines
H
H
H
active low
column
scan lines
H
H
L
75
Scanned Keypad
H
row
return
lines
H
H
H
active low
column
scan lines
H
L
H
76
Scanned Keypad
H
row
return
lines
H
H
H
active low
column
scan lines
L
H
H
77
Scanned Keypad
H
row
return
lines
H
H
H
active low
column
scan lines
H
H
L
78
Scanned Keypad
H
row
return
lines
H
H
H
active low
column
scan lines
H
L
H
79
Scanned Keypad
H
row
return
lines
L
H
H
active low
column
scan lines
L
H
H
80
Keypad Encoder
81
Rotary Pulse Generator (RPG)
82
RPGs

Determine direction of rotation by concatenating
“previous” and “current” codes, and using as
look-up table index
83
Switch (SPST Contact) Debouncer
84
Position Control

Position control
– required in many applications
– complications
• inertia/mechanical loading
• startup torque different than run torque
• gear backlash
– stepping actuators are a good solution for many
positioning problems
• rotational
• linear
– why steppers are a good choice
•
•
•
•
•
high resolution without gearing
fast positioning (up to 1000 steps/sec)
position error (usually) does not accumulate
wide range of high and low torque (large/small) available
simple/efficient drive circuitry
85
Stepper Motor
Interface
86
L
L
Y
01
11
H
00
H
X
10
VMOTOR
FULL STEP MODE
87
H
L
Y
01
11
L
00
H
X
10
VMOTOR
FULL STEP MODE
88
H
H
Y
01
11
L
00
L
X
10
VMOTOR
FULL STEP MODE
89
L
H
Y
01
11
H
00
L
X
10
VMOTOR
FULL STEP MODE
90
L
L
Y
01
11
H
00
H
X
10
VMOTOR
FULL STEP MODE
91
L
H
Y
01
11
H
00
H
X
10
VMOTOR
HALF STEP MODE
92
L
L
Y
01
11
H
00
H
X
10
VMOTOR
HALF STEP MODE
93
H
L
Y
01
11
H
00
H
X
10
VMOTOR
HALF STEP MODE
94
H
L
Y
01
11
L
00
H
X
10
VMOTOR
HALF STEP MODE
95
H
H
Y
01
11
L
00
H
X
10
VMOTOR
HALF STEP MODE
96
H
H
Y
01
11
L
00
L
X
10
VMOTOR
HALF STEP MODE
97
H
H
Y
01
11
H
00
L
X
10
VMOTOR
HALF STEP MODE
98
L
H
Y
01
11
H
00
L
X
10
VMOTOR
HALF STEP MODE
99
L
H
Y
01
11
H
00
H
X
10
VMOTOR
HALF STEP MODE
100
Hobbyist Servos

