da Vinci Technologies EE Pro User Manual Download Page 127

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EE Pro for TI-89, 92 Plus
Equations – RLC Circuits

0 2

⋅ + ⋅

−

Eq. 22.6.6

0 1

−

⋅ + ⋅

−

Eq. 22.6.7

Example 22.6 -

An overdamped RLC circuit consists of a 10

Ω

resistor in series with a 40 mH inductor and a 1

F capacitor. If the initial inductor current is 0 mA and the capacitor is charged to a potential of 5 V, find the

voltage across the capacitor after 1 ms.

Calculated Results (Upper display)

Calculated Results (Lower display)

Solution -

All of the equations need to be selected to solve this problem. Press

„

and enter the known

variables followed by a second press of

„

to solve for the unknowns. The solver takes about five minutes

to solve this problem.

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Summary of Contents for EE Pro

Page 1: ...ional Series EE Pro UQHVYCTG RRNKECVKQP 1P 6 CPF 6 2NWU User s Guide June 1999 da Vinci Technologies Group Inc Rev 1 0 da Vinci Technologies Group Inc 1600 S W Western Blvd Suite 250 Corvallis OR 97333 www dvtg com ...

Page 2: ...with the furnishing performance or use of this manual or the examples herein Copyright da Vinci Technologies Group Inc 1999 All rights reserved PocketProfessional and EE Pro is a registered trademarks of da Vinci Technologies Group Inc We welcome your comments on the software and the manual Forward your comments preferably by e mail to da Vinci at support dvtg com Acknowledgements The EE Pro softw...

Page 3: ...Analysis 3 2 4 Special Function Keys in Analysis Routines 4 2 5 Data Fields Analysis Functions and Sample Problems 5 Example 2 1 7 Example 2 2 7 Example 2 3 8 2 6 Session Folders Variable Names 9 3 AC Circuits 10 3 1 Impedance Calculations 10 Example 3 1 11 3 2 Voltage Divider 11 Example 3 2 11 3 3 Current Divider 12 Example 3 3 12 3 4 Circuit Performance 13 Example 3 4 14 4 Polyphase Circuits 15 ...

Page 4: ...e 36 Example 9 2 37 9 3 Connected Two Ports 37 Example 9 3 38 10 Transformer Calculations 39 10 1 Open Circuit Test 39 Example 10 1 39 10 2 Short Circuit Test 40 Example 10 2 40 10 3 Chain Parameters 41 Example 10 3 42 11 Transmission Lines 43 11 1 Line Properties 43 Example 11 1 44 11 2 Line Parameters 44 Example 11 2 45 11 3 Fault Location Estimate 45 Example 11 3 46 11 4 Stub Impedance Matching...

Page 5: ...ion of session folders 5 15 9 solve nsolve and csolve and user defined functions UDF 6 15 10 Entering a guess value for the unknown using nsolve 6 15 11 Why can t I compute a solution 6 15 12 Care in choosing a consistent set of equations 7 15 13 Notes for the advanced user in troubleshooting calculations 7 16 Resistive Circuits 9 Variables 9 16 1 Resistance Formulas 9 Example 16 1 10 16 2 Ohm s L...

Page 6: ...27 Example 19 1 28 19 2 Thermionic Emission 29 Example 19 2 29 19 3 Photoemission 29 Example 19 3 29 20 Meters and Bridge Circuits 31 Variables 31 20 1 A V Ω Meters 32 Example 20 1 32 20 2 Wheatstone Bridge 33 Example 20 2 33 20 3 Wien Bridge 34 Example 20 3 34 20 4 Maxwell Bridge 35 Example 20 4 35 20 5 Owen Bridge 35 Example 20 5 36 20 6 Symmetrical Resistive Attenuator 36 Example 20 6 37 20 7 U...

Page 7: ... 53 Variables 53 23 1 RL Series Impedance 54 Example 23 1 54 23 2 RC Series Impedance 55 Example 23 2 55 23 3 Impedance Admittance 56 Example 23 3 56 23 4 Two Impedances in Series 56 Example 23 4 57 23 5 Two Impedances in Parallel 57 Example 23 5 58 24 Polyphase Circuits 59 Variables 59 24 1 Balanced Network 59 Example 24 1 60 24 2 Balanced Wye Network 60 Example 24 2 60 24 3 Power Measurements 61...

Page 8: ... 82 Example 27 2 1 83 Example 27 2 2 84 27 3 PN Junction Currents 84 Example 27 3 85 27 4 Transistor Currents 86 Example 27 4 87 27 5 Ebers Moll Equation 87 Example 27 5 88 27 6 Ideal Currents pnp 88 Example 27 6 89 27 7 Switching Transients 89 Example 27 7 90 27 8 MOS Transistor I 91 Example 27 8 91 27 9 MOS Transistor II 92 Example 27 9 93 27 10 MOS Inverter Resistive Load 93 Example 27 10 94 27...

Page 9: ...fier 114 Example 29 1 115 29 2 Power Transistor 115 Example 29 2 116 29 3 Push Pull Principle 117 Example 29 3 117 29 4 Class B Amplifier 118 Example 29 4 119 29 5 Class C Amplifier 119 Example 29 5 120 30 Transformers 121 Variables 121 30 1 Ideal Transformer 121 Example 30 1 122 30 2 Linear Equivalent Circuit 122 Example 30 2 123 31 Motors and Generators 124 Variables 124 31 1 Energy Conversion 1...

Page 10: ...ng the Reference Section 1 32 3 Reference Screens 2 32 4 Viewing Reference Tables 3 33 Resistor Color Chart 5 33 1 Using the Resistor Color Chart 6 Example 33 1 6 34 Standard Component Values 7 Example 34 1 7 35 Semiconductor Data 8 36 Boolean Expressions 12 36 1 Using Boolean Expressions 12 Example 36 1 12 37 Boolean Algebra 14 37 1 Using Boolean Algebra Properties 14 Example 37 1 14 38 Transform...

Page 11: ... 2 How to contact Customer Support 6 Appendix C Bibliography 7 Appendix D TI 89 and TI 92 plus Display and Keystroke Differences 9 D 1 Display Property Differences between the TI 89 and TI 92 plus 9 D 2 Keyboard Differences Between TI 89 and TI 92 Plus 9 Appendix E Error Messages 14 E 1 General Error Messages 14 E 2 Analysis Error Messages 15 E 3 Equation Error Messages 16 E 4 Reference Error Mess...

Page 12: ...ion Summary 1 1 Key Features of EE Pro The manual is organized into three sections representing the main menu headings of EE Pro Analysis Equations Reference AC Circuits Resistive Circuits Resistor Color Chart Polyphase Circuits Capacitors and Electric Fields Standard Component Values Ladder Network Inductors and Magnetism Semiconductor Data Filter Design Electron Motion Boolean Expressions Gain a...

Page 13: ... algebraic expressions Added features are the ability to perform simple computations such as estimating standard or preferred manufacturer component values for inductors resistors and capacitors in addition to a resistor color chart guide which can compute resistance and tolerance from a resistor s color band sequence 1 2 Purchasing Downloading and Installing EE Pro The EE Pro software can only be...

Page 14: ...e Accesses the Reference section of the software Info Helpful hints on EE Pro A selection on any menu can be entered by moving the highlight bar to the item with the arrow key D and pressing alternatively the number or letter of the selection can be typed in The Analysis Equation and Reference menus are organized in a directory tree of topics and sub topics The user can return to a previous level ...

Page 15: ...splays may appear slightly different due to final changes made in the software while the manual was being completed 1 9 Summary The designers of EE Pro have attempted to maintain the following features Easy to use menu based interface Computational efficiency for speed and performance Helpful hints and context sensitive information provided in the status line Advanced EE analysis routines equation...

Page 16: ...iable press and Type to show a brief description of a variable and its entry parameters A variable name cannot be entered which is identical to the variable name ie C for capacitance If a symbolic calculation using the variable name leave the entry blank Variables which accept complex entries ex 115 23 i are followed by an underscore _ ex ZZ1_ 2 1 Introduction Analysis routines have been organized...

Page 17: ...cs in AC Circuits Press z for topics in Gain and Frequency Once an Analysis topic is accessed a pop up menu lists the sub topics available for in the section For example when Two Port Networks is selected the pop up menu shows three sub topics 1 Convert Parameters 2 Circuit Performance 3 Connected Two Ports Each of these sub topics are tagged with a number 1 2 3 as shown in the screen display Thes...

Page 18: ...z Choose Input parameter type Prm 1 z11_ Enter P1 Impedance V1 I1 I2 0 Prm 2 z12_ Enter P1 Impedance V1 I2 I1 0 Prm 3 z21_ Enter P2 Impedance V2 I1 I2 0 Prm 4 z22_ Enter P2 Impedance V2 I2 I1 0 Output Type y Choose Output parameter type The status line contents change if h parameters were chosen for Input Type Prm 1 h11_ Enter P1 Impedance V1 I1 V2 0 Prm 2 h12_ Enter P1 Parameter V1 V2 I1 0 Prm 3 ...

Page 19: ... performed by pressing the 3 followed by Use 2 and O to toggle between the data entry screen and graph window Normally labeled as View This enables the information content highlighted by the cursor to be displayed using the entire screen in Pretty Print format An example of such a screen is shown in the screen displays shown This function is also duplicated by pressing the B key If there is no con...

Page 20: ...ircuit Performance as a topic displays a screen that has all the inputs and output variables Input screen for circuit Press to display a Input variables change when Impedance for Load Type Pop up menu for Load Type Admittance is selected for Load Type Use the cursor bar to highlight Load Type press to display a pop up menu for Impedance or Admittance as a load type Selecting a different load type ...

Page 21: ...isplay of Bode Transfer Function Transfer Function Diagram of Gain Function Hs The Computer Engineering section under the Analysis menu performs calculations involving numbers represented in binary decimal octal or hexadecimal formats within the constraints of parameters defined by the user In any topic of the Computer Engineering section the function key opens a dialog box allowing the user to sp...

Page 22: ...s for Zs_ and 70 89 i for load impedance ZL_ 5 Press the Solve key 6 The results of the calculations are displayed in the data screen as shown Example 2 2 Symbolic Results Find the parameters of a transmission line given the open circuit impedance is Z0_ the short circuit impedance is Z1_ distance is d1 and frequency of measurement is f1 Pull down menu for Analysis Menu for Transmission Lines Inpu...

Page 23: ...1000 10000 50000 for Poles List 5 Press the Solve key 6 The results of the calculations are displayed in the data screen as shown 7 Press N key to revert to the pop up display for Gain and Frequency and press to access Bode Diagram input screen 8 Begin entering parameters for graphing the Gain of the Transfer Function The minimum and maximum values for the horizontal axis show the default settings...

Page 24: ...Y and 6 PASTE feature All inputs and calculated results from Analysis and Equations section are saved as variable names Previously calculated or entered values for variables in a folder are replaced when equations are solved using new values for inputs Overwriting of variable values in graphing When an equation or analysis function is graphed EE Pro creates a function for the TI grapher which expr...

Page 25: ...lgebraic expression of previously defined terms which can be simplified to a numerical result upon entry such as 1 5 x 3 y where x 1 2 i and y 4 Field Descriptions Config Circuit Configuration Press and select Series or Parallel configuration by using D After choosing press to display the input screen updated for the new configuration Elements Element Combination Pressing displays the following ci...

Page 26: ...e Admittance loads in series Load Type Type of Load Press to display the choices Impedance Z or Admittance Y The choice made determines whether the third field is ZZ_ impedance chain or YY_ admittance chain is displayed Vs_ Source Voltage in V Enter a real or complex number variable name or algebraic expression of defined terms ZZ_ Impedances in Series Enter a list of real or complex numbers or al...

Page 27: ... variable name or algebraic expression of defined terms ZZ_ Impedance in ohms Enter a list of real or complex numbers or algebraic expression of defined terms YY_ Admittances in Siemens Enter a list of real or complex numbers or algebraic expressions of defined terms VL_ Load Voltage in V Returns a real complex number or algebraic expression I_ Currents in A Returns a list of real or complex numbe...

Page 28: ...ariable name or algebraic expression of defined terms YL_ Load Admittance in Siemens Enter a real or complex number variable name or algebraic expression of defined terms Field Descriptions Output Screen VL_ Load Voltage in V Returns a real complex number or algebraic expression IL_ Load Current in A Returns a real complex number or algebraic expression P Real Power in W Returns a real number or a...

Page 29: ...Upper Half Output Lower Half 1 Choose Admittance for Load Type 2 Enter the value 10 5 i for Is_ 3 Enter the value 0025 0012 i for Ys_ and 0012 0034 i for a load of YL_ 4 Press to calculate the performance parameters 5 The input and results of computation are shown above ...

Page 30: ... of defined terms ZZC_ Impedance Real or complex number variable name or algebraic Fig 4 1 Wye Network expression of defined terms ZZ1_ Y Impedance Real or complex number variable name or algebraic expression of defined terms ZZ2_ Y Impedance Real or complex number variable name or algebraic expression of defined terms ZZ3_ Y Impedance Real or complex number variable name or algebraic expression o...

