This document contains solutions to problems from Chapter 14. Problem 14.1 calculates the maximum and rms input voltages for a circuit. Problem 14.2 calculates output current and minimum load resistance. Problem 14.3 lists voltage values for another circuit. Problem 14.4 calculates closed-loop gain. The remaining problems involve calculating various voltage and current values, closed-loop gains, and input and feedback resistances for multiple stage amplifier circuits using provided equations. Lengthy calculations are shown for some problems.
This document contains solutions to exercises from Chapter 14. The solutions involve calculations for amplifier circuits using op amp specifications like gain bandwidth, input offset voltage, input bias current, slew rate, and more. Some key results include:
- The input offset voltage for a circuit is calculated as 1.27 mV.
- The input resistance of a voltage follower circuit with a large resistor is calculated as 5000 Mฮฉ.
- The maximum frequency of a circuit before the gain drops by 3 dB is calculated as 20 kHz.
This document contains solutions to examples from Chapter 12 of an exercise book. It includes 12 examples covering topics like:
1) Calculating voltage gain, input resistance, and output resistance of non-inverting op-amps.
2) Analyzing the bandwidth, voltage gain, and input/output resistances of inverting op-amps.
3) Determining the voltage transfer characteristics, transconductance, and voltage gain of BJT common emitter amplifiers.
The examples provide step-by-step workings and calculations to arrive at the key parameters and specifications for different amplifier circuit configurations.
This document contains solutions to problems from Chapter 9. It provides worked examples of calculating voltages and currents in circuits containing operational amplifiers. Key steps and results are shown for multiple circuit configurations, including calculating voltage gains, identifying resistor values, and determining output voltages and currents given input signals. Operational amplifier circuits with one, two and multiple stages are analyzed using relevant equations.
This document contains the solutions to various exercises related to circuit analysis and electronics. Exercise problems cover topics such as diode circuits, rectifiers, filters, and transistor circuits. The solutions involve calculating voltages, currents, power, capacitance values, and other circuit parameters through applying circuit laws and relationships. PSpice simulations are also used to verify some of the circuit responses.
The document contains 30 algebra word problems involving addition and subtraction equations with one or two variables. The equations include adding, subtracting, multiplying, or dividing variables and numbers.
The document discusses different number systems including binary, decimal, hexadecimal, and octal. It covers the basics of each system including their radix value and digit sets. The key topics covered include:
1) Converting between different number systems such as decimal to binary and vice versa.
2) Performing arithmetic operations like addition, subtraction, multiplication in binary. This includes using binary complements for subtraction.
3) Representing signed binary numbers and arithmetic on signed binary numbers.
Capรญtulo 07 falha por fadiga resultante de carregamento variรกvelJhayson Carvalho
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The document provides calculations and solutions to example problems related to fatigue design and analysis. It includes determining endurance limits, fatigue stress concentrations, Goodman diagrams, and calculating fatigue life. Key equations from chapters 3, 7, 8, and the appendix are applied to examples involving shafts, beams, and other mechanical components made from various steel alloys. Material properties, load conditions, and geometric factors are considered to iteratively size components and check designs for sufficient fatigue life.
This document contains sample problems and solutions related to semiconductor physics. Problem EX1.1 calculates intrinsic carrier concentrations ni for GaAs, Ge, and Si at 300K using the intrinsic carrier concentration equation. Problem EX1.2 calculates minority and majority carrier concentrations. Problem EX1.3 calculates drift current density in n-type material.
This document contains solutions to exercises from Chapter 14. The solutions involve calculations for amplifier circuits using op amp specifications like gain bandwidth, input offset voltage, input bias current, slew rate, and more. Some key results include:
- The input offset voltage for a circuit is calculated as 1.27 mV.
- The input resistance of a voltage follower circuit with a large resistor is calculated as 5000 Mฮฉ.
- The maximum frequency of a circuit before the gain drops by 3 dB is calculated as 20 kHz.
This document contains solutions to examples from Chapter 12 of an exercise book. It includes 12 examples covering topics like:
1) Calculating voltage gain, input resistance, and output resistance of non-inverting op-amps.
2) Analyzing the bandwidth, voltage gain, and input/output resistances of inverting op-amps.
3) Determining the voltage transfer characteristics, transconductance, and voltage gain of BJT common emitter amplifiers.
The examples provide step-by-step workings and calculations to arrive at the key parameters and specifications for different amplifier circuit configurations.
This document contains solutions to problems from Chapter 9. It provides worked examples of calculating voltages and currents in circuits containing operational amplifiers. Key steps and results are shown for multiple circuit configurations, including calculating voltage gains, identifying resistor values, and determining output voltages and currents given input signals. Operational amplifier circuits with one, two and multiple stages are analyzed using relevant equations.
This document contains the solutions to various exercises related to circuit analysis and electronics. Exercise problems cover topics such as diode circuits, rectifiers, filters, and transistor circuits. The solutions involve calculating voltages, currents, power, capacitance values, and other circuit parameters through applying circuit laws and relationships. PSpice simulations are also used to verify some of the circuit responses.
The document contains 30 algebra word problems involving addition and subtraction equations with one or two variables. The equations include adding, subtracting, multiplying, or dividing variables and numbers.
The document discusses different number systems including binary, decimal, hexadecimal, and octal. It covers the basics of each system including their radix value and digit sets. The key topics covered include:
1) Converting between different number systems such as decimal to binary and vice versa.
2) Performing arithmetic operations like addition, subtraction, multiplication in binary. This includes using binary complements for subtraction.
3) Representing signed binary numbers and arithmetic on signed binary numbers.
Capรญtulo 07 falha por fadiga resultante de carregamento variรกvelJhayson Carvalho
ย
The document provides calculations and solutions to example problems related to fatigue design and analysis. It includes determining endurance limits, fatigue stress concentrations, Goodman diagrams, and calculating fatigue life. Key equations from chapters 3, 7, 8, and the appendix are applied to examples involving shafts, beams, and other mechanical components made from various steel alloys. Material properties, load conditions, and geometric factors are considered to iteratively size components and check designs for sufficient fatigue life.
This document contains sample problems and solutions related to semiconductor physics. Problem EX1.1 calculates intrinsic carrier concentrations ni for GaAs, Ge, and Si at 300K using the intrinsic carrier concentration equation. Problem EX1.2 calculates minority and majority carrier concentrations. Problem EX1.3 calculates drift current density in n-type material.
This multiple choice question document tests understanding of current liabilities and contingencies. It includes questions about the definition and classification of current liabilities, contingencies, and specific current liability accounts such as accounts payable, notes payable, unearned revenue, and payroll liabilities. It also addresses the accounting treatment for compensated absences, contingencies, and the disclosure of short-term debt refinancing arrangements.
