PocketProfessional™ 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.
Notice This manual and the examples contained herein are provided “as is” as a supplement to EE•Pro application software available from Texas Instruments for TI-89, and 92 Plus platforms. Da Vinci Technologies Group, Inc. (“da Vinci”) makes no warranty of any kind with regard to this manual or the accompanying software, including, but not limited to, the implied warranties of merchantability and fitness for a particular purpose.
Table of Contents 1 Introduction to EE•Pro.. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . i 1.1 Key Features of EE•Pro . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . i 1.2 Download/Purchase Information . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ii 1.3 Manual Ordering . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6 Filter Design . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25 6.1 Chebyshev Filter . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25 Example 6.1. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26 6.2 Butterworth Filter . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
12.5 Binary Conversions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Example 12.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12.6 Binary Comparisons . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Example 12.6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12.7 Karnaugh Map . . . . . .
17.5 Parallel Wires . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Example 17.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17.6 Coaxial Cable . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Example 17.6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17.7 Sphere . . . . . .
21.4 RC Step Response . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Example 21.4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.5 RL Series-Parallel . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Example 21.5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.
Variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68 26.1 Basic Inverter . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69 Example 26.1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69 26.2 Non-Inverting Amplifier . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Example 28.2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 104 28.3 BJT (Common Collector) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 105 Example 28.3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 105 28.4 FET (Common Gate) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 106 Example 28.4 . . .
31.8 DC Series Motor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Example 31.8 . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.9 Permanent Magnet Motor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Example 31.9 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.10 Induction Motor I. . . . . . . . . . . . . .
Appendix Appendix A Frequently Asked Questions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 A.1 Questions and Answers. . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . .1 A.2 General Questions . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . .1 A.3 Analysis Questions . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 A.
Chapter 1 Introduction to EE•Pro Thank you for your purchase of EE•Pro, a member of the PocketProfessional® Pro software series designed by da Vinci Technologies to meet the computational needs of students and professionals in the engineering and scientific fields. Many long hours and late nights have been spent by the designers of this software to compile and organize the subject material in this software.
Analysis-(Chapters 2-14) Analysis is organized into 12 topics and 33 sub-topics. The tools in this section incorporate a wide variety of analysis methods used by electrical engineers.
1.5 Differences between TI-89 and TI-92 plus EE•Pro is designed for two models of graphing calculators from Texas Instruments, the TI-92 Plus and the TI-89. For consistency, keystrokes and symbols used in the manual are consistent with the TI-89. Equivalent key strokes for the TI-92 plus are listed in Appendix D. 1.6 Beginning EE•Pro • To begin EE•Pro, start by pressing the O key. This accesses a pull down menu. Use the D key to move the cursor bar to EE Pro Elec. Eng. and press ¸.
• correspond to the topic headings in the software menus. The chapters list all the equations used and explains their physical significance. These chapters also contain example problems and screen displays of the computed solutions. The Appendices contain trouble-shooting information, commonly asked questions, a bibliography used to develop the software, and warranty information provided by Texas Instruments. 1.
Chapter 2 Introduction to Analysis The Analysis section of the software is able to perform calculations for a wide range of topics in circuit and electrical network design. A variety of input and output formats are encountered in the different topics of Analysis. Examples for each of the input forms will be discussed in some detail. • • • • • • The unit management feature in Equations is not present in Analysis due the variety of computation methods used in this section.
• Pressing „ will access the Analysis section of the software and display a pull down menu listing the topics available. There are 12 sections under Analysis. The sections are accessed by using the D key to move the highlight bar to the desired section and pressing ¸. Alternatively, any section can be accessed by entering the number associated with each section. Thus pressing ¨ will display a pop up menu for AC Circuits, while pressing z will list topics in Gain and Frequency.
Prm 1: z11_ Prm 2: z12_ Prm 3: z21_ Prm 4: z22_ Output Type y4 Parameter z11_; when h parameter is selected this changes to h11_. Parameter z12_; when h parameter is selected this changes to h12_. Parameter z21_; when h parameter is selected this changes to h21_. Parameter z22_; when h parameter is selected this changes to h22_. The right arrow4indicates additional choices for this parameter. Select this using the cursor bar.
Note: If the calculator is turned off automatically or manually while a results screen is being displayed, when EE•Pro is accessed again via the O key, the software automatically bypasses the home screen of EE•Pro and returns to the screen result display. 2.4 Special Function Keys in Analysis Routines When Analysis functions are selected, the function keys in the tool bar access or activate features which are specific to the context of the section.
ˆ ‰ Š • • • 6: (know)- Not active 7: Want - Not active “Edit” - Brings in a data entry line for the highlighted parameter. “Choose” in Capital Budgeting enabling the user to select from one of nine projects. “√ √ Check” requesting the user to press this key to select a highlighted parameter for use in an Analysis computation. Appears only when solving problems in the Ladder Network section and is labeled "Add"; this displays the input screen allowing the user to add new elements to a ladder network.
The Gain and Frequency section under the Analysis menu offers an example where a problem is set up in one topic area (Transfer Function) and graphed under another topic heading (Bode Diagram). A Transfer Function is set up as in the screen shown below. It is important to note that data for Zeroes and Poles is entered as a list, e.g., numbers entered within curly brackets separated by commas.
Example 2.1 (Numeric Results) Find the electrical Circuit Performance of an AC circuit consisting of a voltage source 110+15*i volts and an impedance of 25-12*i ohms. The load for the example is a capacitive impedance 70-89*i. Pop up menu in AC Circuits 1. 2. 3. 4. 5. 6. Input entry complete Computed output From the home screen of EE•Pro press the „ key labeled Analysis to display the pull down menu listing all the sections available under Analysis.
Input entered symbolically Calculated Output also symbolic Example 2.3 (Graphical Results) Construct a Bode diagram for a system with pole locations at 1000, 10000, 50000, a zero at 5000, and a proportionality constant of 1000000. From the home screen of EE•Pro, press the „ key to display the pull down menu listing all the sections available under Analysis. 2. Press z to access the Gain and Frequency section to view a pop up menu of available topics. 3.
2.6 Session Folders, Variable Names EE•Pro automatically stores its variables in the current folder specified by the user in 3 or the HOME screens. The current folder name is displayed in the lower left corner of the screen (default is “Main”). To create a new folder to store values 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).
Chapter 3 AC Circuits This chapter describes the software in the AC Circuits section and is organized under four topics. These topics form the backbone of AC circuit calculations. Impedance Calculations Voltage Divider Current Divider Circuit Performance 3.1 Impedance Calculations The Impedance Calculations topic computes the impedance and admittance of a circuit consisting of a resistor, capacitor and inductor connected in Series or Parallel.
Example 3.1 Compute the impedance of a series RLC circuit consisting of a 10 ohm resistor, a 1.5 Henry inductor and a 4.7 Farad capacitor at a frequency of 100 Hertz. 1. 2. 3. 4. 5. Input Screen Output Screen Choose Series for Config and RLC for Elements using the procedure described above. Enter 100 for Freq. Enter 10 for R, 1.5 for L, and 4.7 for C. Press „to calculate ZZ_ and YY_. The output screen shows the results of computation. 3.
1. 2. 3. 4. 5. Choose Impedance for Load Type Enter the value 110 + 25*i for Vs_. Enter {50, 75 + 22*i, 125 - 40*i} for ZZ_. Press „ to calculate IL_ and V_. The results of the computation are shown in the screen display above. 3.3 Current Divider This section demonstrates how to calculate individual branch currents in a load defined by a set of impedances or admittances connected in parallel. In addition, the voltage across the load is calculated.
3.4 Circuit Performance This section shows how to compute the circuit performance of a simple load connected to a voltage or current source. Performance parameters computed include load voltage and current, complex power delivered, power factor, maximum power available to the load, and the load impedance required to deliver the maximum power. Field Descriptions - Input Screen Load Type: Vs_: Zs_ ZL_: Is_: Ys_: YL_: (Type of Load) Press ¸ to select load impedance (Z) or admittance (Y).
Input Screen 1. 2. 3. 4. 5. Output: Upper Half Output: Lower Half Choose Admittance for Load Type. Enter the value 10 - 5*i for Is_. Enter the value .0025 - .0012*i for Ys_, and .0012 + .0034*i for a load of YL_. Press „ to calculate the performance parameters. The input and results of computation are shown above. EE Pro for TI-89.
Chapter 4 Polyphase Circuits This chapter describes Wye and ∆ arrangements in Polyphase Circuits. Wye ↔ ∆ Conversion Balanced Wye Load Balanced ∆ Load 4.1 Wye ↔ ∆ Conversion The Wye ↔ ∆ Conversion converts three impedances connected in Wye or ∆ form to its corresponding ∆ or Wye form, i.e., . Wye ↔ ∆ or ∆ ↔ Wye Input Fields Input Type: ZZA_: (∆ Impedance) ZZB_: (∆ Impedance) ZZC_: (∆ Impedance) ZZ1_: (Y Impedance) ZZ2_: (Y Impedance) ZZ3_: (Y Impedance) Z1 Selection choices are ∆→Wye or Wye→∆.
Example 4.1 - Compute the Wye impedance equivalent of a ∆ network with impedances 75+12*i, 75-12*i, and 125 ohms. 1. Select ∆ →Y for Input Type. . 2. Enter the values 75+12*i, 75 -12*i, 125 for ZZA_, ZZB_ and ZZC_. 3. Press „ to calculate ZZ1_, ZZ2_ and ZZ3_. Input Parameters The computation results are: Calculated Output ZZ1_: 34.0909 - 5.45455⋅i ZZ2_: 34.0909 + 5.45455⋅i ZZ3_: 20.9782 4.2 Balanced Wye Load A balanced Wye load refers to three identical impedance loads connected in a Wye configuration.
1. Select Balanced Wye Load. 2. Enter the value 50 + 25*i for ZZ_. 3. Press „ to calculate performance characteristics of the circuit. 1 I1 Z W12 V12 I3 W13 V31 Z 3 Z V23 I2 N 2 Input Screen Fig. 4.3 Balanced Wye network with 2 wattmeters Output Screen (upper half) Output Screen (lower half) The results of computation are listed below: V23_: V31_: V1N_: V2N_ : V3N_: I1_: I2_: I3_: P: W12: W13: -55 - 95.2628*i -55 + 95.2628*i 55 - 31.7543*i -55 - 31.7543*i 63.5085*i .625966 - .
VCA_: IA_: IB_: IC_: P: WAB: WAC: (Voltage in V across lines C and A) (Line Current A in A) (Line Current B in A) (Line Current C in A) (Phase Power in W) (Wattmeter reading in across lines A and B) (Wattmeter reading in W across lines A and C) A real or complex number, or algebraic exp. A real or complex number, or algebraic exp. A real or complex number, or algebraic exp. A real or complex number, or algebraic exp. A real number or algebraic expression. A real number or algebraic expression.
Chapter 5 Ladder Network This chapter describes ladder network analysis - a circuit reduction method by which branches of the circuit are treated as sides (series connection) or rungs (parallel or shunt connection) of a ladder. 5.1 Elements of a Ladder Network In the examples that follow, the left end of the ladder is the input end and the right end of the ladder is the output end, where the load is connected.
Capacitor - (ideal capacitor) A capacitor can be added as a rung (parallel) or side (series). Choose Series or Parallel for Config, and enter a value for C in Farads. C C Series C Parallel C L RL An RL series circuit can be added as a rung (parallel) or as an RL parallel circuit as a side (series). Choose Series or Parallel for Config, and enter a value for R in ohms and L in henrys.
v Voltage-Controlled I A controlled voltage can be added only in cascade connection. Specify base resistance and transconductance by entering values for rb in ohms and gm in siemens. Current-Controlled I A controlled current can be added only in cascade connection. Specify base resistance and common base current gain by entering values for rb in ohms and β. gm·v rb VCIS I β ·I rb (hie) (hfe) ICIS Transmission Line A transmission line can be added only in cascade connection.
5.2 Using the Ladder Network General instructions for entering the elements and computing the parameters of a ladder network. 1. The initial screen prompts the user for entry of values for Frequency and Load. 2. Build the ladder by adding elements to it. Press ‰ to insert the first element. Choose an element type and press ¸. Enter the appropriate values. Press „ to update the ladder with the new element just added. A second press of the „ key computes the electrical performance of the circuit. 3.
