Download Week 1-2

Principles to Computer
Engineering ENG_4_532:
Introduction
Introduction
• Mr Ya Bao (Semester 1); Dr. Zhanfang Zhao (Semester 2)
– Room T-815, Tower Block
– email: [email protected]
– Tel: 020 7815 7588
• Office Hours
– Mondays 15:00 – 16:00
– Lecture notes
– www.lsbu.ac.uk/vle,
Text books (semester 1)
 Electric Circuits, Nilsson & Riedel, Prentice-Hall, 10E.
2015; ISBN:978-1-292-06054-5
 Digital Electronics, Roger L. Tokheim, Mcgraw hill, 8E,
2014; ISBN:978-1-259-06092-2
Overview
•
•
•
•
•
•
Module Overview
Assessment Methods
Log books
Good practice in labs
Theory for Labs week 3
Summary
Learning Outcomes
• Develop an understanding of the principles of
circuit theory and basic analogue electronics
• Introduce the concept of digital systems
• To develop your ability to manage and perform
the successive steps involved in typical practical
engineering experiments
• To develop your ability to work as a member of a
team
• To develop your ability to communicate technical
information (report writing)
Assessment Methods
• Course split 50% Semester 1 & 50% Semester 2
• Phase Test: 20%
– Scheduled for 14/Nov/16; Week 8 in LR-313
• Formal report: 20%
– Due by Week 13, submit to School Office (T313) by
10th Jan 2017
• Logbooks: 10%
– Submit with reports to School Office by 10th Jan 2017
Lab Schedule
1.
2.
3.
4.
5.
6.
7.
8.
Introduction to Lab Equipment
Voltage, Current, Power & Ohm’s Law
Kirchoff’s Laws of Current and Voltage (*)
Source Transformations
Wheatstone Bridge
Transformers and Rectifiers
Basic TTL Logic Circuits
Timing Circuits and Counters
RULES: While you are in the lab
•
•
•
•
•
•
No drinking
No eating
No smoking
No dangerous activity
Take care with electrical devices
Switch off your mobile phone and ipod,
distractions waste time
Log Books
• You MUST keep a log book! Do not use loose paper for
intermediate recording before transfer to the log book.
• You must use the log book to document your procedure
and results AS YOU GO ALONG
• Draw graphs only on graph paper, use pencil where needed
• Answer all questions from the script in your log
• Add ideas/points of interest too
• Conclude each experiment with the key points
• You MUST get your LOGBOOK stamped at the end of
session
• You MUST leave your logbook in the lab
• Go to stores J205 or
good stationers and
buy a suitable book
with lined paper
and add graph
paper as needed
• Best logbooks are
Chartwell
Laboratory Books
A4-641C
• Page number
• Date
• Names of the group
members
• Title of the experiment
• Brief description of the
procedure
• Carefully record your
results
• Do NOT use ‘Tipp-Ex’
• Graphs should
have a caption
• Make maximum
use of the graph
paper
• Choose an easy
scale
• Label the axes
• Units
• Error margins/bars
Error Margins
• You MUST estimate error margins for any
reading you make
• You should propagate these errors for any
calculations you make
• Common measurement devices
– Oscilloscope ±3%
– Digital Multimeter ±1%
– Analogue Meter ±5%
The Breadboard Close-Up
Vertically
Connecte
d
Brea
k
Vertically
Connecte
d
Horizontally
Connected
Break
Error Margins
• Common components
– E12 Resistor ±5%
– Electrolytic Capacitor ±20%
– Digital Timer ±0.001%
• Example logbook notation
– 1.00kΩ ±5%
– 56.2 s ±0.3s
• Avoid quoting too many significant figures
– 3 is usually about right
Resistor E series tables of values
Finally
• Please keep your Laboratory Manual safe!
• Lab and lecture schedule on VLE
• All related materials will be posted on VLE –
check often
• Please arrive promptly .
• Any questions?
Electrical Energy
• Energy cannot be created or destroyed, however,
it may be converted from one form into another.
• In the next four lectures we are going to
investigate electrical energy and its application,
from basic concepts to electric circuits.
• A cell is a device that can generate electricity,
more precisely, it is a device that converts stored
chemical energy into electrical energy
• An electrolyte causes EMF
or Voltage to appear across
the terminals of the cell
Conventional Current
• In reality the flow of current relates to the movement of
charged particles (i.e. electrons) which are in fact
negatively charged through conductive material (e.g.
metal wires)
• However, historically scientists have considered the flow
of current from high to low potential (voltage)
• This is considered “Conventional Current” and most
scientists and engineers use this and not “electron flow”.
Charge, Voltage and Energy
• Cells have two principle parameters, the Charge
stored, Q and the terminal Voltage, V.
• Charge is measured in Coulombs [C],
• Voltage is measured in Volts [V]
• Voltage is the Energy Stored per Coulomb of Charge
W E
V