Long term Liabilities in Government accountingAlecxz Herrera
ย
This document discusses different types of long-term liabilities and deferred credits accounts. The "OTHER LONG-TERM LIABILITIES" account is used to record liabilities not classified elsewhere. The "DEFERRED CREDITS" account records money received before it is earned, such as unearned revenue. The "OTHER DEFERRED CREDITS" account is similar but for items not in the deferred credits to income account. Deferred income tax is a liability from income earned but not taxed yet due to accounting and tax differences, to be paid later.
GDP is the total value of all final goods and services produced within a country's borders in a year. It is used to measure the health of a country's economy. Intermediate products and secondhand sales are excluded from GDP calculations to avoid double counting and because no new wealth is created through used goods. When considering GDP, factors like delays in reporting, increases in weapons production versus social services, and changes in quality of life versus costs must be accounted for. GDP measures a country's economic performance through production and economic health through public sentiment.
This document contains a quiz on stockholders' equity concepts from chapter 15. It includes 66 multiple choice questions covering topics like common stock, preferred stock, additional paid-in capital, retained earnings, treasury stock, stock dividends, and calculations related to stockholders' equity. The questions are designed to test the reader's understanding of how various transactions impact the accounting for stockholders' equity.
Ch14 - Organisation theory design and change gareth jonesAnkit Kesri
ย
The document discusses organizational conflict, power, and politics. It describes the sources and stages of organizational conflict, including latent, perceived, felt, and manifest conflict. It also discusses methods for managing conflict, such as resolving structural issues or addressing conflicts between individuals. The document also covers sources of power within organizations, such as authority, resources, and information control. Finally, it discusses organizational politics and tactics for gaining influence, such as building coalitions, and the importance of balancing power within organizations.
This document contains 49 true/false questions about investment planning and securities markets. It covers topics like the differences between investing and speculating, mutual funds, investment plans, common and preferred stocks, primary and secondary markets, brokers, margin trading, and stock indexes. The questions test comprehension of key concepts in investment planning and securities markets.
This document contains solutions to 15 exercises related to circuit analysis. The exercises calculate values like capacitance, resistance, cutoff frequency, and gain based on given circuit diagrams and specifications. Equations are set up and solved to find the desired values. For example, exercise 1 calculates the capacitance needed for a given RC circuit to have a cutoff frequency of 40 kHz. The solutions proceed methodically through each problem, showing the work to arrive at the final answers.
The document contains solutions to example problems from Chapter 9 of an exercise book.
The examples calculate resistor values for circuits involving operational amplifiers to achieve specified voltage gains, current values, and time constants. Computer analysis examples also calculate output voltages and currents for various input conditions in operational amplifier circuits.
This document contains solutions to problems in Chapter 13 regarding computer simulations and analyses of bipolar junction transistor circuits. The key points summarized are:
1) Problem 13.1 provides a computer simulation of a circuit and calculates the overall gain as -1.59x106.
2) Problem 13.6 calculates various currents in a BJT reference circuit, including the reference current of 0.5 mA and collector current of 21.1 ฮผA.
3) Problem 13.8 calculates currents in another BJT reference circuit, including the reference current of 0.22 mA, collector current of 14.2 ฮผA, and other node currents.
This document contains solutions to problems from Chapter 12. It provides detailed calculations and explanations for problems involving feedback amplifiers, frequency response, input and output impedances, and circuit configurations. Key points addressed include:
- Calculating voltage gain, feedback factor, and input/output impedances for various amplifier circuits.
- Deriving frequency response expressions and bandwidth calculations.
- Determining circuit types based on input/output impedance characteristics.
- Solving for voltage and current variables in feedback amplifier circuits.
William hyatt-7th-edition-drill-problems-solutionSalman Salman
ย
This document contains solutions to drill problems from Chapter 2 on electrostatics. It includes calculations of electric fields, electric flux densities, and total charge for various charge distributions using Gauss's law and other concepts of electrostatics. Any errors found in the solutions should be reported to the author.
This document provides solutions to example problems from Chapter 11 of an electronics textbook.
The problems cover topics such as: transistor biasing circuits, common emitter amplifier configurations, determining voltage and current values, and calculating resistor values needed for specific circuit behaviors. Detailed calculations are shown for each example problem to derive the requested values.
1) The document contains example problems from Chapter 6 of a textbook on exercise problems involving BJT circuit analysis.
2) Problem EX6.1 involves calculating currents, voltages, and gain for a common-emitter amplifier.
3) Problem EX6.12 involves calculating the bias resistor values needed to produce a specified quiescent current and voltage for a common-emitter amplifier.
This document contains solutions to problems from Chapter 11. Problem 11.1 solves for resistances in a common emitter amplifier circuit. Part c) finds the maximum and minimum voltages. Problem 11.2 calculates the common-mode gain of a differential amplifier as -0.01012. Problem 11.3a) solves for currents and voltages in a multi-emitter transistor. Part b) solves for currents when the collector voltage is changed.
Solutions manual for microelectronic circuits analysis and design 3rd edition...Gallian394
ย
Solutions Manual for Microelectronic Circuits Analysis and Design 3rd Edition by Rashid IBSN 9781305635166
Download at: https://goo.gl/ShMdzK
People also search:
microelectronic circuits analysis and design rashid pdf
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electronic circuit analysis and design
This document contains solutions to exercises from Chapter 8. The summaries are:
1) The first exercise calculates the maximum power transfer for a common emitter amplifier circuit under different voltage supply conditions.
2) The second exercise calculates temperature increases for different components in a circuit based on power dissipation.
3) The third exercise similarly calculates temperature rises for different components based on known power and thermal resistances.
The document provides information about calculating mean, variance, and standard deviation from a data set. It includes a table of values for number of cycles (x) and failure cycles (f) for a sample of bearings. It then shows the calculations to find:
1) The mean number of cycles is 122.9 thousand cycles.
2) The variance is 912.9 thousand cycles squared.
3) The standard deviation is 30.3 thousand cycles.
The document provides information about calculating mean, variance, and standard deviation from a data set. It includes a table of values for number of cycles (x) and failure cycles (f) for a sample. It then shows the calculations to find:
1) The mean number of cycles is 122.9 thousand cycles
2) The variance is 912.9 thousand cycles squared
3) The standard deviation is 30.3 thousand cycles
The intent is to demonstrate calculating statistics from a data set to characterize the distribution and variability. The example uses cycle life data from a fatigue test to find the central tendency and spread.
This document contains solutions to 10 exercises involving bipolar junction transistors and MOSFET circuits. The key points summarized are:
1) Exercise 10.1 calculates the reference and base currents in a BJT circuit. Exercise 10.2 calculates the reference current, zero bias current, and small signal output resistance.
2) Exercise 10.3 calculates the various currents in a multi-emitter BJT. Exercise 10.4 calculates the early effect resistance and voltage difference between two BJTs.