1. 2. 3. 4. 5. 6. 7. Enter 1E6 for Frequency. Enter 50 for Load. Press ‰ to add the first element and move the highlight bar in the pull down menu to Capacitor and press ¸ to display the input screen for the Capacitor. Select Parallel for Configuration of the capacitor and enter the value 50E-12 for C. Press ¸ to accept the element data and press „ to return a listing of the Ladder Network. Move the highlight bar to 1: Capacitor and press ‰ to enter the second element in this circuit.
1. Enter the frequency and load values: Frequency: 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. EE Pro for TI-89.
Chapter 6 Filter Design This chapter covers a description of the software under the heading Filter Design. Three filter designs are included in this section. Design computations result in the value of component elements comprising the filter. v Chebyshev Filter v Butterworth Filter v Active Filter 6.1 Chebyshev Filters This section of the software computes component values for Chebyshev filters between equal terminations.
Field Descriptions - Output Screen Element1: number. Element2: …. ElementN: (First element in parallel) Returns a real (Second element in series) Returns a real number. (nth element in series because N is always odd) Returns a real number. Example 6.1 Design a low-pass Chebyshev filter with a cutoff at 500 Hz, a termination resistance of 50 ohm, 3 dB pass band ripple, and a 30 dB attenuation at 600 Hz. Input Screen 1. 2. 3. Output Screen Enter 50 for R, 500 for f0, and 600 for f1.
Field Descriptions - Input Screen Char: (Bandpass Characteristic) R: f0: Press ¸ to select Low Pass, High Pass, Band Pass, or Band Elimination. (Termination Resistance in Ohms) Enter a real number. (Cutoff Frequency in Hz - for Low Pass and High Pass) Enter a real number. (Center Frequency in Hz - for Band Pass and Band Elimination) Enter a real number. (Center Frequency in Hz - for Band Pass and Band Elimination) Enter a real number.
C4 High Pass Filter R5 C3 C1 + R2 C4 Band Pass Filter R5 C3 R1 + R2 Fig. 6.3 Active Filter Configurations Field Descriptions - Input Screen Type: f0: A: (Filter Type) (Band Cutoff in Hz) (Midband Gain in dB) Q: (Quality Factor: Q = C: (Capacitor in F) Press ¸ to select Low Pass, High Pass, or Band Pass. Enter a real number or algebraic expression of defined terms. Enter a real number or algebraic expression of defined terms.
Chapter 7 Gain and Frequency This chapter covers the basic principles of circuit analysis using a transfer function model and plots the resulting equations using the classical graphical representation often referred to as a Bode diagram for gain or phase: v Transfer Function v Bode Diagrams 7.1 Transfer Function A transfer function is defined as the ratio of an output to its input signal and is generally modified by a network between the two.
FG H FG H PFE_: IJ FG IJ KH K IJ FG IJ KH K s_ s_ 1− ... z1 z2 s_ s_ 1− 1− ... p1 p2 K ⋅ 1− Eq. 7.1.1 (Partial Fraction Expansion) Returns a symbolic expression of the form: FG H K1 K2 K3 + + +... s_ s_ s_ 1− 1− 1− p1 p2 p3 IJ FG K H IJ FG K H IJ K Eq. 7.1.2 Example 7.1 Find the transfer function and its partial fraction expansion for a circuit with a zero located at -10 r/s and three poles located at -100 r/s, -1000 r/s and -5000 r/s. Assume that the multiplier constant is 100000.
ω-Max: Autoscale: Label Graph Full Screen A-Min: A-Max: θ-Min: (Maximum Frequency in r/s - X axis) Enter a real number. (Scales the plot: hides A-min and A-max fields) Press ˆ to select. Attaches labels to the graph. Select to graph the results in a full screen display. (Minimum amplitude in dB (Y axis) - if Gain is chosen for Plot Type) Enter a real number. (Maximum amplitude in dB (Y axis) - if Gain is chosen for Plot Type) Enter a real number.
Chapter 8 Fourier Transforms This section contains software computing discrete “Fast” Fourier transforms and its inverse. v FFT v Inverse FFT 8.1 FFT A physical process can be monitored in two significantly different ways. First, the process can be monitored in time domain in analog or digital form. Second, the data can be collected in the frequency domain in analog or digital form. In a variety of measurement and digital storage devices, data is gathered at regular, discrete time intervals.
8.2 Inverse FFT This section focuses on transforming data from the frequency domain to the time domain. The inverse transform 1 hn = N N −1 ∑H k ⋅ e2π j⋅k⋅n N k =0 Eq. 8.2.1 algorithm uses the relationship displayed in the above equation, 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) Time: (Time Signal) Enter an array or list of real or complex numbers. Returns time signal. Example 8.
Chapter 9 Two-Port Networks This chapter covers the basic properties of two-port network analysis under three topical headings. Parameter Conversion Circuit Performance Interconnected Two-Ports 9.1 Parameter Conversion input output I2 I1 V1 z, y, h, g, a or b V2 Many electrical or electronic systems are often modeled as two-port networks with four variables, namely input voltage V1, input current I1, output voltage V2, and output current I2.
Field Descriptions - Input Screen The software is configured for the entry of two-port parameters; an input of one type and an output of another type. Input Type: Selecting this item (highlighting and pressing ¸ ) displays a menu of input types available to the user. Selecting z, y, h, g, a, or b automatically updates the names of the remaining parameter to reflect the choice of input type.
9.2 Circuit Performance This section computes the circuit performance of a two-port network. Given a voltage source with a finite impedance and a finite load, the software will compute the input and output impedances, the current and voltage gains, the voltage gain with reference to source, the current gain to the source, the power gain, the power available to the load, the maximum power available to the load, and the load impedance for maximum power deliverable to the load.
Input Screen Results: Upper Half Zin_ (Input Impedance) Iout_ (Output Current) Vout_ (Thevenin Voltage) Zout_ (Thevenin Impedance) Igain_ (Current Gain) Vgain_ (Voltage Gain) VgainAbs (Absolute Voltage Gain) GP_ (Power Gain) Pmax: (Maximum Power at ZLopt) ZLopt_ (Optimum Load Impedance) Results: Lower Half 11989.03667 ohms -.033236 amps 1.66666 volts .145833 ohms -199.651 .832638 .830906 664.947 4.76188 watts .145833 ohms 9.
Field Descriptions - Input Screen Connection: (Type of Connection) Press ¸ to select Cascade, Series Series, Parallel Parallel, Series Parallel, or Parallel Series. First Input Type: Press ¸ to select parameter type z, y, h, g, a, or b for entry. ..111_: ..112_: ..121_: ..122_: Enter a real or complex number, variable name, or algebraic expression of defined terms. Enter a real or complex number, variable name, or algebraic expression of defined terms.
Chapter 10 Transformer Calculations This chapter covers the software features used to perform to calculations for electrical transformers. This section is organized under three categories: Open Circuit Test Short Circuit Test Chain parameters 10.1 Open Circuit Test An open circuit test described here is usually performed at rated conditions of the primary or secondary side of a transformer. It is common practice to apply a voltage to the primary side.
Input Screen 1. 2. 3. Output Screen Enter the values 110 for V1 and 440 for V2. Enter 1 for I1 and 45 for PP1. Press „ to calculate and display the results, as shown above. 10.2 Short Circuit Test Short circuit tests are often a quick method used to determine the winding impedance of a transformer and are usually reported at rated kVA values.
1. 2. 3. Enter the values 5 for V1 and 18 for I2. Enter 5 for PP1, 30 kVA rating and 110 for V1R. Press „ to calculate the results, which are displayed in the output screen. 10.3 Chain Parameters Chain parameters (or the so-called ABCD parameters) are convenient problem-solving tools used in solving transmission and distribution problems.
EE Pro for TI -89, 92 Plus Analysis - Transformer Calculations 42
Chapter 11 Transmission Lines This chapter covers the basic principles of transmission line analysis covered under four categories of topics: Line Properties Line Parameters Fault Location Estimate Stub Impedance Matching 11.1 Line Properties This portion of the software computes the characteristic parameters of a transmission line from fundamental properties of the wires forming the transmission line.
β: (Phase Constant in degrees or radians/unit length) Returns a real number, or algebraic expression. λ: (Wavelength in unit length) vp: (Phase Velocity in unit length/s) Zoc_: (Open Circuit Impedance in Ω) Zsc_: (Short Circuit Impedance in Ω) ρ_ : (Reflection Coefficient) Returns a real number or algebraic expression. Returns a real number or algebraic expression. Returns a real or complex number, or algebraic expression. Returns a real or complex number, or algebraic expression.
d: (Distance to Load Location/ unit length) f: (Frequency in Hertz) Enter a real number or algebraic expression of defined terms or variable. Enter a real number or variable or algebraic expression of defined terms.
docmin: dscmin: A real number, variable name, or algebraic expression of defined terms. (Minimum distance to an open circuit fault in unit lengths) Returns a real number, variable name, or algebraic expression. (Minimum distance to short circuit fault in unit lengths) Returns a real number, variable name, or algebraic expression. Example 11.3 A transmission line measures a capacitive reactance of -275 ohms. The characteristic line impedance is 75 ohms, and has a phase constant of 0.025 r/length.
Example 11.4 A transmission line has a characteristic impedance of 50 ohms and a load of 75 ohms. Estimate the shorting stub location for matching purposes. Input Screen 1. 2. Output Screen Enter the values 50 and 75 for RR0, and ZL_ respectively. Press „ to calculate the results as shown in the screen displays above.
Chapter 12 Computer Engineering This chapter covers functions of interest in the design of logic systems and circuits.
Carry and Range Conditions - The shifting, rotating, arithmetic and bit manipulation operations can result in Carry and Range Flags being modified. The specific condition for the flags to be set depend upon the operation being performed. Range flag is set if the correct result of the operation cannot be represented in the current word size and the complement mode. When the result is out of range, the lower order bits that fit the word size are displayed.
† activates mode dialog box. Binary Arithmetic Screen Display Once the parameters have been set in the binary mode screen, pressing ¸ accepts the entries, while pressing N cancels the entries made. If an incorrect choice is made in the data entry, a dialog box pops up to alert the user that an error has been detected in the data entered. 12.
Input Screen 1. 2. 3. 4. Output Screen Press † to select hexadecimal for Base: for data, 128 bits for Wordsize, Unsigned number system and press ¸ to accept the choice. Enter 25a6 for the first Binary 1 number. Enter 128d for the second Binary 2 number. Choose MULT for Operator; press „ to compute the result. The screen displays above show the input screen and the resulting output.
SL SR RL Shifts the input to the left by one bit and shifts a 0 into the right-most position. Shifts the input to the right by one bit and shifts a 0 into the left-most position. Rotates the input to the left by one bit. The left-most bit wraps around to the right-most position. Rotates the input to the right by one bit. The right-most bit wraps around to the left-most position. Rotates bits left by one bit through carry.
Operator: Press ¸ or B to select. (Binary Operation) Note: All bit operations affect the carry flag only. SB CB B? ΣB Result: Sets the bit in the specified position. Clears the bit in the specified position. Tests the bit in the specified position. Returns the number of bits set. (Binary Function Value) Returns an integer result using the number base set in †/Binary Mode. Example 12.4 Find the bit sum of the a hexadecimal number AE34578F which is 32 bits wide. Input Screen Output Screen 1.
(Input Field for R→B16C and →IEEE) (Binary Function) Real: Function: Real to Binary Binary to Real Real to IEEE IEEE to Real Result: Enter a real number. Press ¸ to display choices available. Converts a real number to a binary number (binary, octal, decimal or hexadecimal). If a number with a fractional part is entered, the value is rounded to the nearest integer. Affects the range flag only. Converts a binary number (decimal, binary, hexadecimal or octal) to a real number.
== Compares two binary or real numbers. The result field shows a 1 if they are equal, otherwise 0. Compares two binary or real numbers. The result field shows a 1 if they are NOT equal, otherwise 0. Compares two binary or real numbers. If the number in the first field is less than the number in the second field, then the result field displays a 1, otherwise 0. Compares two binary or real numbers.
Vars: (List of Variables) A character string consisting of one letter variable names such as ABCD with no spaces between variable names. Prime Impl: (Prime Implicant Expression) Returns a logical algebraic expression in Sum of Products form. Example 12.7 Minimize a five input function with minterms at 0, 2, 4, 6, 8, 10, 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 1. 2. 3.
