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First-Order Circuits First-Order Circuits Overview • RC & RL natural response • RC & RL step response ECE 221 • First-order circuits the easy way • Examples Portland State University t=0 + R v - C iC Ver. 1.65 RC Natural Response: Initial Conditions t = 0− V 0 iR For t < 0, the switch has been closed for a long time. Introduction: RL & RC Circuits • First-Order Circuits: Circuits that contain a single capacitor or inductor and a network of DC sources, resistors, and switches • Can also analyze circuits containing multiple capacitors and inductors if we can combine them into a single equivalent capacitor/inductor • Will analyze using KCL & KVL • Generates a ﬁrst-order diﬀerential equation • Simple enough we can use ordinary calculus to solve • Eventually we will discuss the easy way • Circuits which contain no sources are called source-free ECE 221 First-Order Circuits Ver. 1.65 • If there is no source in the circuit, is called the natural response 1 Portland State University t=0 + R v - C iC Ver. 1.65 RC Natural Response: Initial Conditions t = 0+ V 0 iR For t < 0, the switch has been closed for a long time. First-Order Circuits • What is the voltage across the capacitor at t = 0+ ? ECE 221 • What is the voltage across the capacitor at t = 0− ? 3 Portland State University • How much energy is stored in the capacitor at t = 0+ ? Ver. 1.65 • How much energy is stored in the capacitor at t = 0− ? First-Order Circuits • What is the current in the capacitor for t = 0+ ? ECE 221 • What is the current in the capacitor for t = 0− ? Portland State University 2 4 t=0 = = = = = + R v - C iC First-Order Circuits Ver. 1.65 Ver. 1.65 RC Natural Response: Final Conditions t → ∞ V0 iR As t → ∞, the switch has been opened for a long time. • What is the current in the capacitor as t → ∞? • What is the voltage across the capacitor as t → ∞? First-Order Circuits • How much energy is stored in the capacitor as t → ∞? ECE 221 • Where did the energy go? Portland State University v(t) V0 v(t) = ECE 221 v(t) − RC 1 − dt RC 1 − dv RC t 1 − dv RC 0 1 (t − 0) − RC −t RC t V0 e− RC t ≥ 0 t≤0 V0 RC Natural Response Continued (1) dv(t) dt 1 dv(t) dt v(t) dt du u v(t) 1 du u V0 ln ln (v(t)) − ln (V0 ) = Portland State University V0 t=0 iR + R v - = 0 = 0 = 0 First-Order Circuits C iC RC Natural Response General Equation Solve for v(t). V 0 t=0 + R v - C iC V0 = − e−t/RC R = 21 CV02 e−2t/RC First-Order Circuits iR RC Natural Response Summary ECE 221 iC (t) + iR (t) dv(t) v(t) C + dt R dv(t) v(t) + dt RC We know for t ≤ 0, v(t) = V0 . For t ≥ 0, 5 Portland State University For t > 0, v(t) = V0 e−t/RC ECE 221 v(t) i (t) = − C R 2 w(t) = 21 C (v(t)) 7 Portland State University Ver. 1.65 Ver. 1.65 8 6 0 t≤0 t≥0 First-Order Circuits V0 t V0 e− RC 2τ 3τ Time (s) RC Circuit Natural Response 1τ v(t) = ECE 221 4τ RC Natural Response Continued (3) V0 0 V0 0 Portland State University 2 Time (s) 3 4 τ = 1.0 s τ = 2.0 s τ = 0.5 s RC Circuit Natural Response 1 ECE 221 First-Order Circuits 5τ 5 Ver. 1.65 Ver. 1.65 The time constant τ determines how quickly the voltage settles to its ﬁnal value. 