Page 31: ... circuit as shown in the Fig 4 3 Input Fields Input Type V12_ Reference Voltage in V across lines 1 and 2 Enter a real or complex number variable name or algebraic expression of defined terms ZZ_ Phase Impedance in Ω Enter a real or complex number variable name or algebraic expression of defined terms Result Fields V23_ Voltage in V across lines 2 and 3 A real complex number or variable name V31_ ...

Page 32: ...age VAB_ represents the line voltage from line A to line B and is used as the reference voltage throughout this Delta network The voltages across lines B and C and C and A have the same magnitude as VAB_ but are out of phase by 120º and 120º respectively The software computes the currents IA_ IB_ and IC_ in each leg of the Delta network power dissipated in each phase P wattmeter readings WAB and W...

Page 33: ... across lines A and C A real number or algebraic expression Example 4 3 A Delta network consists of three impedances of 50 25 i with a line voltage of 110 volts across line A and B Find the line current and power measured in a two wattmeter measurement system Input Parameters Calculated Output 1 Select Balanced Delta Load 2 Enter the values 50 25 i for ZZ_ and 110_V for VAB_ 3 Press to calculate p...

Page 34: ...nd of the ladder Field Descriptions Input Screen Frequency Frequency Enter a real number or algebraic expression of defined terms Load_ Initial Load Enter a real or complex number variable name or algebraic expression of defined terms 1 Element 1 closest to the load or output N Element N furthest from the load or output Field Descriptions Element Screen Sixteen different element types are availabl...

Page 35: ...t can be added as a rung parallel or as an RC parallel circuit as a side series Choose Series or Parallel for Config and enter a value for R in ohms and C in farads Series RC Parallel RC C C R R RLC An RLC series circuit can be added as a rung parallel or as an RLC parallel circuit as a side series Choose Series or Parallel for Config and enter a value for R in ohms L in henrys and C in farads Ser...

Page 36: ...Caution be sure to enter a value for the electrical length θ0 which is consistent with the chosen frequency Short Circuited Stub Can be added only in cascade connection Specify characteristic impedance and electrical length by entering values for Z0 in ohms and θ0 in radians Caution be sure to enter a value for the electrical length θ0 which is consistent with the chosen frequency Two Port Network...

Page 37: ...ocation desired and pressing The new element will appear after the a highlighted element 4 A circuit element can be deleted from the ladder by moving the highlight bar to the element and pressing Š 5 Press to compute the overall ladder network parameters 6 Previously calculated results are not automatically updated for new element entries the user must press to re solve for the circuit parameters ...

Page 38: ...a 50 ohm resistor in series 6 Press to calculate the results displayed in the output screen as shown above 7 To delete an element from the network highlight it and press Š the delete key To calculate the ladder network parameters at a second frequency of 10 MHz 1 Press N to return to the input screen 2 In the Frequency field type 10E6 3 Press to calculate the results as displayed in the output scr...

Page 39: ...y 10 000 Hz Load 5000 ohms 2 Enter the ladder elements in the following order Capacitor Parallel 318E 12 Current Controlled I Enter 2500 ohms for RB and 100 for β Resistor Parallel 1E6 ohms Capacitor Series 0 638E 6 farads 3 Press to compute the results which are displayed in the output screen above ...

Page 40: ...hebyshev circuit elements are assumed to be ideal and are illustrated below Odd Elements Even Elements Low Pass Cn Ln High Pass Ln Cn Band Pass Lpn Cpn Lsn Csn Band Elimination Lsn Csn Lpn Cpn Fig 6 1 Chebyshev Filter Elements Field Descriptions Input Screen Char Bandpass Characteristic Press to select Low Pass High Pass Band Pass or Band Elimination R Termination Resistance in ohms Enter a real n...

Page 41: ...creen 1 Enter 50 for R 500 for f0 and 600 for f1 2 Enter 30 for dB and 3 for Ripple 3 Press to calculate the results displayed in the output screen above 6 2 Butterworth Filter This section computes the component values for Butterworth filters between equal terminations Inputs are termination resistance pass band characteristics and attenuation at some out of band frequency The basic form of the f...

Page 42: ...ment in parallel Returns a real number Element2 Second element in series Returns a real number Element3 Third element in parallel Returns a real number Element4 Fourth element in series Returns a real number ElementN nth element in series if n is odd or parallel if n is even Returns a real number Example 6 2 Design a 100 Hz wide Butterworth band pass filter centered at 800 Hz with a 30 dB attenuat...

Page 43: ...eal number or algebraic expression of defined terms C Capacitor in F Enter a real number or algebraic expression of defined terms Field Descriptions Output Screen Element1 First element Returns a real number Element2 Second element Returns a real number Element3 Third element Returns a real number Element4 Fourth element Returns a real number Element5 Fifth element Returns a real number Example 6 ...

Page 44: ...include symbolic expressions for the transfer function and its partial fraction expansion Field Descriptions Inputs Type of Input Press to select Roots or Coefficients Determines whether the third and fourth fields are Zeros and Poles or Numer and Denom Constant Constant Multiplier Enter a real number Default is 1 Zeros Numerator Roots if Roots is chosen for input type Enter an array or list of re...

Page 45: ...in and Frequency screen and select Bode Diagrams 7 2 Bode Diagrams The behavior of the transfer function as the frequency of a sinusoidal source varies is of great interest to engineers A very effective way to grasp the relationship between transfer function and frequency is to plot the magnitude and the argument of the transfer function on two separate graphs These plots are often called Bode gai...

Page 46: ...tire example for the Transfer Function section 2 In the Bode Diagram screen the Xfer field contains the Transfer Function H s _ calculated in the previous example A choice is available to the user for Graph Type Gain or Phase 3 Choose s_ for Indep The default is s_ and should rarely be changed because the Transfer Functions screen always generates transfer functions as functions of lowercase s_ 4 ...

Page 47: ...n the frequency domain using the following equation H h e k n n 0 N 1 2 j k n N π Eq 8 1 1 The variable hn is the nth element in the time domain and Hk is the kth element in the frequency domain The FFT algorithm treats the data block provided as though it is one of a periodic sequence If the underlying data is not periodic the resulting FFT created wave is subject to substantial harmonic distorti...

Page 48: ...n where Hk is the kth element in the frequency domain and hn is the nth element in the time domain Field Descriptions Freq Frequency Spectrum Enter an array or list of real or complex numbers Time Time Signal Returns time signal Example 8 2 Find spectral coefficients for the periodic time signal 1 2 3 2 1 1 Enter 1 2 3 2 1 for Freq 2 Press to calculate and display results in the time domain Time 3...

Page 49: ...ere all four variables are identified For instance a two port network characterized by z parameters is defined by the following pair of equations Fig 9 1 Two Port Network Model V I Z I Z 1 1 11 2 12 Eq 9 1 1 V I Z I Z 2 1 21 2 22 Eq 9 1 2 The four components of z parameters are defined as follows Z V I 11 1 1 with I 0 2 Z V I 12 1 2 with I 0 1 Z V I 21 2 1 with I 0 2 Z V I 22 2 2 with I 0 1 The ta...

Page 50: ...Type Press to display z y h g a or b use D to move the highlight bar to select the choice and press Note The help text e g Enter P1 Impedance V1 I1 I2 0 shows whether the ratio is an impedance admittance or dimensionless ratio which port is being described and whether it is the current or the voltage that is open Field Descriptions Output Screen 11_ Returns a real or complex number variable name o...

Page 51: ... or algebraic expression of defined terms Field Descriptions Output Screen Zin_ Input Impedance in Ω Returns a real or complex number or algebraic expression Iout_ Output Current in A Returns a real or complex number or algebraic expression Vout_ Thevenin Voltage in V Returns a real or complex number or algebraic expression Zout_ Thevenin Impedance in Ω Returns a real or complex number or algebrai...

Page 52: ...ort building blocks Assuming that Brune s criteria is valid for these networks the two port subsections can be interconnected in five ways 1 Cascade The output of network 1 is connected directly to the input of network 2 Network 1 Network 2 Cascade 2 Series Series The inputs and outputs of the two networks are both connected in series Network 1 Network 2 Series Series 3 Parallel Parallel The input...

Page 53: ...f defined terms Output Type Press to select z y h g a or b 11_ Returns a real or complex number variable name or algebraic expression of defined terms 12_ Returns a real or complex number variable name or algebraic expression of defined terms 21_ Returns a real or complex number variable name or algebraic expression of defined terms 22_ Returns a real or complex number variable name or algebraic e...

Page 54: ...erms V2 Secondary RMS Voltage in V Magnitude only A real number variable name or algebraic expression of defined terms I1 Primary RMS Current in A Magnitude only A real number variable name or algebraic expression of defined terms PP1 Primary Real Power in W A real number variable name or algebraic expression of defined terms Field Descriptions Output Screen n Primary to secondary turns ratio Retu...

Page 55: ...algebraic expression of defined terms PP1 Primary Real Power in W A real number variable name or algebraic expression of defined terms kVA kVA rating in kVA A real number variable name or algebraic expression of defined terms V1R Primary Voltage Rating in V A real number variable name or algebraic expression of defined terms Field Descriptions Output Screen n Primary to secondary turns ratio Retur...

Page 56: ...ce in Ω A real or complex number variable name or algebraic expression of defined terms n Primary to secondary turns ratio Returns a real number variable name or algebraic expression Gc Primary core conductance in S A real number or algebraic expression of defined terms Gc 0 Bc Primary core susceptance in S A real number or algebraic expression of defined terms Bc 0 Field Descriptions Output Scree...

Page 57: ...EE Pro for TI 89 92 Plus Analysis Transformer Calculations 42 ...

Page 58: ...eries Inductance in H unit Length A real number variable name or algebraic expression of defined terms L 0 R Series Resistance in Ω unit Length A real number variable name or algebraic expression of defined terms R 0 G Shunt Conductance in Siemens unit Length A real number variable name or algebraic expression of defined terms G 0 C Shunt Capacitance in F unit Length A real number variable name or...

Page 59: ...and calculated values are with respect to miles Input Screen Output Screen upper half Output Screen lower half 1 Enter the values 0 001 85 8 and 0015E 6 for L R and G respectively 2 Enter the values 62E 9 75 and 3 for C ZL_ and d respectively 3 Enter 2000 for f 4 Press to calculate the results The input and output screen displays are shown above 11 2 Line Parameters This topic computes fundamental...

Page 60: ...er or algebraic expression vp Phase Velocity in unit length s Returns a real number or algebraic expression Example 11 2 A transmission line is measured to have an open circuit impedance of 103 6255 2 525 i and an impedance under short circuit conditions of 34 6977 1 7896 i at a distance 1 unit length from the load location All measurements are conducted at 10 MHz Compute all the line parameters I...

Page 61: ...on is circular in nature there are two possible stub locations d1 and d2 Field Descriptions Input Screen ZL_ Load Impedance in Ω A real or complex number variable name or algebraic expression of defined terms RR0 Characteristic Resistance in Ω A real number variable name or algebraic expression of defined terms Field Descriptions Output Screen β d1 Electrical length from a stub at location d1 to t...

Page 62: ...n line has a characteristic impedance of 50 ohms and a load of 75 ohms Estimate the shorting stub location for matching purposes Input Screen Output Screen 1 Enter the values 50 and 75 for RR0 and ZL_ respectively 2 Press to calculate the results as shown in the screen displays above ...

Page 63: ...w the user to modify the word size between 1 and 128 bits 12 1 Special Mode Settings the Key The screen allows the user to set the parameters for solving binary problems Press the key to display a user interface dialog box prompting the user to clarify and specify the foundation for digital arithmetic Digital arithmetic operations are allowed in binary octal decimal and hexadecimal number systems ...

Page 64: ... Decimal Interpretation of a 5 bit Binary Binary 1 s Complement Mode 2 s Complement Mode Unsigned Mode 01111 15 15 15 01110 14 14 14 01101 13 13 13 01100 12 12 12 01011 11 11 11 01010 10 10 10 01001 9 9 9 01000 8 8 8 00111 7 7 7 00110 6 6 6 00101 5 5 5 00100 4 4 4 00011 3 3 3 00010 2 2 2 00001 1 1 1 00000 0 0 0 11111 0 1 31 11110 1 2 30 11101 2 3 29 11100 3 4 28 11011 4 5 27 11010 5 6 26 11001 6 7...

Page 65: ... or an integer preceded by the number base in parenthesis b d o or h Operator Binary Operation Press to select ADD Adds binary or real integer Affects the range and carry flags SUB Subtracts the second binary or real integer from the first Affects both the range And carry flags MULT Multiply the two binary or real numbers Affects both the range and the carry flags DIV Divide the first binary integ...

