1. CAPACITANCE AND INDUCTANCE Introduces two passive, energy storing devices: Capacitors and Inductors LEARNING GOALS CAPACITORS Store energy in their electric field (electrostatic energy) Model as circuit element INDUCTORS Store energy in their magnetic field Model as circuit element CAPACITOR AND INDUCTOR COMBINATIONS Series/parallel combinations of elements RC OP-AMP CIRCUITS Integration and differentiation circuits

2. CAPACITORS First of the energy storage devices to be discussed Basic parallel-plates capacitor CIRCUIT REPRESENTATION NOTICE USE OF PASSIVE SIGN CONVENTION Typical Capacitors

3. Normal values of capacitance are small. Microfarads is common. For integrated circuits nano or pico farads are not unusual PLATE SIZE FOR EQUIVALENT AIR-GAP CAPACITOR

4. Basic capacitance lawLinear capacitors obey Coulomb’s law C is called the CAPACITANCE of the device and has units of One Farad(F)is the capacitance of a device that can store one Coulomb of charge at one Volt. EXAMPLE Voltage across a capacitor of 2 micro Farads holding 10mC of charge V Capacitance in Farads, charge in Coulombs result in voltage in Volts Capacitors can be dangerous!!! Linear capacitor circuit representation

5. The capacitor is a passive element and follows the passive sign convention Capacitors only store and release ELECTROSTATIC energy. They do not “create” Linear capacitor circuit representation LEARNING BY DOING

6. If the voltage varies the charge varies and there is a displacement current Capacitance Law One can also express the voltage across in terms of the current Integral form of Capacitance law … Or one can express the current through in terms of the voltage across Differential form of Capacitance law The mathematical implication of the integral form is... Voltage across a capacitor MUST be continuous Implications of differential form?? DC or steady state behavior A capacitor in steady state acts as an OPEN CIRCUIT

7. LEARNING EXAMPLE CAPACITOR AS CIRCUIT ELEMENT The fact that the voltage is defined through an integral has important implications... Ohm’s Law

8. CAPACITOR AS ENERGY STORAGE DEVICE Instantaneous power Energy is the integral of power If t1 is minus infinity we talk about “energy stored at time t2.” If both limits are infinity then we talk about the “total energy stored.” W

9. Energy stored in 0 - 6 msec Charge stored at 3msec “total energy stored?”.... “total charge stored?”... If charge is in Coulombs and capacitance in Farads then the energy is in …. LEARNING EXAMPLE

12. FIND THE ENERGY

13. LEARNING EXTENSION

14. Energy stored at a given time t J Charge stored at a given time C Current through the capacitor A Electric power supplied to capacitor at a given time Energy stored over a given time interval W J WHAT VARIABLES CAN BE COMPUTED? SAMPLE PROBLEM

15. Current through capacitor Voltage at a given time t Voltage at a given time t when voltage at time to<t is also known V Charge at a given time C Voltage as a function of time V Electric power supplied to capacitor Energy stored in capacitor at a given time W J “Total” energy stored in the capacitor J SAMPLE PROBLEM If the current is known...

16. SAMPLE PROBLEM Compute voltage as a function of time At minus infinity everything is zero. Since current is zero for t<0 we have In particular Charge stored at 5ms Total energy stored Total means at infinity. Hence Before looking into a formal way to describe the current we will look at additional questions that can be answered. Now, for a formal way to represent piecewise functions.... Given current and capacitance

17. Formal description of a piecewise analytical signal

18. Flux lines may extend beyond inductor creating stray inductance effects A TIME VARYING FLUX CREATES A COUNTER EMF AND CAUSES A VOLTAGE TO APPEAR AT THE TERMINALS OF THE DEVICE INDUCTORS NOTICE USE OF PASSIVE SIGN CONVENTION Circuit representation for an inductor

19. A TIME VARYING MAGNETIC FLUX INDUCES A VOLTAGE Induction law INDUCTORS STORE ELECTROMAGNETIC ENERGY. THEY MAY SUPPLY STORED ENERGY BACK TO THE CIRCUIT BUT THEY CANNOT CREATE ENERGY. THEY MUST ABIDE BY THE PASSIVE SIGN CONVENTION FOR A LINEAR INDUCTOR THE FLUX IS PROPORTIONAL TO THE CURRENT DIFFERENTIAL FORM OF INDUCTION LAW THE PROPORTIONALITY CONSTANT, L, IS CALLED THE INDUCTANCE OF THE COMPONENT INDUCTANCE IS MEASURED IN UNITS OF henry (H). DIMENSIONALLY LEARNING by Doing Follow passive sign convention