Chapter 13 Error Functions This topic demonstrates the procedure for computing numeric solutions for the Error Function and the Complementary Error Function. Error Functions 13.1 Using Error Functions The definitions of the Error Function and Complementary Error Function are: erf ( x ) = 2 π x z e − t dt 2 Eq. 13.1.1 0 2 erfc( x ) = 1 − erf ( x ) = π ∞ z e − t dt 2 Eq. 13.1.
Chapter 14 Capital Budgeting This chapter covers the four basic measures of capital budgeting: Payback Period Net Present Value Internal Rate of Return Profitability Index 14.1 Using Capital Budgeting This section performs analysis of capital expenditure for a project and compares projects against one another. Four measures of capital budgeting are included in this section: Payback period (Payback), Net Present Value (NPV), Internal Rate of Return (IRR), and Profitability Index (PI).
Multiple Graphs Activation of this feature enables the overlay of each successive graph (projects) on the same axis. Press ¸ to activate. Full Screen Graph Press ¸ to activate. Field Descriptions - Project Edit Screen Name: t0: t1: .... tn: (Project Name) (Investment at t=0) (Cash flow at t=1) Enter the name of the project. Enter a real number. Enter a positive or negative real number. (Cash flow at t=n) Enter a positive or negative real number. Example 14.
Cash Flow Input: plant1 1. Cash Flow Input: plant2 Output Screen With the highlight bar on the Project field, press ¸ to select a project to edit. Select a project that has not been used (this example uses projects 1 and 2). Press ¸ to return to the Capital Budgeting screen. 2. Press † to select Cash option enter the project edit screen and edit the cash flows. 3.
Chapter 15 Introduction to Equations The Equations section of EE•Pro contains over 700 equations organized into 16 topic and 105 sub-topic menus. • • • • • • The user can select several equation sets from a particular sub-topic, display all the variables used in the set of equations, enter the values for the known variables and solve for the unknown variables. The equations in each sub-topic can be solved individually, collectively or as a sub-set.
1. Pressing … displays the 2. Press ¨ to display the 3. Press © to display the ‘Equations’ menu. menu in ‘Resistive Circuits’ equations for ‘Ohm’s Law’. 4. Select equations by high- 5. Press „ to display the 6. Press ˆ to compute the lighting and pressing ¸. variables in the selected unknown variables. Note: equations. Enter the known variable values. Use the unit toolbar to select units. Computed results ‘‘ are distinguished from entered values ‘é‘.
1. Highlight the result to be 2. The unit menu tool bar 3. Press † to convert the. converted (R). Press ‡ to is now displayed at the top of result of ‘R’ from Ω to MΩ. display the Options menu, press y/Conv. the screen. 15.4 Viewing Multiple Solutions The math engine used by EE•Pro is able to manage complex values for variables (where they are permitted) and calculate more than one solution in cases where multiple answers exist for an entered problem.
the selected equations, or if the selected equations do not establish a closed form relationship between all of the entered values and the unknowns. In such a case, only the calculated variables will be displayed. Press „ to select all of the If there are more unknowns ...a partial solution will be equations in Resistive Formulas. than selected equations or relationships between variables are not established from the selected equations...
1*. Graph an equation by pressing …. Press ¸ 2. Select variables for Independent (x) and Variable units reflect settings in EE•Pro. to choose an equation. Dependent (y) variables. 4. Select graphing options by pressing ¸ 5. Split Screen Mode: Toggle between graph and settings 6: Full Screen Mode: Press O, j and Ñ to return by pressing 2 and O. to EE•Pro.
15.9 solve, nsolve, and csolve and user-defined functions (UDF) When a set of equations is solved in EE•Pro, three different functions in the TI operating system (solve, numeric solve, and complex solve) are used to find the most appropriate solution. In a majority of cases, the entered values are adequate to find numeric solutions using either the solve or csolve functions.
15.11 Why can't I compute a solution? If a solution is unable to be computed for an entered problem, you might check the following: 1. 2. 3.
2: Parallel Admittance 3: RLC Natural Response 4: Under-damped case 5: Critical Damping 6: Over-damped Case 10: Electrical Resonance 1: Parallel Resonance I 2: Parallel Resonance II 3: Lossy Inductor 4: Series Resonance 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 Amplifier B: Source Coupled JFET 16: Motors, Generators 1: Energy Conversion 2: DC Generato
Chapter 16 Resistive Circuits This software section performs routine calculations of resistive circuits. The software is organized in a number of topics listed below. Resistance and Conductance Ohm’s Law and Power Temperature Effects Maximum Power Theorem V and I Source Equivalence Variables A complete list of all the variables used, a brief description and applicable base unit is given below.
16.1 Resistance Formulas Four equations in this topic represent the basic relationship between resistance and conductance. The first equation links the resistance R of a bar with a length len and a uniform crosssectional area A with a resistivity ρ. The second equation defines the conductance G of the same bar in terms of conductivity σ, len and A. The third and fourth equations show the reciprocity of conductance G resistance R, resistivity ρ and conductivity σ. R= ρ ⋅ len A Eq. 16.1.
V = I ⋅R Eq. 16.2.1 P =V ⋅I Eq. 16.2.2 P = I2 ⋅R Eq. 16.2.3 V2 R Eq. 16.2.4 P = V 2 ⋅G Eq. 16.2.5 1 G Eq. 16.2.6 P= R= Example 16.2 - A 4.7_kohm load carries a current of 275_ma. Calculate the voltage across the load, power dissipated and load conductance. Entered Values Computed results Solution - Upon examining the problem, several choices are noted. Either Equations 16.2.1, 16.2.2 and 16.2.6, or 16.2.2, 16.2.3 and 16.2.5 or 16.2.2, 16.2.3 and 16.2.6 or 16.2.1, 16.2.2 and 16.2.
User entered Variables Computed results -PQYP 8CTKCDNGU 44 AΩ 44 AΩ 6 Aº( 6 Aº% %QORWVGF 4GUWNVU α A u- 16.4 Maximum DC Power Transfer The equations under this topic are organized to compute load voltage Vl, load current Il, power dissipation in the load P, maximum power available in the load Pmax, and load impedance Rlm needed for maximum power deliverable to the load.
-PQYP 8CTKCDNGU 8U A8 4U AQJO 4N AQJO %QORWVGF 4GUWNVU 2OCZ A9 2 A9 16.5 V and I Source Equivalence The two equations in this topic show the equivalence between a voltage source and a current source. A voltage source Vs with an internal series resistance of Rs is equivalent in all its functionality to a current source Is with a source resistance Rs connected across it. Vs Rs Vs = Is ⋅ Rs Is = Eq. 16.5.1 Voltage and current sources and its equivalence Eq. 16.
Chapter 17 Capacitors, Electric Fields This section covers seven topics to compute electric field properties and capacitance of various types of structures. When the section is accessed, the software displays the topics in a pop up menu shown above. Point Charge Long Charged Line Charged Disk Parallel Plates Parallel Wires Coaxial Cable Sphere Variables A complete list of all the variables used in this section is given below.
17.1 Point Charge The two equations in this topic calculate the radial electric field Er and the potential V at a point located a distance r away from a point change Q. The first equation shows the inverse square relationship between Er and r, while the second equation shows the inverse relationship between the potential V and distance r. The equations have been generalized to include εr, the relative permittivity of the medium. Q 4 ⋅ π ⋅ ε 0 ⋅ εr ⋅ r 2 Q V= 4 ⋅ π ⋅ ε 0 ⋅ εr ⋅ r Er = Eq. 17.1.1 Eq. 17.1.
Entered Values Computed results Solution - Press „ to display the input screen, enter all the known variables, and press „ to solve the selected equation set. The screen display above shows the computed results. -PQYP 8CTKCDNGU ρN ' AEQWNQODU O T AEO εT %QORWVGF 4GUWNVU 'T A8 O 17.3 Charged Disk These two equations describe the electric field and potential along the vertical axis through the center of a uniformly charged disk.
17.4 Parallel Plates The five equations listed in this topic describe the electrical and mechanical forces in a parallel plate capacitor. Two plates are separated by a distance d which is small compared to the lateral dimensions so fringing field effects can be ignored. The first equation computes the electric field E at the plate for a potential difference V between the plates separated by a small distance d.
cl = π ⋅ ε 0 ⋅ εr d cosh −1 2 ⋅ ra FG H Eq. 17.5.1 IJ K Example 17.5 - Compute the capacitance per unit length of a set of power lines 1_cm radius, and 1.5_m apart. The dielectric medium separating the wires is air with a relative permittivity of 1.04. Entered Values Calculated Output Solution - Press „ to display the input screen, enter all the known variables and press „ to solve the selected equation set. -PQYP 8CTKCDNGU εT TC AEO F AO %QORWVGF 4GUWNVU EN ' A( O 17.
Solution - After examining the problem, all the three equations need to be selected to solve the problem. Press „ to display the input variable screen. Enter all the known variables, and press „ to solve the selected equation set. The computed results are shown in the screen display below. Input Values for the example Computed results -PQYP 8CTKCDNGU TC AEO TD AEO εT T AEO ρN ' AEQWNQODU O %QORWVGF 4GUWNVU EN ' A( O 'T A8 O CPF 8 A8 17.
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.
Chapter 18 Inductors and Magnetism Topics in this section focus on electrical and magnetic properties of physical elements. Long Line Long Strip Parallel Wires Loop Coaxial Cable Skin Effect Variables A complete list of all the variables used in the various topics of this section are listed below along with the default units used for those variables.
δ ρ Skin depth Resistivity m Ω*m 18.1 Long Line The magnetic field B from a current I in an infinite wire in an infinitely long line is computed at a distance r from the line. B= µ0 ⋅ I 2⋅π⋅r Eq. 18.1.1 Example 18.1 - An overhead transmission line carries a current of 1200_A, 10_m away from the surface of the earth. Find the magnetic field at the surface of the earth. Input Screen Calculated Results Solution - Since there is only equation, press „ to display the input screen.
Example 18.2 - A strip transmission line 2_cm wide carries a current of 16025_A/m. Find the magnetic field values 1_m away and 2_m from the surface of the strip. Input Screen Calculated Results Solution - Both equations need to be used to compute the solution. 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.
Solution - Upon examining the problem, all the three equations are needed. 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 + A# + A# Z AEO & AO C AEO %QORWVGF 4GUWNVU ( A0 O . A* O $Z A6 18.
Input Screen Calculated Results Solution Upon examining the problem, the last two equations are needed. Select these using the ¸| key and press „to display the input screen. Enter the known variables and press „ to solve the selected equation set. The screen display of the input and calculated results are shown below. -PQYP 8CTKCDNGU C AO DN AO F AO + A# + A# θ AFGI %QORWVGF 4GUWNVU . AJGPT[ 6 A0 O 18.
computes the effect of higher frequencies on resistance, Reff, in ohms. Since the skin effect is a direct consequence of internal magnetic fields in the conductor, the relative permeability, µr, influences these properties. δ= 1 π ⋅ f ⋅ µ 0 ⋅ µr ρ Eq. 18.6.1 Re ff = π ⋅ f ⋅ µ 0 ⋅ µr ⋅ ρ 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.
Chapter 19 Electron Motion This section covers equations describing the trajectories of electrons under the influence of electric and magnetic fields. These equations are divided into three topics. Electron Beam Deflection Thermionic Emission Photoemission Variables The table below lists all the variables used in this chapter.
v = 2⋅ r= Eq. 19.1.1 me ⋅ v q⋅B Eq. 19.1.2 L ⋅ Ls ⋅Vd 2 ⋅ d ⋅Va Eq. 19.1.3 q ⋅Vd ⋅ z2 2 2 ⋅ me ⋅ d ⋅ v Eq. 19.1.4 yd = y= q ⋅Va me Example 19.1- An electron beam in a CRT is subjected an accelerating voltage of 1250_V. The screen target is 40_cm away from the center of the deflection section. The plate separation is 0.75_cm and the horizontal path length through the deflection region is .35 cm. The deflection region is controlled by a 100_V voltage. A magnetic field of 0.
