ECE 222 Electric Circuit Analysis II Chapter 7 First-Order RC Circuits Herbert G. Mayer, PSU Status Status 5/10/2016 For use at CCUT Spring 2016

Syllabus Definition Natural Response Compute i(t), p(t), and w(t) RC Time Constant τ Examples 1, 2, 3 Bibliography

Definition Circuits with only resistors R and capacitors C are known as RC circuits Both RL and RC circuits are known as first-order circuits, since they can be described by a first- order differential equation DE A first-order DE is adequate, since RC circuits have a single type of component that stores energy Here we analyze first-order circuits with capacitors

Definition Interconnected capacitors in a first-order circuit can all be reduced to an equivalent capacitor Ceq Reduction follows the rules of parallel and series capacitors, or uses the Thevénin Equivalent for finding a corresponding voltage source Or the Norton Equivalent for a current source For n parallel capacitors Cn use: Ceq = C1 + C2 + C3 + . . . Cn Ceq = Σ Cn, for n = 1..3 For n series capacitors Cn use: 1 / Ceq = 1 / C1 + 1 / C2 . . . 1 / Cn

Definition Below is an abstracted first-order RC circuit, in which all resistances are substituted by an equivalent resistor Req and all capacitors by an equivalent Ceq

Natural RC Response with Capacitor

Natural RC Response v(t) Analyze the natural response of first-order RC circuits: Assume a constant voltage source vg has been connected to R1 and C in the circuit below for a long time; via connection point a At time t = 0 the connection is switched from point a to point b Now C is in series with R, separated from R1 and disconnected from the source vg When point b is connected at time t = 0, the voltage across the terminals of C is still vg Thus at time t = 0 C starts to discharge, initially vg, causing current i(t) in the loop

Natural RC Response v(t) When C gets connected to R via point b, there can be no instantaneous voltage change Yet a current starts flowing immediately Sanity check students: is R1 necessary? Voltage v(t) begins to drop, starting at t = 0 with vg

Natural RC Response v(t) Summing up all currents in the b loop, using KCL, using v(t) across the cap, and with v(t=0) = v0 = vg C dv / dt + v(t) / R = 0 [A] dv / v(t) = -1 / ( RC ) dt –from v0 at 0 to v(t) at t ln( v(t) / v0 ) = -t / (RC) v(t) = v0 e -t / (RC) v(t) = v0 e -t/τ

Natural RC Response v(t) Use time constant τ, with τ = RC The Natural Response of the voltage v(t) in a first- order RC circuit is an exponential decay from v0 With v(t) = v0 at t=0, and with v(t) = 0 at t -> ∞ v0 v(t) v(t) = v0 e-t/τ t t

Compute i(t), p(t), and w(t) Knowing function v(t) = v0 e-t/(RC) for the voltage, we can compute i(t) across resistor R: i = v / R -- Ohm’s Law i(t) = v0 e -t/(RC) / R -- for t >= 0 i(t) = v0 e -t/τ / R

Compute i(t), p(t), and w(t) Knowing v(t) = v0 e-t/τ and current i(t) = v0 e-t/τ / R we compute the power: p(t) = v(t) * i(t) -- for t >= 0 p(t) = v02 e-2 t/τ / R With energy w(t) the time integral of p(t) from 0 to t w(t) = p(t) dt w(t) = v02 / R e-2 t/τ dt w(t) = v02 / R e-2 t/τ dt = v02 / R e-2 t/(RC) dt w(t) = ½ RC / R v02 ( 1 - e-2 t/(RC) ) w(t) = C v02 / 2 * ( 1 - e-2 t/τ ) -- from 0 to t

RC Time Constant τ

RC Time Constant τ The time constant for RC circuits is R*C, denoted τ τ has SI unit of seconds [s] [τ] = [R C] = time constant = [V A-1][A s V-1] = [s] As for inductors, abbreviate R*C (RC) with τ Analyze the response of an RC circuit, decaying at time t = 0 with voltage v(t) = v0 Compare with linear decay: We had defined vg = v0 The tangent of e-function v(t) has incline -v0/τ, hence the tangent at v0 crosses the time axis at time t = τ