0 First Order Response & Time Constants Portland State University Voltage (V) Voltage (V) RC Natural Response Comments V0 t≤0 t V0 e− RC t ≥ 0 v(t) = • The voltage response of the RC circuit is an exponential decay • Called the natural response of the circuit – Natural response: Behavior of the circuit with no external sources of excitation First-Order Circuits + R v - C iC Ver. 1.65 • The rate at which the voltage decreases is measured by the time constant, τ • τ = RC • v(t) = V0 e−t/τ • In 5 time constants v(t) is within 1% of its ﬁnal value ECE 221 • We will treat this as the steady-state value 9 Portland State University t=0 iR RC Natural Response Energy V 0 wR (t) = First-Order Circuits Ver. 1.65 Solve for the energy stored in the capacitor, wC (t) for t ≥ 0 and the energy dissipated by the resistor, wR (t). wC (t) = 1 2C 2 1 2 Cv(t) 2 V0 e−t/τ = 2 −2t/τ 1 2 CV0 e ECE 221 = 11 Portland State University 10 12 wR (t) = 2 1 2 CV0 First-Order Circuits First-Order Circuits Ver. 1.65 Ver. 1.65 RC Natural Response Energy Key Ideas 1 − e−2t/τ wC (t) = 2 −2t/τ 1 2 CV0 e • How much energy is dissipated in the resistor? • How much energy is initially stored in the capacitor? ECE 221 Example 1: Workspace (1) ECE 221 • How much energy is stored in the capacitor as t → ∞? Portland State University Portland State University 18 mA t=0 0.2 µF 20 kΩ 0.125m v1 First-Order Circuits First-Order Circuits - v1 + Example 1: RC Natural Response 5 kΩ ECE 221 Example 1: Workspace (2) ECE 221 Find the time constant for t > 0. Find v1 (t). 13 Portland State University 15 Portland State University Ver. 1.65 16 14 10 kΩ Ver. 1.65 t=0 + R v - L i RL Natural Response: Initial Conditions t = 0− t=0 I 0 For t < 0, the switches have not changed for a long time. t=0 L I 0 t=0 t=0 di(t) dt v(t) + Ri(t) di(t) + Ri(t) dt L ECE 221 + R v - + - R v = 0 = 0 L i L i = −Ri(t) First-Order Circuits Ver. 1.65 Ver. 1.65 20 18 RL Natural Response: Initial Conditions t = 0+ t=0 I 0 For t < 0, the switches have not changed for a long time. First-Order Circuits • What is the current in the inductor at t = 0+ ? 19 Portland State University We know for t < 0, i(t) = I0 . For t > 0, Solve for i(t). RL Natural Response General Equation ECE 221 • What is the current in the inductor at t = 0− ? + L i 17 Portland State University • What is the voltage across the inductor at t = 0+ ? Ver. 1.65 • What is the voltage across the inductor at t = 0− ? First-Order Circuits • How much energy is stored in the inductor at t = 0+ ? ECE 221 • How much energy is stored in the inductor at t = 0− ? Portland State University t=0 R v - Ver. 1.65 RL Natural Response: Final Conditions t → ∞ t=0 I 0 As t → ∞, the switches have not changed for a long time. • What is the current in the inductor as t → ∞? • What is the voltage across the inductor as t → ∞? First-Order Circuits • How much energy is stored in the inductor as t → ∞? ECE 221 • Where does this energy go? Portland State University = = = = = R − i(t) L R − dt L R − dv L R t − dv L 0 R − (t − 0) L R −t L R I0 e−t L I0 t≥0 t≤0 RL Natural Response Continued (1) di(t) dt di(t) i(t) du u i(t) 1 du u I0 ln = First-Order Circuits t≤0 t≥0 RL Natural Response Comments ECE 221 i(t) i(t) I0 ln (i(t)) − ln (I0 ) = Portland State University For t > 0, I0 i(t) = I0 