3) Exercise 10.5 again calculates reference current and zero bias current. Exercises 10.6 through 10.11 involve various MOSFET and BJT circuit calculations.
Amth250 octave matlab some solutions (4)asghar123456
ย
This document provides the solutions to assignment questions about linear algebra and linear programming. It includes:
1) The matrix and vector representations of a system of linear equations. It finds the numerical solution and compares it to the exact solution.
2) An analysis of the condition number of Hilbert matrices of increasing size, showing the condition number grows exponentially up to a certain point.
3) Formulating a production planning problem as a linear programming problem to maximize profit subject to resource constraints. It solves the problem as both a linear program and integer program.
This document contains solutions to problems from Chapter 15. It provides detailed calculations and examples for various circuit analysis problems involving filters. Some key points:
- It calculates transfer functions, corner frequencies, and component values for low-pass filters, high-pass filters, and bandpass filters.
- It determines the number of poles needed in a filter to achieve a given attenuation level.
- It analyzes the transfer function of a maximally flat high-pass filter and derives the relationship between component values.
- It provides an example of designing a circuit to meet given low-frequency and high-frequency gain specifications using an op-amp.
The document demonstrates analytical techniques for analyzing and designing passive filter
This multiple choice question document tests understanding of current liabilities and contingencies. It includes questions about the definition and classification of current liabilities, contingencies, and specific current liability accounts such as accounts payable, notes payable, unearned revenue, and payroll liabilities. It also addresses the accounting treatment for compensated absences, contingencies, and the disclosure of short-term debt refinancing arrangements.
Long term Liabilities in Government accountingAlecxz Herrera
ย
This document discusses different types of long-term liabilities and deferred credits accounts. The "OTHER LONG-TERM LIABILITIES" account is used to record liabilities not classified elsewhere. The "DEFERRED CREDITS" account records money received before it is earned, such as unearned revenue. The "OTHER DEFERRED CREDITS" account is similar but for items not in the deferred credits to income account. Deferred income tax is a liability from income earned but not taxed yet due to accounting and tax differences, to be paid later.
GDP is the total value of all final goods and services produced within a country's borders in a year. It is used to measure the health of a country's economy. Intermediate products and secondhand sales are excluded from GDP calculations to avoid double counting and because no new wealth is created through used goods. When considering GDP, factors like delays in reporting, increases in weapons production versus social services, and changes in quality of life versus costs must be accounted for. GDP measures a country's economic performance through production and economic health through public sentiment.
This document contains a quiz on stockholders' equity concepts from chapter 15. It includes 66 multiple choice questions covering topics like common stock, preferred stock, additional paid-in capital, retained earnings, treasury stock, stock dividends, and calculations related to stockholders' equity. The questions are designed to test the reader's understanding of how various transactions impact the accounting for stockholders' equity.
Ch14 - Organisation theory design and change gareth jonesAnkit Kesri
ย
The document discusses organizational conflict, power, and politics. It describes the sources and stages of organizational conflict, including latent, perceived, felt, and manifest conflict. It also discusses methods for managing conflict, such as resolving structural issues or addressing conflicts between individuals. The document also covers sources of power within organizations, such as authority, resources, and information control. Finally, it discusses organizational politics and tactics for gaining influence, such as building coalitions, and the importance of balancing power within organizations.
This document contains 49 true/false questions about investment planning and securities markets. It covers topics like the differences between investing and speculating, mutual funds, investment plans, common and preferred stocks, primary and secondary markets, brokers, margin trading, and stock indexes. The questions test comprehension of key concepts in investment planning and securities markets.
This document contains solutions to 15 exercises related to circuit analysis. The exercises calculate values like capacitance, resistance, cutoff frequency, and gain based on given circuit diagrams and specifications. Equations are set up and solved to find the desired values. For example, exercise 1 calculates the capacitance needed for a given RC circuit to have a cutoff frequency of 40 kHz. The solutions proceed methodically through each problem, showing the work to arrive at the final answers.
The document contains solutions to example problems from Chapter 9 of an exercise book.
The examples calculate resistor values for circuits involving operational amplifiers to achieve specified voltage gains, current values, and time constants. Computer analysis examples also calculate output voltages and currents for various input conditions in operational amplifier circuits.
This document contains solutions to problems in Chapter 13 regarding computer simulations and analyses of bipolar junction transistor circuits. The key points summarized are:
1) Problem 13.1 provides a computer simulation of a circuit and calculates the overall gain as -1.59x106.
2) Problem 13.6 calculates various currents in a BJT reference circuit, including the reference current of 0.5 mA and collector current of 21.1 ฮผA.
3) Problem 13.8 calculates currents in another BJT reference circuit, including the reference current of 0.22 mA, collector current of 14.2 ฮผA, and other node currents.
This document contains solutions to problems from Chapter 12. It provides detailed calculations and explanations for problems involving feedback amplifiers, frequency response, input and output impedances, and circuit configurations. Key points addressed include:
- Calculating voltage gain, feedback factor, and input/output impedances for various amplifier circuits.
- Deriving frequency response expressions and bandwidth calculations.
- Determining circuit types based on input/output impedance characteristics.
- Solving for voltage and current variables in feedback amplifier circuits.
William hyatt-7th-edition-drill-problems-solutionSalman Salman
ย
This document contains solutions to drill problems from Chapter 2 on electrostatics. It includes calculations of electric fields, electric flux densities, and total charge for various charge distributions using Gauss's law and other concepts of electrostatics. Any errors found in the solutions should be reported to the author.
This document provides solutions to example problems from Chapter 11 of an electronics textbook.
The problems cover topics such as: transistor biasing circuits, common emitter amplifier configurations, determining voltage and current values, and calculating resistor values needed for specific circuit behaviors. Detailed calculations are shown for each example problem to derive the requested values.
1) The document contains example problems from Chapter 6 of a textbook on exercise problems involving BJT circuit analysis.
2) Problem EX6.1 involves calculating currents, voltages, and gain for a common-emitter amplifier.
3) Problem EX6.12 involves calculating the bias resistor values needed to produce a specified quiescent current and voltage for a common-emitter amplifier.
This document contains solutions to problems from Chapter 11. Problem 11.1 solves for resistances in a common emitter amplifier circuit. Part c) finds the maximum and minimum voltages. Problem 11.2 calculates the common-mode gain of a differential amplifier as -0.01012. Problem 11.3a) solves for currents and voltages in a multi-emitter transistor. Part b) solves for currents when the collector voltage is changed.
