trans1 - ELEC 101 Transient Circuits 1 2009/10 Spring...

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Unformatted text preview: ELEC 101 Transient Circuits 1 2009/10 Spring Transient Circuit Example Study the behavior of a network as a function of time after a sudden change in the network occurs due to the switches opening and closing. Example switch opens at t = 0 iL t=0 2k You have to key in the pass code within a certain time (say 30 seconds) after entering your house. t The alarm will be ON after 30 seconds if any intruder has triggered the sensor of the alarm system. + VL(t) 5A L 2k VL(t) 0 10 kV a high voltage arc across the switch ELEC 101 1 a Transient Circuits 2 2009/10 Spring Switch Operations t=0 t < 0, switch is opened t = 0 and t > 0, switch is closed c t=0 t=0 t < 0, switch is closed b t=0 t = 0 and t > 0, switch is opened t < 0, switch is closed t = 0 and t > 0, switch is opened ELEC 101 2 Transient Circuits 3 Continuity of Capacitor Voltage VC Capacitor voltage VC is unchanged after switching or VC before switching = VC just after switching or VC is continuous after the switching Example Find VC(t) at t < 0, t = 0, and t > 0. t=0 + VC(t) C 1mF 1m 5V Given C has no initial charge 2009/10 Spring ELEC 101 a Transient Circuits 4 Before switching i.e. at t < 0 i. switch is opened + ii. C has no initial VC(t) charge Q Hence VC = Q/C = 0 c 2009/10 Spring t > 0 i.e. switch is closed VC charges up to 5V (will be discussed later) VC(t) C 1mF 1m 5V 5V c a VC ( t 0) 0V 0 b t VC is continuous at t = 0 (after switching) b Just after switching i.e. at t = 0 d + i. switch is closed VC(t) ii. Since using the continuity theorem, VC is unchanged after the switching C 1mF VC (t 0 ) 0V Hence VC just after switching = VC before switching VC (t 0 ) VC (t 0) 0V 1m 5V At t = i.e. the switch closed for a long time i. circuit is at steady state ii. C behaves like an open circuit VC ( ) 5V ELEC 101 3 Transient Circuits 5 C acts like an open circuit at steady state At steady state, the capacitor is fully charged (discharged), no current flows in the capacitor and the capacitor behaves like an open circuit (iC = 0). 4 Capacitor Current Arc Capacitor current iC NOT continuous with time. The capacitor current can “jump” from one value to another value. Sometime it can cause a high current arc. 2009/10 Spring ELEC 101 Transient Circuits 6 2009/10 Spring VC(t) Sometime we use VC(0) to represent the capacitor voltage right before t = 0, and t = 0+ t = 0 VC(0+) to represent the capacitor voltage right after t = 0. Therefore, VC(0+) = VC(0). Continuity theorem of capacitor voltage 0 t= 0 t=0 t t= 0+ iC(t) t = 0+ On the other hand, iC(0+) ≠ iC(0). If a switching occurs at t = 0, capacitor current right after t = 0 may not necessary equals to the capacitor current right before t = 0. t = 0 t = 0 t 0 t = 0+ t=0 ELEC 101 Transient Circuits 7 2009/10 Spring Find iC(t) at t < 0, t = 0+, and t > 0. Example iC(t) t=0 + iC(t) VC(t) 1m C 1mF + C has no initial charge VC(t) C 1mF 1m 5V iC(t 0) 5kA 5V Hence iC is not continuous at t = 