Unformatted text preview: ELEC 101 1 Basic Elements 1 2009/10 Spring Fundamental Multiplier for prefix will be
given in tests and exam
Derived from Basic Quantities 10−21 zepto (z)
10−24 yocto (y)
1015
1018
1021
1024 peta (P)
exa (E)
zetta (Z)
yotta (Y) Can use the unit to check
equation
Example : E = mc2 ? ELEC 101 Basic Elements 2 Example 2009/10 Spring Sometime one Quantity can be rewritten
in another form that can be easily to
understand. 1 fA = 10−15 A
1 THz = 1012 Hz
1 nm = 10− 9 m = 10 Å
1 Å = 10− 10 m Å = Amgstron radius of an atom = few Å Example
Both sides SI unit
E = m × c2
J = kg × (m / s) 2
E=F×l
J = N × m = (kg m / s2) × m Example
1 Ah is the unit of charge
= 1 A current for 1 hour of operation
= 1 C/s * 3600s = 3600C
A Liion rechargeable battery has
2000mAh capacity (=2 Ah) capacity.
i.e. The battery can supply 2000mA (=2A)
for 1 hr. ELEC 101 2 Basic Elements 3 Charge • Charge (q or Q) is the basic quantity in
electrical circuits 2009/10 Spring • Conservation of charge (charge can not be
destroyed or created ; total charge for an
isolated system is constant )
1 Ah = 1000 mAh = 3600 coulomb • Circuit operations (such as energy transfer and
information processing ) are movement of
charge
• Unit of charge is Coulomb C
• 1 C = charge of 6.24×1018 electrons.
• 1 electron has charge −1.6 × 10−19 C
• 1 proton has charge 1.6 × 10−19 C
• Electron mass ~ 9.1 × 10−31 Kg ;
(proton has 1.6×10−19 C and ~1.67×10−27 Kg )
• electron so far is still a " basic " particle
(nothing inside?) • 1450 mAh = 5220 coulomb
• = 1.45 A for 1 hour battery operation
or 145 mA for 10 hours
or 2.9 A for 0.5 hour, etc. ELEC 101 3 Basic Elements 4 Current
d a 2009/10 Spring Current I is the rate of flow (change) of
charge. Unit is Ampere (A). Sign Convention of current:
the current direction is defined as the
direction that positive (+) charges flow. I = 1A = 1C/s b instant current i (t ) = dq(t )
dt Example +++++ → → I = 1A positive (+) ions i(t): value of current i at time t
Example
c T 1
average current i = ∫ i (t )dt
T0
i: average current from time 0 to T −−−−− → ← I = 1A electrons
Hence current (I) flow direction is opposite
to electron flow direction. ELEC 101 4 Basic Elements 5 Voltage Voltage is often called potential b ground potential = 0V Movement of charge is controlled by
voltage (electric field).
a V = dW/dq c voltage between points A and B is
defined as the energy gained by a 1 C
charge moving from B to A.
Unit of voltage is Volt (V).
1 volt = 1 joule / 1 coulomb. 2009/10 Spring VAB = VA – VB = – (VB – VA ) = –VBA
Voltage is a relative quantity (a vector)
and has magnitude and direction.
Voltage is also called voltage difference or
potential difference Example A (3V) A (3V) Example A (1V)
element Is VAB = – VBA ? •1C
B (0V) 1C charge gains 1J
(1J is required to
bring the 1C from B
to A) VBA VAB B (1V) B (1V)
Yes
VAB = VA – VB
= 3V – 1V = 2V VBA = VB – VA = – 2V ELEC 101 Basic Elements 6 Example Find VA , VB and VAB. 2009/10 Spring Now measuring the voltage of a
battery with a voltmeter. A
1V VAB = 1V
but VA and VB can have
many values B For example VA – VB = 1V – 0V
= 2V – 1V = 3V – 2V, etc
Two symbols for voltage d A A Example 1V ≡ 1V
B B +
− Example Find the voltmeter reading. +
Example A Are they the same?
A− A
− 3V 3V
B B
Yes − 3V
B+ 1.5V,
AA
battery voltmeter V
− ( 1.5V ) ELEC 101 Example Basic Elements 7 Find the voltmeter reading. Example −
1.5V,
AA
battery 2009/10 Spring 5×1018 electrons flow uniformly
through the cross section of a
wire in 2s. Find the current i(t). voltmeter V
+ i = 0.4A ( − 1.5V ) 5 × 1018 electrons
Example Find the voltmeter reading.
