Lecture 7: September 19th, The following slides were derived from those prepared by Professor Oldham for EE40 in Fall 01. Version Date 9/19/01

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1 EES Intro. electronics for S Fall Lecture 7: September 9th, Lecture 7: 9/9/ A.. Neureuther Version Date 9/9/ harging and Discharging of ircuits (Transients) A) Mathematical Method B) EE Easy Method ) Logic D) Generalizations E) Pulse Distortion eading: Schwarz and Oldham 8. Hands opyright, egents of University of alifornia The following slides were derived from those prepared by Professor Oldham for EE in Fall

2 EES Intro. electronics for S Fall Lecture 7: 9/9/ A.. Neureuther Version Date 9/9/ harging and discharging in ircuits (continued) Last Time: We learned that simple the simple circuit with a step input has a universal exponential solution of the form: V = A Be t/ Example: = K, = pf, steps from zero to V at t=: ) Initial value of is ) Final value of is V 3) Time constant is 9 sec Input node ) reaches.3 X in 9 sec opyright, egents of University of alifornia.3v Output node ground time nsec

3 EES Intro. electronics for S Fall Lecture 7: 9/9/ A.. Neureuther ESPONSE: ase (cont.) Proof that V = V ( e t / ) Version Date 9/9/ V in V i i i i V = dv = dt (Ohm's law) (capacitance law) But i = i! Thus, V V dv in = dt or d = (V V in dt ) opyright, egents of University of alifornia

4 EES Intro. electronics for S Fall Lecture 7: 9/9/ A.. Neureuther But V in and = V V = ESPONSE ase (cont.) Proof = constant at t = I claim that the solution to this firstorder linear differential equation is: V = V ( e t/ ) that V = V ( e t/ ) dv We have: = (V V ) Proof by substitution: dt in V = at t = Version Date 9/9/ dv ( V V ) dt? = in Exp. Term gives the value of τ. V / ( ( / )) e t? = V V e t // / / onstant gives A. Initial condition gives AB. OK opyright, egents of University of alifornia

5 EES Intro. electronics for S Fall Lecture 7: 9/9/ A.. Neureuther ESPONSE (cont.) Version Date 9/9/ V in V Generalization i i switches at t = ; then for any time interval t >, in which is a constant, is always of the form: We determine A and B from the initial voltage on, and the value of. Assume switches at t= from Vco to V: First, at t = Thus, as V A B = t, V V A = V Vo V B o initial = V o V Thus, voltage opyright, egents of University of alifornia V = A Be t/τ

6 EES Intro. electronics for S Fall Lecture 7: 9/9/ A.. Neureuther Version Date 9/9/ harging and discharging in ircuits For this example: = K, = pf, steps from zero to V at t=:.3v Input node Output node ground nsec time Note that we found this graph with even using the equation (That is we did not try to evaluate A and B). V = A Be t/ We simply used the dc solution for t< and the dc solution for t>> to get the limits and we used the time constant to get the horizontal scale. We only need the equation to remind us the solution is an exponential. So this will be the basis of our easy method. opyright, egents of University of alifornia

7 EES Intro. electronics for S Fall Lecture 7: 9/9/ A.. Neureuther harging and discharging in ircuits Version Date 9/9/ (The official EE/EE Easy Method) Method of solving for any node voltage in a single capacitor circuit. ) Simplify the circuit so it looks like one resistor, a source, and a capacitor (it will take another two weeks to learn all the tricks to do this.) But then the circuit looks like this: ) The time constant of the transient is τ =. 3) Solve the dc problem for the Input node Output node capacitor voltage before the transient. This is the starting value (initial value) for the transient voltage. ) Solve the dc problem for the capacitor voltage after the transient is over. This is the asymptotic value. ground 5) Sketch the Transient. It is 3% complete after one time constant. ) Write the equation by inspection. opyright, egents of University of alifornia

8 EES Intro. electronics for S Fall Lecture 7: 9/9/ A.. Neureuther harging and discharging in ircuits Version Date 9/9/ (Example of the EE Easy Method) Find Vc(t) for the following circuit: (input switches from V to V at t = ) K 9K Vc ) Simplify the circuit : K Vc ) The time constant of the transient is τ = = nsec pf 3) Before the transient = V so Vc =V pf V K Vc ) After the transient is over = V so Vc = V. This is the asymptotic value. V K Vc opyright, egents of University of alifornia

9 EES Intro. electronics for S Fall Lecture 7: 9/9/ A.. Neureuther harging and discharging in ircuits Version Date 9/9/ (Example of the EE Easy Method) Find Vc(t) for the following circuit: (input switches from V to V at t = ) K 9K Vc We have : Initial value of Vc is V, final value is V and τ = nsec 5) Sketch Vc (t) : Vc Initial value 37% of transient remaining at one time constant t (nsec) Final value pf What is the equation for an exponential beginning at V, decaying to V, with τ = nsec? V c (t) = 3e t/τ opyright, egents of University of alifornia