Position control
– Single-wire interface
– Can not rotate more than ~270°
– why servos are a good choice
• low overhead for control – logic-level interface
• simple interface (PWM or TIM)
Pulse width (0.9-2.9µs)
Refresh period (12-20ms)
– why servos may not be a good choice
• limited range of motion
• lower torque
Clicker Quiz
102
1. Given that the motor coil requires 1 A of current to
operate, that the NPN switching transistor has a VBEsat of
0.6 V and an hFE = 200, and that the microcontroller port pin
can source up to 10 mA of current at a VOH of 4.2 V, a
suitable value for R would be:
A.
B.
C.
D.
E.
100 
270 
720 
1000 
none of the above
12 V
MOTOR
9S12C32
Port Pin
NPN
R
103
2. If the microcontroller port pin was only capable of
sourcing up to 2 mA at a VOH of 4.2 V (e.g., based on
changing the pin from “full” drive to “reduced” drive mode),
the minimum hFE required to operate the 1 A motor coil
would be:
12 V
A.
B.
C.
D.
E.
100
500
1000
5000
none of the above
MOTOR
9S12C32
Port Pin
NPN
R
104
3. The purpose of the diode in the circuit is to:
A.
B.
protect the microcontroller port pin
to dissipate energy in the motor
coil when the transistor turns off
C. to dissipate energy in the motor
coil when the transistor turns on
D. to limit the amount of current
drawn from the 12 V motor supply
E. none of the above
12 V
MOTOR
9S12C32
Port Pin
NPN
R
105
4. Use of optically isolated inputs:
A. helps prevent ESD-induced damage
B. eliminates ground loops
C. prevents port pins from being driven over/under
rated input voltage swing
D. all of the above
E. none of the above
106
5.
The following ABEL program does not realize an 8-bit shift register:
MODULE shiftregA
TITLE ‘8-bit Shift Register A’
DECLARATIONS
clock pin;
serial_in pin;
q0..q7 pin istype ‘reg’;
MODULE shiftregB
TITLE ‘8-bit Shift Register B’
DECLARATIONS
clock pin;
serial_in pin;
q0..q7 pin istype ‘reg’;
EQUATIONS
[q1..q7] := [q0..q6];
q0 := serial_in;
[q0..q7].clk = clock;
EQUATIONS
[q0..q7].clk = clock;
q7 := serial_in;
[q0..q6] := [q1..q7];
END
(A)
END
(B)
MODULE shiftregC
TITLE ‘8-bit Shift Register C’
DECLARATIONS
clock pin;
serial_in pin;
q0..q7 pin istype ‘reg’;
MODULE shiftregD
TITLE ‘8-bit Shift Register D’
DECLARATIONS
clock pin;
serial_in pin;
q0..q7 pin istype ‘reg’;
EQUATIONS
[q7..q1] := [q6..q0];
q0 := serial_in;
[q7..q0].clk = clock;
EQUATIONS
q7.d = serial_in;
[q0..q6].d = [q1..q7].q;
[q0..q7].clk = clock;
END
(C)
END
(D)
(E) none of the above
107
Digitally Controlled Potentiometer
108
Digitally Controlled Potentiometer
The AD5220 contains a
single channel, 128
position, digitallycontrolled variable
resistor (VR) device.
This device performs
the same electronic
adjustment function as
a potentiometer or
variable resistor
optimized for portable
instrument and test
equipment "push
button" applications.
109
Digitally Controlled Potentiometer
The AD7376 performs
the same electronic
adjustment function as
mechanical
potentiometers,
variable resistors, and
trimmers with
enhanced resolution,
solid-state reliability,
and programmability.
With digital rather than
manual control, the
AD7376 provides layout
flexibility and allows
closed-loop dynamic
controllability.
110
Digital Thermometer
111
Humidity and Temperature Sensor
112
Digital/Analog Compass
Digital Compass
Digital Compass
115
Accelerometer
116
Hall Effect Sensor
117
Pressure (Force) Sensor
118
Ultrasonic Range Finder
119
IR Remote
120
RF Link (Transmitter)
121
RF Link (Receiver)
122
2014 Edition © by D. G. Meyer
Microcontroller-Based
Digital System Design
Module 4-C
Basic Regulated Power Supply Design
123
Outline







