Lecture #4 Capacitors; I Vs. V • Capacitors and Energy Stored • Current Versus Voltage for Circuits • Equivalent Straight

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EECS 42 Intro. Digital Electronics, Fall 2003 Lecture 4: 09/04/03 A.R. Neureuther EECS 42 Intro. Digital Electronics, Fall 2003 Lecture 4: 09/04/03 A.R. Neureuther Version Date 09/01/03 Version Date 09/01/03 EECS 42 Introduction to Electronics for BASIC CIRCUIT ELEMENTS Computer Science Andrew R. Neureuther (always supplies some constant given • Voltage Source voltage - like ideal battery) Lecture #4 Capacitors; I vs. V • Current Source (always supplies some constant given current) • Capacitors and Energy Stored • Resistor (Ohm’s law) • Current versus Voltage for Circuits • Wire (“short” – no voltage drop ) • Capacitor (capacitor law – based on energy storage in • Equivalent Straight Line Circuits electric field of a dielectric S&O 5.1-5.2) • Thevenin; Norton Schwarz and Oldham: 5.1 – 5.2; 3.1 Lecture #4 • Inductor (inductor law – based on energy storage in http://inst.EECS.Berkeley.EDU/~ee42/ magnetic field in space S&O 5.1-5.2)

EECS 42 Intro. Digital Electronics, Fall 2003 Lecture 4: 09/04/03 A.R. Neureuther EECS 42 Intro. Digital Electronics, Fall 2003 Lecture 4: 09/04/03 A.R. Neureuther Version Date 09/01/03 Version Date 09/01/03 CAPACITOR CIRCUIT ELEMENT CAPACITOR CURRENT/VOLTAGE Any two conductors a and b separated by an insulator with a difference in voltage V will have an equal and opposite charge on their surfaces whose ab i a value is given by Q = CVab, where C is the capacitance of the structure, and + the + charge is on the more positive electrode. Constitutive relationship: Q = C (Va - Vb). Vab (Q is positive on plate a if Va > Vb) A simple parallel-plate capacitor is b - shown. If the area of the plate is A, A d the separation d, and the dielectric Charge and Current are related: Charge is the time integral of Current constant of the insulator is e, the capacitance equals C = A e/d. dQ dv So i = a so i = C equivalent to Q = C v Symbol or i a dt dt + where we use the associated reference directions. Constitutive relationship: Q = C (Va - Vb). Vab (Q is positive on plate a if Va > Vb) Example: A 1 mA current into a 1 pF capacitance b - produces a voltage change rate of 1mA/1pF = 1 V/ns

EECS 42 Intro. Digital Electronics, Fall 2003 Lecture 4: 09/04/03 A.R. Neureuther EECS 42 Intro. Digital Electronics, Fall 2003 Lecture 4: 09/04/03 A.R. Neureuther Version Date 09/01/03 Version Date 09/01/03 ENERGY STORED IN A CAPACITOR ENERGY STORED IN A CAPACITOR (cont.) More rigorous derivation: During charging, the power flow is You might think the energy (in Joules) is QV, which has the dimension of v × i into the capacitor, where i is into + terminal. We joules. But during charging the average voltage was only half the final integrate the power from t = 0 (v = 0) to t = end (v = V). The i value of V. integrated power is the energy + t = t Final v = VFinal dq v = VFinal v E = v ×i dt = v dt = v dq ò ò ò - 1 1 2 dt Thus, energy is QV = CV . t = t Initial v = VInitial v = VInitial 2 2 but dq = C dv. (We are using small q instead of Q to remind us that it is time varying . Most texts use Q.)

Example: A 1 pF capacitance charged to 5 volts has v = V Final 1 2 1 2 ½(5V)2 (1pF) = 12.5 pJ E = ò Cv dv = CV - CV 2 Final 2 Initial v = VInitial

1 EECS 42 Intro. Digital Electronics, Fall 2003 Lecture 4: 09/04/03 A.R. Neureuther EECS 42 Intro. Digital Electronics, Fall 2003 Lecture 4: 09/04/03 A.R. Neureuther Version Date 09/01/03 Version Date 09/01/03 CAPACITORS IN SERIES CAPACITORS IN PARALLEL + - + - V1 + V2 - Veq | ( | ( | ( Add Currents C1 C2 Equivalent to Ce + i(t) i(t) | ( | ( q i(t) C1 C2 V dV dV i(t) = C1 + C2 Add Voltages - dt dt Equivalent capacitance defined by

dVeq d(V1 + V2 ) Veq = V1 + V2 and i = Ceq = Ceq Equivalent capacitance defined by | ( + dV1 dV2 dt dt i(t) C V(t) i = C1 = C2 dV eq - dt dt i = Ceq dV 1 1 i dt dV i dV i so eq = i( + ) º So 1 = , 2 = , dt C1 C2 Ceq dt C1 dt C2