Page 66: ...e 8641FDB9 Retention of the least significant digits mimics the usual operation of microprocessors 12 3 Register Operations This section allows the user to perform operations on any bit of a binary integer Such operations include shifting bits to the left or right rotating bits to the left or right and shifting or rotating bits through the carry flag The input screen is updated to reflect the choi...

Page 67: ...ion SLN Shifts left by N bits SRN Shifts right by N bits RLN Rotate left by N bits RRN Rotate right by N bits RLCN Rotate left by N bits through the carry RRCN Rotate right by N bits through the carry Result Binary Function Value Returns an integer result using the number base set in Binary Mode Example 12 3 For a binary word 16 bits long shift left by 6 bits the octal number 2275 Input Screen Out...

Page 68: ...pic demonstrates the process of converting real numbers to binary numbers and vice versa The software allows for the conversion of real numbers to the IEEE format In 1985 the Institute of Electrical and Electronic Engineers IEEE a professional association developed standards for Binary Floating Point Arithmetic This Standard referred to Standard 754 specifies two basic forms of floating point form...

Page 69: ...5 Convert a real decimal number 42355 to its IEEE standard View the number in binary base two mode Input Screen Output Screen Result in Pretty Print B 1 Use key to select Binary for Base and 32 bits for Wordsize Press to accept the choices made 2 Enter 42355 for the Real Num 3 Choose Real to IEEE for the Function press to compute the result The screen display above shows the input screen and the r...

Page 70: ...oftware will immediately convert the number to the base setting binary in this case 2 Enter o 11215 and h 128D for Binary 1 and Binary 2 respectively 3 Choose the Operator and press to compute the result The screen display above shows the input and the resulting output 12 7 Karnaugh Map The program provides a symbolic representation of a minimization method in a sum of products form The variable n...

Page 71: ... 11 12 13 14 16 19 29 30 where minterms 4 and 6 are Don t Cares The input variables are V W X Y and Z Find the prime implicant expression Input Screen Output Screen Partial view of the Prime Implicant Expression 1 Enter 0 2 8 10 11 12 13 14 16 19 29 30 for Minterms and 4 6 for Don t Care 2 Enter variable names VWXYZ 3 Press to compute the result The screen displays above show the input screen and ...

Page 72: ...he Error Function and Complementary Error Function are erf x e dt t x z 2 2 0 π Eq 13 1 1 erfc x erf x e dt t x z 1 2 2 π Eq 13 1 2 Field Descriptions X Value Enter a real number global name or algebraic expression Func Error Function type Press to select erf or erfc Result Error Function value Returns a real number or algebraic expression Example 13 1 What is the value of erf 25 Input Screen Outp...

Page 73: ...Flow at time t usually years Payback The number of time periods usually years it takes a firm to recover its original investment NPV The present values of all future cash flows discounted at the selected rate minus the cost of the investment IRR The discount rate that equates the present value of expected cash flows to the initial cost of the project PI The present value of the future cash flows d...

Page 74: ...nter a positive or negative real number tn Cash flow at t n Enter a positive or negative real number Example 14 1 The following projects have been proposed by ACME Consolidated Inc What is the Payback period Net Present Value Internal Rate of Return and Profitability Index of each project Which is the more viable project Table 14 1 Cash Flow for two projects Name of Project Plant 1 Plant 2 Investm...

Page 75: ...ove the highlight bar to Multiple Graphs and press to enable overlaying of successive graphs of each project 8 Press to graph the curvilinear relationship between the Net Present Value and the Discount Rate 9 Press 2 followed by O to enable the graph editing toolbar The curve indicates where k 0 the Net Present Value is simply cash inflows minus cash outflows The IRR is shown at the point where NP...

Page 76: ... displays a pull down menu listing all the topics as shown in the screen display below An arrow to the left of the bottom topic ï indicates more items are listed Pressing 2 D jumps to the bottom of the menu Scroll the highlight bar to an item using the arrow key D and press or type the subject number appearing next to subject heading Resistive Circuits is selected for this example A second menu wi...

Page 77: ...nged 15 2 Viewing an Equation or Result in Pretty Print Sometimes equations and calculated results exceed the display room of the calculator The TI 89 and TI 92 plus include a built in equation display feature called Pretty Print which is available in many areas of EE Pro and can be activated by highlighting a variable or equation and pressing the right arrow key B or pressing the function key whe...

Page 78: ... need to determine which result is most useful to the application 1 Select an equation by 2 Enter known values for 3 If multiple solution exists highlighting and pressing each variable using the tool a dialogue box will appear Press to display bar to designate units Press requesting the user to enter variables to compute the results the number of a solution to view Solution 1 To view another Solut...

Page 79: ...e list to graph Use the same steps as above to select the independent and dependent variables Indep and Depnd from the equation Note all pre existing values stored in the variables used for Indep and Depnd will be cleared when the graphing function is executed The graphing unit scale for each variable reflect the settings in the Equations section of EE Pro Scrolling down the list specify the graph...

Page 80: ...s for a particular session of EE Pro press ƒ TOOLS ª NEW and type the name of the new folder see Chapter 5 of the TI 89 Guidebook for the complete details of creating and managing folders There are several ways to display or recall a value The contents of variables in any folder can be displayed using the moving the cursor to the variable name and pressing ˆ to display the contents of a particular...

Page 81: ...e solution converges It should be noted that the solution generated by nsolve is not guaranteed to be unique i e this solving process cannot determine if multiple solutions exist Table 15 1 User Defined Functions User defined Function Topic Sub topic erf ts τp Solid State PN Junction Current erfc x D t Solid State Semiconductor Basics eegalv Rx RR2 RR3 RR4 Rg Rs Vs Meters and Bridges Wheatstone Br...

Page 82: ...ulating values for the same variable 15 13 Notes for the advanced user in troubleshooting calculations When there are no solutions possible EE Pro provides important clues via key variables eeinput eeprob eeans and eeanstyp These variables are defined during the equation setup process by the built in multiple equation solver EE Pro saves a copy of the problem its inputs its outputs and a character...

Page 83: ... Currents 5 Ebers Moll Equations 6 Ideal Currents pnp 7 Switching Transients 8 MOS Transistor I 9 MOS Transistor II A MOS Inverter R Load B MOS Inverter Sat Load C MOS Inverter Depl Ld D CMOS Transistor Pair E Junction FET 13 Linear Amplifiers 1 BJT CB 2 BJT CE 3 BJT CC 4 FET Common Gate 5 FET Common Source 6 FET Common Drain 7 Darlington CC CC 8 Darlington CC CE 9 EC Amplifier A Differential Ampl...

Page 84: ...ables A complete list of all the variables used a brief description and applicable base unit is given below Variable Description Unit A Area m2 G Conductance S I Current A Il Load current A Is Current source A len Length m P Power W Pmax Maximum power in load W R Resistance Ω Rl Load resistance Ω Rlm Match load resistance Ω RR1 Resistance T1 Ω RR2 Resistance T2 Ω Rs Source resistance Ω T1 Temperat...

Page 85: ...ctance Solution Upon examining the problem two choices are noted Equations 16 1 1 16 1 2 and 16 1 4 or 16 1 1 and 16 1 3 can be used to solve the problem The second choice was made here Press to display the input screen enter all the known variables and press to solve the selected equation set The computed results are shown in the screen display shown here Entered Values Computed results PQYP 8CTK...

Page 86: ...he last option press to open the input screen enter all the known variables and press to solve PQYP 8CTKCDNGU AOC 4 AMΩ QORWVGF 4GUWNVU 8 A8 2 A9 AUKGOGPU 16 3 Temperature Effect This equation models the effect of temperature on resistance Electrical resistance changes from RR1 to RR2 when the temperature change from T1 to T2 is modulated by the temperature coefficient of resistance α RR RR T T 2 ...

Page 87: ...The power dissipation in the load is defined by the equation relating P with Il and Vl The next equation links Pmax to Vs and Rs The last equation represents load resistance needed for a maximum power Vl Vs Rl Rs Rl Eq 16 4 1 Il Vs Rs Rl Eq 16 4 2 P Il Vl Eq 16 4 3 P Vs Rs max 2 4 Eq 16 4 4 Rlm Rs Eq 16 4 5 Example 16 4 A 12_V car battery has a resistive load of 52_ohm The battery has a source imp...

Page 88: ...all its functionality to a current source Is with a source resistance Rs connected across it Is Vs Rs Eq 16 5 1 Vs Is Rs Eq 16 5 2 Example 16 5 Find the short circuit current equivalent for a 5_V source with a 12 5_ohm source resistance Entered Values Computed results Solution Either form of the equation can be used to solve the equation Press to display the user interface enter the values of all ...

Page 89: ...s Parallel Wires Coaxial Cable Sphere Variables A complete list of all the variables used in this section is given below Variable Description Unit A Area m2 C Capacitance F cl Capacitance per unit length F m d Separation m E Electric field V m Er Radial electric field V m Ez Electric field along z axis V m F Force on plate N Q Charge C r Radial distance m ra Inner radius wire radius m rb Outer rad...

Page 90: ...tential Entered Values Computed results Solution Both equations are needed to solve this problem Press to display the input screen enter all the known variables and press to solve for the unknown values Note that ε0 the permittivity of free space does not appear as one of the variables that needs to be entered It is entered automatically by the software However εr the relative permittivity must be...

Page 91: ...charge density of ρs a distance z from the plane of the disk The second equation computes the electrostatic potential Vz at an arbitrary point along the z axis Ez s r z ra z F HG I KJ ρ ε ε 2 0 1 2 2 Eq 17 3 1 Vz s r ra z z ρ ε ε 2 0 2 2 e j Eq 17 3 2 Example 17 3 A charged disc 5 5_cm in radius produces an electric field of 2_V cm 50_cm away from the surface of the disc Assuming that relative per...

Page 92: ...e plates and energy W stored in the capacitor E V d Eq 17 4 1 C r A d ε ε 0 Eq 17 4 2 Q C V Eq 17 4 3 F V C d 1 2 2 Eq 17 4 4 W V C 1 2 2 Eq 17 4 5 Example 17 4 A silicon dioxide insulator forms the insulator for the gate of a MOS transistor Calculate the charge electric field and mechanical force on the plates of a 5_V MOS capacitor with an area of 1250_µ2 and a thickness of 15_µ Use a value of 3...

Page 93: ...oltage between the two conductors of the cable carrying a charge of ρl per unit length and an insulator with a relative permittivity εr The inner conductor has a radius ra while the outer conductor has a radius rb The last equation computes the cable capacitance cl per unit length based on mechanical properties of the cable and insulator V l r rb ra F HG I KJ ρ π ε ε 2 0 ln Eq 17 6 1 Er V r rb ra ...

Page 94: ...a medium with a relative permittivity of εr The second equation computes the electric field outside a sphere at a distance r from the center of the sphere The last equation computes the capacitance between the spheres V Q r ra rb F HG I KJ 4 0 1 1 π ε ε Eq 17 7 1 Er Q r r 4 0 2 π ε ε Eq 17 7 2 C r ra rb rb ra 4 0 π ε ε Eq 17 7 3 Example 17 7 Two concentric spheres 2_cm and 2 5_cm radius are separa...

Page 95: ...itors Electric Fields 20 screen Enter all the known variables and press to solve the selected equation set The computed results are shown in the screen display shown here PQYP 8CTKCDNGU TC AEO TD AEO εT CPF 3 AEQWNQODU QORWVGF 4GUWNVU 8 A8 A ...

Page 96: ...elative permeability unitless a Loop radius or side of a rectangular loop m B Magnetic field T bl Width of rectangular loop m Bx Magnetic field x axis T By Magnetic field y axis T D Center center wire spacing m d Strip width m f Frequency Hz Fw Force between wires unit length N m I Current A I1 Current in line 1 A I2 Current in line 2 A Is Current in strip A m L Inductance per unit length H m L12 ...

Page 97: ...here is only equation press to display the input screen Enter all the known variables and press to solve the equation The computed results are shown in the screen display above PQYP 8CTKCDNGU A T AO QORWVGF 4GUWNVU A6 18 2 Long Strip A thin conducting ribbon strip of width d is infinitely long and carrying a current Is amperes per meter The x and y component of the magnetic field Bx and By are dep...

Page 98: ...th carrying currents I1 and I2 separated by a distance D exert a force F newtons meter between them The second equation computes the magnetic field Bx between the two parallel wires at a distance x from the line carrying current I1 The final equation in this set computes the inductance L from the two wires of diameter a with a spacing of D Eq 18 3 1 Eq 18 3 2 Eq 18 3 3 Example 18 3 A pair of alumi...

Page 99: ...long wire carrying a current I1 The rectangular loop has a width bl parallel to its axis of rotation and a length a perpendicular to the axis of rotation The loop axis of rotation and the infinitely long wire intersect at a 90 degree angle The loop s angle of tilt θ is relative to the plane containing the loop plane and the infinite wire In the side of the loop closest to the infinite wire the cur...

