W. G. Oldham EECS 40 Fall 2001 Lecture 5 Copyright Regents of University of California 1 Review of charging and discharging in RC Circuits Lectures 2 and 3: Gate Delay Ideas Lecture 4: Basic properties of circuit elements Today: Finish up circuit elements How to find delays in circuits which have to charge capacitors (which every circuit effectively has)

W. G. Oldham EECS 40 Fall 2001 Lecture 5 Copyright Regents of University of California 2 BASIC CIRCUIT ELEMENTS Voltage Source Current Source Resistor Capacitor Inductor (like ideal battery) (always supplies some constant given current) (Ohm’s law) (capacitor law – based on energy storage in electric field of a dielectric) (inductor law – based on energy storage in magnetic field produced by a current

W. G. Oldham EECS 40 Fall 2001 Lecture 5 Copyright Regents of University of California 3 DEFINITION OF IDEAL VOLTAGE SOURCE  Special cases: upper case V  constant voltage … called “DC” lower case v  general voltage, may vary with time Symbol   Note: Reference direction for voltage source is unassociated (by convention) Examples: 1) V = 3V 2) v = v(t) = 160 cos 377t Current through voltage source can take on any value (positive or negative) but not infinite

W. G. Oldham EECS 40 Fall 2001 Lecture 5 Copyright Regents of University of California 4 IDEAL CURRENT SOURCE Actual current source examples – hard to find except in electronics (transistors, etc.), as we will see + note unassociated v direction  upper-case I  DC (constant) value lower-case implies current could be time-varying i(t) “Complement” or “dual” of the voltage source: Current though branch is independent of the voltage across the branch

W. G. Oldham EECS 40 Fall 2001 Lecture 5 Copyright Regents of University of California 5 CURRENT-VOLTAGE CHARACTERISTICS OF VOLTAGE & CURRENT SOURCES Describe a two-terminal circuit element by plotting current vs. voltage V i Assume unassociated signs Ideal voltage source   If V is positive and I is only positive  releasing power absorbing power (charging)  But this is arbitrary; i might be negative so we extend into 2 nd quadrant But this is still arbitrary, V could be negative; all four quadrants are possible absorbing power releasing power

W. G. Oldham EECS 40 Fall 2001 Lecture 5 Copyright Regents of University of California 6 CURRENT-VOLTAGE CHARACTERISTICS OF VOLTAGE & CURRENT SOURCES (con’t) V i absorbing power  releasing power  Remember the voltage across the current source can be any finite value (not just zero) If i is positive then we are confined to quadrants 4 and 1: And do not forget i can be positive or negative. Thus we can be in any quadrant. +v_+v_

W. G. Oldham EECS 40 Fall 2001 Lecture 5 Copyright Regents of University of California 7 RESISTOR If we use associated current and voltage (i.e., i is defined as into + terminal), then v = iR (Ohm’s law) Question: What is the I-V characteristic for a 1K  resistor? Slope = 1/R  Answer: V = 0  I = 0 V = 1V  I = 1 mA V = 2V  I = 2 mA etc Note that all wires and circuit connections have resistance, though we will most often approximate it to be zero.

W. G. Oldham EECS 40 Fall 2001 Lecture 5 Copyright Regents of University of California 8 CAPACITOR Any two conductors a and b separated by an insulator with a difference in voltage V ab will have an equal and opposite charge on their surfaces whose value is given by Q = CV ab, where C is the capacitance of the structure, and the + charge is on the more positive electrode. We learned about the parallel-plate capacitor in physics. If the area of the plate is A, the separation d, and the dielectric constant of the insulator is , the capacitance equals C = A  /d. Symbol But so where we use the associated reference directions. Constitutive relationship: Q = C (V a  V b ). (Q is positive on plate a if V a > V b )

W. G. Oldham EECS 40 Fall 2001 Lecture 5 Copyright Regents of University of California 9 ENERGY STORED IN A CAPACITOR You might think the energy (in Joules) is QV, which has the dimension of joules. But during charging the average voltage was only half the final value of V. Thus, energy is.

W. G. Oldham EECS 40 Fall 2001 Lecture 5 Copyright Regents of University of California 10 ENERGY STORED IN A CAPACITOR (cont.) More rigorous derivation: During charging, the power flow is v  i into the capacitor, where i is into + terminal. We integrate the power from t = 0 (v = 0) to t = end (v = V). The integrated power is the energy but dq = C dv. (We are using small q instead of Q to remind us that it is time varying. Most texts use Q.)

W. G. Oldham EECS 40 Fall 2001 Lecture 5 Copyright Regents of University of California 11 INDUCTORS Inductors are the dual of capacitors – they store energy in magnetic fields that are proportional to current. Switching properties: Just as capacitors demand v be continuous (no jumps in V), inductors demand i be continuous (no jumps in i ). Reason? In both cases the continuity follows from non-infinite, i.e., finite, power flow.

W. G. Oldham EECS 40 Fall 2001 Lecture 5 Copyright Regents of University of California 12 SWITCHING PROPERTIES OF L, C Rule: The voltage across a capacitor must be continuous and differentiable. For an inductor the current must be continuous and differentiable. Basis: The energy cannot “jump” because that would require infinite energy flow (power) and of course neither the current or voltage can be infinite. For a capacitor this demands that V be continuous (no jumps in V); for an inductor it demands i be continuous (no jumps in i ).