Basic half-wave power supply
Half-wave ripple analysis
Basic full-wave supply
Full-wave ripple analysis
Basic linear regulator circuits
Heat sinks and thermal resistance
Low dropout linear regulator circuits
124
2
Basic Half-Wave Supply
1
5
3
+
C
RL
8
1
4
A.C. “wall wart”
Question: Why is this called a “half-wave” circuit?
Answer: Only half of the cycles are delivered to the load
125
Basic Half-Wave Dual Polarity Supply
+
CT
-
126
ripple
Analysis
no-load = VN
loaded = VL
3/240 sec = 12.5 ms
Question: What transformer secondary voltage is
required to obtain VN = 5.0 VDC?
5.0/2  3.5 VAC
Note: Amount of ripple is proportional to load and
inversely proportional to size of filter capacitor
Analysis: Can calculate value of C based on amount
of ripple willing to tolerate at a particular load
127
ripple  0.01 V @ 0.5A load
Analysis
no-load = VN
loaded = VL
3/240 sec = 12.5 ms
Example: Have 500 mA load  RL = VL/IL = 5/0.5 = 10
Assume we want no more than 10 mV of ripple  solve for
C given that VL @ t=12.5 ms is ≥ 4.99 V with RL = 10
VL = VN x e-t/RC
 t = -RL x C x ln(VL/VN)
0.0125 = -10 x C x ln(4.99/5.00) = -10 x C x (-2 x 10-3)
= C x (2 x 10-2)  C = 0.0125/(2 x 10-2) = 0.625 F = 625,000 F128
2
1
Basic Full-Wave Supply
1
5
3
2
4
-
+
8
4
+C
3
1
RL
A.C. “wall wart”
Question: Why is this called a “full-wave” circuit?
Answer: All cycles (+/-) are delivered to the load
129
ripple
Analysis
no-load = VN
loaded = VL
1/240 sec = 4.2 ms
Question: What transformer secondary voltage is
required to obtain VN = 5.0 VDC?
5.0/2  3.5 VAC
Note: Amount of ripple is proportional to load and
inversely proportional to size of filter capacitor
Analysis: Can calculate value of C based on amount
of ripple willing to tolerate at a particular load
130
ripple  0.01 V @ 0.5A load
Analysis
no-load = VN
loaded = VL
1/240 sec = 4.2 ms
Example: Have 500 mA load  RL = VL/IL = 5/0.5 = 10
Assume we want no more than 10 mV of ripple  solve for
C given that VL @ t=4.2 ms is ≥ 4.99 V with RL = 10
VL = VN x e-t/RC
 t = -RL x C x ln(VL/VN)
0.0042 = -10 x C x ln(4.99/5.00) = -10 x C x (-2 x 10-3)
= C x (2 x 10-2)  C = 0.0042/(2 x 10-2) = 0.21 F = 210,000 F131
ripple  0.1 V @ 0.5A load
Analysis
no-load = VN
loaded = VL
1/240 sec = 4.2 ms
Example: Still have 500 mA load  RL = VL/IL = 5/0.5 = 10
Now assume we are willing to tolerate 100 mV of ripple 
solve for C given that VL @ t=4.2 ms is ≥ 4.9 V with RL = 10
VL = VN x e-t/RC
 t = -RL x C x ln(VL/VN)
0.0042 = -10 x C x ln(4.90/5.00) = -10 x C x (-2 x 10-2)
= C x (2 x 10-1)  C = 0.0042/(2 x 10-1) = 0.021 F = 21,000 F132
Thought Questions
Question: What’s inside a basic (unregulated) D.C.
“wall-wart”?
2
1
Answer: Not much…
1
5
3
2
+
4
+
8
C
3
1
4
-
Unregulated D.C. “wall-wart”
133
Thought Questions
Question: What value of C is typically used in a
garden-variety D.C. wall-wart?
Answer: 2000F (or less)
Question: If you purchase an “N-volt” unregulated
D.C. wall-wart, what is VN (the “no-load output”)
likely to be?
Answer: N·2 (“not good for anything sensitive”)
134
Thought Question Analysis
Assume still have 500 mA load  RL = VL/IL = 5/0.5 = 10
and that the D.C. wall-wart has C = 2000 F
Also assume VN = 5.0 VDC
Solve for VL @ t=4.2 ms given RL = 10 and C = 2000 F
VL = VN x e- t/RC
 VL = 5 x e- 0.0042/0.02 = 5 x e 0.21
= 5 x 0.81 = 4.05  0.95 V of ripple
Question: Is such a supply (unregulated D.C. wall-wart)
suitable for operating a microcontroller?
Answer: NO – way too much ripple
135
Basic Positive Voltage Linear Regulator
2
1
filter cap
1
5
