ELL 100 - Introduction to Electrical Engineering LECTURE 9: TRANSIENT RESPONSE OF FIRST ORDER CIRCUITS (NATURAL RESPONSE) SOURCE-FREE RC CIRCUITS EXAMPLE Fluid-flow analogy: Electrical circuit: water tank emptying Capacitor discharging through a small pipe through resistance 2 SOURCE-FREE RC CIRCUITS EXAMPLE Mechanical system Electrical equivalent (velocity decays (charge dissipated through damping) through resistance) 3 SOURCE-FREE RC CIRCUITS APPLICATIONS High pass filter Low pass filter 4 SOURCE-FREE RC CIRCUITS APPLICATIONS Timers Camera Flash 555 Timer Oscillators circuits 5 SOURCE-FREE RC CIRCUITS APPLICATIONS Delay Circuits Warning Blinkers 6 SOURCE FREE RC CIRCUITS APPLICATIONS Computer Circuits Digital and Time delay circuits 7 SOURCE FREE RC CIRCUITS APPLICATIONS Pacemakers Timing device in automobile intermittent wiper system 8 SOURCE-FREE RL CIRCUIT APPLICATIONS Pulse Generators Electronic filter Tubelight choke 9 TRANSIENT RESPONSE OF FIRST ORDER CIRCUITS • A first-order circuit is characterized by a first-order differential equation. • Example : • a circuit comprising a resistor and capacitor (RC circuit) • a circuit comprising a resistor and an inductor (RL circuit) Applying Kirchhoff’s laws to RC or RL circuit results in differential equations involving voltage or current, which are first-order. 10 TRANSIENT RESPONSE OF FIRST ORDER CIRCUITS EXCITATION There are two ways to excite the circuits. • Initial conditions of the storage elements– Source-Free Circuits (Energy stored in the capacitor, Energy stored in the inductor) • Independent sources – Forced Excitation circuits (DC sources, Sinusoidal sources, Exponential Sources) 11 TRANSIENT RESPONSE OF FIRST ORDER CIRCUITS NATURAL RESPONSE • The natural response of a circuit refers to the behavior (in terms of voltage or current) with no external sources of excitation. • The circuit has a response only because of the energy initially stored in the energy storage elements (i.e. capacitor or inductor). 12 SOURCE-FREE RC CIRCUIT • A source-free RC circuit occurs when its DC source is suddenly disconnected. • The energy already stored in the capacitor(s) is released to the resistor(s) & dissipated. • RC source-free circuit is analyzed from its initial voltage v(0) = V0 and time constant τ 13 SOURCE-FREE RC CIRCUIT DERIVATION • Assume the voltage v(t) across the capacitor. • Since the capacitor is initially charged, Assume that at time t = 0, the initial voltage is, vV(0) 0 with the corresponding value of the energy stored as 1 w(0) CV 2 2 0 14 SOURCE-FREE RC CIRCUIT DERIVATION Applying KCL at the top node of the circuit, yields iCR + i = 0 By definition, iC = C dv∕dt and iR = v ∕ R. Thus, dv v dv v C 0 or 0 dt R dt RC This is a first-order differential equation. 15 SOURCE-FREE RC CIRCUIT DERIVATION dv 1 => dt v RC t Integrating both sides, we get lnv ln A RC vt => ln A RC t/ RC dv v => v( t ) A e 0 dt RC But from the initial conditions, v(0) = A = V0. t/ RC Hence, v( t ) V0 e (Exponentially Decaying) 16 SOURCE-FREE RC CIRCUIT VOLTAGE RESPONSE • As t increases, the voltage decreases exponentially towards zero. The rapidity with which the voltage decreases is expressed in terms of the time constant, denoted by τ. 17 SOURCE-FREE RC CIRCUIT TIME CONSTANT The time constant of a circuit is the time required for the response to decay to a factor of 1/e or 36.8 percent of its initial value. - /RC -1 V0 e = V 0 e 0.368V 0 RC v( t ) V et/ t/ RC0 v( t ) V0 e 18 SOURCE-FREE RC CIRCUIT TIME CONSTANT t v(t)/V0 τ 0.36788 2τ 0.13534 3τ 0.04979 4τ 0.01832 5τ 0.00674 Graphical determination of the time constant τ from the response curve. 