e−t/τ Ver. 1.65 Ver. 1.65 • The current response of the RL circuit is an exponential decay First-Order Circuits • Called the natural response of the circuit L R ECE 221 • Time constant: τ = Portland State University For t > 0, t=0 + R v - L i RL Natural Response Summary t=0 I0 R i(t) = I0 e−t L R = −RI0 e−t L + R v(t) = −i(t)R First-Order Circuits = 21 LI02 e−2t L ECE 221 w(t) = 21 L (i(t))2 21 Portland State University t=0 R v - L i RL Natural Response Energy t=0 I 0 wL (t) = wR (t) = ECE 221 2 −2t/τ 1 2 LI0 e First-Order Circuits Ver. 1.65 Ver. 1.65 Solve for the energy dissipated by the resistor, wR (t), for t ≥ 0. 23 Portland State University 22 24 First-Order Circuits First-Order Circuits Ver. 1.65 Ver. 1.65 RL Natural Response Energy Key Idea wR (t) = 21 LI02 1 − e−2t/τ wL (t) = 21 LI02 e−2t/τ • How much energy is dissipated in the resistor? • How much energy is initially stored in the inductor? ECE 221 Example 2: Workspace (1) ECE 221 • How much energy is stored in the inductor as t → ∞? Portland State University Portland State University 54 V 3Ω 4.5 Ω t=0 + v - 200 mH iL 50 i1 Example 2: RL Natural Response 9Ω ECE 221 First-Order Circuits First-Order Circuits Example 2: Workspace (2) ECE 221 Find iL (t) for t ≥ 0, v(t) for t > 0, and i1 (t) for t > 0. 25 Portland State University 27 Portland State University 26 200 Ω Ver. 1.65 28 100 Ω i1 Ver. 1.65 t=0 R + v(t) - i(t) t = 0 C V 0 RC Step Response: Initial Conditions t = 0− V s For t < 0, the switches have not changed for a long time. t=0 R R + v(t) - + v(t) - i(t) t = 0 C i(t) t = 0 C V0 Ver. 1.65 RC Step Response: Initial Conditions t = 0+ V s For t < 0, the switches have not changed for a long time. First-Order Circuits • What is the current in the capacitor for t = 0+ ? t=0 RC Step Response Solution ECE 221 • What is the current in the capacitor for t = 0− ? V 29 Portland State University • What is the voltage across the capacitor at t = 0+ ? Ver. 1.65 • What is the voltage across the capacitor at t = 0− ? First-Order Circuits i(t) t = 0 V s Solve for v(t). 31 Portland State University Vs − v(t) − v(t) RC Vs − v(t) Vs ECE 221 = Ri(t) dv(t) = RC dt dv(t) dt = First-Order Circuits Ver. 1.65 We know v(t) = V0 for t ≤ 0. We need to solve for v(t) for t ≥ 0. V0 • How much energy is stored in the capacitor at t = 0+ ? ECE 221 R + C First-Order Circuits Ver. 1.65 • How much energy is stored in the capacitor at t = 0− ? Portland State University t=0 v(t) - 0 RC Step Response: Final Conditions t → ∞ V s As t → ∞, the switches have not changed for a long time. • What is the current in the capacitor as t → ∞? • What is the voltage across the capacitor as t → ∞? ECE 221 • How much energy is stored in the capacitor as t → ∞? Portland State University 30 32 dv V0 = ln v(t) − Vs V0 − Vs = ln (v(t) − Vs ) − ln (V0 − Vs ) = dv(t) 1 dt = RC V − v(t) s dv(t) −1 dt = RC v(t) − Vs du −1 dv = RC u − Vs t v(t) 0 −t RC −t RC First-Order Circuits = Vs + (V0 − Vs ) e−t/RC ECE 221 = (V0 − Vs ) e−t/τ = Vs = Vs + (V0 − Vs ) e−t/τ = vf + vn (t) Natural & Forced Response v(t) du u − Vs RC Step Response Solution Continued −1 RC Portland State University For t > 0 v(t) vf vn (t) Ver. 1.65 • Forced response is what the response eventually is