Understanding First-Order Circuits: RL and RC Models

1. EE42/100 Lecture 9 Topics: More on First-Order Circuits Water model and potential plot for RC circuits A bit on Second-Order Circuits

2. First-Order Circuits • A circuit which contains only sources, resistors and an inductor is called an RL circuit. • A circuit which contains only sources, resistors and a capacitor is called an RC circuit. • RL and RC circuits are called first-order circuits because their voltages and currents are described by first-order differential equations. R R i i vs – + vs – + L C

5. The natural response of an RL or RC circuit is its behavior (i.e. current and voltage) when stored energy in the inductor or capacitor is released to the resistive part of the network (containing no independent sources). The step response of an RL or RC circuit is its behavior when a voltage or current source step is applied to the circuit, or immediately after a switch state is changed.

6. t = 0 i + v – Io Ro L R Natural Response of an RL Circuit • Consider the following circuit, for which the switch is closed for t < 0, and then opened at t = 0: Notation: 0– is used to denote the time just prior to switching 0+ is used to denote the time immediately after switching • The current flowing in the inductor at t = 0– is Io

7. Recall: The current flowing in an inductor cannot change instantly, and the voltage across a capacitor, which is proportional to the charge stored in the capacitor, cannot change instantly. For a first-order circuit these are called initial values of current and voltage. A long time after the circuit configuration changes, the currents and voltages achieve their final, or steady-state values. Later when we talk about second-order circuits – ones that consist of resistors and the equivalent of two energy storage elements, like an L and a C or two Cs – we’ll take a look at the initial and final values of these quantities and their time derivatives.

8. Solving for the Current (t 0) i + v – • For t > 0, the circuit reduces to • Applying KVL to the LR circuit: • Solution: Io Ro L R

9. What Does e-t/t Look Like? e-t/t with t = 10-4 • t is the amount of time necessary for an exponential to decay to 36.7% of its initial value. • -1/t is the initial slope of an exponential with an initial value of 1.

10. Solving for the Voltage (t > 0) • Note that the voltage changes abruptly: + v – Io Ro L R

11. Time Constant t • In the example, we found that • Define the time constant • At t = t, the current has reduced to 1/e (~0.37) of its initial value. • At t = 5t, the current has reduced to less than 1% of its initial value.

12. Transient vs. Steady-State Response • The momentary behavior of a circuit (in response to a change in stimulation) is referred to as its transient response. • The behavior of a circuit a long time (many time constants) after the change in voltage or current is called the steady-state response.

13. Review (Conceptual) • Any first-order circuit can be reduced to a Thévenin (or Norton) equivalent connected to either a single equivalent inductor or capacitor. • In steady state, an inductor behaves like a short circuit • In steady state, a capacitor behaves like an open circuit RTh VTh – + ITh RTh L C

14. Natural Response of an RC Circuit • Consider the following circuit, for which the switch is closed for t < 0, and then opened at t = 0: Notation: 0– is used to denote the time just prior to switching 0+ is used to denote the time immediately after switching • The voltage on the capacitor at t = 0– is Vo t = 0 Ro + v – +  Vo R C

15. Solving for the Voltage (t 0) i • For t > 0, the circuit reduces to • Applying KCL to the RC circuit: • Solution: + v – Ro +  Vo C R

16. Solving for the Current (t > 0) i + v – Ro • Note that the current changes abruptly: +  Vo C R

17. Time Constant t • In the example, we found that • Define the time constant • At t = t, the voltage has reduced to 1/e (~0.37) of its initial value. • At t = 5t, the voltage has reduced to less than 1% of its initial value. (with R in ohms and C in farads, t is in seconds)

18. RL Circuit Inductor current cannot change instantaneously time constant RC Circuit Capacitor voltage cannot change instantaneously time constant Natural Response Summary i + v – L R C R

19. Transient Response of 1st-Order Circuits • We saw that the currents and voltages in RL and RC circuits decay exponentially with time, with a characteristic time constant t, when an applied current or voltage is suddenly removed. • In general, when an applied current or voltage suddenly changes, the voltages and currents in an RL or RC circuit will change exponentially with time, from their initial values to their final values, with the characteristic time constant t as follows: where x(t) is the circuit variable (voltage or current) xf is the final value of the circuit variable t0 is the time at which the change occurs This is a very useful equation!

20. Procedure for Finding Transient Response • Identify the variable of interest • For RL circuits, it is usually the inductor current iL(t) • For RC circuits, it is usually the capacitor voltage vc(t) • Determine the initial value (at t = t0+) of the variable • Recall that iL(t) and vc(t) are continuous variables: iL(t0+) = iL(t0)and vc(t0+) = vc(t0) • Assuming that the circuit reached steady state before t0 , use the fact that an inductor behaves like a short circuit in steady state or that a capacitor behaves like an open circuit in steady state

21. Procedure(cont’d) • Calculate the final value of the variable (its value as t  ∞) • Again, make use of the fact that an inductor behaves like a short circuit in steady state (t ∞)or that a capacitor behaves like an open circuit in steady state (t ∞) • Calculate the time constant for the circuit t= L/Rfor an RL circuit, where R is the Thévenin equivalent resistance “seen” by the inductor t= RCfor an RC circuit where R is the Thévenin equivalent resistance “seen” by the capacitor

22. Example: RL Transient Analysis R = 50 W t= 0 Find the current i(t) and the voltage v(t): i + v – +  L = 0.1 H Vs = 100 V 1. First consider the inductor current i 2. Before switch is closed, i = 0 --> immediately after switch is closed,i = 0 3. A long time after the switch is closed, i = Vs / R = 2 A 4. Time constantL/R = (0.1 H)/(50 W) = 0.002 seconds

