First Order Circuits: Simple RC and RL Circuits Electric Circuits 2 Chapter Objectives • RL and RC circuits are called first-order circuits. In this chapter we will do the following: – Develop a vocabulary that will help us talk about the response of a first-order circuit. – Analyze first-order circuits with inputs that are constant after some particular time, to. – Introduce the notion of a stable circuit and use it to identify stable first-order circuits. – Analyze first-order circuits that experience more than one abrupt change. – Introduce the step function and use it to determine the step response of a first-order circuit. – Analyze first-order circuits with inputs that are not constant. First-Order Circuits • The order of a differential equation is the order of the highest order derivative. • The order of the differential equation is usually equal to the number of capacitors plus the number of inductors in the circuit. • Circuits that contain only one inductor and no capacitors or only one capacitor and no inductors can be represented by a first-order differential equation. • These circuits are called first-order circuits. Source – Free RC Circuit • Consider the circuit, KCL, dv v + =0 C dt R or dv 1 + v=0 dt RC Rearranging the terms and simplifying, dv 1 =− v dt RC dv 1 =− dt v RC Source – Free RC Circuit • From the given, Taking the indefinite integral, dv 1 ∫ v = − RC ∫ dt or t +K ln v = − RC ln v(0) = ln V0 = K v t ln v − ln V0 = ln = − V0 RC v −t = e RC V0 Finally, v(t ) = V0 e −t RC Source – Free RC Circuit • Graph of the voltage response in the simple RC circuit, v(t ) = V0 e −t RC • The rate at which the voltage decays is determined solely by the product of the resistance and the capacitance. Source – Free RC Circuit • Since the response is characterized by the circuit elements and not by an external voltage or current source, the response is called the natural response of the circuit. • Generally, v(t ) = V0 e − ( t −t 0 ) / RC , t ≤ t0 TIME CONSTANTS Time Constants • Considering the circuit, • Where the voltage response is, v = V0 e − t / RC Time Constants • Graphs of v for different values for RC, Time Constants • The current in i the circuit, is given as V0 −t / RC i= e R • The current decreases in the same manner as the voltage Time Constants • The time required for the natural response to decay by a factor of 1/e is defined as the time constant, τ V0 e − ( t +τ ) / RC • Which yields, V0 −t / RC − ( t + RC ) / RC = e = V0 e e τ = RC • Which has the units Ω-F = (V/A)(C/V)=(C/A) = s Time Constants • The voltage response in terms of the time constant, τ − t /τ v = V0 e • Knowledge of the time constant allows us to predict the general form of the response. • Another definition is that it is the time required for the natural response to become zero if it decreases at a constant rate equal to the rate of decay. • A first-order circuit is stable if the time constant is not negative, τ ≥ 0. SOURCE-FREE RL CIRCUIT Source – Free RL Circuit • Consider the circuit, The energy stored at t = 0, 1 2 wL (0) = LI 0 2 KVL, di L + Ri = 0 dt or di R + i=0 dt L Rearranging the terms and simplifying, di R =− i dt L di R = − dt i L Source – Free RL Circuit • From the circuit, Taking the indefinite integral, di R ∫ i = − L ∫ dt or R ln i = − t + K L ln i (0) = ln I 0 = K Or , Finally, i (t ) = I 0 e − Rt L i (t ) = I 0 e −t τ i R ln i − ln I 0 = ln = − t I0 L i − Rt =e L I0 Source – Free RL Circuit • Time Constant: τ = L / R H / Ω = (V ⋅ s / A)(V / A) = s • Units: • Current Response: Source – Free RL Circuit • Instantaneous Power: p(t ) = Ri (t ) = RI 0 e • Energy absorbed by resistor: w(∞) = 1 LI 0 2 2 −2 Rt / L 2 2 RESPONSE TO A CONSTANT FORCING FUNCTION Response to a constant forcing function • Consider the driven RC network: Response to a constant forcing function • For t > 0: Nodal Equation: dv v C + = I0 dt R I0 1 dv + v= dt RC C Separating the variables v − RI 0 dv =− dt RC dv 1 ∫ v − RI 0 = − RC ∫ dt t ln(v − RI 0 ) = − +K RC Response to a constant forcing function • For t > 0: K is the constant