Solutions manual for microelectronic circuits analysis and design 3rd edition...Gallian394
ย
Solutions Manual for Microelectronic Circuits Analysis and Design 3rd Edition by Rashid IBSN 9781305635166
Download at: https://goo.gl/ShMdzK
People also search:
microelectronic circuits analysis and design rashid pdf
microelectronics circuit analysis and design 3rd edition pdf
microelectronic circuits analysis and design pdf
microelectronic circuits analysis and design rashid solution manual
electronic circuit analysis and design
This document contains solutions to exercises from Chapter 8. The summaries are:
1) The first exercise calculates the maximum power transfer for a common emitter amplifier circuit under different voltage supply conditions.
2) The second exercise calculates temperature increases for different components in a circuit based on power dissipation.
3) The third exercise similarly calculates temperature rises for different components based on known power and thermal resistances.
The document provides information about calculating mean, variance, and standard deviation from a data set. It includes a table of values for number of cycles (x) and failure cycles (f) for a sample of bearings. It then shows the calculations to find:
1) The mean number of cycles is 122.9 thousand cycles.
2) The variance is 912.9 thousand cycles squared.
3) The standard deviation is 30.3 thousand cycles.
The document provides information about calculating mean, variance, and standard deviation from a data set. It includes a table of values for number of cycles (x) and failure cycles (f) for a sample. It then shows the calculations to find:
1) The mean number of cycles is 122.9 thousand cycles
2) The variance is 912.9 thousand cycles squared
3) The standard deviation is 30.3 thousand cycles
The intent is to demonstrate calculating statistics from a data set to characterize the distribution and variability. The example uses cycle life data from a fatigue test to find the central tendency and spread.
This document contains solutions to 10 exercises involving bipolar junction transistors and MOSFET circuits. The key points summarized are:
1) Exercise 10.1 calculates the reference and base currents in a BJT circuit. Exercise 10.2 calculates the reference current, zero bias current, and small signal output resistance.
2) Exercise 10.3 calculates the various currents in a multi-emitter BJT. Exercise 10.4 calculates the early effect resistance and voltage difference between two BJTs.
3) Exercise 10.5 again calculates reference current and zero bias current. Exercises 10.6 through 10.11 involve various MOSFET and BJT circuit calculations.
Amth250 octave matlab some solutions (4)asghar123456
ย
This document provides the solutions to assignment questions about linear algebra and linear programming. It includes:
1) The matrix and vector representations of a system of linear equations. It finds the numerical solution and compares it to the exact solution.
2) An analysis of the condition number of Hilbert matrices of increasing size, showing the condition number grows exponentially up to a certain point.
3) Formulating a production planning problem as a linear programming problem to maximize profit subject to resource constraints. It solves the problem as both a linear program and integer program.
This document contains solutions to problems from Chapter 15. It provides detailed calculations and examples for various circuit analysis problems involving filters. Some key points:
- It calculates transfer functions, corner frequencies, and component values for low-pass filters, high-pass filters, and bandpass filters.
- It determines the number of poles needed in a filter to achieve a given attenuation level.
- It analyzes the transfer function of a maximally flat high-pass filter and derives the relationship between component values.
- It provides an example of designing a circuit to meet given low-frequency and high-frequency gain specifications using an op-amp.
The document demonstrates analytical techniques for analyzing and designing passive filter
This document contains a tutorial on applied statistics and mathematics in economics and business. It covers various concepts such as mutually exclusive events, probability, distributions, expected value, and variance. Examples include the probability of employees coming from different countries in an accountancy firm, the likelihood of a boy liking different foods, accident rates by age, and train arrival times. Key formulas and calculations are shown for these examples.
This document summarizes solutions to problems involving circuit analysis using Laplace transforms.
1) The first problem analyzes a simple RC circuit and calculates the transfer function, cutoff frequency, and output response to a step input.
2) The second problem analyzes a similar RC circuit with different component values and calculates the transfer function and cutoff frequency.
3) Additional circuit examples are provided involving resistors, capacitors, and inductors. Transfer functions are derived and cutoff frequencies are calculated.
4) A multiple time constant circuit is analyzed and its frequency response is characterized.
5) Circuits involving operational amplifiers are analyzed to derive transfer functions and calculate bandwidth parameters.
1) The document presents the solution to a structural problem involving forces and moments acting on a beam structure.
2) Equations are set up and solved for the reactions at supports A, C, and E, as well as internal forces FAB, FCB, FCD, FDB, and FEB.
3) The final internal forces are FAB = 3.13 N, FCB = 6.87 N, FCD = -28.34 N, FDB = 9.36 N, FEB = 15.04 N, FA = 12.5 N, FC = 27.5 N, and FE = -12.51 N.
The document summarizes the answers to two economic micro theory questions. For the first question, it presents four isoquant curves showing different combinations of the factors labor (L) and capital (K) that produce the same output levels. For the second question, it calculates average revenue (AR), total revenue (TR), and marginal revenue (MR) for varying quantities of a homogeneous good. It presents the results in a table and graph showing AR, TR, and MR at each quantity from 0 to 15 units.
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Ramey Soft provides software consulting services to assist clients with software design, strategy, tools, and processes in order to ensure project and product success. Their expertise includes web services, database design, web development, management systems, desktop and Android application development, and programming languages such as Java, ASP.net, C/C++, PHP, C#, SQL, and MySQL. They also provide counseling for graduate and post-graduate projects and aim to transform data into knowledge to help clients solve problems and better serve customers. Contact details are provided for Skype, email, and phone numbers.
Ramey Soft provides software consulting services to assist clients with software design, strategy, tools, and processes in order to ensure project and product success. Their services go beyond just coding to transform data into knowledge and help clients solve problems and better serve customers. They offer expertise in web services, database design, web development, management systems, desktop and Android application development, and programming languages like Java, ASP.net, C/C++, PHP, C#, SQL, and MySQL.
Ramey Soft provides software consulting services to assist clients with software design, strategy, tools, and processes in order to ensure project and product success. Their expertise includes web services, database design, web and desktop application development, and management systems for various industries. They also provide counseling for graduate and post-graduate projects, with the goal of transforming data into knowledge to help clients solve problems and better serve customers. Ramey Soft can be contacted via email, Skype, or phone for web development, database design, and application development services in languages like Java, ASP.NET, C/C++, PHP, C#, and technologies like SQL, MySQL, and graphics.
The document contains solutions to example problems from Chapter 17 of an exercise book. Problem EX17.1 involves calculating currents in a circuit with a voltage source and two resistors. Part a calculates the currents with different input voltages. Part b calculates the node voltages. EX17.2 calculates power dissipation for a logic gate circuit with different input voltages.
This document contains the solutions to exercises from Chapter 16. The first exercise solves for the threshold voltage (VTN) of a NMOS transistor. The second exercise calculates the output voltage and current of a common source amplifier. The third exercise analyzes the output characteristics of an inverter. The remaining exercises involve additional circuit analysis problems related to topics in Chapter 16 such as CMOS inverters, logic gates and power calculations.