Page 100: ...adius of the outer conductor of rb is characterized by the inductance L per unit length L rb ra F HG I KJ µ π µ π 0 8 0 2 ln Eq 18 5 1 Example 18 5 A coaxial cable has an inner conductor radius of 2_mm and the outer conductor radius of 15_in Find its inductance per meter Input Screen Calculated Results Solution Press to display the input screen enter all the known variables and press to solve the ...

Page 101: ...Eq 18 6 1 Re ff f r π µ µ ρ 0 Eq 18 6 2 Example 18 6 Find the effect on depth of signal penetration for a 100 MHz signal in copper with a resistivity of 6 5E 6 _ohm cm The relative permeability of copper is 1 02 Input Screen Calculated Results Solution Press to display the input screen enter all the known variables and press to solve the selected equation set The computed results are shown in the ...

Page 102: ...urface area m2 T Temperature K v Vertical velocity m s Va Accelerating voltage V Vd Deflecting voltage V y Vertical deflection m yd Beam deflection on screen m z Distance along beam axis m φ Work function V 19 1 Electron Beam Deflection An electron beam that is subjected to an accelerating voltage Va achieves a velocity v as defined by the first equation The second equation calculates the radius o...

Page 103: ...5 cm The deflection region is controlled by a 100_V voltage A magnetic field of 0 456_T puts the electrons in the beam in a circular orbit What is the vertical deflection distance of the beam when it reaches the CRT screen Display of Input Values Calculated Results Solution The first three equations are needed to solve this problem Select these by highlighting and pressing Press to display the inp...

Page 104: ...lated Results Solution Since there is only equation press to display the input screen enter all the known variables and press to solve the equation The computed results are shown in the screen display shown here PQYP 8CTKCDNGU A O 5 AEO 6 A φ A8 QORWVGF 4GUWNVU A 19 3 Photoemission The two equations in this topic represent the behavior of electrons when excited by photon energy A light beam with a...

Page 105: ...ution sets exist for the entered data in which case the user needs to select a solution which is meaningful to the application When viewing one of a multiple solution set the number of a solution should be entered 1 2 3 etc followed by pressing twice The process of solving and choosing a solution set should be repeated each time the user wishes to view a different solution In this case a positive ...

Page 106: ...Unit a Resistance multiplier unitless b Resistance Multiplier unitless c Resistance Multiplier unitless CC3 Capacitance arm 3 F CC4 Capacitance arm 4 F Cs Series capacitor F Cx Unknown capacitor F DB Attenuator loss unitless f Frequency Hz Ig Galvanometer current A Imax Maximum current A Isen Current sensitivity A Lx Unknown inductance unitless Q Quality Factor unitless Radj Adjustable resistor Ω ...

Page 107: ...tion extends the range of a series ohmmeter with an internal resistance Rm and internal voltage Vs with an adjustable resistor Radj In a practical setup Radj is usually set at its midpoint to compensate for variations in the component values resulting in a systematic error in the measured result Eq 20 1 1 Eq 20 1 2 Eq 20 1 3 Example 20 1 What resistance can be added to a voltmeter with a current s...

Page 108: ...nd two 1000 Ω resistors connected on the known side of the bridge A resistor of 99 Ω was connected to the bridge in the location where the unknown resistor would normally be present The bridge is supplied by a 10 V source with a resistance of 2 5 Ω The galvonemetric resistance is 1 MΩ Find the voltage across the meter and the galvanometric current Entered Values Calculated Results Lower portion of...

Page 109: ...radian frequency ω Eq 20 3 1 Eq 20 3 2 Eq 20 3 3 Eq 20 3 4 Example 20 3 A set of measurements obtained using a Wien bridge is based on the following input All measurements are carried out at 1000_Hz The known resistors RR1 and RR3 are 100 Ω each the series resistance is 200 Ω and Cs is 1 2_µF Find the values of the unknown RC circuit components and the radian frequency Entered Values Calculated Re...

Page 110: ... bridge The bridge are resistors are 1000_Ω each with a 22 µF capacitor and 470 Ω parallel resistance Compute Lx and Rx Entered Values Calculated Results Solution The first two equations are needed to solve the problem Press to display the input screen enter all the known variables and press to compute the solution The calculated results are shown in the screen display above PQYP 8CTKCDNGU U Aµ 44...

Page 111: ... 44 AΩ 44 AΩ QORWVGF 4GUWNVU Z AJGPT 4Z AΩ 20 6 Symmetrical Resistive Attenuator Tee Pad Pi Pad Bridged Pad Balanced Pad Balanced resistive networks are commonly used as attenuators in transmission line circuits The three equations below form the basis of the design for resistive attenuators in a Tee pad a Pi pad a bridged Tee pad or a balanced pad configuration These design equations compute the ...

Page 112: ...uted the values of the multipliers a and b a Tee pad can be constructed using the values Ro a or 16 6395_ohms for the two sides of the Tee and a calue Ro b or 66 931_ohms for the vertical leg of the network 20 7 Unsymmetrical Resistive Attenuator This section contains equations for unsymmetrical resistive attenuator design These equations compute the resistance values for a minimum loss L pad of a...

Page 113: ...e the network to the left of the attenuator possesses an impedance of 100 Ω What is the expected loss in dB Entered Values Calculated Results Solution The last equation is needed to compute the signal attenuation Press to display the input screen enter all the known variables and press to solve the equation The computed results are shown in the screen display above PQYP 8CTKCDNGU 4N AΩ 4T AΩ QORWV...

Page 114: ... chapter along with a brief description and appropriate units Variable Description Unit C Capacitor F Cs Series capacitance F Cp Parallel capacitance F f Frequency Hz iC Capacitor current A iL Inductor current A I0 Initial inductor current A L Inductance H Lp Parallel inductance H Ls Series inductance H Qp Q parallel circuit unitless Qs Q series circuit unitless R Resistance Ω Rp Parallel resistan...

Page 115: ...it consists of a 400_mH inductor and a 125_Ω resistor With an initial current of 100_mA find the inductor current and voltage across the inductor 1_ms and 10_ms after the switch has been closed Entered Values Calculated Results 1 ms Solution Upon examining the problem the first three equations are needed to solve the problem Select these equations using the highlight bar and pressing press to disp...

Page 116: ...e problem all of the equations are needed to solve the problem Press to display the input screen Enter all the known variables and press to solve the set of equations PQYP 8CTKCDNGU Aµ 4 AΩ 8 A8 V AµU QORWVGF 4GUWNVU X A8 K A 9 A τ AU 21 3 RL Step Response These equations describe the response of an inductive circuit to a voltage step stimulus The first equation calculates the time constant τ in t...

Page 117: ...t two equations compute the capacitor voltage vC and current iC in terms of the step stimulus voltage Vs the initial capacitor voltage V0 time t and time constant τ τ τ R C Eq 21 4 1 vC Vs V Vs e t 0 b g τ Eq 21 4 2 iC Vs V R e t 0 b g τ Eq 21 4 3 Example 21 4 A 10_V step function is applied to an RC circuit with a 7 5_Ω resistor and a 67_nfarad capacitor The capacitor was charged to an initial po...

Page 118: ... of Rp Lp and ω The sixth and seventh equations calculate Rs and Ls in terms of Rp Lp and ω The eighth and ninth equations compute Rp and Lp in terms of Rs Ls and Qs The last two of equations calculate Rs and Ls in terms of Rp Lp and Qp ω π 2 f Eq 21 5 1 Qs Ls Rs ω Eq 21 5 2 Rp Rs Ls Rs 2 2 2 ω Eq 21 5 3 Lp Rs Ls Ls 2 2 2 2 ω ω Eq 21 5 4 Qp Rp Lp ω Eq 21 5 5 Rs Lp Rp Rp Lp ω ω 2 2 2 2 2 Eq 21 5 6 ...

Page 119: ...rcuit with values Rp and Cp The first equation converts frequency f to its radian equivalent radian frequency ω The second equation computes the quality factor Qs in terms of ω Rs and Cs The next two equations compute the parallel equivalent values as a function of Rs Cs and ω The fifth equation defines Qp in terms of Rp Cp and ω The sixth and seventh equations compute Rs and Cs in terms of Rp Cp ...

Page 120: ...0000_Ω at 120000_Hz Find its series equivalent Entered Values Calculated Results Solution Upon examining the problem use equations 21 6 1 21 6 3 21 6 4 21 6 6 21 6 7 are needed to solve the problem Select these by highlighting each equation and pressing the key Press to display the input screen enter all the known variables and press to solve PQYP 8CTKCDNGU R _P 4R 150000_Ω H A QORWVGF 4GUWNVU 4U ...

Page 121: ...it α Neper s frequency rad s A1 Constant V A2 Constant V B Susceptance S B1 Constant V B2 Constant V BC Capacitive susceptance S BL Inductive susceptance S C Capacitance F D1 Constant V s D2 Constant V f Frequency Hz G Conductance S I0 Initial inductor current A L Inductance H θ Phase angle rad R Resistance Ω s1 Characteristic frequency rad s s2 Characteristic frequency rad s s1i Characteristic fr...

Page 122: ... reactance XXC is calculated from the frequency ω and capacitance C In the final equation ω is expressed in terms of the oscillation frequency f Zm R X c h2 2 2 Eq 22 1 1 θ F HG I KJ tan 1 X R Eq 22 1 2 X XL XXC Eq 22 1 3 XL L ω Eq 22 1 4 XXC C 1 ω Eq 22 1 5 ω π 2 f Eq 22 1 6 Example 22 1 A circuit consists of a 50 ohm resistor in series with a 20_mhenry inductor and 47_µfarad capacitor At a frequ...

Page 123: ...farads Find the circuit admittance parameters at a frequency of 10 MHz Input and calculated results Input and calculated results upper half lower half Solution All of the equations need to be used to solve this problem Press to display the input screen and enter the values of all known variables Press to compute the unknown parameters PQYP 8CTKCDNGU H A 4 AQJO AµJGPT AµHCTCF QORWVGF 4GUWNVU O A UK...

Page 124: ...DNGU AOJGPT 4 AQJO AµHCTCF QORWVGF 4GUWNVU U T ATCF U U K ATCF U ω ATCF U U T ATCF U U K ATCF U 22 4 Underdamped Case The equations in this section represent the transient response of an underdamped RLC circuit The classical radian frequency ω0 is calculated from the inductance L and the capacitance C in Equation 22 4 1 The Neper s frequency α is shown by the second equation The condition for an u...

Page 125: ... to be selected to solve this problem Press and enter the known variables followed by a second press of to solve for the unknowns PQYP 8CTKCDNGU Aµ AO AO 4 AΩ V AµU QORWVGF 4GUWNVU α α AT U A8 A8 X A8 ωF AT U ω AT U 22 5 Critical Damping The equations in this section represent the RLC transient response of a critically damped circuit The condition for a critically damped system is ω0 α The first t...

Page 126: ...to solve for the unknowns PQYP 8CTKCDNGU Aµ AO AO 4 AΩ V AµU 8 A8 QORWVGF 4GUWNVU A8 U A8 X A8 ω AT U α AT U 22 6 Overdamped Case These equations show the transient performance of an overdamped RLC circuit The condition for an overdamped system is that α ω0 The first two equations define the characteristic frequencies s1 and s2 in terms of the Neper s frequency α and the classical resonant frequen...

Page 127: ...tial inductor current is 0 mA and the capacitor is charged to a potential of 5 V find the voltage across the capacitor after 1 ms Calculated Results Upper display Calculated Results Lower display Solution All of the equations need to be selected to solve this problem Press and enter the known variables followed by a second press of to solve for the unknowns The solver takes about five minutes to s...

Page 128: ...escription Unit C Capacitance F f Frequency Hz I Instantaneous current A Im Current amplitude A L Inductance H θ Impedance phase angle rad θ1 Phase angle 1 rad θ2 Phase angle 2 r R Resistance Ω RR1 Resistance 1 Ω RR2 Resistance 2 Ω t Time s V Total voltage V VC Voltage across capacitor V VL Voltage across inductor V Vm Maximum voltage V VR Voltage across resistor V ω Radian frequency r s X Reactan...

Page 129: ...quation relates the radian frequency and frequency f Note I VR and VL are functions of time I t Im sin ω b g Eq 23 1 1 abs Zm R L b g2 2 2 2 ω Eq 23 1 2 VR Zm t Im sin cos ω θ b g b g Eq 23 1 3 VL Zm t Im cos sin ω θ b g b g Eq 23 1 4 V VR VL Eq 23 1 5 Vm Zm Im Eq 23 1 6 θ ω F HG I KJ tan 1 L R Eq 23 1 7 ω π 2 f Eq 23 1 8 Example 23 1 An RL circuit consists of a 50 Ω resistor and a 0 025 henry ind...

