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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 Li-ion 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 two-terminals (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 two-terminals 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 Two-terminal 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 : Ω-m) 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 μΩ-cm 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 μΩ-cm, Al ≈ 2.7 μΩ-cm Ag ≈ 1.6 μΩ-cm, Au ≈ 2.4 μΩ-cm Silver (Ag) wire, ρ is lowest Example 1cm 2cm Find the resistance. Given ρ = 1.6 μΩ-cm Example copper (Cu) wire : ρ = 1.7 μΩ-cm, L = 1cm, A = 1mm2. Find R. R = ρL/A = (1.7 = 170 μΩ μΩ-cm)(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 = cross-sectional 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 = cross-section 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 i-v 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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