ECE 222 Electric Circuit Analysis II Chapter 10 Natural Response in Series RLC Circuits Herbert G. Mayer, PSU Status 4/19/2016 For use at CCUT Spring 2016

Syllabus Definitions Initial State RLC Circuit i(t) in Series RLC Example 2 Bibliography

a * x + b * dx/dt + c * d2x/dt2 + d = 0 Definitions Natural Response is current discharged in a series RLC circuit by releasing energy stored in the circuit’s capacitor and inductor In a First-Oder Differential Equation (DE), the characteristic variable x = x(t) would occur only in its simple form x, and as first derivate x’(t) or dx/dt. E.g. for variable x and constants a, b, c: a * x + b * dx/dt + c = 0 In a Second-Order DE, the characteristic variable x occurs in original form x, and first derivative, as well as a second derivative Example for variable x with constants a, b, c, d: a * x + b * dx/dt + c * d2x/dt2 + d = 0

Initial State In a series circuit at time t < 0 an electric source (to be removed at t = 0) may charge components C and L If so, there exists residual energy in the series circuit At t = 0 capacitor C in the series circuit holds an initial charge, creating its electric field between the 2 plates And at t = 0 inductor L holds a magnetic field between its terminals This energy discharges, when the circuit is closed and the source is removed exactly at time t = 0

Series RLC Circuit

RLC Circuit Below is a series circuit, analyzed for second-order DE Assume V0 across C in the circuit, no current at t = 0

RLC Circuit – Recall from First-Order SI unit of [H] = [V s A-1] SI unit of [F] = [A s V-1] KVL for Natural Response in RL circuit with single R, L (i.e. first-order), and current i(t) = i: L di / dt + R i = 0 KCL for Natural Response in RC circuit with single R, C (i.e. first-order), and voltage v(t) = v: C dv / dt + v / R = 0

RLC Circuit – Second-Order KVL for Natural Response in series R L C circuit with single R, L, C, with i = i(t) at time t > 0: R i + 1/C i dt + L di / dt + V0 = 0 Once we have current i, we can compute the voltages in all 3 series elements (RLC) Eliminate integral by differentiating: R di/dt + i/C + L d2i/dt2 + 0 = 0 Divide by L, re-order by powers of derivative, and we have a second-order DE for i(t) to solve (a.): d2i / dt2 + R / L * di / dt + i / (LC) = 0 i’’ + i’ R / L + i / LC = 0

RLC Circuit – Characteristic Equation Rewrite as characteristic equation for a series circuit: s2 + s R / L + 1 / (LC) = 0 The roots for the characteristic equation are: s1/2 = -R/2L ± ( (R/2L)2 - 1/LC )½ Abbreviate with α and ω0 as before: s1 = -α + ( α2 - ω02 )½ s2 = -α - ( α2 - ω02 ) ½ α = R / ( 2L ) ω0 = ( LC )-½ Verify SI units [V] for integral-form: R i + 1/C i dt + L di / dt + V0 = 0

RLC Circuit – Second-Order Verify the SI units (dimension) for KCL equation R i + 1/C i dt + L di / dt + V0 = 0 All summands must be of units [V] in the end SI units of R i = [V A-1 A] = [V] SI units of 1/C i dt = [A-1 s-1 V A s] = [V] SI units of L di / dt = [V s A-1 A s-1] = [V] Confirmed!

Series Electric Circuit Damping in Series Electric Circuit

Damping, Like for Parallel RLC

Functions for Damping Classes

Overdamped 1.) Root types: real, distinct roots s1, s2 2.) α vs. ω0: α2 > ω02 3.) Voltage v(t): i(t) = A1 es1 t + A2 es2 t 4.) Root Functions s1, s2: s1 = -α + ( α2 - ω02 )1/2 s2 = -α - ( α2 - ω02 )1/2

Underdamped 1.) Root types: complex, distinct roots s1, s2 2.) α vs. ω0: α2 < ω02 3.) Voltage v(t): i(t) = B1 e-α t cos( ωd t ) + B2 e-α t sin( ωd t ) 4.) Root Functions s1, s2: s1 = -α + ( α2 - ω02 )1/2 s2 = -a - ( α2 - ω02 )1/2 s1 = -α + j ωd s2 = -α - j ωd ωd = ( ω02 - α2 )1/2

Critically Damped 1.) Root types: real, equal roots s1, s2 2.) α vs. ω0: α2 = ω02 3.) Voltage v(t): i(t) = D1 t e-α t + D2 e-α t 4.) Root Functions s1, s2: s1 = -α s2 = -α

RLC Circuit – Second-Order Once current i(t) is found in the series circuit, all voltages vRLC(t) across the RLC components can be computed Given that we know the units for R, L, and C

(See also Example 1 in Parallel RLC Circuit) Series RLC Circuit (See also Example 1 in Parallel RLC Circuit)

Example 2 – Series RLC Circuit Given the circuit for Example 2 below, we compute the following parts of the characteristic equations: Neper frequency α Square of Neper frequency α2 Square of Resonant Radian frequency ω02 Determine, whether the circuit is overdamped, underdamped, or critically damped If applicable , find ωd2 Find the current equation i(t)

Example 2 – Series RLC Circuit Series RLC circuit below has components: R = 560 Ω, C = 0.1 μF, and L = 100 mH C is charged with 100 V at t<=0

Example 2 – Series RLC Circuit 1. Compute Neper frequency α: α = R / 2L α = 560 / ( 2 * 100 * 10-3 ) = 5600 / 2 α = 2800 -- [Radians / s] 2. Compute α2: α2 = 7.84 106 -- [Radians2 / s2] 3. Compute square of Resonant Radian frequency ω02: ω02 = 1 / LC = 1 / ( 100 * 10-3 * 0.1 * 10-6 ) ω02 = 108 -- [Radians2 / s2]

Example 2 – Series RLC Circuit 4. Determine the class of damping: We classify as follows ω02 < a2 - overdamped ω02 > a2 - underdamped ω02 = a2 - critically damped ω02 = 108 -- [Radians2 / s2] α2 = 7.84 106 = 0.0784 108 We see that ω02 > α2 Hence this circuit is underdamped

Example 2 – Series RLC Circuit 5. Compute root ωd: ωd = ( ω02 - α2 )½ ωd = ( 108 – 0.0784 108 )½ ωd = 0.96 * 104 -- [Radians / s] 6. Compute current equation i(t): i(t) = B1 e -α t cos( ωd t ) + B2 e -α t sin( ωd t )

Bibliography Nilsson, James W., and Susan A. Riedel, Electric Circuits, © 2015 Pearson Education Inc., ISBN 13: 9780-13-376003-3 Differentiation rules: http://www.codeproject.com/KB/recipes/Differentiation.aspx Euler’s Identities: https://en.wikipedia.org/wiki/Euler%27s_identity Table of integrals: http://integral-table.com/downloads/single-page-integral-table.pdf