Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Lecture 13 RC/RL Circuits, Time Dependent Op Amp Circuits ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. RL Circuits The steps involved in solving simple circuits containing dc sources, resistances, and one energy-storage element (inductance or capacitance) are: ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. 1. Apply Kirchhoff’s current and voltage laws to write the circuit equation. 2. If the equation contains integrals, differentiate each term in the equation to produce a pure differential equation. 3. Assume a solution of the form K1 + K2est. ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. 4. Substitute the solution into the differential equation to determine the values of K1 and s . (Alternatively, we can determine K1 by solving the circuit in steady state) 5. Use the initial conditions to determine the value of K2. 6. Write the final solution. ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. RL Transient Analysis Find i(t) and the voltage v(t) i(t)= 0 for t < 0 since the switch is open prior to t = 0 Apply KVL around the loop: VS i(t ) R v(t ) 0 ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. RL Transient Analysis VS i(t ) R v(t ) 0 di (t ) i(t ) R L VS dt ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. RL Transient Analysis i (t ) R L di (t ) VS dt di(t ) i (t ) R L VS dt Try i(t ) K1 K 2e st Try i(t ) K1 K 2e st RK 1 ( RK 2 sLK 2 )e VS st VS 100V RK 1 VS K1 2A R 50 L RK 2 sLK 2 0 s R ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. RL Transient Analysis i(t ) 2 K 2e tR / L i(0 ) 0 2 K 2e 2 K 2 K 2 2 0 i(t ) 2 2e tR / L ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. RL Transient Analysis L Define R i(t ) 2 2e t / ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. RL Transient Analysis v(t ) VS i (t ) R 100 50i (t ) 100 50(2 2e t / ) 100e t / ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. RC and RL Circuits with General Sources First order differential equation with constant coefficients di (t ) L Ri (t ) vt (t ) dt vt (t ) L di(t ) i (t ) R dt R dx(t ) Forcing x(t ) f (t ) function dt ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. RC and RL Circuits with General Sources The general solution consists of two parts. ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. The particular solution (also called the forced response) is any expression that satisfies the equation. dx(t ) x(t ) f (t ) dt In order to have a solution that satisfies the initial conditions, we must add the complementary solution to the particular solution. ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. The homogeneous equation is obtained by setting the forcing function to zero. dx(t ) x(t ) 0 dt The complementary solution (also called the natural response) is obtained by solving the homogeneous equation. ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. Step-by-Step Solution Circuits containing a resistance, a source, and an inductance (or a capacitance) 1. Write the circuit equation and reduce it to a first-order differential equation. ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. 2. Find a particular solution. The details of this step depend on the form of the forcing function. 3. Obtain the complete solution by adding the particular solution to the complementary solution xc=Ke-t/ which contains the arbitrary constant K. 4. Use initial conditions to find the value of K. ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. Transient Analysis of an RC Circuit with a Sinusoidal Source q (t ) Ri (t ) 2 sin( 200t ) C t 1 Ri (t ) i (t ) vc (0) 2 sin( 200t ) C0 ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. Transient Analysis of an RC Circuit with a Sinusoidal Source Take the derivative: di (t ) 1 R i (t ) 400 cos( 200t ) dt C di (t ) RC i (t ) 400C cos( 200t ) dt Try a particular solution : i p (t ) A cos( 200t ) B sin( 200t ) ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. Transient Analysis of an RC Circuit with a Sinusoidal Source di(t ) i (t ) 400 x10 6 cos(200 t ) dt A sin(200 t ) B cos(200 t ) A cos(200 t ) B sin(200 t ) 400 x10 6 cos(200 t ) equating the coefficients for sin terms : A B 0 A B equating the coefficients for cos terms : 5 x10 3 B A 400 x10 6 Solving : A 200 x10 6 200 A B 200 x10 6 200 A i p (t ) 200 cos(200 t ) 200 sin(200 t ) ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. Transient Analysis of an RC Circuit with a Sinusoidal Source The complementary solution is given by: ic (t ) Ket / RC (5k)(1F ) 5ms The complete solution is given by the sum of the particular solution and the complementary solution: i(t ) 200 cos( 200t ) 200 sin( 200t ) Ket / A ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. Transient Analysis of an RC Circuit with a Sinusoidal Source Initial conditions: 2sin(0+) = 0 vC(0+) = 1V vR(0+) + vC(0+) = 0 vR(0+) = -1V i(0+) = vR/R = -1V/5000 = -200A = 200cos(0)+200sin(0)+Ke0 = 200 + K K= -400A ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. Transient Analysis of an RC Circuit with a Sinusoidal Source i(t ) 200 cos(200t ) 200 sin( 200t ) 400e t / A ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. Transient Analysis of an RC Circuit with an Exponential Source What happens if we replace the source with 10e-t and the capacitor is initially charged to vc(0)=5? q(t ) Ri (t ) 10e t C t 1 Ri (t ) i (t ) vc (0) 10e t C0 ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. Transient Analysis of an RC Circuit with an Exponential Source Take the derivative: di (t ) 1 t R i (t ) 10e dt C di (t ) t RC i (t ) 10Ce dt Try a particular solution : i p (t ) Ae t ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. Transient Analysis of an RC Circuit with an Exponential Source RCAe t Ae t 10Ce t A RCA 10C A(1 RC ) 10C 6 10C (10 )( 2 x10 ) A 20 A RC 1 2 1 ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. Transient Analysis of an RC Circuit with an Exponential Source The complementary solution is given by: ic (t ) Ke t / RC (1M)( 2F ) 2s The complete solution is given by the sum of the particular solution and the complementary solution: t i(t ) 20e Ke t / 2 A ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. Transient Analysis of an RC Circuit with an Exponential Source Initial conditions: 10e0+ = 10 vC(0+) = 5V vR(0+) + vC(0+) = 10 vR(0+) = 5 i(0+) = vR/R = 5V/1M = 5A = 20e0 + Ke0 = 20 + K K= -15A ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. Transient Analysis of an RC Circuit with an Exponential Source t i(t ) 20e 15e t / 2 A ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. Integrators and Differentiators Integrators produce output voltages that are proportional to the running time integral of the input voltages. In a running time integral, the upper limit of integration is t . ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. vin iin R t t q 1 1 vc iin dt vin dt C C0 RC 0 1 v o v c 0 v o v c vin dt RC 0 t ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. 1 vo t RC t vin t dt 0 ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. If R = 10 k, C = 0.1F RC = 0.1 ms t vo t 1000 vin t dt 0 ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. t t vo t 1000 vin t dt 1000 5dt 5000 t 0 0 t 1ms 1000 5dt 5dt 0 1ms for 0 t 1ms for 1ms t 3ms 10 5000 t ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. Differentiator Circuit dvin 0 vo dvin dq iin C vo RC dt dt R dt ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc. Differentiator Circuit dvin vo t RC dt ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.