Week 4b, Slide 1EECS42, Spring 2005 Prof. WhiteNotes1. Midterm 1 – Thursday February 24 in class.Covers through text Sec. 4.3, topics of HW 4. GSIs will review material in discussion sections prior to the exam. No books at the exam, no cell phones, you may bring one 8-1/2 by 11 sheet of notes (bothsides of page OK), you may bring a calculator, and you don’t need a blue book.Week 4b, Slide 2EECS42, Spring 2005 Prof. WhiteLecture Week 4bOUTLINE– Transient response of 1st-order circuits– Application: modeling of digital logic gateReadingChapter 4 through Section 4.3Week 4b, Slide 3EECS42, Spring 2005 Prof. WhiteTransient Response of 1st-Order Circuits• In Lecture Week 4a, 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)[]τ/)(00 )()(+−−+−+=ttffextxxtxxfis the final value of the circuit variablet0is the time at which the change occursWeek 4b, Slide 4EECS42, Spring 2005 Prof. WhiteProcedure 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 stateWeek 4b, Slide 5EECS42, Spring 2005 Prof. WhiteProcedure (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 capacitorWeek 4b, Slide 6EECS42, Spring 2005 Prof. WhiteExample: 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−−−−=−+=Week 4b, Slide 7EECS42, Spring 2005 Prof. Whitet = 0i+v–R = 50 ΩVs= 100 V+−L = 0.1 HNow solve for v(t), for t > 0:From KVL,()()5022100100)(500teiRtv−−−=−== 100e-500t volts`Week 4b, Slide 8EECS42, Spring 2005 Prof. WhiteExample: 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 kΩWeek 4b, Slide 9EECS42, Spring 2005 Prof. Whitet = 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,A= 5 x 10-4e-100t AWeek 4b, Slide 10EECS42, Spring 2005 Prof. WhiteWeek 4b, Slide 11EECS42, Spring 2005 Prof. WhiteWeek 4b, Slide 12EECS42, Spring 2005 Prof. WhiteWhen 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)Week 4b, Slide 13EECS42, Spring 2005 Prof. White• 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 SignalsWeek 4b, Slide 14EECS42, Spring 2005 Prof. WhiteCircuit 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 statesWeek 4b, Slide 15EECS42, Spring 2005 Prof. WhiteTransition 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/)(−=0Week 4b, Slide 16EECS42, Spring 2005 Prof. WhiteWhat if we step up the input,wait for the output to respond,then bring the input back down? timeVin00timeVin00VouttimeVin00VoutSequential SwitchingWeek 4b, Slide 17EECS42, Spring 2005 Prof. WhiteThe input voltage pulse width must be large 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 = RCWeek 4b, Slide 18EECS42, Spring 2005 Prof. WhiteVinRVoutCSuppose 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 µsWeek 4b, Slide 19EECS42, Spring 2005 Prof.
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