1Lecture 15, Slide 1EECS40, Fall 2003 Prof. KingLecture #15ANNOUNCEMENTS• Prof. King’s office hour today is cancelled• Farhana’s office hours this week are cancelled• HW#5 will be available on Friday 10/3 (due 10/10)• Pick up graded midterms in discussion sectionsOUTLINE– Transient response of 1st-order circuits– Application: modeling of digital logic gateReadingChapter 7.3-7.5(with Prof. Sanders)Lecture 15, Slide 2EECS40, Fall 2003 Prof. KingTransient Response of 1st-Order Circuits• In Lectures 13 and 14, we saw that the currents and voltages in RL and RC circuits decay exponentially with time, with a characteristic time constant τ, 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 τ:where x(t) is the circuit variable (voltage or current)xfis the final value of the circuit variablet0is the time at which the change occurs[]τ/)(00 )()(+−−+−+=ttffextxxtx2Lecture 15, Slide 3EECS40, Fall 2003 Prof. KingProcedure for Finding Transient Response1. 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)2. 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 stateLecture 15, Slide 4EECS40, Fall 2003 Prof. KingProcedure (cont’d)3. Calculate the final value of the variable (its value as t Æ ∞)• Again, make use of the fact that an inductor behaves like a short circuit in steady state (t Æ ∞)or that a capacitor behaves like an open circuit in steady state (t Æ ∞)4. Calculate the time constant for the circuitτ= L/R for an RL circuit, where R is the Théveninequivalent resistance “seen” by the inductorτ= RC for an RC circuit where R is the Théveninequivalent resistance “seen” by the capacitor3Lecture 15, Slide 5EECS40, Fall 2003 Prof. KingExample: RL Transient AnalysisFind the current i(t) and the voltage v(t):t = 0i+v–R = 50 ΩVs= 100 V+−L = 0.1 H1. First consider the inductor current i2. Before switch is closed, i = 0--> immediately after switch is closed, i = 03. A long time after the switch is closed, i = Vs/ R = 2 A4. Time constant L/R = (0.1 H)/(50 Ω) = 0.002 seconds[]Amperes 22 202)(500002.0/)0( tteeti−−−−=−+=Lecture 15, Slide 6EECS40, Fall 2003 Prof. Kingt = 0i+v–R = 50 ΩVs= 100 V+−L = 0.1 HNow solve for v(t), for t > 0:From KVL,()()5022100100)(500teiRtv−−−=−=4Lecture 15, Slide 7EECS40, Fall 2003 Prof. KingExample: RC Transient AnalysisFind the current i(t) and the voltage v(t):t = 0i+v–R2= 10 kΩVs= 5 V+−C = 1 µF1. First consider the capacitor voltage v2. Before switch is moved, v = 0--> immediately after switch is moved, v = 03. A long time after the switch is moved, v = Vs= 5 V4. Time constant R1C = (104Ω)(10-6F) = 0.01 seconds[]Volts 55 505)(10001.0/)0( tteetv−−−−=−+=R1= 10 kΩLecture 15, Slide 8EECS40, Fall 2003 Prof. Kingt = 0i+v–R2= 10 kΩVs= 5 V+−C = 1 µF()4100110555)()(tseRtvVti−−−=−=R1= 10 kΩNow solve for i(t), for t > 0:From Ohm’s Law,5Lecture 15, Slide 9EECS40, Fall 2003 Prof. KingWhen 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).The output of the digital circuit changes between logic 0and logic 1 as computations are performed.Application to Digital Integrated Circuits (ICs)Lecture 15, Slide 10EECS40, Fall 2003 Prof. King• Every node in a real circuit has capacitance; it’s the charging of these capacitances that limits circuit performance (speed)We compute with pulses. We send beautiful pulses in:But we receive lousy-looking pulses at the output:Capacitor charging effects are responsible!timevoltagetimevoltageDigital Signals6Lecture 15, Slide 11EECS40, Fall 2003 Prof. KingCircuit Model for a Logic Gate• Recall (from Lecture 1) that 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:+Vout–RVin(t)+−Cswitches between “low” (logic 0) and “high” (logic 1) voltage statesLecture 15, Slide 12EECS40, Fall 2003 Prof. KingTransition from “0” to “1”(capacitor charging)timeVout0VhighRC0.63VhighVoutVhightimeRC0.37VhighTransition from “1” to “0”(capacitor discharging)(Vhighis the logic 1 voltage level)Logic Level Transitions()RCthighouteVtV/1)(−−=RCthighouteVtV/)(−=07Lecture 15, Slide 13EECS40, Fall 2003 Prof. KingWhat if we step up the input,wait for the output to respond,then bring the input back down? timeVin00timeVin00VouttimeVin00VoutSequential SwitchingLecture 15, Slide 14EECS40, Fall 2003 Prof. KingThe 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.)0123456012345TimeVoutPulse width = 0.1RC0123456012345TimeVout01234560 5 10 15 20 25TimeVoutPulse Distortion+Vout–RVin(t)C+–Pulse width = 10RCPulse width = RC8Lecture 15, Slide 15EECS40, Fall 2003 Prof. KingVinRVoutCSuppose a voltage pulse of width5 µs and height 4 V is applied to theinput of this circuit beginning at t = 0:R = 2.5 kΩC = 1 nF• First, Voutwill increase exponentially toward 4 V.• When Vingoes back down, Voutwill decrease exponentially back down to 0 V.What is the peak value of Vout?The output increases for 5 µs, or 2 time constants.Æ It reaches 1-e-2or 86% of the final value.0.86 x 4 V = 3.44 V is the peak valueExampleτ = RC = 2.5 µsLecture 15, Slide 16EECS40, Fall 2003 Prof. King00.511.522.533.540 2 4 6 8 10Vout(t) =4-4e-t/2.5µsfor 0 ≤ t ≤ 5 µs3.44e-(t-5µs)/2.5µsfor t > 5
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