19.2 Thermionic Emission When certain electronic materials are heated to a high temperature T, the free electrons gain enough thermal energy, forcing a finite fraction to escape the work function barrier φ and contribute to the external current I. The current also depends directly on the surface area S and the so-called Richardson’s constant A0. I = A0 ⋅ S ⋅ T 2 ⋅ e − q ⋅φ k ⋅T Eq. 19.2.1 Example 19.2 - A cathode consists of a cesium coated tungsten with a surface area of 2.45_cm2.
Display of Solution 1 Display of Solution 2 Solution - Both equations are needed to solve the problem. Press „ to display the input screen, enter all the known variables and press „ to solve the equation. EE•Pro displays a notice that multiple solution 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.
Chapter 20 Meters and Bridge Circuits This section covers a variety of topics on meters, commonly used bridge and attenuator circuits. These equations are organized under seven titles. Amp, Volt, and Ohmmeter Owen Bridge Wheatstone Bridge Symmetrical Resistive Attenuator Wien Bridge Unsymmetrical Resistive Attenuator Maxwell Bridge Variables The following is a list of all the variables used in this section.
Rsh Rx Vm Vmax Vs Vsen ω Shunt resistance Unknown resistance Voltage across meter Maximum voltage Source voltage Voltage sensitivity Radian frequency A Ω V V V V r/s 20.1 Amp, Volt, and Ohmmeter The three equations in this section describe the use of resistors in extending the range of ammeters, voltmeters and ohmmeters. A shunt resistor Rsh increases the range of an ammeter with a current sensitivity Isen and a maximum range Imax. A series resistance Rse can extend the range of a voltmeter.
20.2 Wheatstone Bridge A Wheatstone bridge with four resistor elements Rx, RR2, RR3 and RR4 is the foundation of modern measuring systems. When the bridge is balanced, there is no current in the galvanometer circuit. The first equation defines the requirement for a balanced bridge. The voltage across the bridge Vm and the galvanometer current Ig are calculated in as follows. A special function GALV calculates the voltage across the bridge, and is a complex function of Vs, Rx, RR2, RR3, RR4, Rg and Rs. Eq.
20.3 Wien Bridge A Wien bridge circuit is designed to measure an unknown capacitance Cx using a bridge arrangement shown here. The circuit has two methods of varying parameters of the bridge to achieve null. In the first approach, Cx can be computed in terms of RR3, RR1, Rs and Rx while in the second equation, the source frequency can be used as a key controlling parameter. The bridge can also be used to measure the frequency f with the third equation after setting Cx=Cs, Rx=Rs, and RR3=2*RR1.
20.4 Maxwell Bridge A Maxwell bridge is designed to measure the inductance Lx and its series resistance Rx in a bridge circuit. The input stimulus to the bridge circuit is an AC source with a variable frequency and an AC meter detecting a null. The first two equations measure the unknown inductance Lx and its resistance Rx by varying the capacitance Cs, and its parallel resistance Rs and bridge arm resistances RR1 and RR2.
Lx = CC3⋅ RR1⋅ RR4 Rx = Eq. 20.5.1 CC3⋅ RR1 − RR2 CC4 Eq. 20.5.2 Example 20.5- A lossy inductor is plugged into an Owen bridge to measure its properties. The resistance branch has 1000_Ω resistors and a capacitor of 2.25µF on the non-resistor leg and 1.25_µF capacitor on the resistor leg of the bridge. A series resistance of 125_Ω connects the CC4 leg to balance the inductive element.
FG10 H a= 10 b= DB 20 DB 20 2 ⋅ 10 10 FG H DB 10 IJ K −1 Eq. 20.6.1 +1 DB 20 Eq. 20.6.2 −1 DB IJ K c = 10 20 − 1 Eq. 20.6.3 Example 20.6- Design a symmetrical and Bridges Tee attenuator for a 50 Ω load and a 6 DB loss. Entered Values Calculated Results Solution - All three equations are needed. Press „ to display the input screen, enter all the known variables and press „ to solve the equation. The computed results are shown above.
DB = 20 ⋅ log10 ⋅ F GH Rl − Rr Rl + Rr Rr I JK Eq. 20.7.3 Example 20.7- A network needs to be patched by an unsymmetrical attenuator. The network to the right of the attenuator presents a resistive load of 125 Ω, while 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.
Chapter 21 RL and RC Circuits This chapter covers the natural and transient properties of simple RL and RC circuits. The section is organized under six topics. RL Natural Response RC Natural Response RL Step Response RC Step Response RL Series to Parallel RC Series to Parallel Variables The table below lists all the variables used in this chapter, along with a brief description and appropriate units.
21.1 RL Natural Response These four equations define all the key properties for the natural response of an RL circuit with no energy sources. The first equation shows the characteristic time constant τ in terms of the resistance R and the inductance L. The second equation computes the decay of the voltage vL across the inductor with an initial current I0. The third equation displays the decay of the inductor current iL.
capacitance C. The second equation computes the decay of the voltage vC across the capacitor with an initial voltage of V0. The third equation shows the decay of the capacitor current iC. The final equation computes the energy dissipation W. τ = R ⋅C Eq. 21.2.1 t − τ vC = V 0 ⋅ e V 0 − τt ⋅e iC = R 2⋅t − 1 W = ⋅ C ⋅V 0 2 ⋅ 1 − e τ 2 FG H Eq. 21.2.2 Eq. 21.2.3 IJ K Eq. 21.2.4 Example 21.2- An RC circuit consists of a 1.2_µfarad capacitor and a 47_Ω resistor. The capacitor has been charged to 18_V.
Entered Values Calculated Results Solution - Upon examining the problem, all three equations are need to be solve the problem. Press „ to display the input screen, enter all the known variables and press „ to compute the solution. -PQYP 8CTKCDNGU . AO* 4 AΩ + AO# V AU 8U A8 %QORWVGF 4GUWNVU K. A# X. A8 21.4 RC Step Response These three equations describe the step response properties of an RC circuit.
21.5 RL Series to Parallel The equations in this topic show the equivalence relationship between a series RL circuit (Rs and Ls) and its parallel equivalent circuit, Rp, Lp. The first equation translates the frequency, f, to its radian frequency, ω. The second equation specifies the quality factor Qs for Rs and Ls. The third and fourth equations define the values of Rp and Lp in terms of Rs, Ls and ω. The fifth equation computes Qp in terms of Rp, Lp and ω.
Entered Values Calculated Results Solution - Upon examining the problem, the first six equations need to be solved as a set. Select these equations, and press „ to display the input screen enter all the known variables and press „ to solve. The computed results are shown in the screen display shown here. -PQYP 8CTKCDNGU .U AO* 3U H A*\ %QORWVGF 4GUWNVU .R A* 4U AΩ 4R AΩ ω T U 21.
Cs 1 1+ 2 Qs Rp Rs = 1 + Qp 2 Cp = Cs = Eq. 21.6.9 Eq. 21.6.10 c Cp ⋅ 1 + Qp 2 Qp h Eq. 21.6.11 2 Example 21.6 - A parallel RC Circuit consists of a 47_µfarad and 150000_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.
Chapter 22 RLC Circuits The essential equations for computing impedance and admittance, natural response, and transient behavior of RLC circuits are organized into six topics. Series Impedance Parallel Admittance RLC Natural Response Underdamped Transient Critically-Damped Transient Overdamped Transient Variables All the variables used here are listed along with a brief description and units.
ω0 X XXC XL Ym Zm Classical radian frequency Reactance Capacitive reactance Inductive reactance Admittance – magnitude Impedance – magnitude rad/s Ω Ω Ω S S 22.1 Series Impedance The series impedance of an RLC circuit is calculated using the first two equations. The magnitude |Zm| and phase angle θ of the impedance is calculated from the resistance the R and reactance X in the first two equations. The reactance X in the third equation is calculated in terms of the inductive reactance XL.
22.2 Parallel Admittance The admittance of a parallel RLC circuit consists of a magnitude, Ym, and phase angle θ. Both can be calculated in terms of the conductance G and susceptance B. The conductance G is expressed in terms of resistance R, while the susceptance B is expressed in terms of the inductive and capacitive components BL and BC. c Ym h 2 = G2 + B2 θ = tan −1 Eq. 22.2.1 FG G IJ H BK Eq. 22.2.2 1 R B = BL + BC −1 BL = ω⋅L G= Eq. 22.2.3 Eq. 22.2.4 Eq. 22.2.5 BC = ω ⋅ C Eq. 22.2.
s1r = real −α + α 2 − ω 02 e s1i = image −α + s2r = reale −α − s2i = image −α − j α −ω0 j α −ω0 j α −ω0 j Eq. 22.3.1 2 2 Eq. 22.3.2 2 2 Eq. 22.3.3 2 2 Eq. 22.3.4 1 L⋅C 1 α= 2⋅ R⋅C ω0 = Eq. 22.3.5 Eq. 22.3.6 Example 22.3 – A series RLC circuit of Example 22.1 is used to compute the circuit parameters. Input and calculated results (upper half) Input and calculated results (lower half) Solution - All of the equations are needed to solve the parameters from these given set of variables.
ωd = ω 02 − α 2 Eq. 22.4.3 b g b v = B1⋅ e −α ⋅t ⋅ cos ωd ⋅ t + B2 ⋅ e −α ⋅t ⋅ sin ωd ⋅ t B1 = V 0 B2 = − g Eq. 22.4.4 Eq. 22.4.5 α ⋅ V 0 − 2 ⋅ I 0⋅ R ωd b g Eq. 22.4.6 Example 22.4 - A parallel RLC circuit is designed with a 1000 Ω resistor, a 40 mH inductor and a 2 µF capacitor. The initial current in the inductor is 10 mA and the initial charge in the capacitor is 2.5 V. Calculate the resonant frequency and the voltage across the capacitor 1 µs after the input stimulus has been applied.
D1 = I0 + α *V 0 C Eq. 22.5.4 D2 = V 0 Eq. 22.5.5 Example 22.5 - A critically damped RLC circuit consists of a 100 Ω resistor in series with a 40 mH inductor and a 1 µF capacitor. The initial inductor current is 1 mA and the initial capacitor charge is 10 V. Find the voltage across the capacitor after 10 µs. Calculated Results (Upper display) Calculated Results (Lower display) Solution - All of the equations need to be selected to solve this problem.
FG H A1 = FG H FG H Eq. 22.6.6 IJ IJ KK 1 V0 ⋅ + I0 C R s2 − s1 − V 0 ⋅ s1 + A2 = IJ K 1 V0 ⋅ + I0 C R s2 − s1 V 0 ⋅ s2 + 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 µF 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.
Chapter 23 AC Circuits This chapter covers equations describing the properties of AC circuits. ♦ ♦ ♦ ♦ ♦ RL Series Impedance RC Series Impedance Impedance ↔ Admittance Two Impedances in Series Two Impedances in Parallel Variables All the variables here are listed with a brief description and appropriate units.
23.1 RL Series Impedance The equations in this section describe the relationships of an RL series circuit. The first equation shows the sinusoidal behavior of current, I, defined by the amplitude, Im, radian frequency ω, and time t. The second equation defines the magnitude of impedance Zm in terms of the resistance R, the inductance L and ω. The voltages VR and VL across the resistor and inductor are defined by the next two equations.
23.2 RC Series Impedance The equations in this section describe the key relationships involved in an RC series circuit. The first equation shows the sinusoidal behavior of current, I, defined by the amplitude Im, the radian frequency ω, and time t. The magnitude Zm of the impedance of the circuit is calculated in terms of the resistance R, capacitance C, and ω in the second equation. The voltage VR across R and the voltage VC across C are given in the next two equations.
23.3 Impedance ↔ Admittance The equation is designed to convert impedances to admittances with unit management built-in. As shown in the figure to the right, an impedance Z consists of a real and reactive components (R and X) to describe it. The admittance Y consists of real and reactive components (G and B) to describe it. The variables Z and Y have an _ attached to them to add emphasize that they are complex in general and have units attached. Y_ = 1 Z_ Eq. 23.3.1 Example 23.
FG XX 1IJ H RR1 K F XX 2 IJ θ 2 = tan G H RR2 K θ 1 = tan −1 Eq. 23.4.7 −1 Eq. 23.4.8 Example 23.4 - Two impedances consisting of resistances of 100 Ω and 75 Ω and reactive components 75 and 145, respectively are connected in series. Find the magnitude and phase angle of the combination. Input Variables Computed results Solution - Select the first four equations to solve the problem. Press „ to display the input screen, enter all the known variables and press „ to solve the equation.
