Download Transient Response of First-Order Circuits: RL and RC Examples and more Study notes Electrical and Electronics Engineering in PDF only on Docsity! Week 4b, Slide 1EECS42, Spring 2005 Prof. White 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 2EECS42, Spring 2005 Prof. White Lecture Week 4b OUTLINE – Transient response of 1st-order circuits – Application: modeling of digital logic gate Reading Chapter 4 through Section 4.3 Week 4b, Slide 5EECS42, Spring 2005 Prof. White 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 6EECS42, Spring 2005 Prof. White Example: RL Transient Analysis Find the current i(t) and the voltage v(t): t = 0 i + v – R = 50 Ω Vs = 100 V + − L = 0.1 H 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 constant L/R = (0.1 H)/(50 Ω) = 0.002 seconds [ ] Amperes 22 202)( 500002.0/)0( tt eeti −−− −=−+= Week 4b, Slide 7EECS42, Spring 2005 Prof. White t = 0 i + v – R = 50 Ω Vs = 100 V + − L = 0.1 H Now solve for v(t), for t > 0: From KVL, ( )( )5022100100)( 500teiRtv −−−=−= = 100e-500t volts`
EECS42, Spring 2005 Week 4b, Slide 10 Prof. White
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EECS42, Spring 2005 Week 4b, Slide 11 Prof. White
Week 4b, Slide 12EECS42, Spring 2005 Prof. White 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). The output of the digital circuit changes between logic 0 and logic 1 as computations are performed. Application to Digital Integrated Circuits (ICs) Week 4b, Slide 15EECS42, Spring 2005 Prof. White Transition from “0” to “1” (capacitor charging) time Vout 0 Vhigh RC 0.63Vhigh Vout Vhigh time RC 0.37Vhigh Transition from “1” to “0” (capacitor discharging) (Vhigh is the logic 1 voltage level) Logic Level Transitions ( )RCthighout eVtV /1)( −−= RCthighout eVtV /)( −= 0 Week 4b, Slide 16EECS42, Spring 2005 Prof. White What if we step up the input, wait for the output to respond, then bring the input back down? time Vi n 0 0 time Vi n 0 0 Vout time Vi n 0 0 Vout Sequential Switching Week 4b, Slide 17EECS42, Spring 2005 Prof. White 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.) 0 1 2 3 4 5 6 0 1 2 3 4 5 Time Vo ut Pulse width = 0.1RC 0 1 2 3 4 5 6 0 1 2 3 4 5 Time Vo ut 0 1 2 3 4 5 6 0 5 10 15 20 25 Time Vo ut Pulse Distortion + Vout – R Vin(t) C + – Pulse width = 10RCPulse width = RC