20. Differential form of induction law Integral form of induction law A direct consequence of integral form Current MUST be continuous A direct consequence of differential formDC (steady state) behavior Power and Energy stored W Energy stored on the interval Can be positive or negative “Energy stored at time t” Must be non-negative. Passive element!!! J Current in Amps, Inductance in Henrys yield energy in Joules

21. LEARNING EXAMPLE FIND THE TOTLA ENERGY STORED IN THE CIRCUIT In steady state inductors act as short circuits and capacitors act as open circuits

22. L=10mH. FIND THE VOLTAGE THE DERIVATIVE OF A STRAIGHT LINE IS ITS SLOPE ENERGY STORED BETWEEN 2 AND 4 ms J THE VALUE IS NEGATIVE BECAUSE THE INDUCTOR IS SUPPLYING ENERGY PREVIOUSLY STORED LEARNING EXAMPLE

23. L=0.1H, i(0)=2A. Find i(t), t>0 Initial energy stored in inductor “Total energy stored in the inductor” Energy stored between 0 and 2 sec ENERGY COMPUTATIONS SAMPLE PROBLEM Energy stored on the interval Can be positive or negative

24. FIND THE VOLTAGE ACROSS AND THE ENERGY STORED (AS FUNCTION OF TIME) FOR ENERGY STORED IN THE INDUCTOR NOTICE THAT ENERGY STORED AT ANY GIVEN TIME IS NON NEGATIVE -THIS IS A PASSIVE ELEMENT- LEARNING EXAMPLE

26. L=200mH FIND THE CURRENT LEARNING EXAMPLE

27. ENERGYPOWER L=200mH NOTICE HOW POWER CHANGES SIGN ENERGY IS NEVER NEGATIVE. THE DEVICE IS PASSIVE FIND THE POWER FIND THE ENERGY

28. L=5mH FIND THE VOLTAGE LEARNING EXTENSION

29. CAPACITOR SPECIFICATIONS LEARNING EXAMPLE GIVEN THE VOLTAGE WAVEFORM DETERMINE THE VARIATIONS IN CURRENT

30. CURRENT WAVEFORM INDUCTOR SPECIFICATIONS LEARNING EXAMPLE GIVEN THE CURRENT WAVEFORM DETERMINE THE VARIATIONS IN VOLTAGE

32. IDEAL AND PRACTICAL ELEMENTS IDEAL ELEMENTS CAPACITOR/INDUCTOR MODELS INCLUDING LEAKAGE RESISTANCE MODEL FOR “LEAKY” CAPACITOR MODEL FOR “LEAKY” INDUCTORS

33. SERIES CAPACITORS NOTICE SIMILARITY WITH RESITORS IN PARALLEL Series Combination of two capacitors

34. ALGEBRAIC SUM OF INITIAL VOLTAGES POLARITY IS DICTATED BY THE REFERENCE DIRECTION FOR THE VOLTAGE LEARNING EXAMPLE DETERMINE EQUIVALENT CAPACITANCE AND THE INITIAL VOLTAGE OR WE CAN REDUCE TWO AT A TIME

35. SAME CURRENT. CONNECTED FOR THE SAME TIME PERIOD SAME CHARGE ON BOTH CAPACITORS LEARNING EXAMPLE Two uncharged capacitors are connected as shown. Find the unknown capacitance

36. PARALLEL CAPACITORS LEARNING EXAMPLE

37. LEARNING EXTENSION

38. FIND EQUIVALENT CAPACITANCE SAMPLE PROBLEM

39. IF ALL CAPACITORS HAVE THE SAME CAPACITANCE VALUE C DETERMINE THE VARIOUS EQUIVALENT CAPACITANCES SAMPLE PROBLEM

40. All capacitors are equal with C=8 microFarads Examples of interconnections

41. SERIES INDUCTORS LEARNING EXAMPLE

42. PARALLEL INDUCTORS INDUCTORS COMBINE LIKE RESISTORS CAPACITORS COMBINE LIKE CONDUCTANCES LEARNING EXAMPLE

43. LEARNING EXTENSION WHEN IN DOUBT… REDRAW! IDENTIFY ALL NODES PLACE NODES IN CHOSEN LOCATIONS CONNECT COMPONENTS BETWEEN NODES ALL INDUCTORS ARE 4mH