This document contains solutions to exercises from Chapter 13. The solutions involve calculating various currents and voltages in BJT amplifier circuits. Equations are provided and values are calculated for currents, voltages, gains, and other circuit parameters. Detailed steps are shown for each calculation.
1) This document provides solutions to problems involving bipolar junction transistors. It analyzes various circuit configurations to determine values like current, voltage, and resistance.
2) Key steps involve writing equations for the relationships between currents, applying Kirchhoff's laws, and solving the equations simultaneously to find the unknown values.
3) The analysis considers factors like the effect of beta on collector and base currents, and how resistance values must be set for desired output currents and voltages.
The document provides solutions to various problems involving the analysis and design of transistor amplifiers and power supplies. It examines concepts such as maximum power transfer, voltage gain, efficiency, heat dissipation, and voltage/current characteristics. Diagrams and calculations are presented to determine operating voltages and currents, power outputs, temperature rises, and efficiency for different circuit configurations under varying conditions.
This document contains solutions to exercises from Chapter 7. The exercises involve analyzing RC circuits, operational amplifiers, and transistors. The solutions calculate values like capacitance, frequency response, gain, and bias points by setting up and solving equations that model the circuit behavior. Circuit diagrams are included with some of the solutions. The exercises cover topics like low-pass filters, amplifiers, and basic transistor circuits.
This document contains solutions to problems involving BJT amplifier circuit analysis. Problem 6.1 calculates key transistor parameters like transconductance (gm) and output resistance (ro) for two common-emitter amplifiers. Problem 6.2 repeats this calculation for different bias currents. Subsequent problems analyze additional amplifier circuits, determining voltage gains and component values needed to meet specified conditions. Calculations involve setting up and solving equations relating bias currents, voltages, transistor parameters and circuit resistances.
This document contains solutions to problems involving BJT circuit analysis. Key points include:
1) Problem 5.1 analyzes two common emitter amplifier circuits and calculates currents and voltages.
2) Problem 5.2 calculates alpha and collector current values for two common emitter circuits with different beta values.
3) The remaining problems involve calculating various parameter values like currents, voltages, and gains for additional common emitter, common base, and common collector BJT circuits. Circuit analysis methods like Kirchhoff's laws and transistor biasing equations are applied.
4) Graphs and calculations of operating points are presented for some circuits. Effects of component value variations are considered, such as the impact on collector
The document contains solutions to 20 example problems (labeled EX5.1 through EX5.20) relating to bipolar junction transistors. Each problem involves calculating various transistor parameters such as bias currents, voltages, and resistor values given information such as supply voltages, transistor betas, and resistor/collector resistor values. The problems cover topics such as calculating bias points, determining operating modes, and designing transistor circuits.
1) The document contains examples of calculations for MOSFET circuit analysis, including determining operating point values like I_D, V_GS, and g_m.
2) Equations are provided and used to solve for transistor parameters like threshold voltage, transconductance, and output resistance.
3) Circuit examples include common source, common gate, and common drain configurations to determine voltage gain and output resistance.
This document contains solutions to exercises from Chapter 4. The summaries are:
1) Exercise 4.1 calculates the transconductance (gm) and output resistance (ro) for a MOSFET with given parameters.
2) Exercise 4.2 calculates the voltage gain (Av) of a common-source amplifier using the gm from Exercise 4.1.
3) Exercise 4.3 calculates various voltage and current values for a common-source amplifier using given resistor values. It also calculates the voltage gain (Av) of the amplifier.
