Parallel RLC Network Objective of Lecture Derive the equations that relate the voltages across and the currents flowing through a resistor, an inductor, and a capacitor in parallel as: the unit step function associated with voltage or current source changes from 1 to 0 or a switch disconnects a voltage or current source in the circuit. Describe the solution to the 2nd order equations when the condition is: Overdamped Critically Damped Underdamped RLC Network A parallel RLC network where the current source is switched out of the circuit at t = to. Boundary Conditions You must determine the initial condition of the inductor and capacitor at t < to and then find the final conditions at t = ∞s. Replace the capacitor with an open circuit and the inductor with a short circuit. Since the current source has a magnitude of Is at t < to iL(to-) = Is and v(to-) = vC(to-) = 0V vL(to-) = 0V and iC(to-) = 0A Once the steady state is reached after the current source been removed from the circuit at t > to and the stored energy has dissipated through R. iL(∞s) = 0A and v(∞s) = vC(∞s) = 0V vL(∞s) = 0V and iC(∞s) = 0A Selection of Parameter Initial Conditions iL(to-) = Is and v(to-) = vC(to-) = 0V vL(to-) = 0V and iC(to-) = 0A Final Conditions iL(∞s) = 0A and v(∞s) = vC(∞s) = oV vL(∞s) = 0V and iC(∞s) = 0A Since the current through the inductor is the only parameter that has a non-zero boundary condition, the first set of solutions will be for iL(t). Kirchoff’s Current Law iR (t ) iL (t ) iC (t ) 0 v(t ) vR (t ) vL (t ) vC (t ) dvC (t ) vR (t ) iL (t ) C 0 R dt diL (t ) vL (t ) v(t ) L dt d 2iL (t ) L diL (t ) LC iL (t ) 0 2 dt R dt d 2iL (t ) 1 diL (t ) iL (t ) 0 2 dt RC dt LC General Solution 1 1 s s 0 RC LC 2 2 1 1 1 s1 2 RC 2 RC LC 2 1 1 1 s2 2 RC 2 RC LC s1 2 2 o s 2 2 1 2 RC 1 o LC 2 o s 2s 0 2 2 o Note that the equation for the natural frequency of the RLC circuit is the same whether the components are in series or in parallel. Overdamped Case > o implies that L > 4R2C s1 and s2 are negative and real numbers iL1 (t ) A1e s1t iL 2 (t ) A2 e s 2 t t t to iL (t ) iL1 (t ) iL 2 (t ) A1e s1t A2 e s2 t Critically Damped Case o implies that L = 4R2C s1 = s2 = - = -1/2RC iL (t ) A1e t A2 te t Underdamped Case < o implies that L < 4R2C s1 2 o2 jd s2 jd 2 2 o d o2 2 iL (t ) e t [ A1 cos d t A2 sin d t ] Solve for Coefficients A1 and A2 Use the boundary conditions at to- and t = ∞s to solve for A1 and A2. iL (to ) I S Since the current through an inductor must be a continuous function of time. iL (to ) iL (to ) iL1 (to ) iL 2 (to ) I S A1e s1 0 s A2e s2 0 s A1 A2 I S Also know that diL (to ) d vL (to ) L iL1 (to ) iL 2 (to ) 0 dt dt s1 A1e s1 0 s s2 A2e s2 0 s s1 A1 s2 A2 0 Other Voltages and Currents Once current through the inductor is known: diL (t ) vL (t ) L dt vL (t ) vC (t ) vR (t ) dvC (t ) iC (t ) C dt iR (t ) vR (t ) / R Summary The set of solutions when t > to for the current through the inductor in a RLC network in parallel was obtained. Selection of equations is determine by comparing the natural frequency o to . Coefficients are found by evaluating the equation and its first derivation at t = to-. The current through the inductor is equal to the initial condition when t < to Using the relationships between current and voltage, the voltage across the inductor and the voltages and currents for the capacitor and resistor can be calculated.