Page 130: ...ote I VR and VC are functions of time I t Im sin ω b g Eq 23 2 1 abs Zm R C b g2 2 2 2 1 ω Eq 23 2 2 VR Zm t Im sin cos ω θ b g b g Eq 23 2 3 VC Zm t Im cos sin ω θ b g b g Eq 23 2 4 V VR VC Eq 23 2 5 Vm Zm Im Eq 23 2 6 θ ω F HG I KJ tan 1 1 C R Eq 23 2 7 ω π 2 f Eq 23 2 8 Example 23 2 An RC circuit consists of a 100 Ω resistor in series with a 47 µF capacitor At a frequency of 1500 Hz the current...

Page 131: ...rt of 475_ Ω Entered Value Calculated Result Solution Press to display the input screen enter all the known variables and press to solve the equation The computed results are shown in the screen displays above PQYP 8CTKCDNGU K AΩ QORWVGF 4GUWNVU K A5KGOGPU 23 4 Two Impedances in Series These equations combine two impedances Z1 and Z2 in series with real and imaginary parts RR1 and XX1 RR2 and XX2 ...

Page 132: ...mpedances in Parallel Two impedances Z1 and Z2 when connected in parallel result in an equivalent parallel impedance Zp represented by a real part R and a reactive part X The impedances Z1 and Z2 have real and reactive parts RR1 XX1 and RR2 XX2 respectively The first two equations show a numeric expression for magnitude Zm and phase θ θ Simpler versions of R and X are shown in the third and fourth...

Page 133: ...agnitude and phase of the combination Entered Values Calculated Results Solution Use the first and second equations to compute the solution for this problem Select these by highlighting each equation and pressing the key Press to display the input screen enter all the known variables and press to solve the equation The computed results are shown in the screen displays above PQYP 8CTKCDNGU 44 AΩ 44...

Page 134: ...24 1 Balanced Network These equations describe the essential features of a Balanced Network The line voltage VL is defined in terms of phase voltage Vp The second equation expresses the line current IL using the phase current Ip The third equation computes the power in each phase P from Vp Ip and the phase delay θ between voltage and current The last two equations represent the total power PT deli...

Page 135: ...ulates the line current IL from the phase current Ip The power phase P is defined in terms of Vp Ip and phase delay θ The last two equations are used to estimate the total power PT delivered to the system from the parameters VL IL or Vp and Ip VL Vp 3 Eq 24 2 1 IL Ip Eq 24 2 2 P Vp Ip cos θ b g Eq 24 2 3 PT Vp Ip 3 cos θ b g Eq 24 2 4 PT VL IL 3 cos θ b g Eq 24 2 5 Example 24 2 Using the known par...

Page 136: ...sents the total power PT delivered to the three phase load W VL IL 1 6 F HG I KJ cos θ π Eq 24 3 1 W VL IL 2 6 F HG I KJ cos θ π Eq 24 3 2 PT VL IL 3 cos θ b g Eq 24 3 3 Example 24 3 Given a line voltage of 110 V and a line current of 25 A and a phase angle of 0 1 rad find the wattmeter readings in a 2 wattmeter meter system Input variables Computed results Solution All of the equations are needed...

Page 137: ... Resistance Ω Rg Generator resistance Ω Vm Maximum voltage V ω Radian frequency r s ω0 Resonant frequency r s ω1 Lower cutoff frequency r s ω2 Upper cutoff frequency r s ωd Damped resonant frequency r s ωm Frequency for maximum amplitude r s Yres Admittance at resonance S Z Impedance Ω Zres Impedance at resonance Ω 25 1 Parallel Resonance I These ten equations describe resonance properties in para...

Page 138: ...ω1 1 2 1 2 1 2 R C R C L C b g Eq 25 1 4 ω2 1 2 1 2 1 2 R C R C L C b g Eq 25 1 5 β ω ω 2 1 Eq 25 1 6 Q ω β 0 Eq 25 1 7 Q R C L Eq 25 1 8 Q R C ω0 Eq 25 1 9 Example 25 1 Calculate the resonance parameters of a parallel resonant circuit containing a 10 000 Ω resistor a 2 4 µF capacitor and a 3 9 mH inductor The amplitude of the current is 10 mA at a radian frequency of 10 000 rad s Input Values Cal...

Page 139: ... terms of ω0 and Q The fourth and fifth equations compute α the damping coefficient either from R and C or ω0 and Q The final two equations express the damped resonant frequency ωd in terms of α and ω0 or ω0 and Q Q ω β 0 Eq 25 2 1 ω ω 1 0 1 2 1 2 1 2 F HG I KJ Q Q b g Eq 25 2 2 ω ω 2 0 1 2 1 2 1 2 F HG I KJ Q Q b g Eq 25 2 3 α 1 2 R C Eq 25 2 4 α ω 0 2 Q Eq 25 2 5 ω ω α d 02 2 Eq 25 2 6 ω ω d Q 0...

Page 140: ...dance and admittance of the circuit at resonance ω ωm represents the frequency that occurs when the amplitude of the voltage across the resonant circuit is at maximum ω0 1 2 F HG I KJ L C R L Eq 25 3 1 Yres L Rg R C L Rg Eq 25 3 2 Zres Yres 1 Eq 25 3 3 ωm L C R Rg R L L C R L F HG I KJ F HG I KJ F HG I KJ F HG I KJ 1 1 2 2 2 2 2 Eq 25 3 4 Example 25 3 A power source with an impedance Rg of 5 Ω is ...

Page 141: ...the upper and lower cutoff frequencies beyond resonance where the impedance is half the impedance at resonance The expressions for β β represents the bandwidth of the resonant circuit The last equation determines the quality factor Q in terms of R C L and ω ω0 ω0 1 L C Eq 25 4 1 Z R L C F HG I KJ 2 2 1 ω ω Eq 25 4 2 θ ω ω F H GGG I K JJJ tan 1 1 L C R Eq 25 4 3 ω1 2 2 1 2 F HG I KJ R L R L L C Eq ...

Page 142: ...on Upon examining the problem all equations are needed to solve the problem Press to display the input screen enter all the known variables and press to solve the set of equations The computed results are shown in the screen displays above PQYP 8CTKCDNGU Aµ 4 AΩ Aµ 4 AΩ ω ω AT U QORWVGF 4GUWNVU ω ω AT U ω ω AT U ω ω AT U AΩ ...

Page 143: ...ator Differential Amplifier Variables All the variables used here are listed with a brief description and proper units Variable Description Unit Acc Common Mode current gain unitless Aco Common Mode gain from real OpAmp unitless Ad Differential mode gain unitless Agc Transconductance S Aic Current gain unitless Av Voltage gain unitless CC1 Input capacitor F Cf Feedback capacitor F CMRR CM rejectio...

Page 144: ...rrent The first pole frequency fcp is defined by the third equation Small signal rise time tr 10 to 90 is defined by the fourth equation Av Rf RR 1 Eq 26 1 1 Rp RR Rf RR Rf 1 1 Eq 26 1 2 fcp fop Av RR Rf F HG I KJ 1 Eq 26 1 3 tr Rf fop Av RR 35 1 Eq 26 1 4 Example 26 1 Find the gain of an inverter and its optimum value for bias resistance given an input resistance of 1 kΩ and a feedback resistance...

Page 145: ...culated Results Solution Use the first and second equations to compute the solution for this problem Select these by highlighting each equation and pressing the key Press to display the input screen enter all the known variables and press to solve the equation The computed results are shown in the screen displays above PQYP 8CTKCDNGU 4H A Ω 44 AMΩ QORWVGF 4GUWNVU X 4R AΩ 26 3 Current Amplifier Thi...

Page 146: ...the screen displays above PQYP 8CTKCDNGU X 4H AMΩ 4N AMΩ 4Q AΩ 4U AMΩ QORWVGF 4GUWNVU KE 4KP AΩ 4QWV A Ω 26 4 Transconductance Amplifier The two equations in this section specify a closed loop transconductance Agc and output resistance Rout in terms of the resistance Rs and voltage gain Av Agc Rs 1 Eq 26 4 1 Rout Rs Av 1 b g Eq 26 4 2 Example 26 4 Find the transconductance and output resistance fo...

Page 147: ...b g Eq 26 5 2 VU VR Rf Rp Vz Rf Rp 1 Eq 26 5 3 VL VR Rf Rp Vz Rf Rp 2 Eq 26 5 4 Example 26 5 An inverting level detector possesses two zener diodes to set the trip level The setting levels are 4 V and 3 V respectively for the first and second diodes The reference voltage is 5 V the OpAmp is supported by a 10 kΩ bias resistor and a 1 MΩ feedback resistor Find the hysteresis the upper and lower dete...

Page 148: ...lculated Results Solution Use the first three equations to compute the solution for this problem Select these by highlighting each equation and pressing the key Press to display the input screen enter all the known variables and press to solve the equation The computed results are shown in the screen displays above PQYP 8CTKCDNGU 4H A Ω 4R AMΩ 84 A8 8 A8 8 A8 QORWVGF 4GUWNVU 8 A8 44 AΩ 87 A8 26 7 ...

Page 149: ...ghlighting each equation and pressing the key Press to display the input screen enter all the known variables and press to solve the equation The computed results are shown in the screen displays above PQYP 8CTKCDNGU HF AM 8TCVG A8 µU 8QOCZ A8 4H AMΩ QORWVGF 4GUWNVU A 44 AMΩ 26 8 Differential Amplifier These four equations describe the primary relationships used in designing of a differential ampl...

Page 150: ...ridge resistors RR1 RR2 RR3 and RR4 of 10 kΩ 3 9 kΩ 10 2 kΩ and 4 1 kΩ respectively Assume a voltage gain of 90 Entered Values Calculated Results Solution Use the third and fourth equations to compute the solution for this problem Select these by highlighting each equation and pressing the key Press to display the input screen enter all the known variables and press to solve the equation The compu...

Page 151: ...on set to determine which equations in a sub topic form a consistent subset before attempting to compute a solution Semiconductor Basics PN Junctions PN Junction Currents Transistor Currents Ebers Moll Equations Ideal Currents pnp Switching Transients MOS Transistor I MOS Transistor II MOS Inverter Resistive MOS Inverter Saturated MOS Inverter Depletion CMOS Transistor Pair Junction FET Variables ...

Page 152: ...urrent A I0 Saturation current A IB Base current A IC Collector current A ICB0 CB leakage E open A ICE0 CE leakage B open A ICsat Collector I at saturation edge A ID Drain current A IDmod Channel modulation drain current A ID0 Drain current at zero bias A IDsat Drain saturation current A IE Emitter current A IIf Forward current A Ir Reverse current A Ir0 E M reverse current component A IRG G R cur...

Page 153: ...nction potential V pno p density in n material 1 m3 Qtot Total surface impurities unitless Qb Bulk charge at bias C m2 Qb0 Bulk charge at 0 bias C m2 Qox Oxide charge density C m2 Qsat Base Q transition edge C ρn n resistivity Ω m ρp p resistivity Ω m Rl Load resistance Ω τB lifetime in base s τD Time constant s τL Time constant s τo Lifetime s τp Minority carrier lifetime s τt Base transit time s...

Page 154: ...om surface m xd Depletion layer width m xn Depletion width n side m xp Depletion width p side m Z JFET width m 27 1 Semiconductor Basics The nine equations listed under this sub topic describe the basic properties used semiconductor technology The first four equations are a subset which describes the basic properties of free carriers in semiconductors such as resistivity mobility and diffusion pro...

Page 155: ...the seventh equation in terms of the conduction and valence band levels Ec and Ev temperature TT the effective mass of electrons mn and holes mp Ei EF k TT Na ni TT F HG I KJ ln b g Eq 27 1 5 EF Ei k TT Nd ni TT F HG I KJ ln b g Eq 27 1 6 Ei Ec Ev k TT mp mn F HG I KJ b g 2 3 4 ln Eq 27 1 7 The final two equations are diffusion properties of dopants in silicon based on two distinctly different con...

Page 156: ... at the desired depth is 8 x 1017 cm 3 while the surface density is 4 x 1019 cm 3 Entered Values Notice of nsolve routine Calculated Results Solution Equation 21 1 8 is needed to compute the solution for this problem Select it by highlighting and pressing the key Press to display the input screen enter all the known variables and press to solve the equation The nsolve routine is used since x is an...

Page 157: ...depletion layer width xd for a linearly graded junction with a gradient parameter aLGJ Vbi k TT q aLGJ xd ni TT F HG I KJ 2 2 ln Eq 27 2 6 xd s q aLGJ Vbi Va F HG I KJ 12 0 1 3 ε ε Eq 27 2 7 Example 27 2 1 A PN step junction is characterized by an acceptor doping density of 6 x 1016 cm 3 and a donor doping density of 9 x 1017 cm 3 The junction area is 100 µm2 at room temperature For an applied vol...

Page 158: ... Junction Currents These equations characterize the relationships for computing currents in PN junctions They can be classified into four categories The first three equations define the junction currents First the junction current I is expressed in terms of the junction area Aj diffusion coefficients Dn and Dp diffusion lengths LLn and Lp equilibrium densities of minority carriers npo and pno appl...