+ 1.5V,
AA
battery V
−
0V dq(t ) Δq
=
Δt
dt
5 ×1018 × −1.6 ×10 −19 C
=
= −0.4A
2s i( t ) =
voltmeter
( floating i.e. voltmeter
has no reading (or
reading fluctuates) , or
0V for some meters)
Note Reason is
VAB = 1.5V,
but VA and VB are
unknowns.
Voltmeter measures VA + A Voltage is usually defined as
energy per charge ( = dE/dQ ), and
1 Volt = 1 Joule / Coulomb (1 V = 1 J/C). V
−
B
0V Voltage is also defined as
power per current ( = dP/dI ) , and
1 Volt = 1 Watt / Ampere (1 V = 1 W/A). ELEC 101 5 Basic Elements 8 c Circuit and Network
a Circuit elements are models of devices circuit model of 1.5V battery +
1.5V
_ Network is a circuit of high complexity
(can be open circuit).
Example of
a network devices: wires, lamps, batteries, diodes,
transistors, and computer chips, etc.
Example 2009/10 Spring 6 +
9V
_ Lumped Model In this course, we assume all elements and
circuits are lumped.
b Circuit is an interconnection of circuit
elements ( closed circuit )
terminals Example A lumped element is assumed to be physically
small (size << wavelength of signal) such that
currents (signals) can pass through it instantly.
Properties of a lumped element is lumped
(concluded) into a point. Circuit model of
a battery and a
light bulb.
+
9V
_ Lumped element has no physical dimension
(only a mathematical symbol).
Circuit composed of lumped elements is a
lumped circuit. ELEC 101 Basic Elements 9 lumped elements Example C 7 Example Wire A
lamp 9V
battery No dimension Wire B not required in exam Lamp is modeled by several lumped elements.
For example, at high frequency, lamp can not be
simply modeled by 1 lumped element R.
Example circuit with 4 lumped elements L R Distributed Model 2009/10 Spring lamp no dimension Lamp in lumped
model (or lumped
parameter model) A wire (short circuit) is also a lumped element
(zero resistance).
lamp We can also ignore wire A and B
and draw circuit with only 2
lumped elements.
battery 8
1 lumped element Terminal and Port
A twoterminals (or 1 port) element is
an electric device with two external
electrical connections called terminals.
Terminal A Lamp in distributed
model (or distributed
parameter model) port
lamp
3 lumped elements Terminal B element ELEC 101 Basic Elements 10 Example Current and Voltage 9 999mV
1000mV 15V
−15V Vout = −15V Op amp: 5 terminals element 2009/10 Spring Properties of a twoterminals element can be
characterized (described) by the
1. current entering the terminal, and the
2. voltage across the terminals.
Current I D
G B
S Example Common Collector
BJT amplifier Input port
Base
Terminal MOS Transistor:
4 terminals element Collector
Terminal
output port Emitter
Terminal
Ports are defined by application
3 terminals and 2 ports element 10 Voltage V Power in Twoterminal Elements
I A
V P=V×I B
Power P in the element is defined as V × I
(note: V I vectors have opposite directions)
power = rate of change of energy
Power is measured in watts (W) or Joule/sec (J/s).
1 watt = 1W = (1 volt )( 1 ampere ) = 1VA ELEC 101 10 Basic Elements 11 A Passive Element B Element consumes (or dissipates, absorbs)
power from the circuit.
v and i vectors have opposite directions.
Power is defined as positive.
Example Find power information
of the element. 2009/10 Spring Active Element
Element supplies (or delivers) power to
the circuit. v and i vectors have same
directions. Power is defined as negative.
Example Find power information of the
element .
2A 2A
2A 2A
6V or +
6V
_ +
6V
_ or
6V (power supplied = 12W) (power consumed = 12W)
2A and 6V have opposite directions.
Hence passive element.
From definition, P = VI = 6V×2A = 12W.
Or power consumed by element is 12W.
Example of
passive element
(resistance R) +
6V
− 2A 2A and 6V have same directions.
Hence active element.
From definition, P = VI = −(6V × 2A) = −12W.
Power supplied by element is 12W. Example of
active element
(battery ) +
6V
_ 2A ELEC 101 Basic Elements 12
Example Instant Power p(t) C 2009/10 Spring Pav = 0
or called instantaneous power p(t) = dE(t) dE(t) dq
=
× = v(t) ×i(t)
dt
dq dt Example T
11 Summary of Circuit Elements
Circuit
elements Find p(t) at t = 3 sec. Passive Active 12V
4A Resistor Independent
Constant D Average Power Pav Pav = time average of p(t) = 1
T T ∫0 p ( t ) dt Dependent Function
of time Function of
another
current Function
of another
voltage Capacitor Voltage
sources Current
sources instant power p(t) at t = 3s = v(3)×i(3)
= 12V × 4A = 48W (passive element) Inductor Independent
Constant Dependent Function
of time Function
of another
current Function
of another
voltage ELEC 101 12
A Basic Elements 13 Resistance and Resistor
Ohm's law i = v / R or v = i × R Time form of Ohm’s law B v(t) = i(t) × R i +
v 2009/10 Spring i(t)
slope = di/dv = 1/R R v(t) −
R is resistance.