0. a t < 0 switch is opened for a long time C = open circuit c + iC(t) VC(t) C 1mF i C (t 0) i C (t 0 ) 0A b t = 0 switch is closed i. C has no initial charge ii. Since VC (at t = 0+) = VC (at t < 0) = VC(0) = 0V iC(0) 5V V (0 ) 5V 0V C 5kA 1m 1m 1m t>0 i.e. switch is closed VC charges up to 5V 5V d t iC() Since VC() = 5V 5VV () 5V5V C 0A 1m 1m + iC(t) i C () 0 A VC(t) C 1mF 1m 5V ELEC 101 Transient Circuits 8 2009/10 Spring VC(t) 5V b t d VC is continuous after switching (at t = 0, VC = 0 and t = 0+, VC = 0) t>0 c t=0 t<0 a iC(t) t 0 t=0 + iC(t) VC(t) high current arc C 1mF 1m 5V 5kA c t>0 d a t<0 t t 0 b t=0 iC is not continuous after switching (at t = 0, iC = 0 and t = 0+, iC = 5 kA) ELEC 101 5 Transient Circuits 9 Continuity of Inductor Current iL Inductor current iL is unchanged after switching or iL before switching = iL just after switching or iL is continuous after the switching 6 L acts like a short circuit at steady state At steady state, a constant current flows through the inductor. The voltage across the inductor is zero and the inductor behaves like an short circuit (VL = 0). 2009/10 Spring ELEC 101 Transient Circuits 10 Example Find iL(t) at t < 0, t = 0+, and t > 0. t=0 circuit is at steady state for t < 0 iL(t) 1A and switch is L 1k 1k opened at t = 0 a t<0 c 2009/10 Spring t>0 d t switch is closed iL(t) circuit is at steady state L acts like a short circuit iL(t) 1A L switch is opened 1A L 1k 1k i L () 0A 1k 1k i. Since the circuit is at steady state, ii. L behaves like a short circuit Hence the current transient iL(t) iL(t) i L (t 0) 1A b t=0 a 1A switch is opened c iL is discharged to 0A t<0 iL(t) 1A L 1k 1k i. Since iL is unchanged after the switching Hence iL (t = 0+) = iL (t < 0) = 1A d 0 b t iL is continuous at t = 0 (after switching) i L ( t ) 0 A ELEC 101 7 Transient Circuits 11 Inductor voltage Arc Inductor voltage VL is NOT continuous with time. The inductor voltage can “jump” from one value to another value. Sometime it can cause a high voltage arc. Example Find VL(t) at t < 0, t = 0+, and t > 0 . t=0 + iL(t) VL(t) L 1A 1k 1k 2009/10 Spring Example: Fluorescent Lamp Starter ELEC 101 a Transient Circuits 12 Since iL(t < 0) = iL (t = 0+) = iL(0) = 1A t < 0 switch is closed for a long time VL(0) iL(0) 1k L = short circuit Hence VL is not continuous at t = 0 since 1A + VL(0+) 1k 1k VL (t 0) 0V t>0 d t switch is opened circuit is at steady state L acts like a short circuit Since iL() = 0A V () iL() 2 k02 k 0 V L t = 0 switch is opened + VL(t) Since iL(0+) = iL(<0) = 1A + VL(t) L 1k iL(0+) = 1A VL (t 0 ) VL (t 0) c b L 1A1k1kV iL(<0) = 1A + VL(t) L 2009/10 Spring 1A L 1k 1k VL(t 0 ) 1kV a 0 1A 1k 1k VL(t) t c d b iL is NOT continuous 1 kV high voltage arc ELEC 101 8 Transient Circuits 13 iL(t) equation 2009/10 Spring To find time constant after switching, kill V and I sources (short V source and open I source) Step 2 This equation will be given in exam I L (t ) I L ( ) [I L (0 ) I L ( )] e t / is time constant = L/R after switching means t > 0, switch is open iL(t) 1A current source is