3
2
+
4
+C1
8
78XX
1
IN
OUT
2
+
C2
VL
3
1
4
-
VN
-
A.C. “wall wart”
HF noise suppression cap
78XX-series linear regulator IC (XX= 05,12,15 typ)
 can withstand a lot of ripple on input and still
produce a low noise, well-regulated output
 needs input voltage VN  2.5 V higher than
regulated output (VL) to work properly
 VL + 2.5 VDC = Regulator DROPOUT VOLTAGE
136
Source: http://www.nationalsemiconductor.com
137
138
Basic Negative Voltage Linear Regulator
2
1
filter cap
1
5
3
4
4
+
-
2
79XX
VN
1
IN
OUT
2
-
8
3
1
C1
+
C2
VL
+
A.C. “wall wart”
HF noise suppression cap
79XX-series linear regulator IC (XX= 05,12,15 typ)
 used when negative supply voltage needed
 both positive and negative supply regulators can
“co-exist” (need CENTER-TAP A.C. transformer)
 same dropout voltage as positive regulators
139
ripple that can be tolerated
Analysis
dropout = 2.5 V min
no-load = VN
regulated = VL
1/240 sec = 4.2 ms
Design constraints: Need to pick transformer
secondary voltage and filter capacitor C1 such that
VN – ripple voltage ≥ DROPOUT VOLTAGE = VL + 2.5
Note: The larger VN is, the greater the voltage drop
will be across the regulator  larger power (and
heat!) dissipation…but the smaller VN is, the bigger
C1 will need to be to minimize the ripple voltage
140
ripple = 1V
Analysis
dropout = 2.5 V min
no-load = VN
regulated = VL
Example: For a regulated 5 VDC supply, pick VN  8.5 V, so
amount of ripple that can be tolerated is 1 V  transformer
secondary required is 6 VAC
No load: quiescent current is approximately 6 mA 
power dissipation is 8.5 x 0.006 = 51 mW
Full load (1A): 3.5 V @ 1A drops across regulator
(assuming no ripple)  power dissipation = 3.5 X 1 = 3.5 W
Efficiency: 5V @ 1A (5W) delivered to load, but 3.5W
dissipated by regulator  5.0/8.5  59% efficient
141
ripple = 1V
Analysis
dropout = 2.5 V min
VN = 8.5
VL = 5.0
1/240 sec = 4.2 ms
Calculate C1: Have 8.5V @ 1A load (voltage dropped
across regulator is part of load)  RL = VL/IL = 8.5
Assume we want no more than 1 V of ripple  solve for
C1 given that VNmin @ t=4.2 ms is ≥ 7.5 V with RL = 8.5
VL = VN x e-t/RC
 t = -RL x C1 x ln(VNmin/VNmax)
0.0042 = -8.5 x C1 x ln(7.5/8.5) = -8.5 x C x (-0.125)
 C1  3960 F
142
Thought Question
Question: What will happen if we increase the
transformer secondary voltage significantly (e.g.,
to 12 VAC)?  VN  17 VDC
 should be able to tolerate more ripple and “get
by” with a much smaller C1, but power dissipation
will increase and efficiency will decrease
ripple = 9.5V
dropout = 2.5 V min
VN = 17 VDC
VL = 5 VDC
143
Thought Question Analysis
No load: quiescent current is (still) approximately 6 mA 
power dissipation is 17 x 0.006 = 102 mW
Full load (1A): 12 V @ 1A drops across regulator
(assuming no ripple)  power dissipation = 12 X 1 = 12 W
Efficiency: 5V @ 1A (5W) delivered to load, but 12W
dissipated by regulator  5.0/17.0  29% efficient
Calculate C1: Have 17V @ 1A load (voltage dropped
across regulator is part of load)  RL = VL/IL = 17
Assume we want no more than 9.5 V of ripple  solve for
C1 given that VNmin @ t=4.2 ms is ≥ 7.5 V with RL = 17
VL = VN x e-t/RC
 t = -RL x C1 x ln(VNmin/VNmax)
0.0042 = -17 x C1 x ln(7.5/17) = -17 x C1 x (-0.82) = 13.9 x C1
 C1  300 F much smaller than previous case
144
Another Thought Question (or Two)
Question: Can you think of any applications where
we might be willing to tolerate this kind of
inefficiency?
Answer: Application with 12 VDC motor and 5 VDC
microcontroller
Question: How much heat is 5-10 W? What do we