FG XX 1IJ H RR1 K F XX 2 IJ θ 2 = tan G H RR2 K θ 1 = tan −1 Eq. 23.5.7 −1 Eq. 23.5.8 Example 23.5 - For two impedances in parallel possessing values identical to the previous example, calculate the magnitude 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.
Chapter 24 Polyphase Circuits This chapter covers equations used in Polyphase circuits. The equations have been divided into three sections. Power Measurements Balanced ∆ Network Balanced Wye Network Variables All the variables are listed below with a brief description and units. Variable IL Ip P PT θ VL Vp W1 W2 Description Line current Phase current Power per phase Total power Impedance angle Line voltage Phase voltage Wattmeter 1 Wattmeter 2 Unit A A W W rad V V W W 24.
Input variables Computed results Solution - Upon examining the problem, all equations are needed. 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 +. A# θ ATCF 8R A8 %QORWVGF 4GUWNVU +R A# 2 A9 26 A9 8. A8 24.2 Balanced Wye Network These equations describe the relationship for a Balanced Wye Network.
-PQYP 8CTKCDNGU +. A# θ ATCF 8R A8 %QORWVGF 4GUWNVU +R A# 2 A9 26 A9 8. A8 24.3 Power Measurements These three equations for a two-watt meter connection are used to measure the total power of a balanced network. The first two equations determine the watt meter readings W1 and W2 are expressed from the line current IL, line voltage VL, and phase delay θ between the voltage and the current.
Chapter 25 Electrical Resonance The equations in this section describe the electrical properties of resonance in circuits composed of ideal circuit elements. The section is organized under four topics: Parallel Resonance I Parallel Resonance II Resonance in Lossy Inductor Series Resonance Variables The following variables are listed with a brief description and their appropriate units.
reactive parameters L and C. The terms ω1 and ω2 represent the lower and upper cutoff frequencies beyond ω0, where the impedance is half the impedance at resonance. The bandwidth, β, is the difference between ω1 and ω2. The last three equations calculate the quality factor Q in terms of R, C, L and ω0. Im Vm = F1 F 1 I I GGH R + GHω ⋅ C − bω ⋅ LgJK JJK 1 I I FF θ = tan G G ω ⋅ C − H H ω ⋅ L JK ⋅ RJK Eq. 25.1.1 2 2 −1 ω0 = 1 L⋅C ω1 = −1 + 2⋅ R⋅C ω2 = 1 + 2⋅ R⋅C Eq. 25.1.2 Eq. 25.1.
-PQYP 8CTKCDNGU % Aµ( +O AO# 4 AΩ . O* ω AT U %QORWVGF 4GUWNVU β AT U θ ATCF 3 8O A8 ω AT U ω AT U ω AT U 25.2 Parallel Resonance II These equations represent an alternative method of expressing the properties of a resonant circuit in terms of the quality factor Q. The first equation links Q with the resonant frequency ω0 and the bandwidth β.
Solution - All of the equations are needed to compute the solution for this 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( 3 4 AΩ %QORWVGF 4GUWNVU α β AT U ω AT U ω AT U ω AT U ωF AT U 25.
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Entered Values Calculated Results (Upper and Lower Screens) Solution - 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 .
Chapter 26 OpAmp Circuits Seven commonly-used OpAmp circuits are presented in this section. An OpAmp is a direct-coupled high-gain amplifier that can be configured with the use of feedback to circuit elements to achieve overall performance characteristics. There are two inputs labeled ‘+’ and ‘-’. The manner in which input signals are connected to these terminals defines the inverting or non-inverting properties of the circuit.
Rs tr ∆VH VL Vomax VR Vrate VU Vz1 Vz2 Ω s V V V V V/s V V V Voltage divide resistor 10-90% rise time Hysteresis Detection threshold, low Maximum circuit output Reference voltage Maximum voltage rate Detection threshold, high Zener breakdown 1 Zener breakdown 2 26.1 Basic Inverter These equations define the properties of a basic inverter. The first equation relates the voltage gain Av to the feedback resistance Rf and input resistance RR1.
26.2 Non-Inverting Amplifier These equations define the properties of a non-inverting amplifier. The first equation expresses the voltage gain Av in terms of the feedback resistor Rf and resistor RR1. The second equation gives the value of Rp needed in the input circuit to minimize offset current effects. Av = 1 + Rf RR1 Eq. 26.2.1 Rp = RR1⋅ Rf RR1 + Rf Eq. 26.2.2 Example 26.2 - Find the DC gain of a non-inverting amplifier with a feedback resistance of 1 MΩ and a resistance to the load of 18 kΩ.
b Rout = Rs ⋅ 1 + Av g Eq. 26.3.3 Example 26.3 - A current amplifier with a 200 kΩ feedback resistance has a voltage gain of 42. If the source resistance is 1 kΩ, the load resistance is 10 kΩ and the output resistance of the OpAmp is 100 Ω, find the current gain, input and output resistances. Entered Values Calculated Results Solution - Use all of the equations to compute the solution for this problem.
%QORWVGF 4GUWNVU #IE AUKGOGPU 4QWV A Ω 26.5 Level Detector (Inverting) The first equation in this section computes the value of the resistor RR1 attached to an OpAmp inverting input. The second equation calculates the hysteresis (or memory) ∆VH of the level detector circuit. The third and fourth equations define the upper and lower trip voltages VU and VL for an ideal inverting level detector, assuming a reference voltage VR and breakdown voltages Vz1 and Vz2, and in terms of Rp and Rf.
26.6 Level Detector (Non-Inverting) This section computes the value of the resistor RR1 attached to an OpAmp non-inverting input. The next equation calculates the hysteresis (or memory) ∆VH of the level detector circuit. The next two equations define the upper and lower trip voltages VU and VL for an ideal inverting level detector, a reference voltage VR and breakdown voltages Vz1 and Vz2, in terms of Rp and Rf. RR1 = Rp ⋅ Rf Rp + Rf ∆VH = bVz1 + Vz2g ⋅ Rp Eq. 26.6.1 Eq. 26.6.
Rf = Vo max IIf Eq. 26.7.1 Rp = Rf Eq. 26.7.2 Vo max Rf ⋅Vrate 1 RR1 = 2 ⋅ π ⋅ fd ⋅ CC1 1 fd = 2 ⋅ π ⋅ Rf ⋅ CC1 CC1 = Eq. 26.7.3 Eq. 26.7.4 Eq. 26 7.5 Cp = 10 2 ⋅ π ⋅ f 0 ⋅ Rp Eq. 26.7.6 Cf = 1 4 ⋅ π ⋅ f 0 ⋅ Rf Eq. 26.7.7 Example 26.7 - A differentiator circuit designed with an OpAmp has a slew rate of 1.5 V/µs.
Ad = RR3 RR1 Aco = Ad = Acc = Eq. 26.8.1 RR 32 RR3 ⋅ RR1 + RR3 ⋅ CMRR b g Av ⋅ RR3 Eq. 26.8.3 RR12 ⋅ Av 2 + RR32 RR4 ⋅ RR1 − RR2 ⋅ RR3 RR1⋅ RR2 + RR4 b Eq. 26.8.2 g Eq. 26.8.4 Example 26.8 - Find the differential mode gain and the current gain for a differential amplifier with bridge 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.
Chapter 27 Solid State Devices This section covers a variety of topics in solid state electronics. Read this! Note: The equations in this section are grouped under topics which describe general properties of semiconductors or devices. Equations for a variety of specific cases and are listed together under a sub-topic heading and are not necessarily a set of consistent equations which can be solved together.
CL Cox D DB DC DE Dn Dp εox εs Ec EF Ei Ev ffmax γ gd gm gmL Go I I0 IB IC ICB0 ICE0 ICsat ID IDmod ID0 IDsat IE IIf Ir Ir0 IRG IRG0 Is λ Is kD kL kn kn1 kN kP KR L LC LD LE LL EE PRO for TI-89, 92 Plus Equations - Solid State Devices Load capacitance Oxide capacitance per unit area Diffusion coefficient Base diffusion coefficient Collector diffusion coefficient Emitter diffusion coefficient n diffusion coefficient p diffusion coefficient Oxide permittivity Silicon Permittivity Conduction band Fermi level
LLn lNN Lp lP µn µp mn mp N Na nnC Nd nE ni N0 npo p pB φF φGC pno Qtot Qb Qb0 Qox Qsat ρn ρp Rl τB τD τL τo τp τt t TT tch tdis tox tr ts tsd1 tsd2 Ttr V1 Va Vbi VBE VCB VCC EE PRO for TI-89, 92 Plus Equations - Solid State Devices Diffusion length, n n-channel length Diffusion length, p p-channel length n (electron) mobility p (positive charge) mobility n effective mass p effective mass Doping concentration Acceptor density n density, collector Donor density n density, emitter Intrinsic density Surface
VCEs VDD VDS VDsat VEB VG VGS VIH Vin VIL VL VM VOH VOL Vo Vp VSB VT VT0 VTD VTL VTL0 VTN VTP W WB WD WL WN WP x xd xn xp Z CE saturation voltage Drain supply voltage Drain voltage Drain saturation voltage EB bias voltage Gate voltage Gate to source voltage Input high Input voltage Input low voltage Load voltage Midpoint voltage Output high Output low Output voltage Pinchoff voltage Substrate bias Threshold voltage Threshold voltage at 0 bias Depletion transistor threshold Load transistor threshold Load tr
Dn = k ⋅ TT ⋅ µn q Eq. 27.1.3 Dp = k ⋅ TT ⋅ µp q Eq. 27.1.4 The Fermi level EF is a measure of the chemical potential in silicon and is used to estimate the doping density. For instance in a n-type semiconductor, equation 27.1.5 is used to determine the location of the Fermi level, while in a ptype material equation 27.1.6 is used to establish the Fermi level. In some cases, when both donor and acceptor levels are specified, one has to choose either equation 27.1.5 or 27.1.6 whether Nd>Na.
Entered Values Computed Results Solution - Since the dopant is a donor, use equations 27.1.6 and 27.1.7 to find compute a solution. Select these equations and press „ to display the input screen, enter all the known variables and press „ to solve the set of equations. To convert the results of Ef and Ei to units of electron volts, highlight each value press ‡/Opts and y/Conv to display the unit menu in the tool bar. The computed results are shown in the screen displays above.
Vbi = F I GH b g JK k ⋅ TT Nd ⋅ Na ⋅ ln 2 q ni TT Eq. 27.2.1 Equations 27.2.2 - 27.2.4 compute the depletion layer widths xn and xp in the p and the n regions of the junction in terms of the dielectric constant εs, doping densities Nd and Na, built-in voltage Vbi and the applied voltage Va; xd is the total depletion region width for a given applied voltage. 2 ⋅ εs ⋅ ε 0 ⋅ Vbi − Va ⋅ Na q ⋅ Nd ⋅ Na + Nd xn = b Eq. 27.2.2 g Nd ⋅ xn Na Eq. 27.2.3 xd = xn + xp Eq. 27.2.4 xp = Equation 27.2.
66 A- 8C A8 %QORWVGF 4GUWNVU %L ' A( 8DK A8 ZF ' AO ZP ' AO ZR ' AO Example 27.2.2 - A linearly graded junction has an area of 100 µ2, a built-in voltage of 0.8578 V, and an applied voltage of -5.V. The relative permittivity of silicon is 11.8. Under room temperature conditions, what is the junction capacitance, depletion layer width, and the linear-graded junction parameter? Entered Values Calculated Results Solution - Use equations 27.
The so-called Generation-Recombination current IRG0 at 0 bias is calculated by the fourth equation in terms of Aj, average recombination time τo, intrinsic density ni, depletion width xd. The fifth equation shows that applying an external voltage Va, the generation recombination current IRG increases exponentially. IRG 0 = b g −q ⋅ Aj ⋅ ni TT ⋅ xd 2 ⋅ τo b g Eq. 27.3.4 IJ K FG H q ⋅Va q ⋅ Aj ⋅ ni TT ⋅ xd 2⋅k ⋅TT ⋅ e −1 IRG = 2 ⋅ τo Eq. 27.3.