Unit is ohm (symbol Ω) i(t)  v(t) (or I  V) characteristics of a resistor 1 ohm = 1 volt / 1 ampere : 1Ω = 1V/1A C Resistivity, conductivity, sheet resistance + Example
Find I . I 4V
− 2Ω metal
plate resistor electron
e
− carbon
v(t) i(t) w
+ t L I = V / R = 4V / 2Ω = 2A ∴R = ρL
wt Given in exam ρ = resistivity of material (unit : Ωm)
σ = 1/ρ is conductivity ELEC 101 Basic Elements 14 If L = w (square R), R = Rs , Rs is
called sheet resistance ρ
∴ R = RS =
unit Ω / square
t
R is only dependent
on t (thickness) Example 2009/10 Spring Find the resistance .
I 1cm t I Given ρ = 1.6 μΩcm 1cm w 1cm V L=w
V ∴R = ρL ρw ρ 1.6μΩcm
=
==
= 1.6μΩ or 1.6μΩ / sq
A w×t t
1cm Example
Cu ≈ 1.7 μΩcm, Al ≈ 2.7 μΩcm
Ag ≈ 1.6 μΩcm, Au ≈ 2.4 μΩcm
Silver (Ag) wire, ρ is lowest Example 1cm
2cm Find the resistance.
Given ρ = 1.6 μΩcm Example
copper (Cu) wire : ρ = 1.7 μΩcm, L = 1cm,
A = 1mm2. Find R.
R = ρL/A = (1.7
= 170 μΩ μΩcm)(1cm)/0.01cm2 I 2cm
V ∴R = ρL ρw ρ 1.6μΩcm
=
==
= 1.6μΩ or 1.6μΩ / sq
A w×t t
1cm ELEC 101 Basic Elements 15 Conductance D 4V Conductance G = 1/R
unit is mhos , symbol Ω −1 ; or siemens S
Sometimes we use conductance
instead of resistance (for example
in parallel circuit) . 2009/10 Spring + 2A power consumed by 2 Ω = 4V*2A = 8W − energy consumed in 2 sec = 8W*2s = 16J Example Find power information of the elements.
1A R = 2Ω
G = 0.5S 1A 1A 5Ω
5V E Power for R 10V Instant power p(t) = v(t) × i(t)
2
= i(t)2 × R = v(t)
R V2
Average power : Pav = I × R =
R For 5V, P = V×I = 5V × 1A = 5W,
power consumed = 5W, passive element 2 F Energy consumed by a resistor
= P × t = VI × t, t is time
Example Find power and energy
consumed in 2 seconds. For 10V, P = V × I = − 10V × 1A = − 10W,
power supplied = 10W, active element
For 5ohm, P = I2 × R = 12 × 5 Ω = 5W,
Power consumed = 5W, passive element
Hence power consumed = power supplied = 10W
Hence conservation of power is valid. ELEC 101 Basic Elements 16 2009/10 Spring For Bulb B Example
Which light bulb (A or B) is brighter? 400W at 200V 200V R2 I
100W Bulb A
at 200V V 2 ( 200V ) 2
∴R 2 =
=
= 100Ω
400W
P 200V
400W
at 200V Bulb B ∴I = V
200V
=
= 0.4A
R1 + R 2 500Ω For Bulb A ∴ P = I R1 = (0.4 A ) × 400Ω = 64 W
2 For Bulb B ∴ P = I 2 R 2 = (0.4 A ) 2 × 100Ω = 16 W For Bulb A
100W at
200V 2 Bulb A is brighter (by 4 times) 200V 0.4A R1 V2
P=
R1 V 2 ( 200V ) 2
∴ R1 =
=
= 400Ω
100W
P R1 = 400Ω Bulb A R 2 = 100Ω Bulb B 200V ELEC 101 Basic Elements 17 I Example Example 2009/10 Spring I 5Ω Find I . 120V 5V R
I = 1A Example I
5Ω 5V i(t) = When bulb is just turned on
I = −1A q(t) q(t) = 5t C. Find i(t). Some bulbs have different resistances when
they are cold and hot.
