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1 회로 이론 (2014) EMLAB Review of circuit theory I 2 1. Linear system 2. Kirchhoff’s law 3. Nodal & loop analysis 4. Superposition 5. Thevenin’s and Norton’s theorem 6. Resistor, Inductor, Capacitor 7. Operational amplifier 8. First and second order transient circuit EMLAB 3 Linear system i1 (t) Linear system L 1 (t) i2 (t) Linear system L 2 (t) Ai1 Bi2 (t) Linear system L A1 (t) B 2 (t) A system satisfying the above statements is called as a linear system. Resistors, Capacitors, Inductors are all linear systems. An independent source is not a linear system. All the circuits in the circuit theory class are linear systems! EMLAB 4 Examples of Linear system R1 output Resistor R (t ) 1k R1 i1 (t ) R (t ) i1 (t ) R1 input C (t ) - Capacitor C1 C1 i2 (t ) 1n input 1 t C (t ) i2 (t ) dt C0 EMLAB 5 Inductor L (t) i2 (t ) L (t ) L di2 dt EMLAB 6 Kirchhoff’s Voltage law E dr 0 n ( t ) 0 n C Sum of voltage drops along a closed loop should be equal to zero! 1 (t ) 2 (t ) - s (t ) 0 1 (t) R1 2 (t) C1 s (t) - EMLAB 7 Kirchhoff’s Current law J da 0 I n (t ) 0 n S Sum of outgoing(incoming) currents from any node should be equal to zero! 1 (t ) i2 (t ) i1 (t ) 2 (t ) 0 (t) R2 R1 R3 0 - 1 R1 , i2 (t ) 0 - 2 R2 , i3 (t ) 0 -3 R3 i3 (t ) Current definition i n i1 (t ) 3 (t ) a I n (t ) i1 (t ) i2 (t ) i3 (t ) 0 R b a b R 0 - 1 0 - 2 R1 R2 0 -3 R3 0 • To define a current, a direction can be chosen arbitrarily. • The value of a current can be obtained from a voltage drop along the direction of current divided by a resistance met. EMLAB 8 Nodal analysis • Unknowns : node voltages • Kirchhoff’s current law is utilized to form matrix equations. • For each node, the sum of out-going currents become zero. 1 - 0 1 - 0 6k 12k 6m 0 6m 2 - 0 2 - 0 4k 4k 0 2 1 EMLAB 9 Loop analysis • Unknowns : loop currents • Matrix equations are formed by Kirchhoff’s voltage law. • For each loop, the sum of voltage drops are equal to zero. i1 i2 i3 4k (i3 i2 ) 4k i3 0 6k i1 12k (i1 i2 ) 0 i2 6m EMLAB 10 Superposition • Superposition is utilized to simplify the original linear circuits. • If a voltage source is eliminated, it is replaced by a short circuit connected to the original terminals. • If a current source is eliminated, it is replaced by an open circuit. L L1 L 2 Circuit L = Circuit L1 + L2 Circuit Circuit with current source set to zero(OPEN) Circuit with voltage source set to zero (SHORT CIRCUITED) EMLAB 11 Example 02 01 02 4 24 20V 01 01 6k 12V 4V 6k 6k 6k 02 02 1 6k 1 1 6k 12k 6m 6k 2 36 24V 2 1 EMLAB 12 Thevenin and Norton equivalent Resistance obtained with voltage source shorted and current source open V Open circuited voltage measured by voltmeter A Short circuited current measured by ammeter EMLAB 13 Example RTh 2k || 2k 1k VTh RTh 2k || 2k 1k 2k 12V 6V 2k 2k IN 12V 6mA 2k EMLAB Resistor – Input output relationship 14 i(t) I_Probe1.i, A 8 i(t) 6 4 2 0 0 2 4 6 8 10 6 8 10 time, sec v(t) v(t) 20 Vout, V (t ) R i (t ) 15 10 5 0 0 2 4 time, sec EMLAB Capacitor – Input output relationship 15 i(t) 8 I_Probe1.i, A 6 4 2 0 i(t) 0 2 4 6 8 10 time, sec t v(t) v(t) 25 1 t (t ) 0 i ( )d C Vout, V 20 15 10 5 0 0 2 4 6 8 10 time, sec EMLAB Inductor – Input output relationship 16 i(t) I_Probe1.i, A 8 6 4 2 0 0 i(t) 2 4 6 8 10 6 8 10 time, sec v(t) v(t) 4 di (t ) L dt Vout, V 3 2 1 0 -1 0 2 4 time, sec EMLAB 17 First order transient circuit The voltage drop across a capacitor cannot change instantaneously. vC iR 0 dvC 0, vC (t 0) VS dt vC Ae st , Ae st (1 RCs ) 0 vC RC 1 t 1 s , vC Ae RC RC The current through an inductor cannot change instantaneously. di VS (t 0), i (t 0) 0 dt R t V di iR L 0, i (t ) Ae L S dt R R t VS i (t ) [1 e L ] R iR L EMLAB 18 Second order transient circuit R L s C out di 1 t L iR i (t )dt C (0) s dt C0 0 d 2i R di i 2 0 dt L dt LC ( t 0) ( t 0) out 1t i (t )dt C0 Normalized form d 2i di 2 2 0 0i 0 dt 2 dt 2 1 R 0 , LC 2 L 0 2 1 R 0 , LC 20 L i (t ) e st s 2 20 s 02 0 s1, 2 0 0 2 1 EMLAB 19 Solution Over-damped : ζ > 1 Under-damped : ζ <1 i (t ) K1e s t K 2 e s t i (t ) e t A1 cos d t A2 sin d t i ( 0 ) K1 K 2 0 i (0 ) A1 0 1 L 2 di (0 ) s1 K1 s2 K 2 s dt 0 L 1 2 0 d di (0 ) 0 A1 d A2 s dt Critically damped : ζ = 1 i (t ) B1 B2 t e t 0 i (0 ) B1 L di (0 ) 0 B1 B2 s dt EMLAB 20 Critically damped: ζ=1인 경우 d 2i di 2 ( t ) 2 ( t ) 0 0 i (t ) 0 2 dt dt 1 0t [e0t i (t )] 0 [e0t i (t )] 0 e e0t i (t ) B1t B2 i (t ) ( B1t B2 )e 0t EMLAB 21 Transient response Under-damped 0.25 1.6 1 Critically damped 1.4 1.2 Vout, V 1.0 0.8 0.6 1.25 Over-damped 0.4 0.2 0.0 0 5 10 15 20 25 30 time, sec EMLAB 22 Ringing in digital logics in s R s L C out EMLAB Contents : Circuit theory 2 23 1. AC steady-state analysis : 60Hz sinusoidal input signal • Power factor 2. Magnetically coupled networks : transformer 3. Poly-phase circuits : power distribution • Single phase two wire • Three phase 4 wire power distribution 4. Arbitrary input signal • Fourier series and Fourier transform • Laplace transform 5. Two-port network : black box EMLAB 24 Chapter 8 AC steady-state analysis EMLAB 25 Sinusoidal input signal t0 s R L C out 0 ( t 0) d 2x dx 2 (t ) 20 (t ) 0 x (t ) dt 2 dt Vs cos s t (t 0) 입력 전압의 형태 EMLAB 26 Sinusoids x (t ) X M sin t Dimensionless plot As function of time xlead (t ) X M sin (t t0 ) X M amplitude or maximum value angular frequency (rads/sec) t argument (radians) T f 2 Period x(t ) x(t T ), t 1 frequency in Hertz (cycle/sec ) T 2 2 f “Lead by t0” t0 t0 xlag (t ) X M sin (t t0 ) “Lag by t0” EMLAB 27 AC (Alternating Current) • Easy to generate (교류 전압은 만들기 쉽다.) • Easy to change voltage levels. (전압을 변화하기도 쉽다.) • Less damage on human compared with DC (직류에 비해 덜 위험하다.) 60Hz, 220 Vrms EMLAB 28 Solution of Differential Eq. di KVL : L (t ) Ri (t ) VS (t ), i (0 ) 0 dt To solve a differential equation, initial conditions must be specified. L d [ih (t ) i p (t )] dt L V0 cos t (t 0) VS (t ) (t 0) 0 L R [ih (t ) i p (t )] VS (t ) dih (t ) Ri h (t ) 0 dt di p dt (t ) Ri p (t ) VS (t ) i(t ) ih (t ) i p (t ) ih (t ) : Homogeneou s solution i p (t ) : Particular solution EMLAB 29 Solution method #1 di L h (t ) Ri h (t ) 0 dt L di p dt ih (t ) Ae Ae st R t L (t ) Ri p (t ) VS (t ) i p (t ) B cos(t ) For a particular solution, choose a trial function that might produce VS(t) on entering the differential Eq. LB sin( t ) RB cos(t ) V0 cos t ( R cos L sin ) B cos t ( R sin L cos ) B sin t V0 cos t B V0 L , tan R cos L sin R i p (t ) i(t ) Ae L cos(t ), tan 2 2 R R (L) V0 R t L 1 L cos(t ), tan 1 2 2 R R (L) V0 EMLAB 30 Simpler method for sinusoidal source case L di p dt j jt jt (t ) Ri p (t ) VS (t ) i p (t ) B cos(t ) Re{Be e } Re{Ie } d Re{Ie jt } R Re{Ie jt } V0 Re{e jt } dt d Re L Ie jt RIe jt Re{V0 e jt } dt L d Re L Ie jt RIe jt V0 e jt 0 dt d jt Ie RIe jt V0 e jt 0 dt jLIe jt RIe jt V0 e jt L e jt cos t j sin t Complex Polar x jy re j r x 2 y 2 , tan 1 y x x r cos , y r sin V0 e jt V0 jt I , i p (t ) Re{Ie } Re jL R R j L j jt V0e jt V0e e i p (t ) Re Re 2 2 R j L R (L) V0 R (L) 2 2 cos(t ) EMLAB 31 Phasor j V0e jt V0e jt i(t ) Re e Re I e jt Re 2 2 R jL R (L) V0 Phasor : I R jL V0e j R (L) 2 2 V0 R (L) 2 2 • With sinusoidal source function, it is simpler to use a trial solution ~ Re{I ejwt}. • The complex coefficient of the exponential function is called as a phasor. EMLAB 32 Phasor - resistor R i(t) VS (t) I_Probe1.i, A Vout, V 2 1 0 Relationship between sinusoids -1 (t ) -2 0 2 4 6 8 10 12 time, sec 14 16 18 20 i (t ) t EMLAB 33 Phasor - inductor Relationship between sinusoids v(t ) L di (t ) dt (t ) d Re{Ve } L Re{ Ie j t } dt V j L I i (t ) j t t 2 di (t ) v(t ) L dt Re{Ve j t } L 0 d Re{ Ie j t } dt V j L I v(t ) Re{Ve jt } Re{ j L Ie jt } L Re{ Ie j (t / 2) } LI cos(t / 2) EMLAB 34 Phasor - capacitor Relationship between sinusoids i (t ) (t ) i (t ) C dv (t ) dt d Re{Ie } C Re{Ve j t } dt I j C V j t V 0 t 2 I j C EMLAB 35 Examples L 0.05H , I 4 30( A), f 60 Hz C 150 F , I 3.6 145, f 60 Hz Find the voltage across the inductor Find the voltage across the inductor 2 f 120 V jLI V 120 0.05 190 4 30 V 2460 v (t ) 24 cos(120 60) 2 f 120 I jCV V V I jC 3.6 145 120 150 106 190 200 V 235 v (t ) 200 cos(120 t 235) EMLAB 36 Impedance and Admittance I M i Z z VM v AC circuit Impedance : Admittance : Element V VM v VM ( v i ) Z M z R jX I I M i I M I Y YY Y G jB V Z Impedance Admittance R ZR R YR L Z L j L YL C ZC 1 j C 1 R 1 j L YC j C EMLAB 37 Series / parallel combination Z1 Z2 Y1 1 Z1 Y2 1 Z2 Y3 1 Z3 Z3 Yeq Y1 Y2 Y3 Z eq Z1 Z 2 Z 3 Yeq 1 1 1 1 Y1 Y2 Y3 Z eq 1 1 1 1 Z1 Z 2 Z 3 EMLAB 38 Example Find the current i(t) in the network in Fig. E8.8. Y1 Y2 Y1 j 377 50 10 6 j 0.0189 Y2 1 0.0319 j 0.024 3 20 j 377 40 10 Yeq Y1 Y2 0.0319 j 0.0052 0.0323 9.25 I YeqV 0.0323 9.25 120 30 3.88 39.25 i (t ) 3.88 cos(377t 39.25) EMLAB 39 Example 8.15 Super node V1 V2 60 V1 V V 20 2 2 0 1 j1 1 1 j V2 6 V2 V 2 2 1 j1 1 1 j 1 1 V2 1 1 1 j1 6 2 j 1 j1 V2 2.5 j1.5 I0 V2 2.5 j1.5 [ A] 1 EMLAB