19 SOURCE-FREE RC CIRCUIT TIME CONSTANT 20 SOURCE-FREE RC CIRCUIT POWER DISSIPATION The power dissipated in the resistor is V 2 p( t ) vi 0 e2/t R R The energy absorbed by the resistor up to time t is ttV 2 w()() t p d 0 e2/ d R 00R V 2 1 0 e2 / |t CV 2 (1(1 – ee- 22 t/ /τ ) ), RC 22R 00 1 t ,() w CV 2 R 2 0 21 SOURCE-FREE RL CIRCUIT • A circuit with series connection of a resistor and inductor • Current i(t) through the inductor is considered as response of this system. At t = 0, assume that the inductor has an initial current I0, or iI(0) 0 1 Initial energy stored in the inductor w(0) LI 2 2 0 22 SOURCE-FREE RL CIRCUIT RESPONSE OF THE CIRCUIT Applying KVL around the loop, vvLR0 vL = L di/dt and vR = iR. Thus, di di R L Ri 00 i dt dt L it()dit R => dt iL I0 0 i() t Rt Rt/ L => ln i ( t ) I0 e IL0 23 SOURCE FREE RL CIRCUIT RESPONSE OF THE CIRCUIT • Current through inductor decays Rt /L exponentially i() t I0 e • Time constant for the RL circuit is L R t/ i() t I0 e 24 SOURCE FREE RL CIRCUIT POWER DISSIPATION -/t Voltage across the resistor is vR () t iR I0 Re 2 -2t / The power dissipated in the resistor is p vR i I0 Re The energy absorbed by the resistor is tt w()() t p d I2 Re 2 / d R 0 00 1 I2 Re 2 / |t LI 2 (1 e 2 / ), L / R 0 02 0 1 t ,() w LI 2 R 2 0 25 SOLVING NUMERICALS Points to remember : Elements DC steady state Continuous quantity (from t=0- to t=0+) R R - L Short-circuit Current i (v = 0) C Open-circuit Voltage v (i = 0) 26 SOURCE-FREE RC CIRCUIT Q1. Consider the circuit below. Let vC (0)=15 V. Find vc , vx and ix for t > 0. 27 SOURCE-FREE RC CIRCUIT Solution : • We first convert the given circuit into a simple R-C circuit. • Find the equivalent resistance or the Thevenin resistance at the capacitor terminals. 28 SOURCE-FREE RC CIRCUIT 20 5 R4 eq 20 5 The time constant is Req Cs 4(0.1) 0.4 29 SOURCE-FREE RC CIRCUIT t/0.4 2.5t C v( t ) 15 ev( t V ) 15 e V we can use voltage division to get vx 12 2.5t 2.5t vx vC 0.6(15 e ) 9 e V 12 8 t/ v( t ) V0 e v ix 0.75 e2.5t A x 12 30 SOURCE-FREE RC CIRCUIT Q2. The switch in the circuit below is closed for a long time, and then opened at t = 0. Find v(t) for t ≥ 0. Also calculate the energy stored in the capacitor before opening of the switch. 31 SOURCE-FREE RC CIRCUIT Solution: For t < 0, the switch is closed and the capacitor is an open circuit in steady state, as represented in Fig.(a). Using voltage division 9 v( t ) (20)=15 V, t 0 C 93 Since the voltage across a capacitor cannot change instantaneously, the voltage across the capacitor at t = 0− is the same at t = 0+, or vVC (0) V0 15 32 SOURCE-FREE RC CIRCUIT Solution: For t > 0, the switch is open, and we have the RC circuit shown in Fig. (b), Req 1 + 9 10 The time constant is 3 = Req C 10 20 10 0.2s Thus, the voltage across the capacitor for t ≥ 0 is t/ t/0.2 5t v( t ) vC (0)e 15e V 15 e V The initial energy stored 11 w (0)Cv2 (0) 20 10 3 15 2 2.25 J in the capacitor is: C 22C 33 SOURCE-FREE RL CIRCUIT Q3. Assuming that i(0) = 10 A, calculate i(t) and ix(t) in the circuit below 34 SOURCE-FREE RL CIRCUIT Solution: There are two ways we can solve this problem Method -1: The equivalent resistance is the same as the Thevenin resistance at the inductor terminals. Because of the dependent source, we insert a voltage source with vo = 1 V at the inductor terminals a-b, as shown below 35 SOURCE-FREE RL CIRCUIT Applying KVL to the two loops, 1 2(i i ) 1 0 i i (1) 1 2 1 2 2 5 6i 2 i 3 i 0 i i (2) 2 1 1 26 1 Substituting Eq. (2) into Eq. (1) gives vo 1 i = -3A, i = - i =3A => R eqR Th 1 0 1 i 3 1 o L 3 2 s The time constant is 1 R2eq 3 The current through the inductor is i( t ) i (0) ett/ 10 e ( 2/3) A , t 0 36 SOURCE-FREE RL CIRCUIT Method-2: Applying KVL to the circuit For loop 1, 1 di 1 2(ii ) 0 (3) 2 dt 12 For loop 2, 5 6i 2 i 3 i 0 i i 2 1 1 26 1 di 2 Substituting above into Eq. (3) gives 1 i 0 dt 3 1 i(t) 2 => ln t |t i ()(0) t i e(2/3) t 10 e (2/3) t A , t 0 i(0) 3 0 37 SOURCE-FREE RL CIRCUIT The voltage across the inductor is di 2(2/3)tt 10 (2/3) v L 0.5(10) e e V dt 33 Since the inductor and the 2-Ω resistor are in parallel, v i( t ) 1.6667 e(2/3)t A , t 0 x 2 38 SOURCE-FREE RL CIRCUIT Q4. The switch in the circuit below is closed for a long time.