forced to • Also called the steady-state response • Natural response is the part of the response due to the change ECE 221 First-Order Circuits Ver. 1.65 • Also called the transient response because it is temporary Portland State University V s 1 ECE 221 R + v(t) - = i(t) t = 0 C First-Order Circuits 3 RC Step Response 2 Time (s) Forced Natural Total 4 First-Order Circuits 5 Ver. 1.65 Ver. 1.65 36 34 1 = C − (V − Vs ) e−t/τ 0 τ V0 Vs − e−t/τ R R V0 RC Step Response General Equations t=0 v(t) = Vs + (V0 − Vs ) e−t/τ ECE 221 dv i(t) = C dt 1 (V0 − Vs ) e−t/τ R =− 0 RC Step Response Graphed = i(0+ )e−t/τ 33 Portland State University Vs V0 35 Portland State University Voltage (V) Vs R + L i(t) t=0 I0 t=0 RL Step Response: Initial Conditions t = 0− t=0 v - For t < 0, the switches have not changed for a long time. Vs R + L i(t) t=0 I0 t=0 RL Step Response: Initial Conditions t = 0+ t=0 v - For t < 0, the switches have not changed for a long time. R + t=0 I0 First-Order Circuits • What is the current in the inductor at t = 0+ ? t=0 v - L i(t) RL Step Response Solution ECE 221 • What is the current in the inductor at t = 0− ? t=0 37 Portland State University • What is the voltage across the inductor at t = 0+ ? Ver. 1.65 • What is the voltage across the inductor at t = 0− ? First-Order Circuits t=0 V s Solve for i(t). 39 Portland State University Vs − Ri(t) Vs − Ri(t) Vs − i(t) R ECE 221 = v(t) di(t) = L dt L di(t) R dt = First-Order Circuits Ver. 1.65 Ver. 1.65 We know i(t) = I0 for t ≤ 0. We need to solve for i(t) for t ≥ 0. t=0 • How much energy is stored in the inductor at t = 0+ ? ECE 221 R + I 0 First-Order Circuits Ver. 1.65 • How much energy is stored in the inductor at t = 0− ? t=0 v - L i(t) RL Step Response: Final Conditions t → ∞ Portland State University V s As t → ∞, the switches have not changed for a long time. • What is the current in the inductor as t → ∞? • What is the voltage across the inductor as t → ∞? ECE 221 • How much energy is stored in the inductor as t → ∞? Portland State University 38 40 = I0 du u − Vs /R RL Step Response Solution Continued dv i(t) − Vs /R I0 − Vs /R 0 di(t) R dt = L Vs /R − i(t) R di − dt = L i − Vs /R R du − dv = L u − Vs /R t i(t) −R L R −t L R −t L = Vs /R + (I0 − Vs /R) e−t L = ln ECE 221 R First-Order Circuits First-Order Circuits Ver. 1.65 Ver. 1.65 = ln (i(t) − Vs /R) − ln (I0 − Vs /R) i(t) Portland State University = Vf + (V0+ − Vf )e−t/τ RC & RL Circuits the Easy Way v(t) i(t) = If + (I0+ − If )e−t/τ • Every circuit we have seen follows this pattern • It is true, in general ECE 221 • This leads to a general approach to analysis Portland State University = I0 R Vs /R + (I0 − Vs /R) e−t L t≤0 t≥0 RL Step Response General Equations i(t) v(t) t≤0 t≥0 First-Order Circuits di(t) = L dt 0 R (Vs − RI0 ) e−t L = ECE 221 There is a pattern here . . . 41 Portland State University t→∞ xf ≡ lim x(t) Ver. 1.65 →0 x0+ ≡ lim x(||) First-Order Circuits the Easy Way x(t) = xf + (x0+ − xf )e−t/τ 1. Combine any networks of inductors (capacitors) into their single inductor (capacitor) equivalents. 