EECE310_F05_Lecture_Notes__wk_of_9_26_05_

# EECE310_F05_Lecture_Notes__wk_of_9_26_05_ - College of...

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Unformatted text preview: College of Engineering, Architecture, and Computer Sciences Department of Electrical and Computer Engineering Chapter 3 Transient Analysis EECE 310 Preston D. Frazier, Ph.D., P.E., PMP College of Engineering, Architecture, and Computer Sciences Department of Electrical and Computer Engineering Chapter 3-Transient Analysis Transient analysis is the examination of the behavior of a network as a function of time. In the previous chapter, time-invariant circuit elements (resistors) were studied. In this chapter time-varying circuit elements (capacitors and inductors) will be introduced and analyzed. EECE 310 Preston D. Frazier, Ph.D., P.E., PMP College of Engineering, Architecture, and Computer Sciences Department of Electrical and Computer Engineering Chapter 3-Transient Analysis Capacitors (C) are passive elements which are constructed by separating two sheets of conductor, which is usually metallic, by a thin layer of insulating material. EECE 310 Preston D. Frazier, Ph.D., P.E., PMP College of Engineering, Architecture, and Computer Sciences Department of Electrical and Computer Engineering Chapter 3-Transient Analysis Capacitance is equal to quotient of the amount of charge in the capacitor (at a particular point in time) by the amount of voltage across the capacitor (at the same exact point in time). Capacitance is formulated as: q( t ) C= v (t ) EECE 310 Preston D. Frazier, Ph.D., P.E., PMP College of Engineering, Architecture, and Computer Sciences Department of Electrical and Computer Engineering Chapter 3-Transient Analysis The current in a capacitor is equated as: dv (t ) i (t ) = C dt Important Note about Capacitors: If the voltage (v) across a capacitor is constant, then the current (i) within the capacitor is zero and it acts like an open circuit. EECE 310 Preston D. Frazier, Ph.D., P.E., PMP College of Engineering, Architecture, and Computer Sciences Department of Electrical and Computer Engineering Chapter 3-Transient Analysis The current-voltage relationship in a capacitor is derived as: 1 t v (t ) = 1 t0 1 t v (t ) = i (t )dt + i (t )dt C - C t0 1 t v (t ) = v (t0 ) + i (t )dt C t0 EECE 310 Preston D. Frazier, Ph.D., P.E., PMP C - i ( t ) dt College of Engineering, Architecture, and Computer Sciences Department of Electrical and Computer Engineering Chapter 3-Transient Analysis w(t ) = The energy (w) in a capacitor is derived as follows: t - p ( t ) dt = t - v ( t )i ( t ) dt t dv (t ) w(t ) = v (t )C dt = C v (t )dv (t ) - - dt t Cv (t ) w(t ) = 2 EECE 310 Preston D. Frazier, Ph.D., P.E., PMP 2 College of Engineering, Architecture, and Computer Sciences Department of Electrical and Computer Engineering Chapter 3-Transient Analysis Equivalent capacitance (Ceq) for a series combination is given in the following equation: Ceq ( series ) 1 -1 = ( ) j =1 C j n n is equal to the number of capacitors EECE 310 Preston D. Frazier, Ph.D., P.E., PMP College of Engineering, Architecture, and Computer Sciences Department of Electrical and Computer Engineering Chapter 3-Transient Analysis Equivalent capacitance (Ceq) for a parallel combination is stated as the following equation: n Ceq ( parallel ) = C j j =1 EECE 310 Preston D. Frazier, Ph.D., P.E., PMP n is equal to the number of capacitors College of Engineering, Architecture, and Computer Sciences Department of Electrical and Computer Engineering Chapter 3-Transient Analysis Inductors (L) are passive elements which consist of a conducting wire that is wound around a core material, which is usually some nonmagnetic to ferro-magnetic material. EECE 310 Preston D. Frazier, Ph.D., P.E., PMP College of Engineering, Architecture, and Computer Sciences Department of Electrical and Computer Engineering Chapter 3-Transient Analysis The voltage (v) in an inductor is stated as: di (t ) v(t ) = L dt Important Note about Inductors: If the current (i) flowing in an inductor is constant, then the voltage (v) across the inductor is zero and it mimics a short circuit. EECE 310 Preston D. Frazier, Ph.D., P.E., PMP College of Engineering, Architecture, and Computer Sciences Department of Electrical and Computer Engineering Chapter 3-Transient Analysis The current-voltage relationship in an inductor is derived as: 1 t i (t ) = 1 t0 1 t i (t ) = v(t )dt + v(t )dt L - L t0 1 t i (t ) = i (t0 ) + v(t )dt L t0 EECE 310 Preston D. Frazier, Ph.D., P.E., PMP L - v (t ) dt College of Engineering, Architecture, and Computer Sciences Department of Electrical and Computer Engineering Chapter 3-Transient Analysis w(t ) = The energy (w) in an inductor is derived as follows: t - p ( t ) dt = t - v ( t )i ( t ) dt t di (t ) w(t ) = L i (t )dt = L i (t )di (t ) - - dt t Li (t ) w(t ) = 2 EECE 310 Preston D. Frazier, Ph.D., P.E., PMP 2 College of Engineering, Architecture, and Computer Sciences Department of Electrical and Computer Engineering Chapter 3-Transient Analysis Total inductance (LT) for a series configuration is given in the following equation: n LT ( series ) = L j j =1 EECE 310 Preston D. Frazier, Ph.D., P.E., PMP n is equal to the number of inductors College of Engineering, Architecture, and Computer Sciences Department of Electrical and Computer Engineering Chapter 3-Transient Analysis Total inductance (LT) for a parallel combination is stated as the following equation: LT ( parallel ) 1 -1 = ( ) j =1 L j n n is equal to the number of inductors EECE 310 Preston D. Frazier, Ph.D., P.E., PMP College of Engineering, Architecture, and Computer Sciences Department of Electrical and Computer Engineering Chapter 3-Transient Analysis Configuration: Series n Parallel Req ( parallel ) = ( j =1 n Resistors Capacitors Inductors R eq ( series ) = n j =1 Rj 1 -1 ) Rj n n is equal to the number of resistors n is equal to the number of resistors C eq ( series ) 1 -1 = ( ) Cj j =1 C eq ( parallel ) = C j =1 j n is equal to the number of capacitors n is equal to the number of capacitors L eq ( series ) = L j =1 n j Leq ( parallel ) = ( j =1 n 1 -1 ) Lj n is equal to the number of inductors . n is equal to the number of inductors EECE 310 Preston D. Frazier, Ph.D., P.E., PMP College of Engineering, Architecture, and Computer Sciences Department of Electrical and Computer Engineering Chapter 3-Transient Analysis Networks which possess time-varying currents and voltages resulting from a sudden application of the (current of voltage) sources, usually due to switching, are called transients. Important Note about Transient Circuits: Applying KCL (for a parallel circuit) or KVL (for a series network) is necessary for understanding their current-voltage relationship. EECE 310 Preston D. Frazier, Ph.D., P.E., PMP College of Engineering, Architecture, and Computer Sciences Department of Electrical and Computer Engineering Chapter 3-Transient Analysis t=0 EECE 310 Preston D. Frazier, Ph.D., P.E., PMP College of Engineering, Architecture, and Computer Sciences Department of Electrical and Computer Engineering Chapter 3-Transient Analysis Examining the node between the resistor and the capacitor, designate it v(t), and set-up the differential equation (D.EQ.) for a series resistor-capacitor (RC) circuit using KCL: V S - v (t ) dv (t ) - C =0 * dt R EECE 310 Preston D. Frazier, Ph.D., P.E., PMP College of Engineering, Architecture, and Computer Sciences Department of Electrical and Computer Engineering Chapter 3-Transient Analysis Series RC circuit (continued): VS - v(t ) dv(t ) = C * R dt First-Order D.EQ.: VS dv (t ) v (t ) =C + * R dt R dv(t ) v(t ) VS + = * dt RC RC EECE 310 Preston D. Frazier, Ph.D., P.E., PMP College of Engineering, Architecture, and Computer Sciences Department of Electrical and Computer Engineering Chapter 3-Transient Analysis The general solution for a first-order D.EQ. is the following: x(t ) = K1 + K 2 e - at Solve for the variables a, K1, and K2 in the general solution: t - RC 1 2 v(t ) = K + K e Volts * * EECE 310 Preston D. Frazier, Ph.D., P.E., PMP College of Engineering, Architecture, and Computer Sciences Department of Electrical and Computer Engineering Chapter 3-Transient Analysis The general solution for a first-order D.EQ. (continued): Plugging ** into * provides *** K2 e - RC t - RC K1 K 2 e + + RC RC t - RC -t RC VS = *** RC Set t = and discover K1 for *** v(t ) = VS + K 2 e Volts * * EECE 310 Preston D. Frazier, Ph.D., P.E., PMP College of Engineering, Architecture, and Computer Sciences Department of Electrical and Computer Engineering Chapter 3-Transient Analysis The general solution for a first-order D.EQ. (continued): t - v(t ) = VS + K 2 e RC Volts * * Set t = 0 and find K2 for ** (v(t = 0) = 0) v(t ) = VS - VS e t - RC Volts * * EECE 310 Preston D. Frazier, Ph.D., P.E., PMP College of Engineering, Architecture, and Computer Sciences Department of Electrical and Computer Engineering Ascertaining a solution for a transient network may be found taking the following steps: 1) Examine the circuit @ t = 0. 2) Remove the inductor and/or capacitor and replace with short and open circuits, respectively. 3) Discover the current and voltage in the circuit (@ t = 0). 4) Determine the transient (t > 0) solution employing KCL or KVL . EECE 310 Preston D. Frazier, Ph.D., P.E., PMP Chapter 3-Transient Analysis ...
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## This note was uploaded on 10/15/2008 for the course EECE 310 taught by Professor Frazier during the Fall '08 term at Howard.

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