of integration: t ln(v − RI 0 ) = − +K RC Rewriting the equation: v − RI 0 = e − t +K RC Solving for v: v = Ae Exponential function − t RC + RI 0 Constant function Response to a constant forcing function From the equation: v = Ae Exponential function vn = Ae − t RC Natural response − t RC + RI 0 Constant function v f = RI 0 Forced response Response to a constant forcing function Natural and Forced response: v = Ae − t RC + RI 0 v f = RI 0 vn = Ae − t RC Forced response Natural response Response to a constant forcing function Complete response: v = Ae − t RC + RI 0 Response to a constant forcing function – Evaluating the constant, A: v = Ae − t RC + RI 0 • At t=0+, observe that the constant A is now determined not only by the initial voltage (or energy) on the capacitor but also by the forcing function Io 0 − v(0) = V0 = Ae V0 = A + RI 0 • Therefore, RC + RI 0 or A = V0 − RI 0 v(t ) = RI 0 + (V0 − RI 0 )e − t RC THE GENERAL CASE The General Case • All equations of the previous sections are special cases of a general expression given by, dy + Py = Q dt Where: y = unknown variable such as v or i P and Q are constants • Source-free RC circuit dv 1 + v=0 dt RC • Forced RC circuit I0 1 dv + v= dt RC C • Source-free RL circuit di R + i=0 dt L The General Case • Solving for y in • We have, dy + Py = Q dt y = e − Pt ∫ Qe Pt dt + Ae − Pt Where: A = constant of integration Q = function of time or a constant • In the important DC case, where Q is a constant: Q − Pt y = Ae + P y = yn + y f Response of a First Order Circuit to a Constant Input • Thevenin and Norton equivalent circuits simplify the analysis of first-order circuits by showing that all first-order circuits are equivalent to one of two simple first-order circuits. A plan for analyzing first-order circuits. (a) First, separate the energy storage element from the rest of the circuit. (b) Next, replace the circuit connected to a capacitor by its Thevenin equivalent circuit, or replace the circuit connected to an inductor by its Norton equivalent circuit. Thevenin Equivalent Circuit Capacitor Norton Equivalent Circuit Inductor Response of a First Order Circuit to a Constant Input (RC) • Given this RC circuit, • At t=0+, the Thevenin equivalent is in Fig. (b) where, R3 Voc = VS R2 + R3 R2 R3 Rt = R2 + R3 d i (t ) = C v(t ) dt Response of a First Order Circuit to a Constant Input (RC) • KVL around the loop, Voc = Rt i (t ) + v(t ) d = Rt C v(t ) + v(t ) dt d v(t ) Voc = v(t ) + dt Rt C Rt C • The First-order D.E. • Thus, the total response is: v(t ) = Voc + (v(0) − Voc )e − t /( Rt C ) Response of a First Order Circuit to a Constant Input (RL) • Given this RL circuit, • At t=0+, the Norton equivalent is shown in Fig. (b) where, VS I sc = R2 R2 R3 Rt = R2 + R3 d v(t ) = L i (t ) dt Response of a First Order Circuit to a Constant Input (RL) • KCL at the top node, v(t ) I sc = + i (t ) Rt d L i (t ) = dt + i (t ) Rt Rt Rt d i (t ) + i (t ) = I sc dt L L • The First-order D.E. • Thus, the total response is: i (t ) = I sc + (i (0) − I sc )e − ( Rt / L )t SINGULARITY FUNCTIONS Singularity functions • Definition: – Forcing functions whose values change abruptly (e.g. voltage or current sources suddenly inserted in switched circuits) • There are many singularity functions that are useful in circuit analysis: – Step function – Pulse The Step Response • the response of a circuit having only one input which is a unit step function. • can be a voltage or current • initial energies are zero - ∞ < t < 0 The Unit Step Function • The unit step function is the function equal to zero for all negative values of its argument and equal to 1 for all positive values of its argument. • Denoted as u(t), u (t ) = 0, t<0 = 1, • Graphically, t>0 Voltage step sources • The unit step function may be used to represent voltages or current with finite discontinuities • Voltage step source of V volts is represented by the product Vu(t) Voltage step source of V volts Equivalent circuit Current step sources • Current step source of V volts is represented by the product Iu(t) Current step