Clearly, 1 C1C2 Ceq = C1 + C2 CAPACITORS IN PARALLEL Clearly, C = = CAPACITORS IN SERIES eq 1 1 + C1 + C2 Example: 1 pF and 2pF in parallel = 3 pF C1 C2 Example: 1 pF and 2pF in series = 0.67 pF

a • Capacitors: two plate example; Store energy in the electric

Vab field Q = CV, I = C dV/dt and V = (1/C) integral of voltage 1 mA current charging 1 pF V(t) = (I/C)t = (10-3 A/10-12 F) t = 109 V/s t iL • Energy = ½(CV2) (because average voltage is V/2) • At D.C. time derivatives are zero => C is open circuit b • C in parallel add; series 1/C = sum (1/Ci); short together Constitutive relationship Magnetic Flux = Li L (infinite current but conserve charge) Time-rate of change of flux = Voltage • Inductors: coil example; Store energy in the magnetic field; Flux = LI, V = L dI/dt and I = (1/L) (integral of voltage) V = L di /dt ab L In EE42 we do not • At D.C. time derivatives are zero => L is short circuit analyze circuits with Energy = ½(L(i )2) L inductance due to • L in parallel 1/L = sum (1/Li); series add; connect in series the time limitations when have different currents => L1I1+L2I2 = (L1+L2)INEW Not used in EE 42

EECS 42 Intro. Digital Electronics, Fall 2003 Lecture 4: 09/04/03 A.R. Neureuther EECS 42 Intro. Digital Electronics, Fall 2003 Lecture 4: 09/04/03 A.R. Neureuther EXAMPLE of I vs V GRAPH Version Date 09/01/03 EXAMPLE of I vs V GRAPH Version Date 09/01/03 Resistors in Series Simple Circuit, e.g. voltage source + resistor. If two resistors are in series the current is the same; clearly the total If two circuit elements are in series the current is the same; clearly the voltage will be the sum of the two IR values i.e. I(R +R ). 1 2 total voltage will be the sum of the voltages i.e. VS + IR.

Thus the equivalent resistance is R1+R2 and the I-V graph of the series We can graph this on the I-V plane. We find the I-V graph of the pair is the same as that of the equivalent resistance. combination by adding the voltages VS and I R at each current I. Of course we can also find the I-V graph of the combination by adding Lets do an example for =2V, R=2K the voltages directly on the I-V axes. Lets do an example for 1K + 4K resistors I (ma) At 1 mA I (ma) Combined 1K 1V + 4V = 5V + 2V 2K + 2V + + 2VVS + 1K V 4 Combined - 4 - 1 I + V 1K + 4K Resistors in series I V 2K V 4K 2 - add voltages 2KI - - 2 4K 2 Resistors in parallel At 2 mA add currents 2V + 4V = 6V V (Volt) V (Volt) 5 5

2 EECS 42 Intro. Digital Electronics, Fall 2003 Lecture 4: 09/04/03 A.R. Neureuther EECS 42 Intro. Digital Electronics, Fall 2003 Lecture 4: 09/04/03 A.R. Neureuther Version Date 09/01/03 Version Date 09/01/03 GRAPHICAL EQUIVALENT CIRCUIT SIMPLEST EQUIVALENT CIRCUITS A (ma) Combined An adequately I I = 1 mA Short Circuit : VOUT = 0 I IN SS II R1 Circuit equivalent circuit is B R = 1kW IIN = -ISS = - 1 mA N I 1 SS 4 one that has an I vs. V R2 = 6 k W VOUT Open Circuit: I = 0 R R IN graph that is identical 2 3 R3 = 3 kW Note R1||R2 = 2 kW 2 to that of the original C VOUT = ISS R1||R2 circuit. Combined Circuit I 5 V I I (ma) = 1 mA 2kW = 2V SC V OUT OC (Volt) IIN IN VOUT RN Third Point I R 4 Th IN = -ISC = - ( -1 mA) = 1mA IIN = 1 mA + Norton VOUT KCL => I2||3 = ISS + IIN VTh - 2 VOUT = I2||3 R1||R2 V = V = 2V TH OC R is the inverse of the slope = 2 mA 2kW = 4V Thevenin TH Key: It will always be the RTH = RN = VOC / (-ISC) = 2V /(- ( -1 mA)) = 2 kW 5 V (Volt) ISC OUT case that for linear circuit VOC elements the I vs. V is a RTH = R1||R2 straight line. Copyright 2003, Regents of University of California Copyright 2003, Regents of University of California

RTH = RN SHORTCUT METHOS I vs. V and Equivalent Circuits A Look at algebraic relation for the example circuit. • I vs. V for ideal voltage source is a vertical line at V = VSV R1 B ISS • I vs. V for ideal current source is a horizontal line at I = ISC VOC = ISS R1||R2

VOUT R2 R3 • I vs. V for a circuit made up of ideal independent sources and ISC = -ISS C resistors is a straight line.

RTH = RN = VOC / (-ISC) • The simplest circuit for a straight line is an ideal voltage source and a resistor (Thevenin) or a current source and a parallel RTH = RN = (ISS R1||R2)/(-(-ISS)) = R1||R2 resistor (Norton) • The easiest way to find the I vs. V line is to find the intercepts In General turn all of the independent sources to where I = 0 (open circuit voltage V ) and where V = 0 (Short zero and find the remaining equivalent resistance T circuit current I ) seen looking into the terminals. N -1 • The short-cut for finding the (slope) = RT = RN is to turn off Currents sources are turned to zero current all of the independent sources to zero and find the remaining (with any voltage) and voltage sources are turned to zero voltage (with any current). equivalent resistance between the terminals of the elements. Copyright 2003, Regents of University of California Copyright 2003, Regents of University of California