When bulb is just turned on, I = 4A. Find R. Find I . Example 60W at
120V i(t) dq(t) d5t
=
= 5A
dt
dt q(t) = 5t, i.e. in 1s, 5C flow through the area
(or 10C in 2s), hence i(t) = 5A. V 120V
=
= 30Ω
(cold, low R)
4A
I
The instantaneous power when it is just turned
on
V2
(120V ) 2
p( 0 ) =
=
= 480W
R ( 0)
30Ω
∴R = But when bulb is at steady state (turns on
for a long time) V 2 (120V ) 2
∴R =
=
= 240Ω
P
60W (hot , high R) That’s why light bulbs often burn out when
you turn them on. ELEC 101 13 Basic Elements 18 Capacitance and Capacitor Capacitor I QA
+++++++++ V
l
−Q A positive ions
metal plate
dielectric
metal plate
electrons 2009/10 Spring Example
Mica core capacitor : ε = 5 × 8.85x10−12 F/m,
l = 2cm, A = 100cm2. Find C. εA kε o A 5 × 8.85 × 10 −12 F / m × 10 −2 m 2
∴C =
=
=
l
l
0.02m
= 22.1×10 −12 F ≅ 22pF Capacitance C εA kε o A
∴C =
=
l
l Given in exam ε = permittivity of material
εo = permittivity of vacuum
k = dielectric constant (relative
permittivity) of material
l = length between plates
A = crosssectional area of plates Capacitance C for a device is a measure of
the capacity to store energy at a fixed
voltage.
Capacitance is measured in farads (F).
1 farad = 1 coulomb / 1 volt (1F = 1C/1V)
1F is a very large unit.
Application examples of C: tuner, varactor,
memory. ELEC 101 B Basic Elements 19 Faraday's Law 2009/10 Spring ∴ i(t ) = C q(t) = C × v(t)
+
v(t)
− q(t) i(t) slope =
dq/dv = C q(t) Example
2F 2sin t V Find i(t) . C
−q(t) v(t) i(t) q  v characteristics of capacitance C d ( 2 tV )
dv ( t )
= 2F ×
= 2F × 2V / s = 4 A
dt
dt ∴ i(t ) = −C dv(t )
d( 2 sin tV )
= −2F ×
= −2F × 2 cos tV / s = −4 cos tA
dt
dt Differential form of Faraday’s Law
D d q( t )
i( t ) =
dt
dCv(t )
d v( t )
=
=C
dt
dt Energy stored in capacitor i(t) 2 CVC ( t )
∴ EC ( t ) =
2 v(t) C Proof
t Find i(t). 2F t 0 0 0 E( t ) = ∫ p( t )dt = ∫ v( t )i( t )dt = ∫ v( t )C i(t) Example t 2tV t v 2 (t)
= ∫ v( t )Cdv( t ) = C
2
0 dv( t )
dt
dt ELEC 101 Basic Elements 20 CVC
2F × ( 2 V ) 2
∴ EC =
=
= 4J
2
2
2 Example Find i(t) and i(1s).
Given v(t) = (20t – 20) V for 0 < t ≤ 2s i(t) E
v(t) 2F The value of the voltage at any time t, v(t),
depends on its initial value v(0) and all the
values of the current between time 0 and time t. dv(t )
dt
d( 20t − 20)
= 2F ×
i(t)
dt
= 2F × ( 20 − 0)V / s = 40 A t 1
v( t ) = v(0) + ∫ i( t ' )dt '
C0 Example ∴ i (1ms) = 40A dv( t )
Δv
20V − (−20V )
=C
= 2F ×
= 40A
dt
Δt
2s Example
Find energy stored
in the capacitor . Find v(t) from i(t) v(t) ∴ i( t ) = C or i( t ) = C 2009/10 Spring 2F 2V Find v(1s) from i(t). Given
v(0) = −20V.
i(t) = 40A for 0 < t ≤ 2s.
1s 1s 1
1
v(1s) = v(0) + ∫ i( t' )dt' = −20V +
∫ 40Adt
C0
2F 0
1s 40t
= −20V +
= 0V
2F 0 ELEC 101 14 Basic Elements 21 Inductance and Inductor
i(t) permeability μ = 0.25 × 10−3 ,
N = 4, l = 0.01 m, A = 0.1 m2. Find L. Example Inductor L +
v(t)
− μN 2 A 0.25 × 10 −3 H / m × 4 2 × 10 −1 m 2
∴L =
=
l
0.01m
= 40mH N B
A 2009/10 Spring Inductance L
N 2μA
∴L =
l A l Given in exam Faraday’s Law
i(t) Applications of L: tuner, coil, antenna .. φ(t) L Magnetic flux φ(t) N = number of turns
μ = permeability of material
l = length of coils
A = crosssection area of coil
Inductance is measured in henries (unit H) .