2. Use DC analysis to solve for the current (voltage) ﬂowing through the inductor (across the capacitor) at t = 0− 3. Use DC analysis to ﬁnd the initial value of the variable of interest at t = 0+ 4. Use DC analysis to ﬁnd the steady state value of the variable of interest as t → ∞ First-Order Circuits Ver. 1.65 5. Find the equivalent resistance seen by the inductor (capacitor) for t>0 6. Solve for the time constant τ ECE 221 7. Plug into the general equation 43 Portland State University 42 44 100 µA 20 kΩ 0.4 µF 10 kΩ i1(t) 5 kΩ Example 3: RC Circuit 5 kΩ i2(t) t=0 5. Find i1 (t). 10 V 1. Find i1 (0− ) and i2 (0− ). 6. Find i2 (t). First-Order Circuits 2. Find i1 (0+ ) and i2 (0+ ). 3. Explain why i1 (0− ) = i1 (0+ ). ECE 221 First-Order Circuits Example 3: Workspace (2) ECE 221 4. Explain why i2 (0− ) = i2 (0+ ). Portland State University Portland State University Ver. 1.65 Ver. 1.65 45 Portland State University 20 mA 48 Ω First-Order Circuits Example 3: Workspace (1) ECE 221 t=0 12 mH i1 First-Order Circuits 8 mH i2 Example 4: RL Circuit 15 Ω ECE 221 Find vo (t), i1 (t), and i2 (t). 47 Portland State University - vo + Ver. 1.65 10 mA Ver. 1.65 46 48 Portland State University i2 0.4 H + v - 100 Ω First-Order Circuits i1 Example 5: RL Circuit ECE 221 Example 4: Workspace (1) 0.6 H At t = 0− , i1 = 1 mA and i2 = −1 mA. 1. Find v(t) for t ≥ 0+ . ECE 221 First-Order Circuits Ver. 1.65 Ver. 1.65 2. What percentage of the energy initially stored in the inductors is dissipated in the 100 Ω resistor? Portland State University 49 Portland State University 51 Portland State University First-Order Circuits Example 4: Workspace (2) ECE 221 First-Order Circuits Example 5: Workspace ECE 221 Ver. 1.65 Ver. 1.65 52 50 0.6 H 0.4 H i1 + v - Example 6: RL Circuit i2 At t = 0− , i1 = 2 mA and i2 = −1 mA. 1. Find v(t) for t ≥ 0+ . 12 kΩ ECE 221 20 V 4 kΩ 100 Ω First-Order Circuits t=0 200 Ω 240 kΩ Example 7: Hybrid Circuit 12 kΩ 800 Ω First-Order Circuits iL 500 mH Ver. 1.65 vL Ver. 1.65 2. What percentage of the energy initially stored in the inductors is dissipated in the 100 Ω resistor? C vC Portland State University i 33.33 µF ECE 221 Solve for the following for t > 0: vC (t), iC (t), vL (t), iL (t). Portland State University 53 Portland State University 55 Portland State University First-Order Circuits Example 6: Workspace ECE 221 First-Order Circuits Example 7: Workspace (1) ECE 221 Ver. 1.65 Ver. 1.65 56 54 Portland State University Portland State University First-Order Circuits Example 7: Workspace (2) ECE 221 First-Order Circuits Example 8: Workspace (1) ECE 221 Ver. 1.65 Ver. 1.65 vL iL 2H + - 5 kΩ 10 V t=0 4 kΩ Example 8: Hybrid Circuit 5 kΩ t=0 20 mA ECE 221 6 kΩ First-Order Circuits First-Order Circuits Example 8: Workspace (2) ECE 221 + 5 mF 58 vc Ver. 1.65 60 ic Ver. 1.65 - Solve for the following for t > 0: vC (t), iC (t), vL (t), iL (t). 57 Portland State University 59 Portland State University t=0 5 kΩ 125 kΩ 8 nF 20 kΩ 10 V -10 V 12 kΩ vo Example 9: First-Order Op Amp Circuit 3V ECE 221 First-Order Circuits Ver. 1.65 There is no energy stored in the capacitor at t = 0. How long does it take for the op amp to saturate? Repeat this question assuming there is an initial voltage of 2 V on the capacitor. Portland State University 61 Portland State University First-Order Circuits Example 9: Workspace ECE 221 Ver. 1.65 62