Electronics I - Circuit Elements and Kirchhoff's Laws, Slides of Basic Electronics

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BME 372 Electronics I – J.Schesser 3 Circuit Analysis Lesson #1 BME 372 Electronics I – J.Schesser 4 Circuit Analysis • Circuit Elements – Passive Devices – Active Devices • Circuit Analysis Tools – Ohms Law – Kirchhoff’s Law – Impedances – Mesh and Nodal Analysis – Superposition • Examples BME 372 Electronics I – J.Schesser 7 Circuit Elements – Linear Passive Devices – Capacitor: supports a current which is proportional to its changing voltage, device stores an electric field between its plates, and is governed by Gauss’ Law, units: capacitance or farads, f capacitor with theassociated ecapacitanc theof value theis C where)( dt (t)dVCtI C C  BME 372 Electronics I – J.Schesser 8 Circuit Elements – Linear Passive Devices – Inductor: supports a voltage which is proportional to its changing current, device stores a magnetic field through its coils and is governed by Faraday’s Law, units: inductance or henries, h inductor with theassociated inductance theof value theis L where)( dt (t)dILtV L L  BME 372 Electronics I – J.Schesser 9 Circuit Elements - Passive Devices Continued • Non-linear: supports a non-linear relationship among the currents and voltages associated with it – Diodes: supports current flowing through it in only one direction BME 372 Electronics I – J.Schesser 12 Circuit Elements - Active Devices Continued – Independent vs Dependent Sources • An independent source is one where the output voltage or current is not dependent on other voltages or currents in the device • A dependent source is one where the output voltage or current is a function of another voltage or current in the device (e.g., a BJT transistor may be viewed as having an output current source which is dependent on the input current) + -- Vab(t) a b vs vovi Ro Ri Rs RL Avovi ioii + + - - + - BME 372 Electronics I – J.Schesser 13 Circuit Elements - Active Devices Continued • Non-Linear – Transistors: three or more terminal devices where its output voltage and current characteristics are a function on its input voltage and/or current characteristics, several types BJT, FETs, etc. BME 372 Electronics I – J.Schesser 14 Circuits • A circuit is a grouping of passive and active elements • Elements may be connecting is series, parallel or combinations of both BME 372 Electronics I – J.Schesser 17 Series Circuits • Resistors 20Ω 50Ω RT = 20+ 50 = 70Ω 70)5020( 5020 II IIVVV bcabac   a b c BME 372 Electronics I – J.Schesser 18 Series Circuits • Inductors 25h 100h LT =25+ 100 = 125h dt dI dt dI dt dI dt dIVVV bcabac 125)10025( 10025   a b c BME 372 Electronics I – J.Schesser 19 Series Circuits • Capacitors 5f 10f fCT 33.3 3 10 15 50 105 105 10 1 5 1 1           IdtIdt IdtIdtVVV bcabac 10 3) 10 1 5 1( 10 1 5 1 a b c BME 372 Electronics I – J.Schesser 22 Parallel Circuits • Resistors • Inductors • Capacitors R1 R2 CT = C1+ C2 L1 L2 C1 C2 21 21 21 11 1 LL LL LL LT     21 21 21 11 1 RR RR RR RT     T ab R VVRR R V R VIII   )11( 21 21 21     VdtLVdtLL VdtLVdtLIII T ab 1)11( 11 21 21 21 dt dVCdt dVCC dt dVCdt dVCIII T ab   )( 21 2121 a b Iab I1 I2 Iab I1 I2 a b a b Iab I1 I2 BME 372 Electronics I – J.Schesser 23 Combining Circuit Elements Kirchhoff’s Laws • Kirchhoff Voltage Law: The sum of the voltages around a loop must equal zero • Kirchhoff Current Law: The sum of the currents