This document provides solutions to problems involving MOSFET circuit analysis, calculating current (ID) and other parameters using equations that relate gate voltage (VGS), drain voltage (VDS), threshold voltage (VT), and transconductance (Kn or Kp) for n-channel or p-channel MOSFETs. Various regions of operation are considered including saturation, cut-off, and non-saturation. Worked examples calculate ID for different bias conditions and device parameters such as width (W) and length (L).
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Ollama Use Case: Learn how Scenario Specialist Dmitri Bagh has utilized Ollama within FME to input data, create custom models, and enhance security protocols. This segment will include demos to illustrate the full capabilities of FME in AI-driven processes.
Custom AI Models: Discover how to leverage FME to build personalized AI models using your data. Whether itโs populating a model with local data for added security or integrating public AI tools, find out how FME facilitates a versatile and secure approach to AI.
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1. Chapter 14
Problem Solutions
14.1
vo
Ad = = โ80
vi
vo (max) = 4.5 โ vi (max) = 56.25 mV
56.25
So vi (max)rms = = 39.77 mV
2
14.2
(a)
4.5
i2 = = 0.028125 mA
160
4.5
iL = = 4.5 mA
1
Output Circuit = 4.528 mA
v โ4.5
vi = โ o = โ vi = โ0.05625 V
A 80
(b)
v 4.5
io โ 15 mA = o =
RL RL
โ RL (min) = 300 ฮฉ
14.3
(1) vo = 2 V
(2) v2 = 12.5 mV
(3) AOL = 2 ร 104
(4) v1 = 8 ฮผ V
(5) AOL = 1000
14.4
From Eq. (14.4)
2. โ R2 / R1
ACL =
1 โ R2 โ
1+ โ1 + โ
AoL โ R1 โ
โ R2 / R1
โ15 =
1 โ R2 โ
1+ โ1 + โ
2 ร 103 โ R1 โ
โR โ
15 โ 2 โ
โ 1 โ โ R1 โ = R2
15 โ 1 + โ+
โ 2 ร 103 โ 2 ร 103 R1
R2
15.0075 = (0.9925)
R1
R2
= 15.12
R1
14.5
vI โ v1 v1 โ v0 v1
= + and v0 = โ A0 L v1
R1 R2 Ri
v0
so that v1 = โ
A0 L
vI v0 โ 1 1 1โ
+ = v1 โ + + โ
R1 R2 โ R1 R2 Ri โ
So
vI โก1 1 โ 1 1 1 โโค
= โv0 โข + โ + + โโฅ
R1 โฃ R2 A0 L โ R1 R2 Ri โ โฆ
Then
v0 โ(1/ R1 )
= = ACL
vI โก 1 1 โ 1 1 1 โโค
โข + โ + + โโฅ
โฃ R2 A0 L โ R1 R2 Ri โ โฆ
From Equation (14.20) for RL = โ and R0 = 0
1 1 1 (1 + A0 L )
= + โ
Rif Ri R2 1
a. For Ri = 1 kฮฉ
โ(1/ 20)
ACL =
โก 1 1 โ 1 1 1 โโค
โข100 + 103 โ 20 + 100 + 1 โ โฅ
โฃ โ โ โฆ
โ0.05
=
[0.01 + 1.06 ร 10โ3 ]
or
4. 0.0999 50
= = 49.95
0.002 1 + 1
(50)
AOL
which yields
AOL = 1000
14.7
From Equation (14.18)
โA 1 โ
โ โ OL โ โ
v
Avf 1 = o1 = โ Ro R2 โ
v1 โ 1 1 1 โ
โ + + โ
โ RL Ro R2 โ
Or
โ 5 ร 103 1 โ
โโ โ โ
1 100 โ โ(4.99999 ร 103 )
vo1 = โ โ v1 = โ v1
โ1 1 1 โ 1.11
โ + + โ
โ 10 1 100 โ
vo1 = โ4.504495 ร 103 โ v1
Now
i1 vi โ v1
= โกK
v1 R1v1
Then
vi โ v1 = KR1v1
which yields
vi
v1 =
KR1 + 1
Now, from Equation (14.20)
โก 1 โค
1 + 5 ร 103 + โฅ
1 1 โข 10
K= + โข โฅ
10 100 โข 1 + 1 + 1 โฅ
โข
โฃ 10 100 โฅ โฆ
โก 5.0011ร 103 โค
= (0.1) + (0.01) โข โฅ = 45.15495
โฃ 1.11 โฆ
Then
vi vi
v1 = =
( 45.15495)(10 ) + 1 452.5495
We find
โก vi โค
vo1 = โ4.504495 ร 103 โข โฅ
โฃ 452.5495 โฆ
Or
v
Avf 1 = o1 = โ9.9536
vi
For the second stage, RL = โ
5. โ 5 ร 103 1 โ
โโ โ โ
1 100 โ
vo 2 = โ โ v1โฒ = โ4.950485 ร 103 โ v1โฒ
โ1 1 โ
โ + โ
โ 1 100 โ
โก โค
1 1 โข1 + 5 ร 103 โฅ
Kโก + โข โฅ = 49.61485
10 100 โข 1 + 1 โฅ
โข
โฃ 100 โฅ โฆ
vo1 vo1 vo1
v1โฒ = = =
KR1 + 1 (49.61485)(10) + 1 497.1485
Then
vo 2 โ4.950485 ร 103
= = โ9.95776
vo1 497.1485
So
v
Avf = o 2 = (โ9.9536)(โ9.95776) โ Avf = 99.12
vi
14.8
a.
v1 โ vI v v โv
+ 1 + 1 0 =0 (1)
R3 + Ri R1 R2
โก 1 1 1โค v vI
v1 โข + + โฅ= 0 +
โฃ R3 + Ri R1 R2 โฆ R2 R3 + Ri
v0 v0 โ A0 L vd v0 โ v1
+ + =0 (2)
RL R0 R2
or
โก1 1 1โค v A v
v0 โข + + โฅ = 1 + 0L d
โฃ RL R0 R2 โฆ R2 R0
โ v โv โ
vd = โ I 1 โ โ Ri (3)
โ R3 + Ri โ
So substituting numbers:
โก 1 1 1โค v vI
v1 โข + + โฅ= 0 + (1)
โฃ10 + 20 10 40 โฆ 40 10 + 20
or
v1[0.15833] = v0 [0.025] + vI [0.03333]
โก1 1 1 โค v (104 )vd
v0 โข + + โฅ= 1 + (2)
โฃ1 0.5 40 โฆ 40 0.5
6. or
v0 [3.025] = v1 [ 0.025] + ( 2 ร 104 ) vd
โ v โv โ
vd = โ I 1 โ โ 20 = 0.6667 ( vI โ v1 ) (3)
โ 10 + 20 โ
So
v0 [3.025] = v1 [ 0.025] + ( 2 ร 104 ) ( 0.6667 )( vI โ v1 ) (2)
or
v0 [3.025] = 1.333 ร10 4 vI โ 1.333 ร 104 v1
From (1):
v1 = v0 ( 0.1579 ) + vI ( 0.2105 )
Then
v0 [3.025] = 1.333 ร 104 vI โ 1.333 ร 104 โก v0 ( 0.1579 ) + vI ( 0.2105 ) โค
โฃ โฆ
v0 โก 2.1078 ร103 โค = vI โก1.0524 ร104 โค
โฃ โฆ โฃ โฆ
or
v
ACL = 0 = 4.993
vI
To find Rif : Use Equation (14.27)
โ 0.5 0.5 โ
iI โ 1 + + โ
โ 1 40 โ
1 โ โ 0.5 0.5 โ 0.5 โซ (10 ) vd
3
โงโ 1
= v1 โจโ + โ โ 1 + + โ โ โฌโ
โฉโ 10 40 โ โ 1 40 โ (40) 2 โญ 40
iI (1.5125) = v1{(0.125)(1.5125) โ 0.0003125} โ 25vd
or
iI (1.5125) = vI {0.18875} โ 25vd
Now
vd = iI Ri = iI (20) and v1 = vI โ iI (20)
So
iI (1.5125) = [vI โ iI (20)] โ [0.18875] โ 25iI (20)
iI [505.3] = vI (0.18875)
or
vI
= 2677 kฮฉ
iI
Now Rif = 10 + 2677 โ Rif = 2.687 Mฮฉ
To determine R0 f : Using Equation (14.36)
โก โค โก โค
1 1 โข A0 L โฅ 1 โข 103 โฅ
= โ โข โฅ= โ โข โฅ
โฒ
R0 f R0 โข R2 โฅ 0.5 โข1 + 40 โฅ
โข1 + R R โฅ โข 10 20 โฅ
โฃ 1 i โฆ โฃ โฆ