Entered Values Calculated Results Solution - Use the fifth equation to solve this problem. Select this by highlighting the 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 #L A µ^ τ APU 66 A- 8C ZF Aµ %QORWVGF 4GUWNVU +4) A# 27.
Example 27.4 - A junction transistor has the following parameters: α is 0.98, the base current is 1.2 µA while ICBO is 1.8 pA. Find the β, emitter and collector currents. Entered Values Calculated Results Solution - A few different choices are available, however the results might differ slightly due to the combination of equations used. The second, third and fourth equations can be used to solve this problem. Select these by highlighting each equation and pressing the ¸ key.
αr ⋅ Ir = Is Eq. 27.5.7 The last three equations define ICE0 and ICB0 in terms of αf , αr, βf and Ir0. b g ICB 0 = 1 − αr ⋅ αf ⋅ Ir 0 b Eq. 27.5.8 g ICE 0 = ICB 0 ⋅ βf + 1 ICE 0 = b Ir 0 ⋅ 1 − αf ⋅ αr 1 − αf Eq. 27.5.9 g Eq. 27.5.10 Example 27.5.1 - A junction transistor has a forward and reverse α of 0.98 and 0.10 respectively. The collector current is 10.8 mA while the forward current is 12.5 mA. respectively.
IB = FG H IJ K FG H IJ K q ⋅VBE q ⋅VCB q ⋅ A2 ⋅ DC q ⋅ A1⋅ DE ⋅ nE ⋅ e k ⋅TT − 1 + ⋅ nnC e k ⋅TT − 1 LC LE DB ⋅ pB WB α= DB ⋅ pB DE ⋅ nE + WB LE Eq. 27.6.3 Eq. 27.6.4 Example 27.6 - Find the emitter current gain α for a transistor with the following properties: base width of 0.75 µm, base diffusion coefficient of 35 cm2/s, emitter diffusion coefficient of 12 cm2/s, and emitter diffusion length of 0.35 µm. The emitter electron density is 30,000 cm-3 and the base density is 500,000 cm-3.
F I 1 G JJ tr = τB ⋅ lnG ⋅ τt GH 1 − ICsat J IB ⋅ τB K Eq. 27.7.3 FG IB ⋅τB IJ H ICsat ⋅τt K F I G JJ 2 ⋅ IB ⋅ τB tsd 2 = τB ⋅ lnG IB ⋅ τB I J GH ICsat ⋅ τt ⋅ FGH1 + ICsat J ⋅ τt K K I F JJ GG IC ⋅ b1 − αr g JJ k ⋅ TT G IB G ⋅ ln 1 + VCEs = q GG F IC ⋅ b1 − αf gI JJ GG αr ⋅ GGG1 − IB αf JJJ JJ KK H H tsd 1 = τB ⋅ ln Eq. 27.7.4 Eq. 27.7.5 Eq. 27.7.6 Example 27.7 - Find the saturation voltage for a switching transistor at room temperature when a base current of 5.
27.8 MOS Transistor I The seven equations in this section form the basic equations of charge, capacitance and threshold voltage for a MOS transistor. The first equation shows the Fermi potential φF defined in terms of temperature TT, the intrinsic carrier density ni, and the hole density p. φF = FG b gIJ H K ni TT k ⋅ TT ⋅ ln q p Eq. 27.8.
Solution - Use the second through last 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.
The last four equations calculate performance parameters transconductance gm, transit time through the channel Ttr, maximum frequency of operation ffmax, and drain conductance gd. b g b g gm = kn ⋅ VGS − VT 4 2 ⋅L 3 Ttr = µn ⋅ VGS − VT Eq. 27.9.7 Eq. 27.9.8 gm 2 ⋅ π ⋅ Cox ⋅ W ⋅ L gd = kn ⋅ VGS − VT ff max = b Eq. 27.9.9 g Eq. 27.9.10 Example 27.9 - An nMOS transistor has a 6 µ width and 1.25µ gate length. The electron mobility is 500 cm2/V/s. The gate oxide thickness is 0.01µm.
VOH = VDD Eq. 27.10.2 FG 1 + VDD − VT IJ ⋅VOL + 2VDD = 0 H kD ⋅ Rl K kD ⋅ Rl kD bVDD − Vog ⋅ c2 ⋅ bVIH − VT g ⋅Vo − Vo h = VOL2 − 2 ⋅ 2 2 Rl Eq. 27.10.3 Eq. 27.10.4 The final equation computes the midpoint voltage VM for which the driver transistor is in saturation. kD ⋅ VM − VT 2 b g = bVDDRl− VM g 2 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.
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 EE PRO for TI-89, 92 Plus Equations - Solid State Devices Lower Display 95
Solution 8: Upper Display Lower Display Solution - Use all of the equations to solve for the problem. Press „ to display the input screen, enter all the known variables and press „ to solve the equation set. Eight complete solutions are generated, in this case, requiring the user to select an answer most relavent to the situation (ie: physical significance, realistic values for components, currents, etc). Mathematically setting up and solving the problem takes some time.
The output high voltage VOH is calculated in terms of the drain supply voltage VDD, the threshold voltage at zero bias VT0, the fermi potential φF and the body coefficient γ in the fourth equation. The fifth equation defines the input voltage Vin in terms of the ratio KR between the load kL and drive kD MOS constants, VDD, the threshold of the load and drive transistors VTL and VTD. The sixth equation defines the threshold voltage of the load transistor, VTL.
an output voltage of 3.0_ V find the output high voltage, the input high voltage, and the threshold of the load device. Assume a input voltage of 2.5_V. Upper Display Middle Display Lower Display Solution - Use Equations 27.11.1-27.11.4 and 27.11.6-17.11.7 to get a complete solution to the problem on hand. 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 charging time tch for the CL is defined next. The current in the depletion load I0 is given by the last equation. tch = CL ⋅VL I0 Eq. 27.12.5 I 0 = kL ⋅VTL2 Eq. 27.12.6 Example 27.12 - A MOS inverter with a depletion mode transistor as the load has a driver transistor 5_µ wide and 1_µ long while the load is a depletion mode device with a 0 bias threshold of -4_V, 3_µ long and 3_µ wide.
kP = µp ⋅ Cox ⋅ WP lP Eq. 27.13.1 kN = µn ⋅ Cox ⋅ WN lNN Eq. 27.13.2 kP ⋅ VDD − VTP VIH = 2 ⋅Vo + VTN + kN kP 1+ kN c h FG 2 ⋅Vo − VDD − VTP + kN ⋅VTN IJ H K kP VIL = kN 1+ kP kN ⋅ Vin − VTN 2 b g 2 = {VDD > VTP } Eq. 27.13.3 {VDD ≤ VTP } Eq. 27.13.4 kP ⋅ VDD − Vin − VTP 2 c h 2 Eq. 27.13.5 Example 27.13 - Find the transistor constants for an N and P MOS transistor pair given: N transistor: WN=4 µm, lNN=2.
constant εs, Nd, Vbi, and the drain saturation voltage VDsat. The last two equations display the relationship for drain voltage and drain current upon saturation.
Chapter 28 Linear Amplifiers This section covers linear circuit models (i.e., small signal models) used in making first order calculations using bipolar or junction transistors in amplifier circuits. These circuit models are referred to by many different names such as small signal circuit model, AC circuit model, linear circuit model. In addition, popular device configurations such as the Darlington pair, emitter-coupled pair, differential amplifier, and a source-coupled pair topics have been included.
Rin Rl Ro Rs Ω Ω Ω Ω Input resistance Load resistance Output resistance Source resistance 28.1 BJT (Common Base) These six equations represent properties of a transistor amplifier connected in the common base configuration at mid frequencies. The first equation relates the common base current gain α0 with the common emitter current gain β0. The second equation computes the input impedance Rin at the input terminals of the amplifier from the emitter and base resistances, re and rb.
-PQYP 8CTKCDNGU α TD AMΩ TTE A/Ω TG A Ω 4N AMΩ 4U A Ω %QORWVGF 4GUWNVU #K #QX #X β 4KP AΩ 4Q ' AΩ 28.2 BJT (Common Emitter) This section contains the equations for an amplifier, at mid frequencies, connected in the common emitter configuration. The first equation displays the current gain α0 in relation to the common current gain β0.
28.3 BJT (Common Collector) These six equations describe the properties of a transistor amplifier connected in a common collector configuration at mid frequencies. The first equation couples the common emitter current gain α0 with the common base current gain β0. The second equation computes the input impedance Rin in terms of base resistance rb, emitter resistance re, β0, and load resistance Rl.
28.4 FET (Common Gate) The equations in this section focus on an FET amplifier in the common gate configuration. The amplification factor µ is described in the first equation in terms of the transconductance gm and the drain resistance rd. In the second equation the input resistance Rin is described as a function of load resistance Rl, rd and µ. The voltage gain Av is defined by the third equation in terms of Rl, rd and µ.
b Rl + rd g µ +1 F rd ⋅ Rl IJ Av = − gm ⋅ G H rd + Rl K Rin = Eq. 28.5.2 Eq. 28.5.3 Ro = rd Eq. 28.5.4 Example 28.5 - Find the voltage gain of an FET configured as a common-source based amplifier. The transconcductance is 2.5 x 10-3 siemens, a drain resistance of 18 kΩ and a load resistance of 100 kΩ. Find all the parameters for this amplifier circuit. Entered Values Calculated Results Solution - Use all of the equations to compute the solution for this problem.
Entered Values Calculated Results Solution - Use all of the equations 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 in the screen display above. -PQYP 8CTKCDNGU IO AUKGOGPU TF AMΩ 4N AMΩ %QORWVGF 4GUWNVU #X µ 4KP AΩ 4Q A Ω 28.
-PQYP 8CTKCDNGU β TD AMΩ TG AΩ 4$# AMΩ 4N AMΩ 4U AMΩ %QORWVGF 4GUWNVU #K 4KP ' AΩ 4Q AMΩ 28.8 Darlington (CC-CE) The Darlington configuration connected as a common collector-common emitter configuration is described in this section. The first two equations define the input resistance Rin and output resistance Ro, in terms of base resistance rb, emitter resistance re, collector resistance rrc, and current gain β0.
28.9 Emitter-Coupled Amplifier Two classes of emitter-coupled amplifiers are covered in this section. The first equation shows the general relationship between β0 and α0; the current gains under common base and common emitter configurations. The next three equations show the input resistance Rin, output resistance Ro, and voltage gain Av for a common collector-common base method of connection.
28.10 Differential Amplifier The gain Ad in the differential mode of operation is given by the first equation. The common mode gain, Ac is defined in terms of the external collector and emitter resistances RCA and REA and the emitter resistance re. The last two equations show input resistance for differential and common mode inputs Rid & Ric. 1 Ad = − ⋅ gm ⋅ RCA 2 −α 0 ⋅ RCA Ac = 2 ⋅ REA + re Rid = 2 ⋅ rb + β 0 ⋅ re b Eq. 28.10.1 Eq. 28.10.2 g Eq. 28.10.3 Ric = β 0 ⋅ REA Eq. 28.10.4 Example 28.
µ = gm ⋅ rd Eq. 28.10.3 CMRR = gm ⋅ Rs Eq. 28.10.4 Example 28.11 - Find the gain parameters of a source-coupled JFET pair amplifier if the external drain resistance is 25 kΩ, and the source resistance is 100 Ω. The drain resistance 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.
Chapter 29 Class A, B and C Amplifiers This chapter covers the section called Class A, B and C Amplifiers. These amplifier circuits forms the basis of a class of power amplifiers used in a variety of applications in the industry. Read this! Note: The equations in this section are grouped under topics which describe general properties of Class A, B, and C amplifiers.
Po PP Q θJA R Rl RR0 RR2 RB Rrc Rxt S TA TJ ∆Tj V0 V1 VBE VCC ∆VCE VCEmx VCEmn Vm VPP XXC XC1 XC2 XL Power output Compliance Quality factor Thermal resistance Equivalent resistance Load resistance Internal circuit loss Load resistance External base resistance Coupled load resistance External emitter resistance Instability factor Ambient temperature Junction temperature Change in temperature Voltage across tank circuit Voltage across tuned circuit Base emitter voltage Collector supply voltage Voltage swing
The compliance PP is defined as the full voltage swing across the emitter and collector and is expressed in terms of the minimum and maximum transformer ratings VCEmx and VCEmn. VPP represents the peak to peak voltage in the secondary transformer. The final two equations compute the output power Po and the conversion efficiency ζ. PP = VCEmx − VCEmn Eq. 29.1.5 VPP = n ⋅ PP Eq. 29.1.6 ∆IC 2 ⋅ R 8 Po ζ= Pdc Po = Eq. 29.1.7 Eq. 29.1.8 Example 29.