killed (replaced by open circuit) L 1mH 1k 1k Example Find iL(t) for t = 0+ and t > 0. t=0 iL(t) 1mH 1A L 1k 1k Step 1 At t = 0+ L 1 mH 1 s R 1 k Find iL(), and substitute into the iL(t) equation iL(0+), i L (0 ) i L ( t 0) 1 A At t = switch is opened, L is short i L ( ) 0 A Step 3 Substitute IL(), IL(0), into IL(t) equation i L (t ) i L ( ) [i L (0 ) i L ( )] e t / 0 A [1 A 0 A]e t / 1s 1 e t / 1s A iL(t) 1A 0 t ELEC 101 Transient Circuits 14 Example Find i(t) for t > 0. Circuit is at steady state for t < 0. t=0 8V a L 2H t<0 VO(t) Neon lamp 100 4k Neon lamp: Need a high voltage to ignite. i(t) Lamp ignites when |VO| > 300V. R of the lamp is very large. Circuit is at steady state. L acts like a short circuit. i(<0) 8V 2009/10 Spring 8V VO(<0) 100 4k 2H i(t 0) i(<0) 8V 0.08 A 100 Neon lamp is OFF as VO(t<0) = 8 V < 300V ELEC 101 b Transient Circuits 15 at t = 0+ switch is connected to 4 k i(t 0 ) i(t 0) 0.08 A VO (0 ) 8V i(0) 4 k 2009/10 Spring i( t ) i() [i(0) i()] e t 2mA [80mA 2mA] e t / 0.5 ms 8V 0.08A 4k 312 V -312V 8V 8V 4k i(0+) 2H c i(t) Magnitude of VO > 300V. Hence neon lamp ignites. 100 -320V i(0+) at t = L acts like a short circuit. i ( ) d VO(0+) 8V 2mA 4100 at t > 0 100 4k 2H L R 2H 0.5 ms 4100 80 mA 2 mA 0 t=0 t ELEC 101 9 Transient Circuits 16 VC(t) equation Step 2 This equation will be given in exam VC (t) VC () [VC (0 ) VC ()] et / 2009/10 Spring To find time constant after switching, kill V and I sources (short V source and open I source) after switching means t > 0 , switch is closed is time constant = CR C 1mF 1m kill V source Find VC(t) for t = 0+ and t > 0. Example CR 1 mF 1 m 1 s t=0 + VC(t) C 1mF 1m 5V Step 3 Substitute VC(), VC(0), into VC (t) equation VC (t ) VC ( ) [VC (0 ) VC ( )] e t / Step 1 C has no initial charge At t = 0+, VC (0 ) VC ( t 0) 0V 5V [0V 5V ]e t / 1s 5 (1 e t / 1s )V VC(t) 5V At t = , switch is closed, C is open VC () 5V 0 t ELEC 101 Transient Circuits 17 11 Proof for continuity of iL and VC 10 + iC VC iL + VL CVC ( t ) 2 EC 2 Li ( t ) EL L 2 2 V VC(0) = A E(t) tangent at t = 0 0 VC (t) A e t VC (t 5 τ) VC () Energy EC and EL must be continuous with time (unchanged after switching) a 2009/10 Spring t VC() 0.368 A VC(5) 0.007 A VC() 0 A Hence VC and iL must be continuous with time. 99.3% b Another short proof iC C If VC dVC ( t ) dt VL L di L (t ) dt (or iL ) is not continuous, infinite iC (or VL ) is required. Hence dVC/dt and diL/dt must be finite, or VC and iL must be continuous with time. ELEC 101 12 Transient Circuits 18 13 Proof of VC(t) equation VT Proof of iL(t) equation I(t) R t=0 C 2009/10 Spring I(t) t=0 + V(t) at t 0 IT R L + V(t) at t 0 dI V (IT I) R L dt dt dI dI L / R IT I I IT VT V dV I C R dt dt dV dV CR VT V V VT V(t) VT A e I(t) IT A etR / L t / CR t 0 t t 0 V() VT 0 V(t) V() [V(0) VT ] e t / CR I(0) IT A t V(0) VT A I() IT 0 I(t) I() [I(0) IT ] etR / L ELEC 101 14 Transient Circuits 19 15 C in parallel i 2009/10 Spring C in series i i i C1 V C1 C2 V Ceff Ceff C1 C2 V V V1 V2 Proof V 1 1 ( ) idt C1 C2 dV dV dV dV C2 (C1 C2 ) Ceff dt dt dt dt Example C1 = 3F and C2 = 6F. Find Ceff . 