need to do with it? What will it cause if we don’t do
with it what we need to do with it??
- relatively speaking, a LOT
- dissipate it
- premature device failure
145
More Thought Question Analysis
Need to pick properly sized HEAT SINK to help device
dissipate heat
Idea of THERMAL RESISTANCE  measured in ºC/W
(temperature rise per watt dissipated)  “lower thermal
resistance is better”
Thermal resistance is ~ inversely proportional to price.
146
Source: http://www.mouser.com
147
More Thought Question Analysis
Need to pick properly sized HEAT SINK to help device
dissipate heat
Idea of THERMAL RESISTANCE  measured in ºC/W
(temperature rise per watt dissipated)  “lower thermal
resistance is better”
Example: For first case analyzed (6 VAC secondary),
let’s say we don’t want heat sink/junction temperature
to rise more than 10º C above ambient temperature 
need a thermal resistance of approx 10º C/ 3.5 W  3
Thermalloy choice “B” comes close (3.4) – see chart
148
149
3.4
150
More Thought Question Analysis
Need to pick properly sized HEAT SINK to help device
dissipate heat
Idea of THERMAL RESISTANCE  measured in ºC/W
(temperature rise per watt dissipated)  “lower thermal
resistance is better”
Example: For first case analyzed (6 VAC secondary),
let’s say we don’t want heat sink/junction temperature
to rise more than 10º C above ambient temperature 
need a thermal resistance of approx 10º C/ 3.5 W  3
Thermalloy choice “B” comes close (3.4) – see chart
What if we’re “cheap” (only want to spend $0.43 for the
heat sink instead of $1.66)? – choice “R” has a thermal
resistance of 24  will experience 24 x 3.5 = 84º C
temperature rise – ouch!!
P.S. Don’t forget to use HEAT SINK COMPOUND
151
152
24
153
Low Dropout (LDO) Linear Regulator
2
1
filter cap
1
5
3
MIC29150X/TO220
2
+
4
+C1
8
1
IN
OUT
3
+
C2
VL
3
1
4
-
VN
-
A.C. “wall wart”
HF noise suppression cap
Example: Micrel MIC2915X-series
 dropout voltage is only 0.3-0.4 V (compared with
2.5 V for the standard 78XX-series regulators)
 tradeoff with LDO regulators is that since the
dropout voltage is lower, the amount of input ripple that
can be tolerated is lower than a “standard” regulator 154
155
156
78xx “Flashback” for
comparison…
157
158
ripple = 0.6V
Analysis
dropout = 0.4 V
VN = 6 V
VL = 5 V
transformer secondary is 4.25 VAC
1/240 sec = 4.2 ms
Calculate C1: Have 6V @ 1A load (voltage dropped
across regulator is part of load)  RL = VL/IL = 6
Assume we want no more than 0.6 V of ripple  solve for
C1 given that VNmin @ t=4.2 ms is ≥ 5.4 V with RL = 6
VL = VN x e-t/RC
 t = -RL x C1 x ln(VNmin/VNmax)
0.0042 = -6 x C1 x ln(5.4/6.0) = -6 x C x (-0.105) = C x 0.632
 C1  6650 F bigger than “standard” case!
159
Analysis
No load: quiescent current is approximately 8 mA 
power dissipation is 6 x 0.008 = 48 mW
Full load (1A): 6 V @ 1A drops across regulator
(assuming no ripple)  power dissipation = 6 X 1 = 6 W
Efficiency: 5V @ 1A (5W) delivered to load, but 1 W
dissipated by regulator  5.0/6.0  83% efficient
Heat sink: Let’s say we don’t want heat sink/junction
temperature to rise more than 10º C above ambient
temperature  need a thermal resistance of approx
10º C/ 1 W  10  use Thermalloy choice “A” ($1.04)
160
161
10
162
Clicker Quiz
163
1
2
1
3
9 VAC
2
-
+
4
VN
8
C1
3
1
4
VF = 0.7
5
1. Given that the transformer secondary is rated at 9 V and
that the forward voltage of each diode is 0.7 V, then VN
(the no-load D.C. output voltage) will be approximately:
A. 7.6 V
B. 8.3 V
C. 10.75 V
D. 11.33 V
E. 12.73 V
F. none of the above