Page 159: ...f ts p τ Eq 27 3 8 Example 27 3 1 A PN Junction is characterized as having a junction area of 100 µm2 an applied voltage of 0 5 V and diffusion coefficients for electrons and holes of 35 cm2 s and 10 cm2 s respectively The diffusion lengths for electrons and holes are 25 µm and 15 µm The minority carrier densities are 5 x 106 cm 3 electrons and 25 cm 3 holes Find the junction current and the satur...

Page 160: ... relationships between the emitter base and collector currents IE IB and IC respectively The first equation defines the common base current gain α as the ratio of IC to IE The second equation defines the common emitter current gain β in terms of α α IC IE Eq 27 4 1 β α α 1 Eq 27 4 2 The third equation represents Kirchoff s current law for the bipolar junction transistor IE IB IC Eq 27 4 3 The next...

Page 161: ...ons These ten equations show a collection of relevant relationships developed by J J Ebers and J L Moll in the mid 1950s recognizing the reciprocal behavior of bipolar function transistors The first three equations connect the emitter collector and base currents IE IC and IB in terms of forward and reverse current gain αf and αr and the forward and reverse currents IIf and Ir IE IIf r Ir α Eq 27 5...

Page 162: ...hown in the screen displays above PQYP 8CTKCDNGU α αH αT AO H AO QORWVGF 4GUWNVU βH βT µ T A U A 27 6 Ideal Currents pnp The four equations in this set form the basis of transistor action resulting in emitter base and collector currents in a pnp transistor The first three equations show the emitter collector and base currents IE IC and IB in terms of emitter base area A1 diffusion coefficients DE ...

Page 163: ...he work of Ebers and Moll was supplemented by Gummel to model transistor behavior in several different ways The concept of charge in the base of the transistor controlling switching times became an important contribution to switching theory The first equation expresses the base charge at the edge of saturation Qsat in terms of the collector saturation current ICsat and the base transit time τ τt T...

Page 164: ...e 27 7 Find the saturation voltage for a switching transistor at room temperature when a base current of 5 1 mA is used to control a collector current of 20 mA The forward and reverse α s are 0 99 and 0 1 respectively Entered Values Calculated Results Solution Use the last equation to solve this problem Select the equation by highlighting and pressing the key Press to display the input screen ente...

Page 165: ...rface charge density Qb is influenced by the substrate bias VSB Qb q Na s F 0 2 0 2 ε ε φ Eq 27 8 3 Qb q Na s F VSB 2 0 2 ε ε φ Eq 27 8 4 A thin oxide layer with a thickness tox on the surface of the semiconductor results in a capacitance Cox per unit area in the fifth equation Cox ox tox ε ε0 Eq 27 8 5 The sixth equation defines the body coefficient γ in terms of Cox Na and εs The final equation ...

Page 166: ... 27 9 1 kn n ox tox 1 0 µ ε ε Eq 27 9 2 kn kn W L 1 Eq 27 9 3 The fourth equation defines IDmod the drain current when the transistor is operating under saturation in terms of kn gate voltage VGS threshold voltage VT modulation parameter λ and drain voltage VDS The basic physics behind the increase in drain current comes from the channel widths being non uniform under the gate because f a finite p...

Page 167: ...solution for this problem Press to display the input screen enter all the known variables and press to solve the equations The computed results are shown in the screen displays above PQYP 8CTKCDNGU εQZ γ γ A 8 λ λ A 8 µ µ µP AEO 8U φ φ A8 VQZ Aµ 8 5 A8 8 5 A8 85 A8 86 A8 9 Aµ QORWVGF 4GUWNVU QZ A O HHOCZ A IF AUKGOGPU IO AUKGOGPU A OQF A MP A 8 MP A 8 6VT AU 86 A8 27 10 MOS Inverter Resistive Load...

Page 168: ... for which the driver transistor is in saturation kD VM VT VDD VM Rl 2 2 b g b g Eq 27 10 5 Example 27 10 1 Find the driver device constant output and mid point voltages for a MOS inverter driving a 100_kΩ resistive load Driver properties include a 3 µm wide gate a length of 0 8 µm Cox of 345313 pF cm2 The electron mobility is 500 cm2 V s VIH 2 8 V VT 1_V and VDD 5_V Solution 1 Upper Display Lower...

Page 169: ...Equations Solid State Devices 95 Solution 3 Upper Display Lower Display Solution 4 Upper Display Lower Display Solution 5 Upper Display Lower Display Solution 6 Upper Display Lower Display Solution 7 Upper Display Lower Display ...

Page 170: ...lts for all four solutions are shown in the screen displays above In this example VOH VOL VIL Vo and VM have to be positive and between 0 and VDD PQYP 8CTKCDNGU QZ AR EO Aµ µ0 AEO 8U 4 AMΩ 8 A8 8 A8 86 A8 9 Aµ QORWVGF 4GUWNVU 5QNWVKQP M A 8 8 A8 81 A8 81 A8 8Q A8 5QNWVKQP M A 8 8 A8 81 A8 81 A8 8Q A8 5QNWVKQP M A 8 8 A8 81 A8 81 A8 8Q A8 5QNWVKQP M A 8 8 A8 81 A8 81 A8 8Q A8 5QNWVKQP M A 8 8 A8 81...

Page 171: ...TL VT0 and KR The last five equations show the performance parameters of the inverter circuit VIH VDD VTL KR VT 2 3 1 0 b g Eq 27 11 7 Vo VDD VTL VT VT KR KR 0 0 1 d i Eq 27 11 8 The equation for gmL defines the transconductance of the load circuit while the equation for τL defines the characteristic time to charge the load capacitance CL gmL kL VDD VTL b g Eq 27 11 9 τL CL gmL Eq 27 11 10 The cha...

Page 172: ...m PQYP 8CTKCDNGU QZ AR EO µP AEO 8 U Aµ Aµ 9 Aµ 9 Aµ γ A 8 φ φ A8 8 A8 8Q A8 86 A8 QORWVGF 4GUWNVU M A 8 M A 8 4 8 A8 81 A8 86 A8 27 12 MOS Inverter Depletion Load This section lists the design equations for a MOS inverter with a depletion load The first two equations compute device constants kL and kD for the load and the driver transistors in terms of their geometries WD WL LD and LL kL n Cox WL...

Page 173: ...lution 2 Upper Display Solution 2 Lower Display Solution The problem can be solved when equations 27 12 1 27 12 4 are selected Highlight each of these and press Press to display the input screen enter all the known variables and press to solve the equations Two complete solutions are computed for this case and are shown in the screen displays above PQYP 8CTKCDNGU QZ AR EO γ A8 Aµ Aµ µP AEO 8 U φ A...

Page 174: ...first four equations Select these equations by highlighting and pressing Press to display the input screen enter all the known variables and press to solve the equations The computed results are shown in the screen display above PQYP 8CTKCDNGU 90 Aµ N00 Aµ µ µP AEO 8U QZ AR EO 860 A8 862 A8 N2 Aµ 8 A8 µ µR AEO 8U 92 Aµ QORWVGF 4GUWNVU M0 A 8 M2 A 8 8 A8 8Q A8 27 14 Junction FET These five equation...

Page 175: ...t VG 2 0 ε ε 1 6 when VG Vp and VDsat Vbi Vp R S T U V W l q 2 3 Eq 27 14 3 VDsat VG Vp when VG Vp and VDsat Vbi Vp R S T U V W l q 2 3 Eq 27 14 4 IDsat ID VG Vp F HG I KJ 0 1 2 Eq 27 14 5 Example 27 14 Find the saturation current when the drain current at zero bias is 12 5 µA the gate voltage is 5 V and the Pinchoff voltage is 12 V Entered Values Calculated Results Solution Use the third equation...

Page 176: ...ton CC CC Darlington CC CE Emitter Coupled Amplifier Differential Amplifier Source Coupled JFET Pair Variables The table lists all variables used in this section along with a description and appropriate units Variable Description Unit α0 Current gain CE unitless Ac Common mode gain unitless Ad Differential mode gain unitless Ai Current gain CB unitless Aov Overall voltage gain unitless Av Voltage ...

Page 177: ...ain Av from re rb α0 β0 and the load resistance Rl The last equation computes the overall voltage gain for the amplifier system Aov from Rin rrc re α0 β0 and the source impedance Rs β α α 0 0 1 0 Eq 28 1 1 Rin re rb β0 Eq 28 1 2 Ro rrc Eq 28 1 3 Ai α0 Eq 28 1 4 Av Rl re rb α β 0 0 Eq 28 1 5 Aov rrc Rin Rin Rs re rb F HG I KJ α β 0 0 Eq 28 1 6 Example 28 1 A common base configuration of a linear am...

Page 178: ... computes the voltage gain Av from β β0 the load source and input resistances Rl Rs and Rin The last equation calculates the overall voltage gain Aov from the source impedance Rs Rl Rin and β β0 β α α 0 0 1 0 Eq 28 2 1 Rin rb re β0 Eq 28 2 2 Ro rrc Eq 28 2 3 Ai β0 Eq 28 2 4 Av Rl re rb β β 0 0 Eq 28 2 5 Aov Rl Rs Rin β0 Eq 28 2 6 Example 28 2 Using the same inputs as in the previous problem with t...

Page 179: ...age gain Av and overall voltage gain Aov for the amplifier system Aov includes the effect of source impedance Rs β α α 0 0 1 0 Eq 28 3 1 Rin rb re Rl β β 0 0 1 b g Eq 28 3 2 Ro re Rs rb b g β0 Eq 28 3 3 Ai rrc rrc Rl re 1 0 α b g Eq 28 3 4 Av Rl re Rl α0 Eq 28 3 5 Aov Rl Rs Rin Rl β β 0 1 0 1 b g b g Eq 28 3 6 Example 28 3 An amplifier in a common collector configuration has a gain α0 of 0 99 The ...

Page 180: ...10 kΩ The external gate resistance is 1 MΩ and the drain resistance is 125 kΩ The transconductance is 1 6 x 10 3 siemens Find the mid band parameters Entered Values Calculated Results Solution All of the equations need to be used to compute the solution for this problem Press to display the input screen enter all the known variables and press to solve the equations The computed results are shown i...

Page 181: ...tions The computed results are shown in the screen display above PQYP 8CTKCDNGU IO AUKGOGPU TF AMΩ 4N AMΩ QORWVGF 4GUWNVU X µ µ 4KP AΩ 4Q A Ω 28 6 FET Common Drain The first equation defines the amplification factor µ µ in terms of transconductance gm and drain resistance rd The second equation computes input resistance Rin as a function of load resistance Rl rd and µ µ Voltage gain Av is defined ...

Page 182: ...tance rb and source resistance Rs The final equation computes overall current gain Ai for the transistor pair in terms of β β0 re Rl and the external base resistance RBA Rin re re Rl β β 0 0 b g c h Eq 28 7 1 Ro re re rb Rs β β 0 02 b g c h Eq 28 7 2 Ai RBA RBA Rl re β β 0 0 2 b g Eq 28 7 3 Example 28 7 Transistors in a Darlington pair having a β β0 value of 100 are connected to a load of 10 kΩ Th...

Page 183: ...nces the source impedance and the current gain β β0 Rin rb re β0 Eq 28 8 1 Ro rrc β0 Eq 28 8 2 Av Rl re Rs β02 Eq 28 8 3 Example 28 8 An amplifier circuit has a base emitter and load resistance of 1 5 kΩ 25 Ω and 10 kΩ respectively The configuration has a value of β β0 equal to 100 The source and collector resistances are 1 kΩ and 100 kΩ Find the voltage gain input and output resistances Upper Scr...

Page 184: ...nding input resistance Rin and output resistance Ro β α α 0 0 1 0 Eq 28 9 1 Av Rl re Rl F HG I KJ β β 0 2 0 Eq 28 9 2 Ai α β 0 0 Eq 28 9 3 Rin re rb β0 Eq 28 9 4 Ro rrc Eq 28 9 5 Example 28 9 An emitter coupled pair amplifier is constructed from transistors with α0 0 98 The emitter base and collector resistances are 25 Ω 2000 Ω and 56 kΩ respectively If the load resistance is 10 kΩ find the mid ba...

Page 185: ...ind the common mode differential resistance and gains Upper Screen Display Lower Screen Display Solution Use all of the equations to compute the solution for this problem Select these by highlighting each equation and pressing the key Press to display the input screen enter all the known variables and press to solve the equations The computed results are shown in the screen displays above PQYP 8CT...

Page 186: ...esistance is 12 kΩ and the transconductance is 6 8 x 10 3 siemens Entered Values Calculated Results Solution Use all of the equations to compute the solution for this problem Select these by highlighting each equation and pressing the key Press to display the input screen enter all the known variables and press to solve the equations The computed results are shown in the screen displays above PQYP...

Page 187: ...ionships could generate erroneous results or no solution at all Read the description of each equation set to determine which equations in a sub topic form a consistent subset before attempting to compute a solution Class A Amplifier Power Transistor Push Pull Principle Class B Amplifier Class C Amplifier Variables The variables used in this section are listed along with a brief description and uni...