leaving (entering) a node must equal zero BME 372 Electronics I – J.Schesser 24 Combining Rs, Ls and Cs • We can use KVL or KCL to write and solve an equation associated with the circuit. – Example: a series Resistive Circuit V(t) = I(t)R1 + I(t)R2 V(t) = I(t)(R1 + R2) R1 + -- V(t) I(t) R2 BME 372 Electronics I – J.Schesser 27 Combining Rs, Ls and Cs – Example: a series RLC circuit – Or to simplify this analysis, we can concentrate on special cases  dttICdt tdILRtItV )(1)()()( 1 11 C1 R1 L1 + -- V(t) I(t) BME 372 Electronics I – J.Schesser 28 Impedances • Our special case, signals of the form: V(t) or I(t) =Aest where s can be a real or complex number • This is only one portion of the solution and does not include the transient response. t t t ttt t t tt t etIA eA eeeAe dtAe dt dAeAee dttI dt tdItIe 5 5 5 555 5 5 55 5 36 10)(; 36 10 )36( ) 5 55510(10 551010 )( 2. 1)(510)(10        t t AetI fChLRetV dttI Cdt tdILRtItV 5 111 5 1 11 )( : trysLet' 2.;5;10;10)( :assume sLet' )(1)()()(     BME 372 Electronics I – J.Schesser 29 Impedances • Since the derivative [and integral] of Aest = sAest [=(1/s)Aest], we can define the impedance of a circuit element as Z(s)=V/I where Z is only a function of s since the time dependency drops out. BME 372 Electronics I – J.Schesser 32 Complex Numbers • Complex numbers: What are they? • What is the solution to this equation? ax2+bx+c=0 • This is a second order equation whose solution is: a acbbx 2 42 2,1  BME 372 Electronics I – J.Schesser 33 What is the solution to? 1. x2+4x+3=0 3,12 24 2 44 2 12164 2 3444 2 2,1  x BME 372 Electronics I – J.Schesser 34 What is the solution to? 2. x2+4x+5=0 ????? 2 44 2 20164 2 5444 2 2,1  x BME 372 Electronics I – J.Schesser 37 Rectangular Form • Rectangular (or cartesian) form of a complex number is given as z = x+jy x = Re{z} is the real part of z y = Im{z} is the imaginary part of z Re{z} Im{z} Rectangular or Cartesian x y z = x+jy (x,y) BME 372 Electronics I – J.Schesser 38 Polar Form • is a complex number where: • r is the magnitude of z • θ is the angle or argument of z (arg z) Re{z} Im{z} Polar x y z = r e jθ (r,θ) θ r z  re j  r BME 372 Electronics I – J.Schesser 39 Relationships between the Polar and Rectangular Forms z = x + jy = r e jθ • Relationship of Polar to the Rectangular Form: x = Re{z} = r cos θ y = Im{z} = r sin θ • Relationship of Rectangular to Polar Form: )arctan( and 22 x yyxr   BME 372 Electronics I – J.Schesser 42 Euler’s Formula e jθ = cos θ + j sin θ • We can use Euler’s Formula to define complex numbers z = r e jθ= r cos θ + j r sin θ = x + j y Re{z} Im{z} θ  BME 372 Electronics I – J.Schesser 43 Complex Exponential Signals • A complex exponential signal is define as: • Note that it is defined in polar form where – the magnitude of z(t) is |z(t)| = A – the angle (or argument, arg z(t) ) of z(t) = (ωot + ϕ) • Where ωo is called the radian frequency and ϕ is the phase angle (phase shift) )()(   tj oAetz BME 372 Electronics I – J.Schesser 44 Complex Exponential Signals • Note that by using Euler’s formula, we can rewrite the complex exponential signal in rectangular form as: • Therefore real part is the cosine signal and imaginary part is a sine signal both of radial frequency ωo and phase angle of ϕ )sin()cos( )( )(      tjAtA Aetz oo tj o BME 372 Electronics I – J.Schesser 47 Complex Exponential Function as a function of time • Let’s look at this Re{z} Im{z}  tjteetz tjtj  2sin2cos1)( 