10 EES Intro. electronics for S Fall Lecture 7: 9/9/ A.. Neureuther Version Date 9/9/ OU METHOD AVOIDS ALL MATH! Sketch waveform (starts at Vco, ends asymptotically at V, initial slope intersects at t = or transient is 3% complete at t=) Write equation: a. constant term A = limit of V as t b. preexponent B = initial value constant term 3 5 in = 3 V o = 3 5 V t /. = 5 5e t = in =.µsec V o = opyright, egents of University of alifornia V time (msec) = 8e t / t in microseconds

11 EES Intro. electronics for S Fall t = in V Examples =.sec Lecture 7: 9/9/ A.. Neureuther V = A Version Date 9/9/ Be t / V o = V o = 5 V V = 8 A= B= 8 A= B= time (sec)...3. time (sec) V = 8 A= B= 8 A= B= time (sec) time (sec) opyright, egents of University of alifornia

12 EES Intro. electronics for S Fall t = µsec in = µ sec V o = 5V opyright, egents of University of alifornia 5 3 Lecture 7: 9/9/ A.. Neureuther OMPLIATION: Event Happens at t (Solution: Shift reference time to time of event) Example: switch closes at µsec We shift the time axis here by one microsecond, i.e. imagine a new time coordinate t* = tµsec so that in the new time domain, the event happens at t* = and our standard solution applies. Of course we replace t* by t µsec in the equations and plots. Thus instead of t* = we have t = µsec, etc. V Version Date 9/9/ ( t ) = 5e 8 time (microseconds)

13 EES Intro. electronics for S Fall Lecture 7: 9/9/ A.. Neureuther FINAL EXAMPLE Version Date 9/9/ Your photo flash charges a µf capacitor from a 5V source through a K resistor. If the capacitor is initially uncharged, how long must you wait for it to reach 95% charged (7.5 V)? Solution: = K 5V 3 F K V 3 = V o sec = V 5 3 V=7.5 at t =?? By inspection: 7.5 = 5( e x ) t V o = 5 5e e t x t so opyright, egents of University of alifornia /, 7.5 = ( ) 5 8 time in seconds t x = sec

14 EES Intro. electronics for S Fall Lecture 7: 9/9/ A.. Neureuther Version Date 9/9/ Generalizations Switching of circuits with multiple resistors and sources use T seen looking back into the circuit from the terminals of the capacitor after the switching changes conditions. Inductor circuits use τ = L / T opyright, egents of University of alifornia

15 EES Intro. electronics for S Fall Input node Pulse Distortion An example of analysis Output node O ground Lecture 7: 9/9/ A.. Neureuther Version Date 9/9/ ule: The pulse width must be wider than to avoid severe pulse distortion. Example: Lets find the shape of the put pulse for pulse width of.,, and Method: Use 5V pulse height. Let = µsec. eplace voltage source with a switch which shorts the input to ground for t<, switches to 5V at t=, and switches back to ground at t=.µsec, or t=µsec, t=µsec. Thus we have two problems: #: V rising from zero and #: V falling back toward zero. ascade solutions! opyright, egents of University of alifornia

16 EES Intro. electronics for S Fall Lecture 7: 9/9/ A.. Neureuther PULSE: Output is ising exponential then Falling exponential Version Date 9/9/ Example: Switch rises at t =, falls at t =., or µsec (Do µsec case) 5V = µ sec V Solution: for = µsec: during the first rise V obeys: t V = 5[ e ] V o = Now starting at µsec we are discharging the capacitor so the form is a falling exponential with initial value 3. V: 5 3 What is equation? Thus at t = µsec, rising voltage reaches 5 [ e ] = 3.V 3 5 time (microseconds) opyright, egents of University of alifornia

17 5V EES Intro. electronics for S Fall = µ sec V V o = Lecture 7: 9/9/ A.. Neureuther PULSE DISTOTION other cases Version Date 9/9/ Switch rises at t =, falls at t =. or µsec (i.e.. or µsec pulse widths) 5 3 Solve for V in the other two cases (. or µsec) just as for µsec 3 5 time (microseconds) At t =.µsec the put has only risen to.5v! Whereas for µsec pulse width, put reaches to within.995 % of 5V. You need to verify the numbers! opyright, egents of University of alifornia time (microseconds)

18 EES Intro. electronics for S Fall Input node Pulse Distortion An example of analysis Output node O ground Lecture 7: 9/9/ A.. Neureuther Version Date 9/9/ ule: The pulse width must be wider than to avoid severe pulse distortion. PW = time (microseconds) PW = time (microseconds) PW = time (microseconds) opyright, egents of University of alifornia