1 henry = 1 weber / 1 ampere : 1H = 1Wb/1A φ(t) = L×i(t)
slope
= dφ/di = L
i(t)
φ  i characteristics
of inductance C Differential form of Faraday’s Law v( t ) = dφ dLi
di
=
=L
dt
dt
dt ELEC 101 Basic Elements 22 φ(t) = ? i(t) = 2tA Example v(t) = ? 2H ∴φ(t) = L×i(t) =2H×2t A= 4t weber
∴ v( t ) = L Find energy stored
in the inductor . 2H 4V LI L (t ) 2H × ( 2t ) 2
∴ EL (t ) =
=
= 4t 2 J
2
2
E Find i(t) from v(t) i(t) = 2tA t 1
∴ i(t ) = i(0) + ∫ v(t' )dt'
L0 v(t) = ? 2H ∴ v( t ) = − L i(t) = 2tA Example 2 d( 2tA )
di
= 2H ×
= 2H × 2 A / s = 4V
dt
dt Example 2009/10 Spring di
d( 2tA )
= −2H ×
= −2H × 2 A / s = −4 V
dt
dt  Example
i(t)
Find v(t). D Energy stored in inductor
t t t 0 0 0 E( t ) = ∫ p( t )dt = ∫ v( t )i( t )dt = ∫ i( t )L
2 t di( t )
dt = L ∫ i( t )di( t )
dt
0 LI ( t )
∴ EL ( t ) = L
2 Given i(t) = 20t − 20 A for 0 < t ≤ 2s. L = 0.5 H .
10V di
d( 20t − 20)
v(t ) = L = 0.5H ×
= 10V
dt
dt ELEC 101 Basic Elements 23 ( VO = 30V ; 30V ; 0V??? ) 15 Active Circuit Element A Independent (or Ideal) Voltage Source
I V +
or V
_ I 2009/10 Spring By definition, an ideal voltage source can
not be connected across a zero R. An independent voltage
source maintains a
constant voltage V across
its terminals, independent
of the current I (load). i(t) = −2A Example
5V A close example of a simple ideal voltage
source is a battery.
2A
Example 5V 5V R 2A
5V voltage source supplies energy to load R.
Example 0Ω
30V load
R R = 1Ω ; 1GΩ ; 0Ω VO = ? R=0 5V 2A is actually flowing into the + terminal of 5V
source. 5V source absorbs energy (charging of
battery). Battery behaves like a passive element. Example iv curve of ideal voltage source v(t) = 3V = constant
for both i(t) > 0 and < 0
(supplying and
consuming power) i(t) 0 v(t)
3V ELEC 101 B I Basic Elements 24 2009/10 Spring (IO = 3A ; 3A ; 0A ???) Independent (or Ideal) Current Source V By definition, an ideal current source can
not be connected across an infinite R. An independent current source
maintains a constant current I
across its terminals, independent
of the voltage V (load). 2A A current source is usually constructed using
complicated electronic components (circuits).
2A Example
I flows out of +ve terminal
of 2A source.
5V
2A current source supplies
energy to R.
Source is an active element. 5V 2A
R
2A R=∞ Example
I flows out of −ve terminal
5V
of the source.
I source absorbs energy.
Source behaves as a passive
element (charging of source). 2A Example
i v Curve of ideal current source Example IO = ?
3A load R R= 0Ω ; 1GΩ ; ∞Ω I = 1A
i(t) = 1A = constant
for both v(t) > 0 and < 0
(supplying and consuming
power) i(t)
0 v(t) ELEC 101 C Basic Elements 25 Real Voltage Source and
Real Current Source
i + R
V i v I R − Real voltage source 30V D Open and Short circuit
Short circuit:
R=0 ideal voltage or
current source
plus internal
resistance R Example Example Example 1mΩ +
v
− 2009/10 Spring I Real current source B− ~3A ~30V ? 3A 10Ω 10Ω A
I 10kΩ 5V ? + Open circuit:
R=∞ By convention: Current
flows from
high voltage (potential)
point A ( + ) to
low voltage point B ( − ).
Voltage at A (VA ) is 5V
higher than voltage at B
(VB) , i.e. if VB = 1V, VA =
6V (or VB is 5V lower
than VA ). ...
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This note was uploaded on 01/28/2011 for the course ELEC 101 taught by Professor Chan during the Fall '09 term at HKUST.
 Fall '09
 CHAN

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