source of V volts Equivalent circuit Voltage and current step generators Network with V applied at t=0 Network with I applied at t=0 The Unit Step Function delayed • Generally, u (t − t0 ) = 0, t < t0 = 1, t > t0 • Graphically, Graph of step function u(t-to) • The step response of a circuit is the response of the circuit to the sudden application of a constant source v= Vou(t-to) when all the initial conditions of the circuit are equal to zero. Pulse Signal • A pulse signal has a constant nonzero value for a time duration of Δt = t – to • Consider the pulse source: t < t0 0 v(t ) = V0 t0 < t < t1 0 t1 < t Pulse Signal • The pulse can be obtained using two-step voltage sources, Vo occurring at t = to and –Vo occurring at t = t1 v1 (t ) = V0u (t − t0 ) v2 (t ) = −V0u (t − t1 ) Pulse Signal • For t = to, v1 (t ) = V0u (t − t0 ) • For t = t1, v2 (t ) = −V0u (t − t1 ) v1 (t ) = V0u (t − t0 ) Thus, v(t ) = v1 (t ) + v2 (t ) v(t ) = V0 [u (t − t0 ) − u (t − t1 )] Checking: For t < to, v(t ) = V0 [0 − 0] = 0 V v2 (t ) = −V0u (t − t1 ) For t0 < t < t1, v(t ) = V0 [1 − 0] = V0 For t > t1, v(t ) = V0 [1 − 1] = 0 V t < t0 0 v(t ) = V0 t0 < t < t1 0 t1 < t Pulse Signal: Equivalent Source • The pulse can be obtained using two-step voltage sources, Vo occurring at t = to and –Vo occurring at t = t1 v(t ) = V0u (t − t0 ) − V0u (t − t1 ) Exercise: • Using unit step functions, write an expression for the current i(t), that satisfies i (t ) = 0, = −2 A, = 4 A, = 0, t < 10 ms 10 < t < 20 ms 20 < t < 40 ms t > 40 ms Exercise: i2 (t ) = 6u (t − 0.02) i1 (t ) = −2u (t − 0.01) i3 (t ) = −4u (t − 0.04) Exercise: • Using unit step functions, write an expression for the current i(t), that satisfies i (t ) = 0, = −2 A, = 4 A, = 0, t < 10 ms 10 < t < 20 ms 20 < t < 40 ms t > 40 ms • Answer: − 2u (t − 0.01) + 6u (t − 0.02) − 4u (t − 0.04) A Exercise: • Square wave First Pulse Second Pulse Third Pulse The Response of a First-Order Circuit to a Nonconstant Source • The differential equation of an RL or RC circuit: dx(t ) + ax(t ) = y (t ) dt • Where y(t) is a constant if the source is constant (DC) and α=1/τ the reciprocal of the time constant. • Using the integrating factor method, which consists of multiplying the equation by a factor that makes the left-hand side a perfect derivative, and then integrating both sides. The Response of a First-Order Circuit to a Nonconstant Source • The forced response was found to be dictated by the form of y(t). dx(t ) + ax(t ) = y (t ) dt x f = e −αt ∫ e bt eαt dt 1 xf = e −αt eα +b α +b x f = e −αt ∫ e (α +b ) t dt e − bt xf = α +b • The forced response of an RL or RC circuit is of the same form as the forcing function itself. The Response of a First-Order Circuit to a Nonconstant Source • Sources can be either: Aperiodic (nonperiodic) – non-repeating e.g. exponential Periodic – repeats itself exactly after a fixed length of time. f (t + Τ) = f (t ) Smallest positive number T that satisfies the equation is called the period. If T has no value, the source is aperiodic. The Response of a First-Order Circuit to a Nonconstant Source Special Case: Damped Exponential Input : y (t ) = e − bt e − bt Response : x f = a −b For the special case where a=b, we have a-b=0, the form of the response is indeterminate. When the natural response contains a term of the same form as the forcing function, we need to multiply the assumed form of the forced response by t. SAMPLE PROBLEMS Ex. 1. Determining the Natural Response of an RL Circuit The switch in the circuit shown has been closed for a long time before it is opened at t = 0. Find a) b) c) d) iL(t) for t ≥ 0, i0(t) for t ≥ 0+, v0(t) for t ≥ 0+, the percentage of the total energy stored in the 2 H inductor that is dissipated in the 10 Ω resistor. Ex. 1. Determining the Natural Response of an RL Circuit Solution: a) Find iL(t) for t ≥ 0, t − iL (t ) = I 0e τ The switch has been closed for a long time prior to t = 0, so we know the voltage across the inductor must be zero at t = 0 . Therefore the initial current in the inductor is 20 A at t = 0-. Hence iL(0+) also