โฒ
or R0 f = 3.5 ฮฉ
Then R0 f = 1 kฮฉ 3.5 ฮฉ
โ R0 f = 3.49 ฮฉ
b. Using Equation (14.16)
dACL โ 5 โ dA
= (โ10) โ 3 โ โ CL = โ(0.05)%
ACL โ 10 โ ACL
14.9
7. v0 โ A0 L vd v0 โ vI
+ = 0 and vd = vI โ v0
R0 Ri
So
v0 A0 L v v
โ โ (vI โ v0 ) + 0 โ I = 0
R0 R0 Ri Ri
โก1 A 1โค โก1 A โค
v0 โข + 0 L + โฅ = vI โข + 0 L โฅ
โฃ R0 R0 Ri โฆ โฃ Ri R0 โฆ
โก 1 (104 ) 1 โค โก 1 (104 ) โค
v0 โข + + โฅ = vI โข + โฅ
โฃ 0.2 0.2 100 โฆ โฃ100 0.2 โฆ
v0 [5.000501ร 104 ] = vI [5.000001ร 10 4 ]
v0
So ACL = = 0.9999
vI
b. Set vI = 0
v0 โ A0 L vd v0
i0 = + and vd = โv0
R0 Ri
โก1 A 1โค
i0 = v0 โข + 0 L + โฅ
โฃ R0 R0 Ri โฆ
Then
1 1 A0 L 1
= + +
R0 f R0 R0 Ri
or
1 1 (104 ) 1
= + +
R0 f 0.2 0.2 100
which yields
R0 f โ 0.02 ฮฉ
14.10
8. vI 1 โ v1 vI 2 โ v1 v1 โ v0
+ =
20 10 40
vI 1 vI 2 v0 โก1 1 1โค
+ + = v1 โข + + โฅ
20 10 40 โฃ 20 10 40 โฆ
v
and v0 = โ A0 L v1 so that v1 = โ 0
A0L
Then
โง1 1 โ 7 โโซ
vI 1 (0.05) + vI 2 (0.10) = โv0 โจ + โ
3 โ โโฌ
โฉ 40 2 ร 10 โ 40 โ โญ
= โv0 [2.50875 ร 10โ2 ]
โ v0 = โ1.993vI 1 โ 3.986vI 2
ฮv0 2 โ 1.993 ฮv
= โ 0 = 0.35%
v0 2 v0
14.11
โ 40 โ โ 4โ
vB = โ โ v2 = โ โ v2 = 0.8v2 (1)
โ 40 + 10 โ โ 5โ
v1 โ v A v A โ v0
=
10 40
v1 v0 โ1 1 โ
+ = vA โ + โ
10 40 โ 10 40 โ
v1 (0.1) + v0 (0.025) = vA (0.125) (2)
v0 = A0 L vd = A0 L (vB โ v A ) (3)
or
v0 = A0 L [0.8v2 โ v A ]
v0
โ 0.8v2 = โv A
A0 L
v0
โ v A = 0.8v2 โ
A0 L
Then
9. โก v โค
v1 (0.1) + v0 (0.025) = (0.125) โข0.8v2 โ 0 โฅ
โฃ A0 L โฆ
โก 0.125 โค
v1 (0.1) โ v2 (0.1) = โv0 โข0.025 +
โฃ 103 โฅโฆ
โ2
= โv0 [2.5125 ร 10 ]
v0
โ Ad = = 3.9801
v2 โ v1
ฮAd 0.0199
โ = โ 0.4975%
Ad 4
14.12
a. Considering the second op-amp and Equation (14.20), we have
โก โค
1 1 1 โข1 + 100 โฅ 101
= + โ โข โฅ = 0.10 +
Rif 2 10 0.1 โข 1 + 1 โฅ (0.1)(11)
โข 0.1 โฅ
โฃ โฆ
So Rif 2 = 0.0109 kฮฉ
The effective load on the first op-amp is then
RL1 = 0.1 + Rif 2 = 0.1109 kฮฉ
Again using Equation (14.20), we have
1
1 + 100 +
1 1 1 0.1109 = 0.10 + 110.017
= + โ
Rif 10 1 1 + 1 + 1 11.017
0.1109 1
so that
Rif = 99.1 ฮฉ
b. To determine R0 f :
For the first op-amp, we can write, using Equation (14.36)
โก โค โก โค
1 1 โข A0 L โฅ 1 โข 100 โฅ
= โ โข โฅ = โ โข โฅ
R0 f 1 R0 โข R2 โฅ 1 โข 40 โฅ
1+ 1+
โข โฅ โข โฅ
โฃ 1 || 10 โฆ
โฃ R1 || Ri โฆ
which yields R0 f 1 = 0.021 kฮฉ
For the second op-amp, then
โก โค
โข โฅ
1 1 โข A0 L โฅ
= โ
R0 f R0 โข R2 โฅ
โข1 + ( R + R ) || R โฅ
โฃ 1 0f1 i โฆ
โก โค
1 โข 100 โฅ
= โ โข โฅ
1 โข1 + 0.10 โฅ
โข (0.121) ||10 โฅ
โฃ โฆ
or R0 f = 18.4 ฮฉ
c. To find the gain, consider the second op-amp.
11. vI (1) + v01 (0.1113)(2.1) = โv01
or vI = โv01 (1.234)
We had v01 = โv02 (1.246)
So vI = v02 (1.246)(1.234)
v02
or = 0.650
vI
v02
d. Ideal =1
vI
So ratio of actual to ideal = 0.650.
14.13
(a) For the op-amp. A0 L โ f 3dB = 106
106
f 3dB = = 50 Hz
2 ร 104
For the closed-loop amplifier.
106
f 3dB = = 40 kHz
25
(b) Open-loop amplifier.
2 ร 104 2 ร 104
A= โ| A | =
f 2
1+ j โ f โ
f3dB 1+ โ โ
โ f3dB โ
2 ร 104
f = 0.25 f 3dB โ A = = 1.94 ร 104
2
1 + (0.25)
2 ร 104
f = 5 f 3โ dB โ A = = 3.92 ร 103
1 + (5) 2
Closed-loop amplifier
25
f = 0.25 f 3dB โ A = = 24.25
1 + (0.25) 2
25
f = 5 f 3โ dB โ A = = 4.90
1 + (5) 2
14.14
The open loop gain can be written as
A0
A0 L ( f ) =
โ f โโ f โ
โ1 + j โ โโ 1 + j โ โ
โ f PD โ โ 5 ร 106 โ
where A0 = 2 ร 105.
The closed-loop response is
A0 L
ACL =
1 + ฮฒ A0 L
At low frequency,
2 ร 105
100 =
1 + ฮฒ (2 ร 105 )
So that ฮฒ = 9.995 ร 10โ3.
Assuming the second pole is the same for both the open-loop and closed-loop, then
โ f โ โ1 โ f โ
ฯ = โ tan โ1 โ โ โ tan โ 6 โ
โ f PD โ โ 5 ร 10 โ
12. For a phase margin of 80ยฐ, ฯ = โ100ยฐ.
So
โ f โ
โ100 = โ90 โ tan โ1 โ 6 โ
โ 5 ร 10 โ
or
f = 8.816 ร 105 Hz
Then
A0 L = 1
2 ร 105
=
2 2
โ 8.816 ร 105 โ โ 8.816 ร 105 โ
1+ โ โ 1+ โ 6 โ
โ f PD โ โ 5 ร 10 โ
or
8.816 ร 105
โ 1.9696 ร 105
f PD
or
f PD = 4.48 Hz
14.15
(a) 1st stage
(10) f3โ dB = 1 MHz โ f 3โ dB = 100 kHz
2nd stage
(50) f3โ dB = 1 MHz โ f 3โ dB = 20 kHz
Bandwidth of overall system โ 20 kHz
(b) If each stage has the same gain, so
K = 500 โ K = 22.36
2
Then bandwidth of each stage
(22.36) f 3โ dB = 1 MHz โ f 3โ dB = 44.7 kHz
14.16
Ao
A=
f
1+ j
f3โ dB
Ao 5 ร 104
A= โ 200 = โ f3โ dB = 40 Hz
2 2
โ f โ โ 104 โ
1+ โ โ 1+ โ โ
โ f3โ dB โ โ f3โ dB โ
Then
fT = (5 ร 104 )(40) โ fT = 2 MHz
14.17
13. (5 ร 104 ) f PD = 106 โ f PD = 20 Hz
(25) f 3โ dB 106 โ f3โ dB = 40 kHz
Avo 25
Av = โ Av =
f 2
1+ j โ f โ
f3โ dB 1+ โ 3 โ
โ 40 ร 10 โ
At f = 0.5 f 3โ dB = 20 kHz
25
Av = = 22.36
1 + (0.5) 2
At f = 2 f 3โ dB = 80 kHz
25
Av = = 11.18
1 + (2) 2
14.18
(20 ร 103 ) โ Avf MAX
= 106 โ Avf MAX
= 50
14.19
From Equation (14.55),
SR 10 ร 106
F P BW = =
2ฯ VP 0 2ฯ (10)
or
F P BW = f max = 159 kHz
14.20
a. Using Equation (14.55),
8 ร 106
VP 0 =
2ฯ (250 ร 103 )
or
VP 0 = 5.09 V
b.