Page 188: ...ircuit parameter Ω XC1 π equivalent circuit parameter Ω XC2 π equivalent circuit parameter Ω XL π series reactance Ω 29 1 Class A Amplifier The eight equations in this section form the basis for analyzing a Class A amplifier with an ideal transformer coupled to a resistive load Rl The first equation specifies the equivalent load resistance R from the load resistance Rl in the secondary winding of ...

Page 189: ...imum current is 110 mA Upper Display Lower Display Solution Use all of the equations to solve this problem Press to display the input screen enter all the known variables and press to solve the equation The computed results are shown in the screen displays above PQYP 8CTKCDNGU O 3 AO OCZ AO P 4N Ω 8 A8 8 OZ A8 8 OP A8 QORWVGF 4GUWNVU 8 A8 ζ 2FE A9 2Q A9 22 A8 4 AΩ 4N A Ω 822 A8 29 2 Power Transist...

Page 190: ...ain of 125 A 750 Ω base resistance is coupled to an external emitter resistance of 10 kΩ The ambient temperature is 75 F and the thermal resistance of the unit is 10 C W The power that needs to be dissipated is 12 5W The base emitter voltage is 1 25V while ICBO is 1_ma Find the junction temperature collector current and the instability factor Input variables Computed results Solution We note from ...

Page 191: ...Use the third equation to compute the solution for this problem Select the equation using the highlight bar and pressing the key Press to display the input screen enter all the known variables and press to solve the equation The computed result is shown in the screen display above PQYP 8CTKCDNGU 0 0 44 AΩ 8 A8 QORWVGF 4GUWNVU 2Q A9 29 4 Class B Amplifier Power transistors that are connected in a p...

Page 192: ... π K 4 Eq 29 4 6 Pd VCC R K K F HG I KJ 2 4 2 2 π π Eq 29 4 7 V gm Rl Vm hOE Rl 1 2 2 1 1 2 F H GGG I K JJJ Eq 29 4 8 IC gm Vm hOE Rl F H GGG I K JJJ π 1 1 2 Eq 29 4 9 Example 29 4 A Class B amplifier provides 5 W to an effective load of 50 Ω The collector voltage is 25 V If the peak current is 500 mA find the average DC current and the efficiency of power conversion Input variables Computed resul...

Page 193: ...he equations to compute XL have two distinct forms If this equation is part of your selection be advised to ensure that the proper inputs are specified ζ I Rrc I Rrc RR 2 2 0 b g Eq 29 5 1 XXC V Q Po 02 Eq 29 5 2 XL XXC Q Q 2 2 1 Eq 29 5 3 XC Rl Q 1 Eq 29 5 4 XL Q Rl Rl RR 1 2 d i Eq 29 5 5 XC RR Q 2 2 Eq 29 5 6 Example 29 5 A Class C amplifier is supplying a tuned circuit with a quality factor of...

Page 194: ...oltage V V2 Secondary voltage V XX1 Primary reactance Ω XX2 Secondary reactance Ω Xin Equivalent primary reactance Ω Xl Reactive part of load Ω Zin Primary impedance Ω ZL Secondary load Ω 30 1 Ideal Transformer Four equations describe the properties of an ideal transformer The first equation relates the primary and secondary voltages V1 and V2 in terms of the primary and secondary turns N1 and N2 ...

Page 195: ...ns to solve this problem Press to display the input screen enter all the known variables and press to solve the equation set The computed results are shown in the screen display above PQYP 8CTKCDNGU AO 0 0 8 A8 AΩ QORWVGF 4GUWNVU A 8 A8 KP AΩ 30 2 Linear Equivalent Circuit The first two equations define the primary voltage and current V1 and I1 in terms of V2 and I2 The last two equations expand t...

Page 196: ... with an impedance of 12 5 kΩ Find the voltage and current on the secondary side in addition to the equivalent impedance on the primary side Upper Screen Display Lower Screen Display Solution All of the equations are used to solve this problem Press to display the input screen enter all the known variables and press to solve the selected equation set The computed results are shown in the screen di...

Page 197: ...at all Read the description of each equation set to determine which equations in a sub topic form a consistent subset before attempting to compute a solution Energy Conversion DC Generator Separately Excited DC Generator DC Shunt Generator DC Series Generator Separately Excited DC Motor DC Shunt Motor DC Series Motor Permanent Magnet Motor Induction Motor I Induction Motor II Single Phase Inductio...

Page 198: ...tance Ω Re Ext shunt resistance Ω Rel Magnetic reluctance A Wb Rf Field coil resistance Ω Rl Load resistance Ω Rr Equivalent rotor resistance Ω Rs Series field resistance Ω Rst Stator resistance Ω s Slip unitless sf Slip for forward flux unitless sm Maximum slip unitless T Internal torque N m Tb Backward torque N m Tf Forward torque N m Tgmax Breakdown torque N m TL Load torque N m Tloss Torque lo...

Page 199: ...ced by Ns turns moving with an angular velocity ωs sweeping a magnetic flux of φ Es Ns s ω φ 2 Eq 31 1 4 Example 31 1 A conductor having a length of 15 cm and a cross sectional area of 0 5 cm2 is subjected to a magnetic induction of 1 8 T and a field intensity of 2 8 A m The magnetic reluctance is 0 46 A Wb The conductor has 32 turns and is moving at a rotational speed of 62 rad s Find the magneti...

Page 200: ...to its electrical counterpart namely the emf and current in the armature Ea and Ia and the voltage and current in the field windings Ef and If The next equation for torque connects T with K φ and the current Ia T m Ea Ia Ef IIf ω Eq 31 2 6 T K Ia φ Eq 31 2 7 The armature resistance is given by the equation for Ra in terms of N ap coil length L area A and its resistivity ρ Ra N L ap A ρ 2 Eq 31 2 8...

Page 201: ...t resistance re and field coil resistance Rf The next equation evaluates armature induced voltage Ea as a function of machine constant K mechanical radian frequency ωm and flux φ IIf Vfs Rf Re Eq 31 3 1 Ea K m ω φ Eq 31 3 2 The third and fourth equations are alternate forms of expressing terminal voltage Vt in terms of load current IL load resistance Rl armature resistance Ra Vt IL Rl Eq 31 3 3 Vt...

Page 202: ...ine constant K the mechanical angular frequency ω ωm and flux φ φ Ea K m ω φ Eq 31 4 1 The second equation defines terminal voltage Vt in terms of the field current IIf external resistance Re and field coil resistance Rf The third equation computes Vt in terms of load current IL and load resistance Rl The fourth equation expresses Vt as the induced emf Ea minus armature IR drop Vt Rf IIf Re Eq 31 ...

Page 203: ... be the same The second equation computes the terminal voltage Vt in terms of the induced emf Ea load current IL armature resistance Ra and series field windings Rs Ia IIf Eq 31 5 1 Vt Ea Ra Rs IL Eq 31 5 2 Example 31 5 Find the terminal voltage of a series generator with an armature resistance of 0 068 Ω and a series resistance of 0 40 Ω The generator delivers a 15 A load current from a generated...

Page 204: ...al power relationship between ω ωm and φ φ by the inverse quadratic relationship The next set of equations show the relationship between T the torque lost due to friction Tloss and the torque load TL The last equation in this set shows relationship of power with torque T and angular velocity ω ωm ω φ φ m Vt K Ra T K b g2 Eq 31 6 6 T Tloss TL Eq 31 6 7 P T m ω Eq 31 6 8 Example 31 6 Find the termin...

Page 205: ...If Re b g Eq 31 7 1 Vt K m Ra Ia φ ω Eq 31 7 2 The third equation refers to the torque available at the load TL due to the current Ia in the armature minus the loss of torque Tloss due to friction and other reasons TL K Ia Tloss φ Eq 31 7 3 The fourth equation gives the definitive relationship between the back emf Ea K φ φ and ω ωm Ea K m ω φ Eq 31 7 4 The next equation displays the reciprocal qua...

Page 206: ...Ra Rs Rd IL φ ω b g Eq 31 8 1 TL K IL Tloss φ Eq 31 8 2 The third equation defines the back emf in the armature Ea in terms of K φ and mechanical frequency ωm The fourth equation shows torque generated at the rotor due the magnetic flux φ and current IL Ea K m ω φ Eq 31 8 3 T K IL φ Eq 31 8 4 The next equation shows a reciprocal quadratic link between ωm Vt K φ Ra Rs Rd and torque T ω φ φ m Vt K R...

Page 207: ...e shown in the screen displays above PQYP 8CTKCDNGU 4C Ω 4F AΩ 4U A Ω 6 A0O 8V A8 ω ωO AT U QORWVGF 4GUWNVU C A8 A8 A A φ φ A9D A9D 31 9 Permanent Magnet Motor These five equations characterize the basic features of a permanent magnet motor The first equation shows the back emf Ea in terms of machine constant K flux φ φ and radian velocity ω ωm The second equation shows the connection between gene...

Page 208: ...key variables used in evaluating the performance of an induction motor The first equation expresses the relationship between the radian frequency induced in the rotor ω ωr the angular speed of the rotating magnetic field of the stator ω ωs the number of poles p and the mechanical angular speed ω ωm ω ω ω r s p m 2 Eq 31 10 1 The second third and fourth equations describe the slip s using ω ωr and ...

Page 209: ...1 10 10 Rr RR KM 1 2 Eq 31 10 11 Example 31 10 Find the mechanical power for an induction motor with a slip of 0 95 a rotor current of 75 A and a resistance of 1 8 Ω Entered Values Calculated Results Solution Choose the next to last equation to compute the solution Select by highlighting and pressing Press to display the input screen enter all the known variables and press to solve the equation PQ...

Page 210: ... ωs Tm p s Va Rst XL Rst max 3 4 2 2 2 ω Eq 31 11 4 The maximum slip sm in the fifth equation represents the condition when dT ds 0 sm Rr Rs XL 2 2 Eq 31 11 5 The sixth equation defines the so called breakdown torque Tgmax of the motor The final equation relates Rr with machine constant KM and the rotor resistance per phase RR1 Tg p s Va Rs XL Rst max 3 4 2 2 2 ω Eq 31 11 6 Rr RR KM 1 2 Eq 31 11 7...

Page 211: ...s Isf Rr sf 2 1 2 2 ω Eq 31 12 2 Tb p s Isb Rr sf 2 1 2 2 2 ω b g Eq 31 12 3 Example 31 12 Find the forward slip for an eight pole induction motor with a stator frequency of 245 rad s and a mechanical radian frequency of 62 5 rad s Entered Values Calculated Results Solution The first equation is needed to compute the solution Select by highlighting and pressing Press to display the input screen en...

Page 212: ...nous machine with a mechanical rotational velocity of 31 rad s The motor has eight poles a field current of 1 8 A and experiences an applied voltage of 130 V Entered Values Calculated Results Solution The first and second equations are needed to compute the solution Select these using the cursor bar and pressing Press to display the input screen enter all the known variables and press to solve the...

Page 213: ...Properties Fourier Laplace and z transforms properties commonly used Fundamental Constants SI prefixes and the Greek Alphabet Pull down menu on TI 89 and TI 92 Plus Unlike Analysis and Equations the screen formats for the topics in the Reference section can differ significantly depending on the information presented Some Reference tables such as the Resistor Color Chart and Standard Component Valu...

Page 214: ...ion When accessing the Semiconductor Data section a dialog box appears listing the available topics Use D key to move the highlight bar to 3 5 and 2 6 Compounds and press This displays the electronic and physical properties of the compound gallium phosphide GaP Sections in Semiconductor Data Properties of GaP Materials available in 3 5 2 6 Compounds section Properties of CdS Note that GaP has an a...

Page 215: ...ided into three sub topics Definitions Properties and Transform Pairs For example navigate from Transforms Fourier Transforms Transform Properties A screen display below shows the lists the equations displaying the fundamental properties of Fourier transform pairs Note that the name of the selected transform equation appears in the status line For example if the highlight bar is moved to the sixth...

Page 216: ...EE Pro for TI 89 92 Plus Reference Introduction to Reference Section 4 Regular View of Laplace Transform Inverse View of Laplace Transform ...

Page 217: ... of the color chart is included in the software and is displayed when the function key is pressed Table 33 1 Description of Colors in resistors Band positions represent 3 Band 4 Band 5 Band Band 1 digit digit digit Band 2 digit digit digit Band 3 multiplier multiplier digit Band 4 N A tolerance multiplier Band positions represent 3 Band 4 Band 5 Band Band 1 digit digit digit Band 2 digit digit dig...

Page 218: ...lerance Resistor Tolerance Returns a percent 33 1 Using the Resistor Color Chart Select this topic from the Reference section and press 1 Select the number of bands on the resistor display automatically updates for the entry 2 In each Band field select a color using B and pressing 3 The results are displayed in the Value and Tolerance lines of the display Example 33 1 Find the value and tolerance ...