is 20 A , because an instantaneous change in the current cannot occur in an inductor. Thus, I0 = 20 A. Ex. 1. Determining the Natural Response of an RL Circuit Solution: a) Find iL(t) for t ≥ 0+, We replace the resistive circuit connected to the terminals of the inductor with a single resistor of 10 Ω. I0 = 20 A. Req = 2 + (40 || 10) = 10 Ω Time constant, τ = L/ Req = 0.2 s; 1/τ = 5 iL(t)=20e-5t A Ex. 1. Determining the Natural Response of an RL Circuit Solution: b) Find i0(t) for t ≥ 0+, Using current division, 10 i0 = −iL 10 + 40 10 −5t i0 = 20e 10 + 40 i0 = 4e −5t A 1 i0 = 20e 5 −5t Ex. 1. Determining the Natural Response of an RL Circuit Solution: c) Find v0(t) for t ≥ 0+, Using Ohm’s Law, v0 = 40i0 ( v0 = 40 4e −5t −5t ) v0 = 160e V Ex. 1. Determining the Natural Response of an RL Circuit Solution: d) the percentage of the total energy stored in the 2 H inductor that is dissipated in the 10 Ω resistor. 2 ( ) −5t 2 v 0 160e p10 Ω (t ) = = 10 10 p10 Ω (t ) = 2560e −10tW ∞ w10 Ω (t ) = ∫ 2560e −10t dt = 256 J 1 0 2 1 w(0) = Li = (2)(400) 2 2 w(0) = 400 J 256 ×100 = 64% 400 Ex. 2. Determining the Natural Response of an RC Circuit The switch in the circuit shown in has been in position x for a long time. At t = 0, the switch moves instantaneously to position y. Find a) vc(t) for t ≥ 0, b) v0(t) for t ≥ 0+, c) i0(t) for t ≥ 0+, d) the total energy dissipated in the 60 kΩ resistor. Ex. 2. Determining the Natural Response of an RC Circuit Solution: a) Find vc(t) for t ≥ 0, −t vc (t ) = V0e τ Because the switch has been in position x for a long time, the 0.5 μF capacitor will charge to 100 V and be positive at the upper terminal. We can replace the resistive network connected to the capacitor at t = 0+ with an equivalent resistance of 80 kΩ. Hence the time constant of the circuit is (0.5 x 10-6)(80 x 103) or 40 ms. Thus, −25t vc (t ) = 100e V ,t ≥ 0 Ex. 2. Determining the Natural Response of an RC Circuit Solution: b) Find v0(t) for t ≥ 0+, Using voltage division, 48 v0 = vc (t ) 80 v0 = 60e −25t V ,t ≥ 0 + Ex. 2. Determining the Natural Response of an RC Circuit Solution: c) Find v0(t) for t ≥ 0+, Using Ohm’s Law, v0 (t ) i0 (t ) = 3 60 ×10 i0 (t ) = e −25t mA, t ≥ 0 Ex. 2. Determining the Natural Response of an RC Circuit Solution: d) total energy dissipated in the 60 kΩ resistor. Power dissipated: Total Energy dissipated: ∞ p60 kΩ (t ) = i 0 (t )(60 × 10 ) 2 3 p60 kΩ (t ) = 60e −50t mW , t ≥ 0 + w60 kΩ (t ) = ∫ i 2 0 (t )(60 ×103 )dt 0 w60 kΩ (t ) = 1.2mJ Ex. 3. Determining the Step Response of an RC Circuit The switch in the circuit shown has been in position a for a long time. At t = 0, the switch moved to position b. Find a) vc(t) for t ≤ 0, b) Final value of vc(t). c) Time constant in position b. d) Expression for vc(t) for t ≥ 0+. e) Expression for i(t) for t ≥ 0+. f) How long after the switch in position b does the capacitor voltage equal zero. g) Plot vc(t) and i(t) versus t. Ex. 3. Determining the Step Response of an RC Circuit Solution: a) Initial value of vc(t) The switch has been in position a for a long t ime, so the capacitor look s like an open circuit. Therefore the voltage across the capacitor is the voltage across the 60 Ω resistor. From the voltage divider rule, the voltage across the 60 Ω resistor is 40 x [60/ (60 + 20)] or 30 V. As the reference for vc is positive at the upper terminal o f the capacitor , we have vc(0) = -30 V. Ex. 3. Determining the Step Response of an RC Circuit Solution: b) Final value of vc(t) After the switch has been in position b for a long time, the capacitor will look like an open circuit in terms o f the 90 V source. Thus the final value of the capacitor voltage is +90 V. Ex. 3. Determining the Step Response of an RC Circuit Solution: c) Time constant in position b. τ = RC τ = (400 ×10 )(0.5 ×10 ) 3 τ = 0.2s −6 Ex. 3. Determining the Step Response of an RC Circuit Solution: d) Expression for vc(t) for t ≥ 0+. v(t ) = Voc + (v(0) − Voc )e − t /( Rt C ) Voc = 90V v(0) = −30V τ = 0.2s −5t vc (t ) = 90 − 120e V , t ≥ 0 vc (t ) = 90 + (− 30 − 90 )e −5t