Q Q
• Where W = Energy Stored in Joules [ J ] and Q =
Charge
• Example: A cell uses 1500 Joules of energy to
generate 1000 Coulombs of charge, what is its
voltage?
Current and Charge
• The smallest charge is a single electron which
has 1.6x10-19 Coulombs
• The rate of flow of electrical charge is termed
‘Current’
Q
I
t
• Where Q = Charge [C] and t = time [s]
• Current is measured in Amps [A]
• Example: If 1000C of electrons travel through
a wire in 100s, what is the current in A and
mA?
Resistance
• Resistance is the property of a material to
“resist” the flow of current
• Conductors have low resistivity per unit area
• Insulators have high resistivity per unit area
• The flow of current through a resistive
material causes a potential difference (or
voltage) to develop across it
• Fixed external resistors are very useful circuit
components and are made from materials
with a known resistivity per unit area
Value of resistor can be calculated by the colour ring printed on.
Resistor colour coding
Voltage Divider Example
• Given that V = 10 and the voltage at VB =
3 what are the voltages VAB and VBC?
-
+
I
V=10V
VBC
C
VC
VAB
A
B
VB
V
B
VA
Figure for Question 1.2 A cell and two resistors in series
Electrical Power
• Electrical power, P, is given by the amount of
electrical energy converted (or absorbed) per unit
time in Watts [W]
E VQ
P


