EE292: Fundamentals of ECE

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1 EE292: Fundamentals of ECE Fall 2012 TTh 10:00-11:15 SEB 1242 Lecture

2 2 Outline Review Steady-State Analysis RC Circuits RL Circuits

3 3 DC Steady-State Analysis Analysis of C, L circuits in DC operation Steady-state non-changing sources A long time after switch event Capacitors i = C dv Voltage is constant no current open circuit Inductors v = L di Current is constant no voltage short circuit

4 4 Example What are initial conditions of this circuit Open switch no current in circuit i x (0 ) = 0 No charge on capacitor no voltage across it v x (0 ) = 0

5 5 Example Solve for the steady-state values Capacitors Open circuit Inductors Short circuit

6 6 Example Solve for the steady-state values Voltage v x by voltage divider R v x = V 2 5 s = 10 = 5 V R 1 +R Current i x by Ohm s Law i x = V = V = 10 = 1 A R R eq 5+5 v x = i x R 2 = 1 5 = 5V

7 7 Transients The study of time-varying currents and voltages Circuits contain sources, resistances, capacitances, inductances, and switches Studied using our basic analysis methods KCL, KVL, node-voltage, mesh-current But, more complex due to differential relationships between current and voltage with capacitors and inductors

8 8 Discharging a Capacitor A C dv c(t) RC dv c(t) + v c(t) R = 0 + v c (t) = 0 i c i R A Differential equation describes the voltage across the capacitor over time Solution is of the exponential form v c t = Ke st

9 9 Discharging a Capacitor RC dv c(t) + v c (t) = 0 Substitute v c t = Ke st into differential equation RCKse st + Ke st = 0 Solve for s RCs + 1 Ke st = 0 RCs + 1 = 0 s = 1 RC RC is known as the time constant v c t = Ke t/(rc)

10 10 Discharging a Capacitor v c t = Ke t/(rc) Solve for K Voltage before and after switch are the same v c 0 = v c (0 + ) v c 0 + = V i = Ke 0 = K V i is initial charge on capacitor K = V i v c t = V i e t/(rc)

11 11 Voltage/Time Characteristics v c t = V i e t/(rc) τ = RC Time constant of the circuit The amount of time for voltage to decay by a factor of e 1 = Decays to 0 in about five time constants (5τ) Large τ longer decay time Larger R less current Larger C more charge

12 12 Charging a Capacitance Assume capacitor is fully discharged no voltage across capacitor v c 0 = 0 i R A i c A C dv c(t) RC dv c(t) + v c t V s R = 0 + v c (t) = V s Assume solution of the form v c t = K 1 + K 2 e st Solve for s, K 1 RCK 2 se st + K 1 + K 2 e st = V s 1 + RCs K 2 e st + K 1 = V s s = 1 RC K 1 = V s

13 13 Charging a Capacitance v c t = V s + K 2 e t/(rc) Solve for K 2 v c 0 + = 0 = V s + K 2 e 0 K 2 = V s Final solution v c t = V s V s e t/(rc) Transient response eventually decays to a negligible value Steady-state response or forced response

14 14 RC Current Voltage v c t = V s V s e t/(rc) Current i c = V s v c (t) i c = C = C dv c(t) R V s RC e t/(rc) i c = V s R e t/(rc) i c

15 15 General 1 st -Order RC Solution Notice both the current and voltage in an RC circuit has an exponential form The general solution for current/voltage is: x represents current or voltage t 0 represents time when source switches x f - final (asymptotic) value of current/voltage τ time constant (RC) Find values and plug into general solution

16 16 Example i c Solve for v c (t) v f = V s steady-state analysis v c 0 + = 0 no voltage when switch open τ = RC equivalent resistance/capacitance v c t = V s + 0 V s e t/(rc) = V s V s e t/(rc) Solve for i c (t) i f = 0 i c 0 + = V s v c (0 + ) R i c t fully charged cap no current = V s 0 R = V s R = 0 + V s R 0 e t/(rc) = V s R e t/(rc)

