Review of charging and discharging in RC Circuits (an enlightened approach)

Transcription

1 EES 42 SPING 21 Lecture 7 opyright egents of University of alifornia eview of charging and discharging in ircuits (an enlightened approach) Before we continue with formal circuit analysis - lets review circuits ationale: Every node in a circuit has capacitance to ground, like it or not, and it s the charging of these capacitances that limits real circuit performance (speed) elevance to digital circuits: We communicate with pulses voltage We send beautiful pulses But we receive lousy-looking pulses and must restore them voltage charging effects are responsible. So lets review them.

2 EES 42 SPING 21 Lecture 7 opyright egents of University of alifornia Simplification for behavior of ircuits Before any input change occurs we have a dc circuit problem (that is we can use dc circuit analysis to relate the put to the input). Long after the input change occurs things settle down. Nothing is changing. So again we have a dc circuit problem. We call the period during which the put changes the transient voltage We can predict a lot ab the transient behavior from the pre- and post-transient dc solutions voltage

3 EES 42 SPING 21 Lecture 7 opyright egents of University of alifornia What environment do pulses face? Every wire in a circuit has resistance. Every junction (called nodes) has capacitance to ground and other nodes. The active circuit elements (transistors) add additional resistance in series with the wires, and additional capacitance in parallel with the node capacitance. Input node + - I Output node O ground A pulse originating at node I will arrive delayed and distorted at node O because it takes to charge through If we focus on the circuit which distorts the pulses produced by, it consists simply of and. ( is just the -varying source which produces the input pulse.)

4 EES 42 SPING 21 Lecture 7 opyright egents of University of alifornia The ircuit to Study (All single-capacitor circuits reduce to this one) Input node Output node ground represents total resistance (wire plus whatever drives the input node) represents the total capacitance from node to the side world (from devices, nearby wires, ground etc)

5 EES 42 SPING 21 Lecture 7 opyright egents of University of alifornia ESPONSE ase 1 ising voltage. apacitor uncharged: Apply + voltage step V1 Input node + - Output node ground jumps at t=, but cannot jump like. Why not? Because an instantaneous change in a capacitor voltage would require instantaneous increase in energy stored (1/2V²), that is, infinite power. (Mathematically, V must be differentiable: I=dV/dt) V does not jump at t=, i.e. V(t= + ) = V(t= - ) Therefore the dc solution before the transient tells us the capacitor voltage at the beginning of the transient.

6 EES 42 SPING 21 Lecture 7 opyright egents of University of alifornia V1 ESPONSE ase 1 apacitor uncharged: Apply voltage step Input node approaches its final value asymptotically (It never quite gets to V1, but it gets arbitrarily close). Why? Output node ground After the transient is over (nothing changing anymore) it means d(v)/dt = ; that is all currents must be zero. From Ohm s law, the voltage across must be zero, i.e. =. That is, V1 as t. (Asymptotic behavior) Again the dc solution (after the transient) tells us (the asymptotic limit of) the capacitor voltage during the transient. + -

7 EES 42 SPING 21 Lecture 7 opyright egents of University of alifornia 5 ESPONSE Example apacitor uncharged: Apply voltage step of 5 V Input node learly starts at V ( at t = + ) and approaches 5V. Output node We know this because of the pre-transient dc solution (V=) and post-transient dc solution (V=5V). + - ground So we know a lot ab during the transient - namely its initial value, its final value, and we know the general shape.

8 EES 42 SPING 21 Lecture 7 opyright egents of University of alifornia V1 V1 ESPONSE: ase 1 (cont.)? Exact form of? Input node Output node + - ground Equation for : Do you remember general form? V1 = V 1 (1-e -t/τ ) Exponential! τ

9 EES 42 SPING 21 Lecture 7 opyright egents of University of alifornia eview of simple exponentials. ising Exponential from Zero = V 1 (1-e -t/τ ) Falling Exponential to Zero = V 1 e -t/τ at t =, at t, 8 =, and V 1 also at t = τ, =.63 V 1 at t =, at t, 8 = V 1, and, also at t = τ, =.37 V 1 V 1 V 1.63V 1.37V 1 τ τ

10 EES 42 SPING 21 Lecture 7 opyright egents of University of alifornia Further eview of simple exponentials. ising Exponential from Zero Falling Exponential to Zero = V 1 (1-e -t/τ ) = V 1 e -t/τ We can add a constant (positive or negative) = V 1 (1-e -t/τ ) + V 2 = V 1 e -t/τ + V 2 V 1 + V 2 V 1 + V 2.63V 1 + V 2.37V 1 + V 2 V 2 V 2 τ τ

11 EES 42 SPING 21 Lecture 7 opyright egents of University of alifornia Further eview of simple exponentials. ising Exponential Falling Exponential = V 1 (1-e -t/τ ) + V 2 = V 1 e -t/τ + V 2 Both equations can be written in one simple form: = A + Be -t/τ Initial value (t=) : = A + B. Final value (t>>τ): = A Thus: if B <, rising exponential; if B >, falling exponential A Here B < A+B Here B > A+B A

12 EES 42 SPING 21 Lecture 7 opyright egents of University of alifornia V in V1 ESPONSE: ase 1 (ising exponential) V in V1 i = V 1 (1-e -t/τ ) i τ How is τ related to and? If is bigger, it takes longer (τ ). If is bigger, it takes longer (τ ). Thus, τ is proportional to. In fact, τ =! Thus, = V 1 (1-e -t/τ )

13 EES 42 SPING 21 Lecture 7 opyright egents of University of alifornia ESPONSE: ase 1 (cont.) Proof that V = V (1 e 1 t / ) V in i i i i V = dv = dt (Ohm's law) (capacitance law) But i = i! Thus, V in dv dt dv = dt or = 1 (V in V )

14 EES 42 SPING 21 Lecture 7 opyright egents of University of alifornia But and V = V in 1 V = ESPONSE ase 1 (cont.) Proof = constant at t = + that V I claim that the solution to this first-order linear differential equation is: V = V (1 e t/ ) 1 = V (1 e t/ ) 1 dv We have: = 1 (V V ) Proof by substitution: dt in dv 1 ( V V ) dt? = in V -- / 1 ( (1 / )) 1 e t? = V V e t // 1 / 1 / clearly V V e t/ 1 e t/ 1 = and + = at t = OK

15 EES 42 SPING 21 Lecture 7 opyright egents of University of alifornia ESPONSE (cont.) V in Generalization i i switches at t = ; then for any 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 V1: First, at Thus, as t = V A + B = t, V V 1 A = V 1 Vo V B o initial = V o V Thus, 1 = A + Be -t/τ voltage