IB = b − IC ⋅ Rxt − VBE Rxt + RB g Eq. 29.2.3 The fourth equation expresses a more exact form for the collector current IC in terms of hFE, Rxt, RB, ICBO, and VBE. In using this equation, care must be taken to ensure that Eq.29.2.3 and Eq. 29.2.4 are not selected at the same time. Such a choice will lead to the inability of the solver engine to perform the computation accurately. IC = b g hFE ⋅ Rxt + RB − hFE ⋅VBE + ⋅ ICBO hFE ⋅ Rxt ⋅ RB hFE ⋅ Rxt + RB Eq. 29.2.
transformer. The secondary winding is connected to a load RR2. The first equation computes an equivalent resistance R based on the maximum current supplied to the load Imax and the collector supply voltage VCC. The power output Po is computed by the second equation in terms of VCC and R. The final equation computes the power Po in terms of the load resistance RR2 and the transformer windings N1 and N2. Care must be exercised in selecting the equations.
2 ⋅ K ⋅ Im ax ⋅VCC π 2 ⋅ K ⋅VCC 2 Pdc = π ⋅R Pdc = Eq. 29.4.3 Eq. 29.4.4 The efficiency of power conversion ζ is given by the fifth and sixth equations. The power dissipated by the circuit Pd is computed in the seventh equation. The eighth equation calculates the voltage V1 across a tuned RLC circuit in terms of the transconductance gm, load resistance Rl, and output conductance hOE.
29.5 Class C Amplifier These six equations outline the properties of a Class C amplifier. The first equation defines the efficiency of conversion ζ in terms of the current I, the coupled-in load Rrc, and the equivalent internal circuit loss resistance RR0. The next equation computes the tuned circuit parameters which have a capacitive reactance of XXC, which is given in terms of the load voltage V0, quality factor Q, and power Po.
Chapter 30 Transformers The contents in this chapter is divided into two topics. Ideal Transformer Linear Equivalent Circuit Variables All variables used in this section are listed here with a brief description and units. Variable I1 I2 N1 N2 RR1 RR2 Rin Rl V1 V2 XX1 XX2 Xin Xl Zin ZL Description Primary current Secondary current # primary turns # secondary turns Primary resistance Secondary resistance Equiv.
V 1 N1 = V2 N2 Eq. 30.1.1 I 1⋅ N 1 = I 2 ⋅ N 2 Eq. 30.1.2 V 1⋅ I 1 = V 2 ⋅ I 2 Eq. 30.1.3 F N 1 IJ Zin = G H N 2K Eq. 30.1.4 2 ⋅ ZL Example 30.1 - An ideal transformer has 10 primary turns and 36 secondary turns. The primary side draws 500 mA when subjected to a 110 V input. If the load impedance is 175 Ω, find the input impedance at the primary side of the transformer in addition to the voltage and current on the secondary end.
F N 1 IJ ⋅ b XX 2 + Xlg Xin = XX 1 + G H N 2K 2 Eq. 30.2.4 Example 30.2 - The transformer in the above problem has a primary and secondary resistance of 18 Ω and 5 Ω, respectively. The corresponding coils have a reactance of 6 and 2.5 Ω. The secondary side is loaded 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.
Chapter 31 Motors and Generators This section has thirteen topics covering various aspects of motors and generators. The topics are organized under these headings. Read this! Note: The equations in this section are grouped under topics which describe general properties of semiconductors or devices. Equations for a variety of specific cases and are listed together under a sub-topic heading and are not necessarily a set of consistent equations which can be solved together.
Isb Isf K Kf KM L θ N Ns ρ φ p P Pa Pma Pme Pr RR1 Ra Rd Re Rel Rf Rl Rr Rs Rst s sf sm T Tb Tf Tgmax TL Tloss Tmmax TTmax Ts Va Vf Vfs Vt ωm ωme ωr ωs Wf XL Backward stator current Forward stator current Machine constant Field coefficient Induction motor constant Length of each turn Phase delay Total # armature coils # stator coils Resistivity Flux # poles Power Mechanical power Power in rotor per phase Mechanical power Rotor power per phase Rotor resistance per phase Armature resistance Adjustable resist
31.1 Energy Conversion The four equations in this section describe the fundamental relationship amongst electrical, magnetic and mechanical aspects of a system. For example, the first two equations show two ways of computing energy density Wf stored in a magnetic field. The first equation uses the field intensity H and flux density B in a magnetic region with length L and area A. The second an electric analogy to the magnetic circuit as it uses the magnetic reluctance Rel and flux φ to compute Wf.
31.2 DC Generator The first equation describes the relation between electrical radian frequency ωme, the mechanical radian frequency ωm, and the number of poles in the generator p. The next equation expresses the emf generated per turn Eta with the relative motion of the coil with respect to the magnetic field φ. p ⋅ ωm 2 p Eta = ⋅ ωm ⋅ φ π ωme = Eq. 31.2.1 Eq. 31.2.
Display: Upper Half Display: Lower Half Solution - Choose the first six equations. Select these by highlighting each equation and pressing ¸. 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 displays above. -PQYP 8CTKCDNGU CR 'H A8 +C A# ++H AO# 0 φ A9D R ωO AT U %QORWVGF 4GUWNVU 'C A8 'VC A8 - 6 A0O ωOG AT U 31.
Display: Upper Half Display: Lower Half Solution - Use all the equations to compute the solution for 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 displays above. -PQYP 8CTKCDNGU - φ A9D 4C AΩ 4G A Ω 4H A Ω 4N AMΩ 8HU A8 ωO AT U %QORWVGF 4GUWNVU 'C A8 ++H AO# +. A# 8V A8 31.
Entered Values Computed Results Solution - Use the first equation to solve this problem. Select this by pressing ¸. Press „ to display the input screen, enter all the known variables and press „ to solve the selected equation. The computed result is shown in the screen display above. -PQYP 8CTKCDNGU 'C A8 φ A9D, ωO AT U %QORWVGF 4GUWNV - 31.5 DC Series Generator The two equations in this section describe the properties of a series DC generator.
31.6 Separately-Excited DC Motor These equations form the working foundation for a separately excited motor. The first equation calculates the field voltage Vf in terms of the field current IIf and field coil resistance Rf. Vf = Rf ⋅ IIf Eq. 31.6.1 The second equation computes the terminal voltage Vt in terms of the machine constant K, magnetic flux φ, mechanical radian frequency ωm, armature current Ia, and armature resistance Ra. Vt = K ⋅ φ ⋅ ωm + Ra ⋅ Ia Eq. 31.6.
-PQYP 8CTKCDNGU 'C A8 +C A# φ A9D 4C A Ω 4H A Ω 8H A8 ωO AT U %QORWVGF 4GUWNVU ++H AO# - 6 A0O 8V A8 31.7 DC Shunt Motor These seven equations describe the principal characteristics of a DC shunt motor. The first equation expresses the terminal voltage Vt in terms of the field current IIf and field resistance Rf along with the external field resistance Re.
Solution - Use the fourth equation to solve this problem. Select the equation with the cursor bar and press ¸. Press „ to display the input screen, enter all the known variables and press „ to solve the selected equation. The computed result is shown in the screen display above. -PQYP 8CTKCDNGU - φ A9D ωO AT U %QORWVGF 4GUWNV 'C A8 31.8 DC Series Motor These eight equations describe the performance characteristics of a series DC motor.
Solution 1: Upper Display Solution 1: Lower Display Solution 2: Upper Display Solution 2: Lower Display Solution - The first, third and fifth equations are needed to compute a solution. Select these by highlighting and pressing ¸. Press „ to display the input screen, enter all the known variables and press „ to solve the selected equation set. There are two possible solutions for this example. Type the number of the solution set to be viewed and press ¸ twice.
ωm = Vt Ra ⋅ T − 2 K ⋅φ K ⋅φ Eq. 31.9.5 b g Example 31.9 – Find the machine constant for a permanent motor rotating at 62.5 rad/s in a magnetic flux field of 1.26 Wb. Assume a 110 V back emf. Entered Values Calculated Results Solution - The first equation is needed to compute the solution. Select it by highlighting and pressing ¸. Press „ to display the input screen, enter all the known variables and press „ to solve the equation.
Pme = 3 ⋅ p ωm ⋅ ⋅ Pma 2 ωs Eq. 31.10.6 Pme = T ⋅ ωm Eq. 31.10.7 The eighth equation expresses torque in terms of p, Pma, and ωs. T = 3⋅ p Pma ⋅ 2 ωs Eq. 31.10.8 The last three equations show an equivalent circuit representation of induction motor action and links the power Pa with rotor resistance Rr, rotor current Ir, slip s, rotor resistance per phase RR1 and the machine constant KM. Pma = Rr ⋅ Ir 2 + 1− s ⋅ Rr ⋅ Ir 2 s Eq. 31.10.9 Pa = 1− s ⋅ Rr ⋅ Ir 2 s Eq. 31.10.10 Rr = RR1 KM 2 Eq.
Pma = Rr 2 ⋅ Ir s Eq. 31.11.1 The second equation shows the expression for torque T in terms of poles p, Pma and radian frequency of the induced voltage in the stator ωs. The third equation is an alternate representation of torque in terms of the applied voltage Va, stator resistance Rst, Rr, inductive reactance XL, and ωs. T= 3 Pma ⋅ p⋅ 2 ωs T= 3 p Rr ⋅ ⋅ ⋅ 2 ωs s Eq. 31.11.2 Va 2 FG Rst + Rr IJ H sK Eq. 31.11.
-PQYP 8CTKCDNGU R 4UV AΩ 8C A8 ωU AT U :. A Ω %QORWVGF 4GUWNV 6OOCZ A0O 31.12 Single-Phase Induction Motor These three equations describe the properties of a single-phase induction motor. The first equation defines the slip for forward flux sf with respect to the forward rotating flux φ, the radian frequency of induced current in the stator ωs, the number of poles p, and the angular mechanical speed of the rotor ωm.
ωm = 2 ⋅ ωs p TT max = 3 ⋅ Eq. 31.13.1 p IIf ⋅Va ⋅ 2 ωs Eq. 31.13.2 bg Pma = Va ⋅ Ia ⋅ cos θ T= Eq. 31.13.3 Pme ωm T = 3⋅ Eq. 31.13.4 p Pma 2 ωs Eq. 31.13.5 Example 31.13 – Find the stator radian frequency and the maximum torque for a synchronous 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.
Chapter 32 Introduction to Reference This chapter guides the user through the Reference Part of EE•Pro. The information in the Reference section of the software is organized in a similar fashion as Analysis and Equations, except the information is generally noninteractive. 32.
• Pressing † accesses the menu for the Reference section listing the topics. EE•Pro is structured with a hierarchy of screens for choosing a specific topic. Select a topic of Reference by moving the highlight bar to the desired section using the D key and pressing 2. ¸. Alternatively, type in the number corresponding to the section desired. For example, press ª to access Semiconductor Data, or press { to access the Transforms section. Semiconductor Data Menu Transforms Menu 32.
• Other properties available in the section include Donor and Acceptor levels in Silicon shown in the screen displays below. Silicon Donor Level Silicon Acceptor Levels The last topic in Semiconductor Properties includes the colors observed for Silicon dioxide and Silicon Nitride thickness. These colors are arranged in order of thickness (µm). Some colors appear multiple times due to multiple diffraction orders. SiO2/Si3N4 Color Choices Properties of SiO2/Si3N4 (thickness in µm) 32.
Regular View of Laplace Transform EE Pro for TI - 89, 92 Plus Reference - Introduction to Reference Section Inverse View of Laplace Transform 4
Chapter 33 Resistor Color Chart This section of EE•Pro allows the user to enter the color sequence of a physical resistor and compute its value and tolerance. Most physical resistors come with a band of colors to help identify its value. There are 3 variations of color bands used in practice: 3, 4 or 5 band of colors. The table below identifies the hierarchy used in practice. A picture of the color chart is included in the software and is displayed when the function key † is pressed.