RC Time Constant τ Example: vb(t) = v0 e-t/τ and the red vr(t) = v0 - t v0 / τ V0 vb (t) = v0 e-t/τ vr (t) = v0-t v0/τ t τ

Example 1 for Capacitor

Example 1 At time t < 0 a capacitor C of 0.5 μF is connected by a 10 kΩ resistor to a 100 V constant voltage source via connection a Hence voltage vC across terminals of C is = 100 V At t = 0 the connection is switched to b C discharges across the purely resistive circuit Compute several time-dependent electric units

Example 1 – The Circuit Parallel Rp & Serial Req Time constant τ Voltage vC(t) at C Voltage vP(t) at 60 kΩ Current i(t) thru 32 kΩ iP(t) through 60 kΩ Power p(t) dissipated in the 60 kΩ

Example 1 – Rp, Req , and τ a.) Rp is two resistors in parallel: 240 // 60 Rp = 240 * 60 / ( 240 + 60 ) = 48 [kΩ] Req is two resistors in series: 48 kΩ + 32 kΩ Req = Rp + 32 kΩ = 80 [kΩ] b.) Time constant τ = R C with C = 0.5 μF, Req = 80 kΩ is: τ = 0.5 * 10-6 * 80 * 103 = 40 [ms]

Example 1 – Voltages vC(t) and vP(t) c.) Compute voltage vC(t) with vC(0) = v0 = 100 [V] vC(t) = v0 e -t/τ vC(t) = 100 e -t/40 * 1000 vC(t) = 100 e- 25 t d.) Compute voltage vp(t) at 60 kΩ vP(t) = vC * Rp / Req vP(t) = 48/80 vC vP(t) = 0.6 vC vP(t) = 60 e-25 t

Example 1 – Currents i(t) and iP(t) e.) Compute current i(t) through 32 kΩ resistor i = v / R -- in general i(t) = vC / Req i(t) = 100 e -25 t / 80 [V / kΩ] i(t) = 1.25 e -25 t [mA] f.) Compute current iP(t) in 60 kΩ iP(t) = i(t) * 240 / ( 240+60) iP(t) = 1.0 e -25 t [mA]

Example 1 – Power p(t) g.) Given the following: p = i2 * R = e -25 t * R [W] R = 60 [kΩ] iP(t) = 1.0 e -25 t [mA] Compute the power p(t) dissipated in the 60 kΩ p(t) = e -25 t -25 t * 60 --10-3 * 10-3 * 103 p(t) = 60 e -50 t [mW]

Example 2 Two Series Capacitors

Example 2 Example 2 circuit has 2 series capacitors C1 and C2, connected via R = 250 [kΩ] C1 is charged with 4 [V] at time t = 0, and C2 with 24 [V] but opposite direction C1 and C2 discharge at t = 0, now compute: Equivalent Ceq τ for Ceq Voltage v(t) at R Current i(t) Voltage v1(t) Voltage v2(t) Initial energy in C1 Initial energy in C2 Final energy in C1 & C2

Example 2 – The Circuit First find v(t), then use it to compute i(t) through R With i(t), compute v1(t) and v2(t): opposite direction! For v(t), we replace C1 and C2 in series with Ceq = 4 [μF]

Example 2 – Equivalent Circuit Find v(t), the series of v1(t) and v2(t), but opposite polarity, so use – signs to compute v1(t) and v2(t) Ceq yields initial v0 = 20 [V] at Ceq of 4 [μF]

Example 2 – Ceq and τ a.) Compute Ceq with C1 = 5 μF and C2 = 20 μF in series Ceq = C1 * C2 / ( C1 + C2 ) Ceq = 5 * 20 / ( 5 + 20 ) = 4 [μF] b.) Compute time Constant τ τ = R * C τ = 250 * 103 * 4 10-6 τ = 1 [s]