44. ALL INDUCTORS ARE 6mH NODES CAN HAVE COMPLICATED SHAPES. KEEP IN MIND DIFFERENCE BETWEEN PHYSICAL LAYOUT AND ELECTRICAL CONNECTIONS SELECTED LAYOUT LEARNING EXTENSION

45. L-C

46. RC OPERATIONAL AMPLIFIER CIRCUITS INTRODUCES TWO VERY IMPORTANT PRACTICAL CIRCUITS BASED ON OPERATIONAL AMPLIFIERS THE IDEAL OP-AMP

47. RC OPERATIONAL AMPLIFIER CIRCUITS -THE INTEGRATOR IDEAL OP-AMP ASSUMPTIONS

48. RC OPERATIONAL AMPLIFIER CIRCUITS - THE DIFFERENTIATOR IDEAL OP-AMP ASSUMPTIONS KVL DIFFERENTIATE IF R1 COULD BE SET TO ZERO WE WOULD HAVE AN IDEAL DIFFERENTIATOR. IN PRACTICE AN IDEAL DIFFERENTIATOR AMPLIFIES ELECTRIC NOISE AND DOES NOT OPERATE. THE RESISTOR INTRODUCES A FILTERING ACTION. ITS VALUE IS KEPT AS SMALL AS POSSIBLE TO APPROXIMATE A DIFFERENTIATOR

49. ABOUT ELECTRIC NOISE ALL ELECTRICAL SIGNALS ARE CORRUPTED BY EXTERNAL, UNCONTROLLABLE AND OFTEN UNMEASURABLE, SIGNALS. THESE UNDESIRED SIGNALS ARE REFERRED TO AS NOISE THE DERIVATIVESIMPLE MODEL FOR A NOISY 60Hz SINUSOID CORRUPTED WITH ONE MICROVOLT OF 1GHz INTERFERENCE. noise signal noisesignal

50. LEARNING EXTENSION

52. USING GROUND WIRE TO REDUCE CROSSTALK APPLICATION EXAMPLE CROSS-TALK IN INTEGRATED CIRCUITS REDUCE CROSSTALK BY Reducing C12 Increasing C2 SMALLER COST? EXTRA SPACE BY GROUND WIRE Simplified Model

53. LEARNING EXAMPLE SIMPLE CIRCUIT MODEL FOR DYNAMIC RANDOM ACCESS MEMORY CELL (DRAM) REPRESENTS CHARGE LEAKAGE FROM CELL CAPACITOR NOTICE THE VALUES OF THE CAPACITANCES THE ANALYSIS OF THE READ OPERATION GIVES FURTHER INSIGHT ON THE REQUIREMENTS SWITCHED CAPACITOR CIRCUIT CELL AT THE BEGINNING OF A MEMORY READ OPERATION

54. CELL READ OPERATION IF SWITCH IS CLOSED BOTH CAPACITORS MUST HAVE THE SAME VOLTAGE ASSUMING NO LOSS OF CHARGE THEN THE CHARGE BEFORE CLOSING MUST BE EQUAL TO CHARGE AFTER CLOSING Even at full charge the voltage variation is small. SENSOR amplifiers are required After a READ operation the cell must be refreshed

55. LEARNING EXAMPLE IC WITH WIREBONDS TO THE OUTSIDE FLIP CHIP MOUNTING GOAL: REDUCE INDUCTANCE IN THE WIRING AND REDUCE THE “GROUND BOUNCE” EFFECT A SIMPLE MODEL CAN BE USED TO DESCRIBE GROUND BOUNCE

56. MODELING THE GROUND BOUNCE EFFECT IF ALL GATES IN A CHIP ARE CONNECTED TO A SINGLE GROUND THE CURRENT CAN BE QUITE HIGH AND THE BOUNCE MAY BECOME UNACCEPTABLE USE SEVERAL GROUND CONNECTIONS (BALLS) AND ALLOCATE A FRACTION OF THE GATES TO EACH BALL

57. LEARNING BY DESIGNPOWER OUTAGE “RIDE THROUGH” CIRCUIT CAPACITOR MUST MAINTAIN AT LEAST 2.4V FOR AT LEAST 1SEC. DESIGN EQUATION http://www.wiley.com/college/irwin/0470128690/animations/swf/6-21.swf

58. Proposed solution DESIGN EXAMPLE DESIGN AN OP-AMP CIRCUIT TO REALIZE THE EQUATION Needs integrator And adder adder Design equationsTwo equations in five unknowns!! Not too large. Not too small Seems a reasonable value If supply voltages are 10V (or less) all currents will be less than 1mA, which seems reasonable ANALYSIS OF THE SOLUTION integrator