164
1
2
1
9 VAC
3
4
VF = 0.7
5
2
-
+
4
78XX
VN
1
IN
OUT
2
+
8
C2
VL
3
1
C1
-
2. Given that the transformer secondary is rated at 9 V and
that the forward voltage of each diode is 0.7 V, calculate the
efficiency operating at full load (1 amp) if a 7805 is used
(ignoring power factor correction):
A.
B.
C.
D.
E.
F.
39%
44%
47%
56%
100%
none of the above
165
1
2
1
3
9 VAC
2
-
+
4
78XX
VN
1
IN
OUT
2
+
8
C1
C2
VL
3
1
4
VF = 0.7
5
-
3. Given that the 7805 has a dropout voltage of 2.5 V,
calculate the amount of ripple that can be tolerated
(rounded to the nearest 0.1 V) when sizing C1:
A. 0.1 V
B. 0.8 V
C. 2.5 V
D. 3.8 V
E. 8.8 V
F. none of the above
166
1
2
1
3
4
VF = 0.7
5
9 VAC
2
-
+
4
78XX
VN
1
IN
OUT
2
+
8
C2
VL
3
1
C1
-
4. Based on similar linear power supply examples, the size
of C1 for the design above should be on the order of:
A.
B.
C.
D.
E.
F.
10 µF
100 µF
1000 µF
10,000 µF
100,000 µF
none of the above
167
1
2
1
3
9 VAC
2
-
+
4
78XX
VN
1
IN
OUT
2
+
8
C1
C2
VL
3
1
4
VF = 0.7
5
-
5. If the junction temperature of the 7805 is to rise no more
than 20° C above ambient temperature, then the thermal
resistance of its heat sink should be no more than:
A. 0.3
B. 1.0
C. 3.2
D. 6.4
E. 12.8
F. none of the above
168
2
1
STANDARD 5V
REGULATOR
Dropout
= 2.5 V
78XX
1
VN = 8 V
Ripple = 0.5 V max
5
3
2
4
-
+
4
VN
1
IN
OUT
2
+
8
C2
VL
3
1
C1
-
6. Assuming a ripple tolerance of 0.5 V, the maximum
efficiency at full load (1 A) that can be achieved with a 5 V
regulator circuit based on a standard linear regulator
(that has a dropout voltage of 2.5 V) is approximately
(again, ignoring power factor correction):
A.
B.
C.
D.
E.
50%
60%
63%
70%
none of the above
169
1
2
1
5
3
2
4
-
+
4
VN = 6 V
Ripple
VN = 0.5 V max
LOW-DROPOUT
5V REGULATOR
Dropout
= 0.5 V
78XX
1
IN
OUT
2
+
8
C2
VL
3
1
C1
-
7. Assuming a ripple tolerance of 0.5 V, the maximum
efficiency at full load (1 A) that can be achieved with a 5 V
regulator circuit based on a low-dropout linear regulator
(that has a dropout voltage of 0.5 V) is approximately:
A.
B.
C.
D.
E.
70%
77%
83%
91%
none of the above
170
1
2
1
5
3
2
4
-
+
4
VN = 6 V
Ripple
VN = 0.5 V max
LOW-DROPOUT
5V REGULATOR
Dropout
= 0.5 V
78XX
1
IN
OUT
2
+
8
C2
VL
3
1
C1
-
8. The minimum value of C1 that should be used for this
low-dropout 5V 1 A power supply design is approximately:
A.
B.
C.
D.
E.
2000 µF
HINT:
Value of RL is based on VN @ 1 amp load
4000 µF
t = -RL x C1 x ln(VNmin/VNmax), where
8000 µF
10,000 µF
none of the above
t = 0.0042
171
1
2
1
5
VF = 1.1
3
2
4
-
+
4
VN = 6 V
Ripple
VN = 0.5 V max
LOW-DROPOUT
5V REGULATOR
Dropout = 0.5 V
78XX
1
IN
OUT
2
+
8
C2
VL
3
1
C1
-
9. Assuming each diode in the bridge rectifier has a
forward voltage of 1.1 V (typical), the transformer (RMS)
secondary voltage should be rated at approximately:
A.
B.
C.
D.
E.
5.0 VAC
5.8 VAC
6.3 VAC
9.0 VAC
none of the above
172
2
1
LOW-DROPOUT
5V REGULATOR
Dropout
= 0.5 V
78XX
1
5
3
VF = 1.1
2
4
-
+
4
VN
1
IN
OUT
2
+
8
C2
VL
3
1
C1
-
10. What if, when doing a parts search, you can only
find suitable transformers that have secondary voltage
ratings of 5.0 VAC, 6.0 VAC or 6.3 VAC?
A.
B.
C.
D.
E.
pick the 5.0 VAC transformer and increase C1 to 16,000 µF
pick the 6.0 VAC transformer and reduce C1 to 5000 µF
pick the 6.0 VAC transformer and reduce C1 to 4000 µF
pick the 6.3 VAC transformer and reduce C1 to 2000 µF
none of the above
173
2014 Edition © by D. G. Meyer
Microcontroller-Based
Digital System Design
Module 4-D
Hardware and Software Development Tips
174
Outline