 VI
• Hence
t
t
• Where E = Energy [ J] absorbed, t = time [s], Q =
Charge [C] and I = Current [A]
• Example: A DC motor consumes 2000J of electrical
energy per second when it is in use. Find:
– a) the power consumed by the motor
– b) given that the motor requires 200V to operate deduce
the electric current flowing through the motor.
Internal Resistance of Cells
• All the materials inside cells have some resistance
• The resistance inside a cell is called its “internal
resistance”, this is denoted by r, and causes a voltage
loss when loaded by an external resistance
V
VL
-
+
+
-
r
-
VR
+
External resistor, R
I
Internal Resistance and Voltage
Drop
• Example: A cell V has an internal voltage
1.8 V and the lost voltage VL is 0.3 V
(dropped across its internal resistance).
What is the terminal voltage V across an
external resistor?
V
=1.8V
-
VL=0.3
+V
-
+
r
-
VR
+
External resistor,
R
I
Connecting Cells in Series
• In order for cells to be connected in series the
positive terminal of Cell 1 must connect to the
negative terminal of Cell 2
Cell 1
VC
-
Cell 2
+
VB
-
VBC
+
VA
VAB
Figure 1.4 Two cells in series
• The voltage across Cell 1 VBC = 15 V, and the
voltage across Cell 2 is VAB = 15 V. If we set VC = 0,
then the voltage across two cells VAC is?
Connecting Cells in Parallel
• In practice two (or more) cells with the
same voltage can be connected in parallel
• The voltage across the connected cells is
the same as that across any single one of
them
• The current available from the connected
cells is then multiplied by the number of
cells (as the internal resistance is
effectively reduced)
Circuits and Resistance
• The simplest circuit is one with an electric source
(a cell here) and a resistor, which are connected
with metal wires, as shown in Figure 2.1.
Cell
-
+
V
Metal
wire
R
I
Figure 2.1 Electric circuit
• As mentioned previously, the flow of electrical
current through a resistance causes a voltage
drop (as with the internal resistance of a cell)
Ohm’s Law
• In the early 19th century Georg Ohm proved that the
amount of current through a resistance is
proportional to the voltage across it and inversely
proportional to its resistance. Hence
V
I
R
• This can also be written as:V  IR
• Thus; if there is 1Volt across a resistor and it has 1A
flowing through it, the resistor = 1 Ohm []
• Example: 220V is supplied to a DC motor which has
a resistance of 22 Ω. Find the current flowing in the
motor?
• Answer: I = V/R = 220/22 = 10Amps
Kirchhoff’s Current Law
• Kirchhoff’s current law states that, “The sum of all
the currents passing through any node in a circuit
equals zero”.
• This is the same as saying, “the sum of currents
flowing into that node is equal to the sum of
V
currents flowing out of that
See figure
- node”.
+
T
R1
I1
R2
IT
I2
A
R3
I3
• Hence at Node A;I T  I1  I 2  I 3  0 or I T  I1  I 2  I 3
Kirchhoff’s Current Law: Example 1
• Given that IT = 6 Amps, I1 = 2 Amps, I2 = 1
Amp
• What is the value of I3?
VT
-
+
R1
I1
R2
IT
I2
A
R3
I3
• Answer: I3 = IT – I1 – I2 = 6 – 2 – 1 = 3 Amps
Kirchhoff’s Current Law: Example 2
• Given that I1 = 1.5 Amps, I2 = 0.5 Amps, I3 = 1 Amp
• What is the value of IT?
VT
-
+
R1
I1
R2
IT
I2
A
R3
I3
• Answer: IT = I1 + I2 + I3 = 1.5 + 0.5 + 1 = 3 Amps
Kirchhoff’s Voltage Law
• Kirchhoff’s voltage law states that, “The sum of all
the voltages around a closed loop equals zero”.
• This is the same as saying, “the sum of voltage
drops in a loop is equal to the sum of voltage rises
in a loop”
+
+
V3
V1
–
–
V2
+
–
 V1  V2  V3  0
equivalent ly
 V1  V2  V3  0
or simply
V1  V2  V3
Kirchhoff’s Voltage Law: Example 1
• Given that V3 = 10 Volts and V2 = 4 Volts
• What is the value of V1?
+
+
V3
V1
–
–
V2
+
–
• Answer: V1 = V3 – V2 = 10 – 4 = 6 Volts
Kirchhoff’s Voltage Law: Example 2
• Given that V1 = 2 Volts and V2 = 7 Volts
• What is the value of V3?
+
+
V3
V1
–
–
V2
+
–
• Answer: V3 = V1 + V2 = 2 + 7 = 9 Volts
Series Resistance
• When resistors are connected end to end
they are said to be in series
• The total resistance can be shown to be the
sum of the individual resistances
I
T
VT
V1
I1
V2
I2
V3
• IT = I1 = I2 = I3 and VT = V1 + V2 + V3
• VT = I1R1 + I2R2 + I3 R3 = IT(R1 + R2 + R3)
• Thus:- RT = VT/IT = R1 + R2 + R3
I3
Series Resistance: Example 1
• Given that, in the circuit below;
• R1 = 10, R2 = 20 and R3 = 30
• What is the total series resistance?
IT
VT
R1
I1
R2
I2
R3
I3
• Answer: RT = 10 + 20 + 30 = 60
Series Resistance: Example 2
• Given that, in the circuit below;
• What is the total series resistance?
• What is the current I?
5V
+
-
2Ω
3Ω
I
• Answer: RT = 2 + 3 = 5
• Answer: I = 5/5 = 1 Amp
Resistors in Parallel
• When components are connected in the same way
at both ends, they are said to be in parallel (e.g.
resistors in figure below)
V
T
-
+
IT
I1
V1
R1
I2
V2
R2
I3
V3
R3
• Applying Kirchoff’s Laws give us IT = I1 + I2 + I3 and
VT = V1 = V2 = V3
Resistors in Parallel (2)
• Applying Ohm’s Law to each resistor gives:
VT
V1 V2 V3
IT 
 I1  I 2  I 3 


RT
R1 R2 R3
• Since VT = V1 = V2 = V3, they can be
cancelled out and hence we have
1
1
1
1



RT R1 R2 R3
• Alternatively this can be re-written specifically
for three resistors as:
RR R
RT 
1
2
3
R1 R2  R2 R3  R3 R1
Resistors in Parallel: Example 1
• Given the circuit below, what is the parallel
resistance?
• Answer: 1  1  1  1  0.1  0.042  0.025  0.167
RT
10
• Alternatively:
RT 
24
40
RT 
1
 6
0.167
R1 R2 R3
10  24  40
9600


 6
R1 R2  R2 R3  R3 R1 10  24  24  40  10  40 1600
Resistors in Parallel: Example 2
• Given the circuit below, what is the parallel
resistance?
• Answer: 1  1  1  1  0.05  0.033  0.017  0.1
RT
• Alternatively:
20
30
60
RT 
1
 10
0.1
R1 R2 R3
20  30  60
36000
RT 


