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Second Order Circuits ES-3 Download: http://www.ece.tufts.edu/~hopwood/downloads/es3/SecondOrderCircuits.ppt Second Order Circuits • Contain two independent reactive components • This results in a second order differential equation containing d2i/dt2 or d2v/dt2 Example: Series RLC circuit L VS(t) i(t) + C KVL: VL + VR + VC = Vs R vL (t ) vR (t ) vC (t ) vS (t ) t di 1 L iR i (t )dt vC (0) vS (t ) dt C0 dvS d 2i di i (t ) L 2 R 0 dt dt C dt d 2i R di i (t ) 1 dvS 2 dt L dt LC L dt d 2i di 1 dvS 2 2 0 i (t ) 2 dt dt L dt R 1 ...where and 0 . 2L LC Find the natural response : vS 0 d 2i di 2 2 0 i (t ) 0 2 dt dt di d 2i 2 2 0 i (t ) 0 2 dt dt try a solution in the form : i (t ) kest substituting : s 2 kest 2skest 02 kest 0 giving : s 2 2s 02 0 Use the quadratic equation : 2 (2 ) 2 4(1)02 2 02 s 2(1) Three Cases • Case 1 - Overdamped: >o large R: R > 2L 1 LC s1 2 02 0 s2 0 2 2 0 The natural response is two decaying exponentia ls : in (t ) K1e s1t K 2 e s2t Three Cases • Case 2 - Underdamped: o small R: R 2L 1 LC s1 2 02 1 02 2 j 02 2 s2 2 02 j 02 2 The natural response is two decaying phasors in (t ) K1e s1t K 2 e s2t t K1e e j ( 02 2 ) t 2 2 t j ( 0 ) t K 2e e Ke t cos(nt )...where n 02 2 Three Cases • Case 3 – Critically damped: o R 2L 1 LC s1 0 2 2 0 s2 0 0 2 2 0 The natural response is in (t ) K1e t K 2tet Examples L VS(t) L = 10uH i(t) + C C = 10nF R o = 3.16x106 rad/sec (504 kHz) R = 10, 63.2, and 1000 ohms = 5x105, 3.16x106, and 5x107 s-1 vs t Underdamped Critically Damped Overdamped Overdamped: R=1000W 10V 5V 0V 0s 0.2us 0.4us 0.6us 0.8us 1.0us V(R1:2) Time 1.2us 1.4us 1.6us 1.8us 2.0us Underdamped: R=10W 10V 5V 0V -5V -10V 1us 0s 2us 3us 4us 5us V(R1:2) Time 6us 7us 8us 9us 10us Critically Damped: R=63.2W 10V 5V 0V -5V 1us 0s 2us 3us 4us 5us V(R1:2) Time 6us 7us 8us 9us 10us Mechanical Analogy Wikipedia.com Automobile: Mass inductor Suspension: Spring capacitor Shock Absorber: Damper resistor Force voltage Velocity current