Memory Models
 Discussion
 Application Code Organization
 Dealing With Random Behavior
 Avoiding Really Stupid Tricks
175
Memory Models - 1

What are the primary differences between generalpurpose processor memory models and embedded
processor memory models?
 “Flat” memory model (typically no memory
hierarchy or virtual memory)
 Limited (fixed, “non-infinite”) SRAM data space and
Flash program space
 “Non-homogeneous” memory types
SRAM – “read/write” (volatile unless battery
backup used)
Flash – “read only” (non-volatile in-circuit, sectorerasable and reprogrammable)
EEPROM – “read mostly” (non-volatile in-circuit,
176
byte-erasable and reprogrammable)
Memory Models - 2

How do these differences in memory models influence
way in which high-level language code is written?
 Don’t use too high a level of abstraction
Avoid use of big library routines (e.g., printf)
Avoid dynamic memory allocation
Avoid complex data structures
Avoid recursive constructs
Watch declarations (char, int, long)
 Treat “C” like a “macro-assembly” language
 Remember that floating point support is emulated by
lengthy software routines
 Remember that using table lookup might be a better
approach for transcendental functions (sin, cos, tan,
log) than calculation via software emulation
177
Useful #define Tricks

Get input from GPIO
#define BUTTON_1_MASK
#define BUTTON_1_PRESSED

0x04
( PORT1_IN & BUTTON_1_MASK )
Drive output pins
#define LED_1_MASK
#define LED_1_ON
#define LED_1_OFF
0x20
( PORT1_OUT |= LED_1_MASK )
( PORT1_OUT &= ~(LED_1_MASK) )
if( BUTTON_1_PRESSED ) {
LED_1_ON;
} else {
LED_1_OFF;
}
Discussion - 1

What does “real time” mean (or, what are the
key characteristics of a “real time” system)?
 there are “mission critical” timing constraints
(usually tied to input/output data sampling
rates and/or data processing overhead)
 service latencies are known and fairly tightly
bounded
 are typically “event-driven”
 require low overhead context switching
179
Discussion - 2

What does “fail safe” mean in the context of
embedded software (firmware) development?



A fail-safe or fail-secure device is one
that, in the event of failure, responds in a
way that will cause no harm, or at least a
minimum of harm, to other devices or
danger to personnel.
cite examples of “fail-safe” device behavior
cite examples of “non fail-safe” device
behavior
180
Application Code - 1

What are some possibilities for organizing
embedded application code?

polled program-driven
“round robin” polling loop
advantage: simple!
disadvantage: large number of devices  big
loop  large latency
181
Application Code - 2

What are some possibilities for organizing
embedded application code?

interrupt- (event-) driven
sometimes called “event driven”  all processing
(after initialization) is in response to interrupts
may want CPU to “sleep” between interrupts to
reduce power consumption
182
Application Code - 3

What are some possibilities for organizing
embedded application code?

command-driven or “flag”-driven (also
referred to as “state machine”)
“hybrid” of program-driven and interrupt-driven
service routines “activated” based on (ASCII string)
commands received
alternately, activities of polling loop can be
controlled by state of various “flags” set by interrupt
service routines (e.g., in response to button presses,
time slice expiration, etc.)
183
Application Code - 4

What are some possibilities for organizing
embedded application code?

real-time OS kernel (timer-interrupt driven)
data structure provides list of currently enabled tasks
(can be dynamically inserted/deleted)
 periodic interrupt (RTI) used to determine when
tasks rolled in/out
 can vary relative priority of enabled tasks by
changing time slice allocation
184
Dealing With Random Behavior - 1

Possible “random” behaviors



digital input values read from a port are “random” –
check pull-up (internal/external), most likely inputs are
floating (also check for cold solder joints)
digital input values read from a port have a consistent
(wrong) pattern – check programming of port pins –
make sure they are configured as inputs – input may
be damaged due to ESD/over-voltage
the same value is read from adjacent port pins – check
for solder bridges, cold solder joints, or floating
adjacent pins
185
Dealing With Random Behavior - 2

Possible “random” behaviors…



digital values sent to an output port don’t appear on the
pins (port pins are not driven) – check to make sure port
pins are programmed to be outputs, and that the “drive”
register (if present) is set correctly (some pins can be
configured to be open-drain)
some output pins are always driven high or always driven
low – this is most likely due to a “stuck at” fault resulting
from a failure in the pad driver circuit (if there are spare
port pins available, fly wire around the problem:
otherwise, it’s time to replace the microcontroller)
the same value is output on adjacent port pins – check
for solder bridges (and possible damage to C)
186
Dealing With Random Behavior - 3