Example 2 – c.) v(t) and d.) i(t) c.) Voltage v(t) is the same as in the equivalent circuit with Ceq = 4 μF, R = 250 kΩ, and τ = 1 [s] Initial voltage v0 in equivalent circuit is given by difference of voltages v1 and v2 = 20 V, direction of v(t) v(t) = v0 e-t/τ v(t) = 20 e-t/τ v(t) = 20 e-t [V] d.) Current i(t) follows Ohm’s Law at R i(t) = v(t) / 250,000 i(t) = 20 / 250 e-t [mA] i(t) = 80 e-t [μA] for t >= 0

Example 2 – e.) v1(t) and f.) v2(t) e.) Given i(t), compute v1(t) i(t) = 80 e-t 10-6 [A] and C1 = 5 [μF] v1(t) = -1/C1 i(t) dt + v1(0) -- from 0 to t v1(t) = -106 / 5 * 80 e-t 10-6 dt - 4 [V] v1(t) = -16 e-t dt - 4 -- from 0 to t v1(t) = 16 e-t - 16 - 4 = 16 e-t - 20 [V] f.) Given i(t), compute v2(t) v2(t) = -1/C2 i(t) dt + v2(0) -- from 0 to t v2(t) = -106 / 20 * 80 e-t 10-6 dt + 24 [V] v2(t) = 4 e-t - 4 + 24 = 4 ( e-t + 5 ) [V]

Example 2 – Initial Energy in C1 & C2 g.) Compute initial energy w1(0) in C1 of 5 μF w(t) = ½ C v02 ( 1 - e-2t/τ ) hence for w1(t) w1(t) = ½ C1 v1(0)2 w1(t) = ½ 5 10-6 42 = 40 [μJ] h.) Compute initial energy w2(0) in C2 of 20 μF w(t) = ½ C v02 -- for w2(t) w2(t) = ½ C2 v2(0)2 w1(t) = ½ 20 10-6 242 w1(t) = 5.769 [mJ]

Example 2 – Final Energy in C1 & C2 i.) Final energy w1(t=∞) in C1 of 5 μF, v1(t=∞) = -20 V w1(∞) = ½ C1 v1(∞)2 w1(∞) = ½ 5 10-6 * 400 w1(∞) = 1 [mJ] Final energy w2(t=∞) in C2 of 20 μF, v2(t=∞) = 20 V w2(∞) = ½ C2 v2(∞)2 w2(∞) = ½ 20 10-6 * 400 w2(∞) = 4 [mJ]

Example 3 Parallel RC Circuits in Series

Example 3 Example 3 has 2 RC circuits in series, plus one 15 kΩ resistor, separable by switch a-b Parallel to constant voltage of 15 V Switch a-b is opened after a long time, at t = 0

Example 3 Compute the following a.) Voltage drop v0 = v1 + v2 for t < 0 across 2 series capacitors, with switch a-b being closed a long time, also v1 and v2 separately b.) Time constants τ1 and τ2 for the 2 RC circuits c.) Voltages v1(t) and v2(t) for the time t > 0, i.e. with switch a-b open

Example 3 a.) Solution for voltage drop v0 For closed switch, R = 15 + 20 + 40 = 75 kΩ total resistance across constant voltage source of 15 V Use voltage division for v0 v0 = 15 * ( 20 + 40 ) / 75 v0 = 12 [V] v1 = 15 * 20 / 75 = 4 [V] v2 = 15 * 40 / 75 = 8 [V] b.) Solution for time constants in parallel RC, τ = R * C τ1 = 20 * 5 [kΩ μF] = 100 [V A-1 A s V-1] 10-3 τ1 = 100 [ms] τ2 = 40 * 1 [kΩ μF] = 40 [ms]

Example 3 c.) Solution for voltages v0(t), v1(t), v2(t) after time t > 0 v0(t) = v1(t) + v2(t) v1(t) = v1(0) e –t/τ1 v1(t) = 4 e –10 t v2(t) = v2(0) e –t/τ2 v2(t) = 8 e –25 t v0(t) = 4 e –10 t + 8 e –25 t

Bibliography Electric Circuits, 10nd edition, Nilsson and Riedel, Pearsons Publishers, © 2015 ISBN-13: 978-0-13-376003-3 Table of integrals: http://integral-table.com/downloads/single-page-integral-table.pdf