Page 219: ...ress to display selection The values range from tolerance 20 down to 0 05 Component Type of Component Press to display component choices Resistor Inductor or Capacitor Output Field Value Closest Standard Value to the Desired Value Returns a Preferred Value Bands Resistor Color Bands if the Component is a Resistor Returns the color bands in a resistor Example 34 1 A design calculation yields 2 6 mi...

Page 220: ...Crystal Structure ρ Density εr Relative permittivity Nc Density of states CB Nv Density of states VB mle Longitudinal e mass mte Transverse e mass mlh Light hole mass mhh Heavy hole mass φ Electron affinity EG Band gap ni Intrinsic Density a Lattice constant αth Thermal expansion coefficient MP Melting Point τ Carrier lifetime µn Electron mobility µp Hole Mobility Raman E Raman Photon Energy Sp Ht...

Page 221: ...C unitless g cm 3 Si Donor Levels Output Field Silicon Donor levels are displayed relative to the conduction band The donor list includes the following Li Sb P As Bi Te Ti C Se Se Cr Ta Ta Cs Ba S Mn Mn VB All values are displayed in electron volts eV relative to the conduction band In some cases there is a VB designation next to the element In these cases the location of the donor level is refere...

Page 222: ...ll values in are in electron volts eV relative to the valence band For some elements there is a CB appended to the element name In these cases the location of the acceptor level is referenced relative to the conduction band For example the acceptor level for Mg is displayed as 0 11_eV CB follows the element name Mg indicating that the acceptor level is 0 11_eV below the conduction band On the othe...

Page 223: ... the following parameters SiO2 Value in µm Si3N4 Value in µm Order Value as a number Silicon Press to display the pull down menu listing the color of silicon with SiO2 or Si3N4 deposited on the surface 1 Silicon 2 Brown 3 Golden brown 4 Red 5 Deep Blue 6 Blue 7 Pale Blue 8 Very Pale Blueœ 9 Silicon A Light Yellow ZnSe B Yellow C Orange Red D Red E Dark red F Blue G Blue Green H Light Green I Orang...

Page 224: ...own in this section and can be viewed by pressing the function key The Boolean algebra rules are sixteen in number and are listed as functions F0 F15 They are also identified by their name as shown in the Table 36 1 Example 36 1 Find the properties of an Exclusive OR XOR 1 Use the arrow key D to move the highlight bar down the list Continue scrolling until the status line reads Exclusive OR XOR 2 ...

Page 225: ...rence Boolean Expressions 13 F6 x y x y Exclusive OR XOR F7 x y OR F8 x y NOT OR NOR F9 x y x y Equivalence XNOR F10 y Complement NOT F11 x y Implication F12 x Complement NOT F13 x y Implication F14 x y NOT AND NAND F15 1 Identity ...

Page 226: ...e Transforms or z Transforms 2 Select Definitions Properties or Transform Pairs 3 Use the D or C keys to move to the transform line desired 4 The forward transforms are displayed by default Pressing toggles between the forward and inverse formats 5 Press B to view the transform property in Pretty Print Information is presented on either side of the colon as shown below The term on the left side of...

Page 227: ...ansforms and press 2 Press a second time to access Definitions 3 Move the highlight bar to the second definition and press B to display the equation in Pretty Print format 4 Press A or B to scroll Press any key to return to the previous screen Fourier Definitions Pretty Print of Fourier Time domain Transform Example 38 2 View the Laplace transform of the time function f t t 1 In the initial Transf...

Page 228: ...tion Const Description π circle ratio µq mass µ mass e e Napier constant φ0 magnetic flux quantum γ Euler constant µB Bohr magneton φ golden ratio µe e magnetic moment α fine structure µN nuclear magneton c speed of light µp p magnetic moment ε0 permittivity µµ µ magnetic moment F Faraday constant a0 Bohr radius G Newton s Gravitational constant R Rydberg constant g acceleration due gravity c1 1st...

Page 229: ...ew a constant use the arrow key D key to move the highlight bar to the value and press the View key The status line at the bottom of the screen gives a verbal description of the constant Example 39 1 Look up the value of π 1 Pi is the first value to appear in the constant sections Make sure it is selected by the highlight bar using the arrow keys 2 Access the View function by pressing key 3 Press ...

Page 230: ... the highlight bar to select a SI prefix multiplier The name of the prefix is displayed in the status line The prefix and multiplier can be viewed by pressing the key Table 40 1 SI Prefix Table Prefix Multiplier Prefix Multiplier Y Yotta Z Zetta E Exa P Peta T Tera G Giga M Mega k Kilo h Hecto da Deka 1E24 1E21 1E18 1E15 1E12 1E9 1E6 1E3 1E2 1E1 d deci c Centi m Milli µ Micro n Nano p Pico f Femto...

Page 231: ...ial keystrokes are listed in the TI 89 manual They are repeated here for convenience of the user Alternatively 2 or followed by will access an internal menu listing several Greek characters Table 40 1 Key stroke Sequence Greek Letter Key stroke Sequence Greek Letter c j Á α c j c β c j b δ c j Î b c j e ε c j Í φ c j m γ c j Î m Γ c j y λ c j z µ c j π c j Î Π c j ρ c j ª σ c j Îª Σ c j Ü τ c j ω ...

Page 232: ...seems to be occurring How can I halt this process A Some computations can take a long time particular if many equations and unknowns are being solved or a complex analysis function has been entered Notice if the message in the status line at the bottom right of the screen reads BUSY This indicates that the TI math engine is attempting to solve the problem Pressing the key usually halts a computati...

Page 233: ...ent variables in the current folder are cleared To preserve data under variable names which may conflict with EE Pro s variables run EE Pro in a separate folder using the guidelines above Q An item which is supposed to be displayed in a menu doesn t appear A Some menus have more than eight items If an arrow ï appears next to the digit 8 use the arrow key D to scroll the menu and view the remaining...

Page 234: ... to be in common SI units F m A Ω etc which are stated in the status line for each parameter In some cases variables will be expressed in units arbitrarily chosen by the user example The variable len in Transmission Lines can be entered in km m miles however the answers will be expressed in units of len If a value is entered that is inconsistent with the expected data type an error dialog will app...

Page 235: ...ra characters in the units to match TI s syntax This causes extra characters to appear or symbols to appear differently in Pretty Print Q There are already values stored in some of my variables How do I clear those values A The values can be accessed via VAR LINK menu To delete variables in VAR LINK use the file management tool provided use ƒ key to access file management tools check the variables...

Page 236: ...e made available AS IS to you are any subsequent user Although no warranty is given for the License Material the media if any will be replaced if found to be defective during the first three 3 months of use For specific instructions contact the TI webstore via www ti com THIS PARAGRAPH EXPRESSES da Vinci s MAXIMUM LIABILITY AND YOUR SOLE AND EXCLUSIVE REMEDY 6 Limitations da Vinci makes no warrant...

Page 237: ...RES 1 800 842 2737 e mail ti cares ti com For questions specific to the use and features of EE Pro contact da Vinci Technologies Group Inc phone 1 541 757 8416 Ext 109 9 AM 3 PM P S T Pacific Standard Time Monday thru Friday except holidays email support dvtg com Customers outside the U S Canada Puerto Rico and the Virgin Islands For questions that are specific to the purchase download and install...

Page 238: ...uit Analysis 3rd Edition Charles E Merrill Publishing Company Columbus OH 1977 9 Jordan Edward C Reference Data for Radio Engineers Radio Electronics Computers and Communications 7th Edition Howard and Sams Indianapolis IN 1985 10 Coughlin Robert F and Driscoll Frederick F Operational Amplifiers and Linear Integrated Circuits Prentice Hall Englewood Cliffs NJ 1991 11 Johnson David E and Jayakumar ...

Page 239: ...n New Jersey 1981 24 Yarborough Raymond B Electrical Engineering Reference Manual Professional Publications Belmont CA 1990 25 Schroeder Dieter K Advanced MOS Devices Addison Wesley Reading MA 1987 26 Neudeck Gerold W The PN Junction Diode Addison Wesley Reading MA 1983 27 Pierret Robert F Semiconductor Fundamentals Addison Wesley Reading MA 1983 28 Pierret Robert F Field Effect Devices Addison We...

Page 240: ...Screen 156 x 39 pixels 25 characters 4 lines 236 x 51 pixels 39 characters 6 lines Vertical Split Screen 77 x 80 pixels 12 characters 10 lines 117 x 104 pixels 19 characters 13 lines Vertical Split Screen 1 3rd Not supported 236 x 33 pixels 39 characters 4 lines Vertical Split screen 2 3rd Not supported 236 x 69 pixels 39 characters 8 lines Horizontal Split Screen 1 3rd Not supported 77 x 104 pixe...

Page 241: ...92 Plus key strokes Representation in the manual Function Keys F1 ƒ ƒ ƒ F2 F3 F4 F5 F6 2 ƒ ˆ ˆ F7 2 F8 2 Š Š Trig Functions Sin 2 Ú W W Cos 2 Û X X Tan 2 Ü Y Y Sin 1 Ú 2 W SIN 1 Cos 1 Û 2 X COS 1 Tan 1 Ü 2 Y TAN 1 Alphabet keys A j Á Ñ A B j c Ò B C j d Ó C D j b D E j e E F j Í Ô F G j m G H j n H I j o I J j p J K j K L j y L M j z M N j Õ N O j Ö O P j P Q j Q R j R S j ª S T Ü T U j U ...

Page 242: ...ation in the manual Log EXP LN 2 Ù x x ex Ù 2 x s Special Characters π 2 Z 2 Z T θ Z Ï Ï Negation i 2 2 I 2 J Graphing Functions Y ƒ W Window E Graph R Editing Functions Cut 2 5 Copy 6 Paste N 7 Delete 0 0 8 Quit 2 N K Insert 2 0 2 0 Recall 2 2 Store Backspace 0 0 0 Parenthesis Brackets c c c d d d 2 c 2 c 2 d 2 d 2 b 2 b g 2 e 2 e h Math Operations Addition Subtraction Multiplication p p p Divisi...

Page 243: ... µ 2 µ Â Operations Greater than 2 2 Ã cont Absolute value Í 2 K Í Angle 2 Square root 2 p 2 p Approximate Tables TblSet T Table Y Modifier Keys 2nd 2 2 2 Diamond Shift Î Î Alphabet j j Alphabet lock 2 j Special Areas Math 2 z 2 z I Mem 2 2 Var Link 2 2 Units 2 ª 9 Char 2 2 Ans 2 2 Entry 2 2 Special Characters Single Quote 2 Á È Double Quote 2 2 L É Back slash 2 2 Á Ì Underscore 3 Colon 2 y 2 Ï Ë ...

Page 244: ...µ 0 Comma b b Decimal point Main Functions Home Q Function Specific Key TI 89 Key strokes TI 92 Plus key strokes Representation in the manual Main Functions Mode 3 3 cont Catalog 2 Clear M M Custom 2 ª Enter ON OFF 2 2 ESCAPE N N N Application O O O Cursor Movement Top C C C Right B B B Left A A A Bottom D D D ...

Page 245: ...occur when the system is low on available memory resources Consult your TI 89 manual on methods of viewing memory status and procedures for deleting variables and folders to make more memory available 4 The message Unable to save EE Pro data will be displayed if EE Pro is unable to save information of its last location in the program before exiting due to low memory availability Consult your TI 89...

Page 246: ...der network that contains no defined elements Add elements to the network by pressing 4 Invalid variable reference Conflict with system variable or reserved name This error is displayed if a variable name is entered as an input which is identical to a variable name used by EE Pro This message is also displayed if an entered variable name is a reserved system variable A complete list of variables r...

Page 247: ...cash flow entries 13 The message Data Error Reinitializing project will occur if the project data has been altered or is inconsistent with current state EE Pro will restart E 3 Equation Messages If a value is entered that is inconsistent with the expected data type an error dialog will appear which lists the entry name the description and the expected data type s and the expected units If an error...

Page 248: ...erges upon may not be unique It may be possible to find a solution starting from a different initial guess To specify an initial guess enter a value for the unknown and then use F5 Opts 7 Want to designate it as the variable to solve for More information on the differences between the solve nsolve and csolve functions is listed in the TI 89 manual One complete useable solution found All of the unk...

Page 249: ...rt Yi1 yi99 z1 x y z99 x y yc θc zfact xrid ygrid zmin eyeφ θmax tstep difto1 fldres plotStep r1 θ r99 θ u1 n u99 n zc nc xmin ymin xres zmax eyeϕ θstep t0 dtime nmin sysMath Graph Zoom Zxmin Zymin Zxres Ztmin Ztmaxde Zmax zeyeϕ zpltstep Zxmax Zymax zθmin ztmax ztstepde zzsc1 znmin Zxscl Zyscl zθmax ztstep ztplotde zeyeθ znmax Zxgrid Zygrid zθstep zt0de zzmin zeyeφ zpltstrt Statistics ü Σx2 Σy2 me...

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