Possible “random” behaviors…



analog input values read from the ADC are always zero
(or, always the same value) – check ADC programming
and device driver (make sure you are waiting for the
“conversion complete” flag to be set)
analog input values read from the ADC are “random”
(or, are inconsistent) – check ADC reference voltages
and input pin solder joints (cold joint can act like a
capacitor); also, check driver routine to make sure it is
handshaking on the “conversion complete” flag
the lower two bits (or so) of values read from the ADC
are “random” – this is “normal” (typically all the lower
two bits are good for is rounding the result!)
187
Dealing With Random Behavior - 4

Possible “random” behaviors…


the SCI/SPI is not receiving or transmitting data – most
likely a configuration/driver problem (look at Tx line on
MSO; when Tx works, loop back to Rx line and check to
make sure the same characters your driver is
transmitting are being received)
the SCI is “alive” (can see waveforms on oscilloscope),
but board will not communicate with a PC “com” port
or a serial LCD – check baud rate and character frame
format of sending/receiving end; check to make sure
RS232 level translators generating correct polarity and
voltage swing (nominally  9 V); check to make sure
side “A” Tx connected to side “B” Rx, and vice-versa 188
Dealing With Random Behavior - 5

Possible “random” behaviors…


applications run on the processor for a few seconds,
and then “crash” – may have random interrupts that
are not implemented or handled correctly, stack may
be overflowing (“creep” due to software error)
microcontroller gets too warm for comfort – pins
programmed as outputs are “fighting”, input signals
not at supply rail potential (weak 1’s), or may be
drawing too much current from port pin (source/sink)
189
Clicker Quiz
190
Q1. The code organization used in Labs 7-10 can
best be categorized as:
A. polled program-driven
B. interrupt-driven (event-driven)
C. flag-driven (state machine)
D. real-time OS kernel (timer-interrupt driven)
E. none of the above
191
Q2. When running the ADC in 10-bit mode with
the analog input connected to a fixed voltage
source, the lower two bits keep changing on
successive samples. This is most likely…
A. due to a programming error
B. due to a power supply issue
C. due to a “stuck at” fault (bad pad driver)
D. what should be expected (i.e., normal)
E. none of the above
192
Q3. If a logic “1” is constantly output on a pin that is
programmed as an input port, this most likely is…
A. due to a programming error
B. due to a power supply issue
C. due to a “stuck at” fault (bad pad driver)
D. what should be expected (i.e., normal)
E. none of the above
193
Q4. If different values are written to a port but the
pins are always “0”, this most likely is…
A. due to a programming error
B. due to a power supply issue
C. due to a “stuck at” fault (bad pad driver)
D. what should be expected (i.e., normal)
E. none of the above
194
Q5. If the same value is read from adjacent input
port pins, this most likely is due to a…
A. solder bridge
B. cold solder joint
C. floating adjacent pin
D. all of the above
E. none of the above
195
“Avoid Really Stupid Tricks (RST)” - 1

Do NOT solder parts, attach wires, connect
probes, etc. while your board is powered up


Temporary short circuits can totally fry
numerous components on your board
Remember that the soldering iron tip is
GROUNDED, and that random points tied to
ground on your board while it is in operation
could be a very bad thing!
196
“Avoid Really Stupid Tricks (RST)” - 2

Do NOT attempt to “probe” the pads of
surface mount parts



Few have been known to do this successfully
Shorts between/among microcontroller pins,
regardless of how temporary, can be disastrous
This is why your kit has a breadboard header!
197
“Avoid Really Stupid Tricks (RST)” - 3

Do NOT connect port pins directly to power
supply rails to obtain a “1” or “0” when
debugging (or, in general)




Pin might be programmed as an output, at
which point its pad driver will be instant toast
If pin programmed as an input, might be overbiased and destroy input buffer circuit
ALWAYS use a resistor (10K is a good value) as
a pull-up or pull-down – ALWAYS
To test ADC (analog) inputs, ALWAYS use a
potentiometer (10K is a good value) – ALWAYS198
“Avoid Really Stupid Tricks (RST)” - 4

Do NOT attempt to power different parts of
your circuit with different (external) power
supplies



Think about what will happen to your
board/parts if two essentially infinite current
sources are trying to drive (even slightly)
different voltages on the same net/trace…
Then, look for “burn marks” on fellow students
Use a SINGLE power supply for all logic
components (including the microcontroller
module)
199
“Avoid Really Stupid Tricks (RST)” - 5

Do NOT power your project using the USB
BDM interface (in dialog box, uncheck the
“supply power to target” option)

Look no further than the “dead” PC in lab with
blown out USB interface module…
200