2)1(2  t=2/8 seconds arg(z(t))=2π x2/8= π/2; z(t)= 0 + j1 t=1/8 seconds arg(z(t))=2π x1/8=π/4; z(t)=0.707+j 0.707 t=3/8 seconds arg(z(t))=2π x3/8 = 3π/4; z(t)= -0.707+ j0.707 t=4/8 seconds arg(z(t))=2π x4/8 = π; z(t)= -1+ j0 t=5/8 seconds arg(z(t))=2π x5/8 = 5π/4; z(t)= -0.707 - j0.707 t=6/8 seconds arg(z(t))=2π x6/8 = 3π/2; z(t) = 0 - j t=7/8 seconds arg(z(t))=2π x7/8= 7π/4; z(t) = 0 .707- j0.707 t=0 seconds arg(z(t))=2π x0=0; z(t)=1+ j0 t=8/8 seconds arg(z(t))=2π x8/8 = 2π ; z(t)= 1+ j0 BME 372 Electronics I – J.Schesser 48 Phasor Representation of a Complex Exponential Signal • Using the multiplication rule, we can rewrite the complex exponential signal as • X is complex amplitude of the complex exponential signal and is also called a phasor   j tjtjjjtjtj Ae eeAeeAeAetz oooo    X X X toequalnumber complex a is where )( )( BME 372 Electronics I – J.Schesser 49 Graphing a phasor • X=A e jϕ can be graphed in the complex plane with magnitude A and angle ϕ: Re Im ϕ  X=A e jϕ BME 372 Electronics I – J.Schesser 52 Sinusoidal Steady State • If V(t) = A cos (ωt+θ), then we can represent V(t) as Re{Aej(ωt+θ )} = Re{Aejθ ejωt} Since from Euler’s formula: A e j(ωt+θ )=Acos(ωt+θ ) + j Asin (ωt+θ ) BME 372 Electronics I – J.Schesser 53 Sinusoidal Steady State • What is Ae j(ωt+θ) ? – First, it is a complex function since it is a function of a complex number. If we plot on the complex plane, it has a magnitude of A and angle of ωt+θ. It can be viewed as a vector which rotates in time around the origin of the complex plane at angular velocity ω and at t=0 is at θ degrees from the real axis. – We can represent this function by a PHASOR in terms of rectangular coordinates or polar coordinates Real Numbers Complex Plane At t=0 Imaginary Numbers θ Real Numbers Complex Plane At t=20 Imaginary Numbers 20ω+θ MAGNITUDEANGLE (phasor notation) or in this case V  A BME 372 Electronics I – J.Schesser 54 Sinusoidal Steady State Continued • We define the Voltage phasor as V and current phasor as I • Define SSS impedance as Z = V / I using Ohm’s Law • Then the impedances become:  For an inductor V= jL I ZL  jL L 2 ; Here we say the voltage across an inductor leads the current through it by 90.  For a capacitor V= 1 jC I ZC  1 jC  1 C   2 ; Here we say the voltage across a capacitor lags the current through it by 90.  For a resistor V= R I ZR  R  R0 ; Here we say the voltage across a resistor is in phase with the current through it. BME 372 Electronics I – J.Schesser 57 Sinusoidal Steady State Continued ]) )1( [tancos( )1( )( tion,representa time theback to Converting ] )1( [tan )1()1( 0 1 0 1 1 1 1 2 1 1 2 1 1 1 1 1 2 1 1 2 1 1 11 1 11 R C L t C LR AtI R C L C LR A C LjR A Cj LjR A                            I BME 372 Electronics I – J.Schesser 58 Homework R3 1k R1 1k a b Find the total resistance R2 R45k 5k BME 372 Electronics I – J.Schesser 59 Homework Find the total resistance Rab where R1 = 3Ω, R2 = 6Ω, R3 = 12Ω, R4 = 4Ω, R5 = 2Ω, R6 = 2Ω, R7 = 4Ω, R8 = 4Ω R1 R7 R2 R8 R3 R4 R6 R5 a b BME 372 Electronics I – J.Schesser 62 HomeworkC3 1f C1 1f a b Find the total capacitance C2 C45f 5f BME 372 Electronics I – J.Schesser 63 Homework Find and plot the impedance Zab(jω) as a function of frequency. Use Matlab to perform the plot. R=1C=1 L=1 b a BME 372 Electronics I – J.Schesser 64 Homework R=1C=1L=1 b a Find and plot the impedance Zab(jω) as a function of frequency. Use Matlab to perform the plot.