3F 9V 6F idt +18C V2 = 3V 6V 9V Ceff = 3F + 6F = 9F 1 Ceff C1 = 3F and C2 = 6F. Find Ceff and V2 . Ceff 3F // 6F 2F Example Ceff V C2 Proof i C1 1 1 1 Ceff C1 C2 18C +18C 18C +18C 2F 18C V2 = 3V ELEC 101 16 17 L in parallel 2009/10 Spring L in series i i i V Transient Circuits 20 i is equivalent to V L1 L2 is equivalent to L1 V 1 1 1 L eff L1 L 2 Proof 1 1 ) Vdt L1 L 2 i 1 L eff V2 Vdt V V1 V2 (L1 L2 ) Example Example 3H V Leff Leff L1 L2 Proof i( V1 L2 Leff di di Leff dt dt Find Leff . Find Leff . 6H Leff = 3H//6H = 2H 3H Leff 6H Leff = 3H + 6H = 9H Leff ELEC 101 Example Transient Circuits 21 Find iO(t) for t 0. Circuit is stable for t < 0. Kill 12V (replace by a short) = CR = 200F(3k //3k) 0.3s 3k V 3k iO C 200F 12V 6k t=0 3 Find VC(t) first for t 0. Then use VC(t) to find iO(t). use VC (t) VC () [VC (0 ) VC ()] e Find VC(0) for t < 0 t / 3k V 3k iO C 12V 2 12V 4 6k VC () 12V for t 0 3k V 3k C 200F 5 Use VC(t) to find iO(t) for t 0 6k 3k 6V 6k Substitute VC(0+), VC() and into VC(t) equation for t 0 9k t/ 9V VC (0 ) VC (0 ) 9V VC(t) VC() [VC(0 ) VC()]e 12k 6 [9 6]et/0.3s V 6 3et/0.3s V iO(<0)=12V/12k = 1mA Find time constant 12V 200F Find VC () VC() 6V VC(0) 9V 0.3s VC ( 0) 12V 3k V 3k C 3k V 3k C 1 2009/10 Spring iO(t) V (t) C 2 1et/0.3s mA 3k t VC(t) 6 3e0.3s V 9V 6V 0 0.3s iO(t) 2 1e t t 0.3s mA 3mA 2mA 1mA 0 0.3s t ELEC 101 Transient Circuits 22 Example Charge and discharge of flash bulb Close S1 to charge and close S2 to discharge. Find iC(t) and VC(t) . camera t=0 flash 3k V (t) S2 S1 C 15V 1 100F iC(t) 3 a b At t < 0, circuit is in steady state. Hence C acts like an open circuit. Therefore VC(<0) = 0V. 3k V (t) S2 C 100F At t = , i.e. S1 is closed for a long time. VC VC () 15V c Find time constant CR 100 F 3 k 0.3 s Charge 15V At t = 0, S1 is closed. VC (0) VC ( 0) 0V d S1 2009/10 Spring iC(t) Substitute into equation and find the voltage across C in the charging process VC (t) V() [V(0 ) V()] et / CR 15V [0 15V] et / 300ms 15V(1 et / 300ms ) ELEC 101 2 Transient Circuits 23 2009/10 Spring Discharge After S1 is closed for a long time, VC = 15V. S1 is then opened. At t = 0, S2 is closed. 3k V (t) S2 C S1 15V 100F iC(t) VC(t) is continuous 1 3 2 charge VC(t) 15 (1 e t / 300ms )V 15V 15 et / 0.3msV a VC (0 ) VC ( 0) 15V b VC ( ) 0 V c CR 100 F 3 0.3 ms d VC ( t ) V() [V(0 ) V()] e t / CR 0 [15V 0] e t / 0.3ms 15V e t / 0.3ms discharge tangent 0 300ms S1 close S2 open 0 0.3ms S2 close S1 open ELEC 101 Transient Circuits 24 2009/10 Spring Example Find iC(t) from VC(t). 1 S1 Charge 15V 3k V (t) S2 C iC(t) is NOT continuous iC(t) 100F 1 iC (0 ) 5 mA i (t) iC() 0A 15V VC(t) iC(t) 5et 300ms mA 3k C 15V 5mA 3 k Discharge 15V S1 discharge 5 et / 300msmA 0.3ms 0 300ms 2 2 charge 0 5 e t / 0.3ms A 3k V (t) S2 C 100F iC (0 ) 5 A iC () 0A V (t) iC(t) C 5et 0.3ms A 3 iC(t) 3 15V 5A 3 High current arc (5A) lights up flash bulb ...
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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.

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