 10
R1 R2  R2 R3  R3 R1 20  30  30  60  20  60 3600
Hybrid connected resistors
Series - Parallel Circuits
Potentiometers
• A three-terminal resistor with a sliding or rotating
contact that forms an adjustable voltage divider.
• These devices are commonly called
“Potentiometers” and are typically used in volume
controls on audio equipment
0V
Vout
Vin
Potentiometers continued
• It can be shown that the voltage across
the bar is a fraction of the input voltage
depending on the ratio of the input and
V
V
I

output resistances.
R
R
• Hence as I is the same for the entire
resistive strip, Ohm’s Law shows us that
-
+
out
in
out
I
Vout
Vin
Rin
Rou
V
t out
in
Sliding
bar
Rout
 Vin
Rin
Potentiometers: Example
• Calculate Rout when we require an output
voltage of 10V from a voltage divider,
which has the total resistance of 100 Ω
and can supply the maximum voltage 50
Rout
V.
Vout  Vin
Rin
• Answer:
Rout
• Hence
Vout
10
 Rin
 100  20
Vin
50
Voltage Dividers
• Consider the circuit shown in Figure 3.2
below containing two discrete resistors.
We can develop an equation that
describes the voltage across each resistor
R1 and R2 in terms of the input voltage and
Vin
I

a ratio of resistors
T
R R
1
Vin
-
V1  IT R1
+
IT
- V1 +
- V2 +
R1
R2
Figure 3.2 Potential Divider
2
V2  IT R2
Vin
V1 V2


R1 R2 R1  R2
 R1 

V1  Vin 
 R1  R2 
 R2 

V2  Vin 
 R1  R2 
Voltage Dividers: Example
• Calculate V1 and V2 when Vin = 24V, R1 =
8Ω and R2 = 40Ω
 R1 

V1  Vin 
 R1  R2 
•
 R2 

V2  Vin 
 R1  R2 
 R1 
 8  24  8


V1  Vin 
 24
 4V


Answer: R1  R2   8  40  48
 R2 
 40  24  40
  24
V2  Vin 
 20V

48
 8  40 
 R1  R2 
The Wheatstone Bridge
• We use an “Ohmmeter” to measure an
unknown resistance
• The heart of the simplest Ohmmeter is a socalled “Wheatstone Bridge” circuit
• If R1 was a variable resistor, we can adjust it
until Vab = 0
The Balanced Wheatstone Bridge
• When Vab = 0, a special condition occurs: the bridge
is said to be “balanced”, i.e. Va = Vb
• This implies that ig = 0, hence from KCL, i4 = i3 and i2
= i1
• Further, from Ohm’s Law; i4R4 = i2R2 and i3R3 = i1R1
The Wheatstone Bridge continued
• Hence
i1 R1 i3 R3

i2 R2 i4 R4
R3
R1 
R2
R4
R1 R4  R3 R2
The Wheatstone Bridge: Example
• Calculate R1 in a Wheatstone bridge when it
is balanced and when R2 = 300Ω, R3 =
200Ω, R4 = 100Ω .
R3
R1 
R2
R4
• Answer:
R3
200
R1 
R2 
300  600
R4
100
Capacitors
Type of Capacitors
Capacitance
• Capacitors are devices which store electrical
charge
• A capacitor consists of two plates separated
by an insulator, as shown in Figure 4.1
• The negative plate is connected to a low
potential and the positive plate to a high
potential
Insulator
Q
-
+
-
+
-
+
V
Negative plate
Positive plate
Figure 4.1
Capacitance continued
• The total amount of the charge stored, is
denoted by Q and the voltage across the
plates by V
• The capacitance then is defined as
Q
• Where C is in Farads
C
[F ]
V
• 1 Farad = 1 Coulomb per Volt
• 1mF, µF, nF (nano), pF(pico)
• 0.1 pF is the smallest available in capacitors
for general use in electronic design
Energy Stored in a Capacitor
• When charged, a capacitor stores electrical
energy
• Recall the formula for electrical energy in a circuit,
which is W = VQ
• However, we need to be careful as the voltage
between the plates in a capacitor varies from 0 to
V
• Hence, to be more accurate we should use the
average voltage
0  Vab Vab
Vm 
Vab
W