W. G. Oldham EECS 40 Fall 2001 Lecture 5 Copyright Regents of University of California 13 The RC Circuit to Study ( All single-capacitor circuits reduce to this one) R represents total resistance (wire plus whatever drives the input) C represents the total capacitance from node to the outside world (from devices, nearby wires, ground etc)

W. G. Oldham EECS 40 Fall 2001 Lecture 5 Copyright Regents of University of California 14 RC RESPONSE EXAMPLE Case 1 Capacitor uncharged: Apply + voltage step of 2V Input nodeOutput node ground R C Vin Vout + - time V in 0 0 The capacitor voltage, Vout, is initially zero and rises in response to the step in input. Its initial value, just after the step, is still zero WHY? Its final value, long after the step is is 5V WHY? 5 time V out We need to find the solution during the transient. Time period during which the output changes: transient

W. G. Oldham EECS 40 Fall 2001 Lecture 5 Copyright Regents of University of California 15 RC RESPONSE Vin “jumps” at t=0, but Vout cannot “jump” like Vin. Why not? Case 1 – Rising voltage. Capacitor uncharged: Apply + voltage step  Because an instantaneous change in a capacitor voltage would require instantaneous increase in energy stored (1/2CV²), that is, infinite power. (Mathematically, V must be differentiable: I=CdV/dt) Input nodeOutput node ground R C Vin Vout + - V does not “jump” at t=0, i.e. V(t=0 + ) = V(t=0 - ) time Vin 0 0 V1 Vout Therefore the dc solution before the transient tells us the capacitor voltage at the beginning of the transient.

W. G. Oldham EECS 40 Fall 2001 Lecture 5 Copyright Regents of University of California 16 RC RESPONSE Case 1 – Capacitor uncharged: Apply voltage step After the transient is over (nothing changing anymore) it means d(V)/dt = 0 ; that is all currents must be zero. From Ohm’s law, the voltage across R must be zero, i.e. Vin = Vout. Vout approaches its final value asymptotically (It never quite gets to V1, but it gets arbitrarily close). Why?  That is, Vout  V1 as t  . (Asymptotic behavior) time Vin 0 0 V1 Vout Input nodeOutput node ground R C Vin Vout + - Again the dc solution (after the transient) tells us (the asymptotic limit of) the capacitor voltage during the transient.

W. G. Oldham EECS 40 Fall 2001 Lecture 5 Copyright Regents of University of California 17 RC RESPONSE: Case 1 (cont.) Exact form of Vout? Equation for Vout: Do you remember general form? ? Exponential ! Vout Input nodeOutput node ground R C Vin Vout + - V out = V 1 (1-e -t/  ) time Vout 0 0 V1 

W. G. Oldham EECS 40 Fall 2001 Lecture 5 Copyright Regents of University of California 18 Review of simple exponentials. Rising Exponential from Zero Falling Exponential to Zero at t = 0, V out = 0, and at t ,V out  V 1 also at t = , V out = 0.63 V 1 8 at t = 0, V out = V 1, and at t ,V out  0, also at t = , V out = 0.37 V 1 8 time V out 0 0 V1V1 .63V 1 V out V1V1 time 0 0 .37V 1 V out = V 1 (1-e -t/  ) V out = V 1 e -t/ 

W. G. Oldham EECS 40 Fall 2001 Lecture 5 Copyright Regents of University of California 19 Further Review of simple exponentials. Rising Exponential from Zero Falling Exponential to Zero time V out 0 0 V 1 + V 2 .37V 1 + V 2 V2V2 V out = V 1 (1-e -t/  ) V out = V 1 e -t/  V out = V 1 (1-e -t/  ) + V 2 V out = V 1 e -t/  + V 2.63V 1 + V 2 V out 0 V 1 + V 2 time 0  V2V2 We can add a constant (positive or negative)

W. G. Oldham EECS 40 Fall 2001 Lecture 5 Copyright Regents of University of California 20 Further Review of simple exponentials. Rising Exponential Falling Exponential Both equations can be written in one simple form: Thus: if B 0, falling exponential Initial value (t=0) : V ou t = A + B. Final value (t>>  ): V ou t = A time V out 0 0 A+B A Here B > 0 V out 0 A time 0 A+B Here B < 0 V out = V 1 (1-e -t/  ) + V 2 V out = V 1 e -t/  + V 2 V out = A + Be -t/ 

W. G. Oldham EECS 40 Fall 2001 Lecture 5 Copyright Regents of University of California 21 RC RESPONSE: Proof RC is “time constant”  ! Large R or large C => slow rise and fall

W. G. Oldham EECS 40 Fall 2001 Lecture 5 Copyright Regents of University of California 22 RC RESPONSE: Proof V out (t) = V in + [V out (t=0)-V in ]e -t/RC We know Is a solution? Substitute:

W. G. Oldham EECS 40 Fall 2001 Lecture 5 Copyright Regents of University of California 23 Example Input nodeOutput node ground R C Vin Vout + - The simple RC circuit with a step input has a universal exponential solution of the form: Example: R = 1K, C = 1pF, Vin steps from zero to 10V at t=0: time Vin Vout 1nsec 6.3V 1)Initial value of Vout is 0 2)Final value of Vout is 10V 3)Time constant is sec 4)Vout reaches 0.63 X 10 in sec V out = A + Be -t/RC V out = e -t/1nsec DRAW THE GRAPH Now just write the equation:

W. G. Oldham EECS 40 Fall 2001 Lecture 5 Copyright Regents of University of California 24 Charging and discharging in RC Circuits (The official EE40 Easy Method ) Input nodeOutput node ground R C Vin Vout + - 4) Solve the dc problem for the capacitor voltage after the transient is over. This is the asymptotic value. 5) Sketch the Transient. It is 63% complete after one time constant. 6) Write the equation by inspection. 1) 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: Method of solving for any node voltage in a single capacitor circuit. 2) The time constant of the transient is  = RC. 3) Solve the dc problem for the capacitor voltage before the transient. This is the starting value (initial value) for the transient voltage.