Notes 1. 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 (both sides of page OK), you may bring a calculator, and you don t need a blue book. Week 4b, Slide 1
Lecture Week 4b OUTLINE Transient response of 1 st -order circuits Application: modeling of digital logic gate Reading Chapter 4 through Section 4.3 Week 4b, Slide 2
Transient Response of 1 st -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) x f is the final value of the circuit variable x( t) = x f + [ ] ( )/ τ ( + t t x t ) x e t is the time at which the change occurs Week 4b, Slide 3 f +
Procedure for Finding Transient Response 1. Identify the variable of interest For RL circuits, it is usually the inductor current i L (t) For RC circuits, it is usually the capacitor voltage v c (t) 2. Determine the initial value (at t = t + ) of the variable Recall that i L (t) and v c (t) are continuous variables: i L (t + ) = i L (t ) and v c (t + ) = v c (t ) Assuming that the circuit reached steady state before t, 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 Week 4b, Slide 4
Procedure (cont d) 3. 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 ) 4. Calculate the time constant for the circuit τ = L/R for an RL circuit, where R is the Thévenin equivalent resistance seen by the inductor τ = RC for an RC circuit where R is the Thévenin equivalent resistance seen by the capacitor Week 4b, Slide 5
Example: RL Transient Analysis Find the current i(t) and the voltage v(t): t = R = 5 Ω i + V s = 1 V + v L =.1 H 1. First consider the inductor current i 2. Before switch is closed, i = --> immediately after switch is closed, i = 3. A long time after the switch is closed, i = V s / R = 2 A 4. Time constant L/R = (.1 H)/(5 Ω) =.2 seconds i( t) = 2 + [ ] ( t )/.2 5t 2 e = 2 2e Amperes Week 4b, Slide 6
t = R = 5 Ω i + V s = 1 V + v L =.1 H Now solve for v(t), for t > : ( 5t From KVL, v( t) = 1 ir = 1 2 2e )( 5) = 1e -5t volts` Week 4b, Slide 7
Example: RC Transient Analysis Find the current i(t) and the voltage v(t): R 1 = 1 kω t = i + V s = 5 V + R 2 = 1 kω v C = 1 µf 1. First consider the capacitor voltage v 2. Before switch is moved, v = --> immediately after switch is moved, v = 3. A long time after the switch is moved, v = V s = 5 V 4. Time constant R 1 C = (1 4 Ω)(1-6 F) =.1 seconds v( t) = 5 + [ ] ( t )/.1 1t 5 e = 5 5e Volts Week 4b, Slide 8
R 1 = 1 kω t = i + V s = 5 V + R 2 = 1 kω v C = 1 µf Now solve for i(t), for t > : From Ohm s Law, i( t) Vs v( t) 5 = = R1 1 = 5 x 1-4 e -1t A ( 1t 5 5e ) 4 A Week 4b, Slide 9
Week 4b, Slide 1
Week 4b, Slide 11
Application to Digital Integrated Circuits (ICs) When we perform a sequence of computations using a digital circuit, we switch the input voltages between logic (e.g., Volts) and logic 1 (e.g., 5 Volts). The output of the digital circuit changes between logic and logic 1 as computations are performed. Week 4b, Slide 12
We compute with pulses. Digital Signals We send beautiful pulses in: voltage time But we receive lousy-looking pulses at the output: voltage time Capacitor charging effects are responsible! Every node in a real circuit has capacitance; it s the charging of these capacitances that limits circuit performance (speed) Week 4b, Slide 13
Circuit 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: R + V in (t) + C V out switches between low (logic ) and high (logic 1) voltage states Week 4b, Slide 14
Logic Level Transitions Transition from to 1 (capacitor charging) Transition from 1 to (capacitor discharging) V out ( t) = V high ( t / RC 1 e ) V out ( t) = V high e t / RC V out V out V high V high.63v high.37v high RC time RC time (V high is the logic 1 voltage level) Week 4b, Slide 15
Sequential Switching What if we step up the input, Vin time wait for the output to respond, Vin Vout time then bring the input back down? Week 4b, Slide 16 Vin Vout time
Pulse Distortion V in (t) + R C + V out The 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.) Vout Pulse width =.1RC 6 5 4 3 2 1 1 2 3 4 5 Time Vout Pulse width = RC 6 5 4 3 2 1 1 2 3 4 5 Time Week 4b, Slide 17 Vout Pulse width = 1RC 6 5 4 3 2 1 5 1 15 2 25 Time
Example Suppose a voltage pulse of width 5 µs and height 4 V is applied to the input of this circuit beginning at t = : τ = RC = 2.5 µs V in R R = 2.5 kω C = 1 nf V out C First, V out will increase exponentially toward 4 V. When V in goes back down, V out will decrease exponentially back down to V. What is the peak value of V out? The output increases for 5 µs, or 2 time constants. It reaches 1-e -2 or 86% of the final value..86 x 4 V = 3.44 V is the peak value Week 4b, Slide 18
V out (t) 4 3.5 3 2.5 2 1.5 1.5 2 4 6 8 1 t (s) V out (t) = { 4-4e -t/2.5µs for t 5 µs 3.44e -(t-5µs)/2.5µs for t > 5 µs Week 4b, Slide 19
Week 4b, Slide 2