17 17 First-Order RL Circuits Contains DC sources, resistors, and a single inductance Same technique to analyze as for RC circuits 1. Apply KCL and KVL to write circuit equations 2. If the equations contain integrals, differentiate each term in the equation to produce a pure differential equation Use differential forms for I/V relationships for inductors and capacitors 3. Assume solution of the form K 1 + K 2 e st 4. Substitute the solution into the differential equation to determine the values of K 1 and s 5. Use initial conditions to determine the value of K 2 6. Write the final solution

18 18 RL Example Before switch i 0 = 0 KVL around loop V s Ri t i t + L R di t L di t = V s R = 0 Notice this is the same equation form as the charging capacitor example Solution of the form i t = K 1 + K 2 e st Solving for K 1, s K 1 + K 2 e st + L R K 2se st = V s R K 1 = V s R 1 + L R s = 0 S = R L Solving for K 2 i 0 + = 0 = V s + K R 2e tr/l 0 = V s R + K 2e 0 K 2 = V s R Final Solution i t i t = V s V s R R e tr/l = 2 2e 500t

19 19 RL Example i t = 2 2e 500t Notice this is in the general form we used for RC circuits τ = L R Find voltage v(t) v f = 0, steady-state short v 0 + = 100 No current immediately through R, v = L di(t) v t = 100e t/τ

20 20 Exercise 4.5 Initial conditions For t < 0 All source current goes through switched wire i R t = i L t = 0 A v t = i R t R = 0 V For t = 0 + (right after switch) i L t = 0 Current can t change immediately through an inductor i R t = 2 A, by KCL v t = i R t R = 20 V Steady-state Short inductor v t = 0 Short circuit across inductor i R = 0 All current through short i L = 2 A By KCL

21 21 Exercise 4.5 Initial conditions For t < 0 All source current goes through switched wire i R t = i L t = 0 A v t = i R t R = 0 V For t = 0 + (right after switch) i L t = 0 Current can t change immediately through an inductor i R t = 2 A, by KCL v t = i R t R = 20 V Steady-state Short inductor v t = 0 Short circuit across inductor i R = 0 All current through short i L = 2 A By KCL

22 22 Exercise 4.5 Can use network analysis to come up with a differential equation, but you would need to solve it Instead, use the general 1 st -order solution Time constant τ τ = L R = 2 10 = 0.2 Voltage v(t) v t = e t/0.2 = 20e t/0.2 V Current i L (t), i R (t) i L t i R t = e t/0.2 = 2 2e t/0.2 A = e t/0.2 = 2e t/0.2 A

23 23 RC/RL Circuits with General Sources Previously, RC dv c(t) + v c (t) = V s What if V s is not constant RC dv c(t) + v c (t) = v s (t) v s (t) Now have a general source that is a function of time The solution is a differential equation of the form τ dx(t) + x(t) = f(t) Where f(t) is known as the forcing function (the circuit source)

24 24 General Differential Equations General differential equation τ dx(t) + x(t) = f(t) The solution to the diff equation is x t = x p t + x h (t) x p t is the particular solution x h (t) is the homogeneous solution

25 25 Particular Solution τ dx p t + x p t = f(t) The solution x p t is called the forced response because it is the response of the circuit to a particular forcing input f t The solution x p t will be of the same functional form as the forcing function E.g. f t = e st x p t = Ae st f t = cos ωt x p t = Acos ωt + Bsin(ωt)

26 26 Homogeneous Solution τ dx h t + x h t = 0 x h t is the solution to the differential equation when there is no forcing function Does not depend on the sources Dependent on initial conditions (capacitor voltage, current through inductor) x h t is also known as the natural response Solution is of the form x h t = Ke t/τ

27 27 General Differential Solution Notice the final solution is the sum of the particular and homogeneous solutions x t = x p t + x h (t) It has an exponential term due to x h (t) and a term x p t that matches the input source

28 28 Second-Order Circuits RLC circuits contain two energy storage elements This results in a differential equation of second order (has a second derivative term) This is like a mass spring system from physics

29 29 RLC Series Circuit Solve for current through capacitor i t = C dv c t i t = C dv s t d2 i t 2 R L di t L d2 i t i t = LC di t R 1 L dv s t KVL around loop v s t L di t Solve for v c t v c t = v s t L Take derivative dv c t = dv s t i t R v c t = 0 di t L d2 i t 2 i t R di t R The general 2 nd -order constant coefficient equation

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