23. R = 50 W t= 0 i + v – +  L = 0.1 H Vs = 100 V Now solve for v(t), for t > 0: From KVL,

24. Example: RC Transient Analysis R1 = 10 kW t= 0 Find the current i(t) and the voltage v(t): i + v – +  C = 1 mF Vs = 5 V R2 = 10 kW 1. First consider the capacitor voltage v 2. Before switch is moved, v = 0 --> immediately after switch is moved,v = 0 3. A long time after the switch is moved, v = Vs= 5 V 4. Time constantR1C = (104W)(10-6 F) = 0.01 seconds

25. R1 = 10 kW t= 0 i + v – +  C = 1 mF Vs = 5 V R2 = 10 kW Now solve for i(t), for t > 0: From Ohm’s Law,

26. When we perform a sequence of computations using a digital circuit, we switch the input voltages between logic 0 (e.g. 0 Volts) and logic 1 (e.g. 5 Volts). Application to Digital Integrated Circuits (ICs) The output of the digital circuit changes between logic 0 and logic 1 as computations are performed.

27. Digital Signals We compute with pulses. We send beautiful pulses in: voltage • Every node in a real circuit has capacitance; it’s the charging of these capacitances that limits circuit performance (speed) time But we receive lousy-looking pulses at the output: voltage time Capacitor charging effects are responsible!

28. Circuit Model for a Logic Gate • Electronic building blocks referred to as “logic gates” are used to implement logical functions (NAND, NOR, NOT) in digital ICs • Any logical function can be implemented using these gates. • A logic gate can be modeled as a simple RC circuit: R + Vout – Vin(t) +  C switches between “low” (logic 0) and “high” (logic 1) voltage states

29. Logic Level Transitions Transition from “0” to “1” (capacitor charging) Transition from “1” to “0” (capacitor discharging) Vout Vout Vhigh Vhigh 0.63Vhigh 0.37Vhigh time time 0 0 RC RC (Vhigh is the logic 1 voltage level)

30. Vin Vin Vout Vout 0 0 time time 0 0 Sequential Switching What if we step up the input, wait for the output to respond, then bring the input back down? Vin 0 time 0

31. 6 6 6 5 5 5 4 4 4 Vout Vout Vout 3 3 3 2 2 2 1 1 1 0 0 0 0 0 5 1 2 10 15 3 4 20 25 5 0 1 2 3 4 5 Time Time Time Pulse Distortion R The input voltage pulse width must be long enough; otherwise the output pulse is distorted. (We need to wait for the output to reach a recognizable logic level, before changing the input again.) + Vout – + – Vin(t) C Pulse width = 0.1RC Pulse width = RC Pulse width = 10RC

32. Example R Suppose a voltage pulse of width 5 ms and height 4 V is applied to the input of this circuit beginning at t = 0: V V out in C R = 2.5 kΩ C = 1 nF t = RC = 2.5 ms • First, Vout will increase exponentially toward 4 V. • When Vin goes back down, Vout will decrease exponentially • back down to 0 V. • What is the peak value of Vout? • The output increases for 5 ms, or 2 time constants. •  It reaches 1-e-2 or 86% of the final value. • 0.86 x 4 V = 3.44 V is the peak value

33. 4 3.5 3 2.5 2 1.5 1 0.5 0 0 2 4 6 8 10 { 4-4e-t/2.5ms for 0 ≤ t ≤5 ms 3.44e-(t-5ms)/2.5ms for t > 5 ms Vout(t) =

34. A Bit on Second-Order Circuits A second-order circuit consists of resistors and the equivalent of two energy storage elements (Ls, Cs). A second-order circuit is characterized by a second-order differential equation (contains second-derivatives of time) Example: A circuit containing R, L and C in series with a voltage source; a circuit with R, L and C in parallel.

35. Initial and final values of v, i, dv/dt, and di/dt Example: The switch in this circuit has been closed for a long time. It opens at t = 0. Find: i(0+), v(0+), di(0+)/dt, dv(0+)/dt, i(infinite time), v(infinite time) • Values for t < 0 • Values for t = 0+ • Values for t = infinity

36. Values for t < 0

37. b. Values for t = 0+

38. c. Values for t = infinity

39. i(t) R + vs(t) C - L A 2nd Order RLC Circuit • Application: Filters • A bandpass filter such as IF amplifier for the AM radio. • A lowpass filter with a sharper cutoff than can be obtained with an RC circuit.

40. vr(t) + - R + + vc(t) vs(t) C - - vl(t) - + L The Differential Equation i(t) KVL around the loop: vr(t) + vc(t) + vl(t) = vs(t)

41. The Differential Equation The voltage and current in a second order circuit is the solution to a differential equation of the following form: xp(t) is the particular solution (forced response) and xc(t) is the complementary solution (natural response). (the forcing function – the driving voltage or current source)

42. The Particular Solution • The particular solution xp(t) is usually a weighted sum of f(t) and its first and second derivatives. • If f(t) is constant, then xp(t) is constant. • If f(t) is sinusoidal, then xp(t) is sinusoidal (with the same frequency as the source, for a circuit of only linear elements)

43. The Complementary Solution The complementary solution has the following form: K is a constant determined by initial conditions. s is a constant determined by the coefficients of the differential equation.

44. Characteristic Equation • To find the complementary solution, we need to solve the characteristic equation: • The characteristic equation has two roots-call them s1 and s2.

45. Overdamped, critically damped and underdamped response of source-free transiently excited 2nd-order RLC circuit Assume circuit is excited by energy stored in C or L of series RLC circuit. Assume i(t) = K1es1t + K2es2t where (with slightly different notation) a = R/2L is called the damping factor and is the undamped natural frequency If a > w0 overdamped case a If a = w0  critically damped case b If a< w0  underdamped case c