V
Q

Q
m
• So
2
2
and we know Q  CVab
• Hence W  Vab CVab  1 CVab2
2
2

2
Energy Stored in a Capacitor:
Example
• Question: A capacitor is supplied with 10
V in a circuit. If its capacitance is 150µF,
what is the electrical energy stored in the
capacitor?
• Answer:
1
1
2
W  CVab  150 10 6 10 2  75 10  4 J  7.5mJ
2
2
Charging and discharging a
capacitor
Charging and discharging a
capacitor
Charging and Discharging a
Capacitor
• Charging and discharging a capacitor from a
DC (direct current) source is shown below
V
• We assume that the voltage source, V, has no
internal resistance
• If the switch was held in position 2 for a long
time, then the capacitor would be completely
discharged, Vc = 0V
Charging a Capacitor
• If the switch is then moved to position 1, current will
start to flow through the resistor R, thereby
charging the capacitor, C
• The voltage across the plates of the capacitor will
rise in time, until after a long time, the capacitor will
have the same voltage as the supply, V
V
VC
Discharging a Capacitor
• If the switch is then moved back to position 2,
current will start to flow through the resistor R,
thereby discharging the capacitor, C
• The voltage across the plates of the capacitor will
fall in time, until after a long time, the capacitor will
have no charge at all and again, Vc = 0V.
V
VC
ESSENTIAL FORMULAS
Engineering Science
Electrics
Essential formulas
W E
V

Q Q
Where W = Energy Stored in Joules [ J ] and
Q = Charge in Coulombs [C]
Voltage is measured in Volts [V]
Q
I
t
Where Q = Charge [C] and t = time [s]
Current I is measured in Amps [A] or mA
E VQ
P 
 VI
t
t
Where E = Energy [ J] absorbed, t = time
[s], Q = Charge [C] and I = Current [A]
Power is measured in Watts [W]
Ohm’s Law
V
I
R
I = Current Amp, [A]
V = Voltage, Volts, [V]
R = Resistance, Ohms, [Ω]
P = Power, Watts, [W]
V
R
I
V  IR
2
V
P  VI  I R 
R
2
KC
L
VT -
+
R1
I T  I1  I 2  I 3  0 or I T  I1  I 2  I 3
R2
I1
I2
R3
IT
A
I3
KVL
 V1  V2  V3  0
equivalent ly
 V1  V2  V3  0
or simply
V1  V2  V3
+
+
V3
–
V1
–
V2 +
–
IT
Series
Resistance
R =R +R +R
T
1
2
3
VT
V1
I1
Resistors in
Parallel
1
1
1
1



RT R1 R2 R3
V2
I2
-
+
V3
VT
IT
I1
V1
R1
I2
V2
R2
I3
V3
R3
I3
-
Potentiometer
s
Vin Vout
I

Rin Rout
I
+
Vin
Rin
Rout
Vout
Vout
Sliding bar
Rout
 Vin
Rin
Vin
-
Voltage
Dividers
 R1 
 V2  Vin  R2 
V1  Vin 
R R 
 R1  R2 
2 
 1
+
IT
- V1 +
- V2 +
R1
R2
Figure 3.2 Potential Divider
The Wheatstone
Bridge
balanced, i.e. Va =
Vb
R3
R1 
R2
R4
R1R4  R2 R3
Capacitance
Q
C
V
[F ]
1 Farad = 1 Coulomb per Volt
Vab
1
W
CVab  CVab2
2
2
Charging and Discharging a
Capacitor
time constant t = RC
VC
When discharging
When charging
VC
Source Transformation




Source transformations are useful method of circuit analysis
It is theoretically possible to replace any given arbitrary linear
circuit containing any number of sources and resistances with
either a Thévenin equivalent or Norton equivalent circuit
A simple source transformation allows a voltage source with
series resistance to be replaced by a current source with a
parallel resistance and vice versa.
Both circuits behave in the same way to external loads.
a
b
Thévenin Equivalent Circuit



All linear circuits can be modelled by an independent
voltage source and a series resistor
The voltage of the source is the open circuit voltage of the
network across ‘a’ and ‘b’
The resistance is determined from the short circuit current
Norton Equivalent Circuit



All linear circuits can be modelled by an independent
current source and a parallel resistor
The current of the source is the short circuit current
through ‘a’ and ‘b’
The resistance is determined from the open circuit voltage
(same as Thévenin resistance)
Example 1

Find both Thévenin and Norton equivalent circuits for the
circuit below
END