BLUE VIOLET GREY WHITE 6 7 8 9 xE6 xE7 xE8 xE9 0.25% 0.1% 0.05% - Field Description Input Fields Num. of Bands: (Number of Bands) Band 1: (see table above) Band 2: (see table above) Band 3: (see table above) Band 4: (see table above) Band 5: (see table above) Pressing ¸ displays choice of 3, 4, or 5 bands. Pressing ¸ displays choice of colors. Pressing ¸ displays choice of colors. Pressing ¸ displays choice of colors. Pressing ¸ displays choice of colors. Pressing ¸ displays choice of colors.
Chapter 34 Standard Component Values In this section, the software computes the inverse of the color chart computation described in the previous chapter (i.e.: given a value and a tolerance for a resistor, a color sequence is generated). As a side benefit the calculation algorithm also allows the user to estimate suitable “off-the-shelf” standard components for needed resistors, inductors and capacitors. These are also referred to as "Preferred Values" of components available from manufacturers.
Chapter 35 Semiconductor Data Physical, chemical, electrical, electronic and mechanical properties of common semiconductors are presented in this section. The information is organized under five (5) topics listed in detail in Table 35-1. All properties are listed at 300 °K, unless otherwise specifically stated. Details of how to access the information is included in the table.
III-V, II-VI Compounds GaP ! Press ¸ to display pull down menu listing the III-V and II-VI semi-conducting compounds. They are: 1:GaP 2:GaSb 3:InAs 4:InP 5:InSb 6:CdS 7:CdSe 8:CdTeœ 9:ZnS A:ZnSe B:ZnTe Input Field: Compound: To select a compound, use the D key to move to highlight bar to the item and press ¸ (or enter the item number).
Acceptor Levels EE Pro for TI-89, 92 Plus Reference - Semiconductor data Ag(VB): Pt(VB): Si(VB): Si(VB): Na(VB): Au(VB): V: Mo: Mo(VB): Mo(VB): Hg(VB): Hg(VB): Sr: Sr(VB): Ge: Ge(VB): K: K(VB): Sn: W: W: W: W(VB): W(VB): Pb: O: O: Fe: Fe: Fe(VB): Output Field: Silicon Acceptor levels are displayed relative to the valence band.
SiO2/Si3N4 Colors Pd: Be: Be: Zn(CB): Zn: Au(CB): Co(CB): Co: Co: V: Ni(CB): Hg(CB): Hg(CB): Cu: Cu: Cu: Sn: Pb: O(CB): O: Input Field: Compound: and appear more than once. Silicon ! Press ¸ to display the pull down menu listing the color of silicon with SiO2 or Si3N4 deposited on the surface.
Chapter 36 Boolean Expressions This section covers the Boolean expressions reference table, which includes 16 commonly-used Boolean expressions. This section also contains a diagram of the most commonly-used logic components. 36.1 Using Boolean Expressions This section displays a list of rules for Boolean algebra for two variables x and y: "+" is used to indicate the OR function. "." is used to indicate AND function. " ' " represents a logical inversion.
F6 F7 F8 F9 F10 F11 F12 F13 F14 F15 EE Pro for TI-89, 92 Plus Reference - Boolean Expressions (x.y')+(x'.y) x+y (x+y)' (x.y)+(x'.y') y' x+y' x' x'+y (x.
Chapter 38 Transforms This section accesses a series of tables containing transforms of common interest to electrical engineers. The transforms are listed as three topics - Fourier, Laplace and z-Transforms. Each topic contains information categorized under three sub-topics - Definitions, Properties and Transform pairs. All formulae can be viewed in Pretty Print equation-display format. These sub-topics are not interactive, i.e.
f (t ): 1 2π +∞ zc b g F w ⋅ e − i⋅ω ⋅t dw h Inverse Fourier Transform −∞ Status Line Message • The status line gives a description of the equation highlighted. The descriptions use standard mathematical terminology such as Modulation, Convolution, Frequency Integration, etc. Example 38.1 -What is the definition of the Inverse Fourier transform? 1. In the main Transforms screen, move the highlight bar to Fourier Transforms and press ¸. 2. Press ¸ a second time to access Definitions. 3.
Chapter 39 Constants This Constants Reference Table section lists the values and units for 43 commonly-used universal constants. These constants are embedded in equations in the Equations section of EE•Pro and are automatically inserted during computations. 39.1 Using Constants The Constants section in Reference is designed to give a quick glance for commonly used constants. It lists values of accuracy available by the standards of measurement established by appropriate international agencies.
These constants were arranged in the following order: universal mathematical constants lead the list followed by universal physical constants, atomic and quantum mechanical constants, radiation constants, standard temperature and pressure, universal gas constant and molar constants. To view 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.
Chapter 40 SI Prefixes The SI Prefixes section displays the prefixes used by the Systeme International [d’Unit[eacute]s] (SI). 40.1 Using SI Prefixes The prefixes are listed in the order shown in Table 40-1. The D key is used to move 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.
Chapter 41 Greek Alphabet This section displays the Greek Alphabet and their names. There are several Greek letters supported by the TI - 89. To enter the Greek letters, the sequential 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.
Appendix A Frequently Asked Questions A complete list of commonly asked questions about the EE•Pro are listed here. Review this list for your questions prior to calling for Technical support. You might save yourself a phone call! The material is covered under four general headings. A.1 Questions and Answers General Questions Analysis Questions Equations Questions Reference Questions A.2 General Questions The following is a list of questions about the general features of EE•Pro: Q.
A. EE•Pro automatically stores its variables in the current folder specified by the user in 3 or the HOME screens. The current folder name is displayed in the lower left corner of the screen (default is “Main”). To create a new folder to store values 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).
Q. The calculated angle or radian frequency result isn’t correct. A. Check to be sure the 3 settings list degrees or radians for angle units and make sure it matches the units of your entered value or desired value. Secondly, if the result is greater than 2π or less than -2π (result ≥ 360o), the TI solve(...) function may be generating a non principal solution. A principal solution is defined as a value between 0 and 2π (or between 0 and 360o).
A. When the Units feature is on, values can be entered and saved in any unit. When units are off, values can be entered in any unit, but the values will automatically be displayed on the screen in the default SI units. This is necessary so that when a series of equations are solved, all the values are consistent. Q. When solving a set of equations “Too many unknowns to finish solving.” is displayed. Why? A. Sometimes the solver doesn’t have enough to solve for all the remaining, unknown variables.
Appendix B Warranty, Technical Support B.1 da Vinci License Agreement By downloading/installing the software and/or documentation you agree to abide by the following provisions. 1. License: da Vinci Technologies Group, Inc. (“da Vinci”) grants you a license to use and copy the software program(s) and documentation from the linked web/CD page (“Licensed Materials”). 2.
Some states do not allow the exclusion or limitation of incidental or consequential damages, so the above limitation may not apply. B.2 How to Contact Customer Support Customers in the U.S., Canada, Puerto Rico, and the Virgin Islands For questions that are specific to the purchase, download and installation of EE•Pro, or questions regarding the operation of your TI calculator, contact Texas Instruments Customer Support: phone: 1.800.TI.CARES (1-800-842-2737) e-mail: ti-cares@ti.
Appendix C Bibliography In developing EE•Pro a number of resources were used. The primary sources we used are listed below. In addition, a large list of publications, too many to list here, were used as additional references. 1. Campsey, B. J., and Eugene F. Brigham, Introduction to Financial Management, Dryden Press, 1985, pp. 392-400 2. Van Horne, James C., Financial management and Policy, 8th Edition, Prentice-Hall, 1989, pp. 126-132 3. Edminister, Joseph A.
20. Elmasry, Mohamed I., Editor, Digital MOS Integrated Circuits, IEEE Press, New York, 1981 21. Kuo, Benjamin C., Automatic Control Systems, Prentice Hall, New jersey, 1991 22. Stevenson Jr., William D., Elements of Power Systems Analysis, McGraw-Hill International, New York, 1982 23. Wildi, Theodore, Electrical Power Technology, John Wiley and Son, New Jersey, 1981 24. Yarborough, Raymond B., Electrical Engineering Reference Manual, Professional Publications, Belmont, CA 1990 25. Schroeder, Dieter K.
Appendix D: TI-89 & TI-92 PlusKeystroke and Display Differences D.1 Display Property Differences between the TI-89 and TI-92 Plus The complete display specifications for both the TI-89 and TI-92 Plus calculators are displayed below. Table D-1 TI-89 and TI-92 Plus display specifications. Property Display size Pixel Aspect ratio Full Screen TI-89 TI-92 Plus 240 x 128 1.88 40 characters/line 13 Lines 236 x 51 pixels 39 characters, 6 lines Vertical Split Screen (1/3rd) 160 x 100 1.
Table D-2 Keyboard Differences, Representation in Manual Function Specific Key Function Keys Trig Functions Alphabet keys TI-89 Key strokes TI-92 Plus key strokes F1 F2 F3 F4 F5 F6 F7 F8 ƒ „ … † ‡ 2ƒ 2„ 2… ƒ „ … † ‡ ˆ ‰ Š ƒ „ … † ‡ ˆ ‰ Š Sin 2Ú W W Cos 2Û X X Tan 2Ü Y Y Sin-1 Cos-1 ¥Ú 2W ¥ Û 2X [SIN-1] [COS-1] Tan-1 ¥Ü 2Y [TAN-1] A j Á Ñ A B jc Ò B C jd Ó C D jb D E je E F jÍ G jm G H jn H I jo I J jp J K j^ jy K L M Ô Representation in t
V j µ W j¶ Ù Ú Û j· X Y Z Space Function Log, EXP Special Characters Specific Key Editing Functions Ù X Y Z LN 2 Ù x x ex ¥ Ù 2 x s π 2Z 2Z T ¥Z · 2½ Ï Ï ¥½ · 2I 2J · ) * Y= ¥ƒ ¥W # Window Graph ¥„ ¥… ¥E ¥R $ % Cut ¥2 5 Copy Paste Delete ¥¤ ¥N Representation in the manual ¥0 2N ¥ 0 6 7 8 20 2§ § 0 2 0 K / 2 § § 0 £ § 0 ( c c ) { } [ ] d 2c 2d 2b 2e d 2 2 2 2 Addition « « « Subtraction Multiplication Division | p e | p e | p e Recall Store Ba
Raise to power Enter Exponent for power of 10 Equal Integrate Differentiate Function Specific Key Math Operations cont.
Nine Zero Comma Decimal point Main Functions Home Function Specific Key Main Functions cont. Mode Catalog Clear Custom Enter ON OFF ESCAPE Application Cursor Movement o µ b ¶ o µ b ¶ ¥ Q 9 0 , .
Appendix E Error Messages General Error Messages Analysis Error Messages Equations Error Messages Reference Error Messages E.1 General Error Messages 1. NOTE: Make sure the settings in the 3 screen do not have the following configuration. Angle: DEGREE Complex Format: POLAR EE•Pro works best in the default mode settings of your calculator (ie. Complex Format: REAL, or Angle: RADIAN). If one of a set of error messages appears which includes “An error has occurred while converting....
8. “storage error...” This message is set to occur if the user attempts to enter a value into a variable which is locked or archived, or a memory error has occured. Check the current status of the variable by pressing ° and scrolling to the variable name, or check the memory parameters by pressing ¯. 9. “Invalid variable reference. Conflict with system variable or reserved name.” This can occur if a variable name is entered which is reserved by the TI operating system.
for the number base. If a number does not begin with a letter prefix in parenthesis, then it is assumed to be in the number base set in the †/Mode dialog. 8. “Entry truncated” will appear if an entered value exceeds the binary word size allocated in the †/Mode dialog. For example, the decimal number 100 is expressed as a 7 bit binary number:1100100. If a word size of 6 bits is designated in the †/Mode dialog, the latter six digits of the number (100100) will be entered into the available word space.
• • • • • • • • • • "Unable to find a solution in the time allowed. Examine variables eeinput and eeprob to see the exact statement of the problem. EE\xB7Pro sets Exact/Approx mode to AUTO during solve.” “No equations have been selected. Please select either a single equation to solve by itself, or several equations to solve simultaneously.
Appendix F: System Variables and reserved names The TI-89 and TI-92 Plus has a number of variable names that are reserved for the Operating System. The table below lists all the reserved names that are not allowed for use as variables or algebraic names.