1 1
Period T = = = 4 ร 10โ6 s
f 250 ร 103
One-fourth period = 1 ฮผ s
VP 0
Slope = = SR = 8 V/ฮผ s
1ฮผ s
โ VP 0 = 8 V
14.21
For input (a), maximum output is 5 V.
14. S R = 1 V/ฮผs
so
For input (b), maximum output is 2 V.
For input (c), maximum output is 0.5 V so the output is
14.22
For input (a), max v01 = 3 V.
Then v02 max
= 3(3) = 9 V
For input (b), max v01 = 1.5 V.
15. Then v02 max
= 3 (1.5 ) = 4.5 V
14.23
f MAX = 20 kHz, SR = 0.8 V / ฮผ s
SR 0.8 ร 106
V po = = โ V po = 6.37 V
2ฯ f MAX 2ฯ (20 ร 103 )
14.24
โV โ โV โ
I1 = I S 1 exp โ BE1 โ , I 2 = I S 2 exp โ BE 2 โ
โ VT โ โ VT โ
Want I1 = I 2 , so
โV โ
5 ร 10โ14 (1 + x) exp โ BE1 โ
I1
=1= โ VT โ
I2 โV โ
5 ร 10โ14 (1 โ x) exp โ BE 2 โ
โ VT โ
(1 + x) โ V โV โ
= exp โ BE1 BE 2 โ
(1 โ x) โ VT โ
Or
1+ x โ V โ VBE1 โ โ VOS โ
= exp โ BE 2 โ = exp โ โ
1โ x โ VT โ โ VT โ
โ 0.0025 โ
= exp โ โ = 1.10
โ 0.026 โ
Now
1 + x = (1 โ x )(1.10) โ
x = 0.0476 โ 4.76%
14.25
From Equation (14.62),
โ vCE1 โ โ vCE 2 โ
โ1+ V โ I โ 1+ V โ
โ AN โ
= S3 โ โ AN โ
โ 1 + vEB โ I S 4 โ 1 + vEC 4 โ
โ V โ โ โ
โ AP โ โ VAP โ
For vCE 2 = 0.6 V, then vEC 4 = 5 V. We have vCE1 = 5 V so
16. โ 5 โ โ 0.6 โ
โ 1 + 80 โ I S 3 โ 1 + 80 โ
โ โ= โ โ โ
โ 1 + 0.6 โ I S 4 โ 1 + 5 โ
โ โ โ โ
โ 80 โ โ 80 โ
or
I S 3 (1.0625) 2
= = 1.112
I S 4 (1.0075) 2
So
I S 3 = (10โ14 ) (1.112 )
or
I S 3 = 1.112 ร 10โ14 A
14.26
By superposition:
R
vo (vi ) = โ 2 โ vi = โ50vi
R1
โ R โ
vo (vos ) = โ 1 + 2 โ โ vos = 51vos
โ R1 โ
So
vo = vo ( vi ) + vo ( vos ) = โ50vi + 51vos
For vi = 20 mV and vos + 2.5 mV
vo = โ50(0.02) + 51(0.0025) = โ0.8725 V
For vi = 20 mV and vos = โ2.5 mV
vo = โ50(0.02) + 51(โ0.0025) = โ1.1275 V
So
โ1.1275 โค vo โค โ0.8725 V
14.27
vo = โ50vi = โ50 โก ยฑ2.5 mV + sin ฯ t ( mV ) โค
โฃ โฆ
vo = [ ยฑ0.125 โ 0.25sin ฯ t ] ( v )
14.28
17. 0.5 ร 10โ3
I= = 5 ร 10โ8 A
104
Also
i
dV 1 I
I = C o โ Vo = โซ Idt = โ t
dt C0 C
Then
5 ร 10โ8
5= t โ t = 103 s
10 ร 10โ6
14.29
โ 100 โ
a. | v01 | = 10 โ 1 + โ or | v01 | = 110 mV
โ 10 โ
Then
โ 50 โ
| v02 | = | v01 | (5) + 10 โ 1 + โ = (110)(5) + (10)(6)
โ 10 โ
or
| v02 | = 610 mV
14.30
v0 due to vI
โ 1 โ
v0 = (0.5) โ1 + โ = 0.9545 V
โ 1.1 โ
Wiper arm at V + = 10 V, (using superposition)
โ R1 || R5 โ โ 0.0909 โ
v1 = โ โ (10) = โ โ (10)
โ R1 || R5 + R4 โ โ 0.0909 + 10 โ
= 0.090
โ 1โ
Then v01 = โ โ โ (0.090) = โ0.090
โ 1โ
Wiper arm in center, v1 = 0 and v02 = 0
Wiper arm at V โ = โ10 V, v1 = โ0.090
So
v03 = 0.090
Finally, total output v0 : (from superposition)
Wiper arm at V + ,
v0 = 0.8645 V
Wiper arm in center,
v0 = 0.9545 V
Wiper arm at V โ ,
18. v0 = 1.0445 V
14.31
a. R1โฒ = R2 = 0.5 || 25 = 0.490 kฮฉ
โฒ
or
R1โฒ = R2 = 490 ฮฉ
โฒ
b. From Equation (14.75),
โ 125 ร 10โ6 โ
(0.026) ln โ โ14 โ
+ (0.125) R1โฒ
โ 2 ร 10 โ
โ 125 ร 10โ6 โ
= (0.026) ln โ โ14 โ
+ (0.125) R2โฒ
โ 2.2 ร 10 โ
0.586452 + (0.125) R1โฒ = 0.583974 + (0.125) R2 โฒ
0.002478 = (0.125)( R2 โ R1โฒ)
โฒ
So R2 โ R1โฒ = 0.0198 kฮฉ โ 19.8 ฮฉ
โฒ
Then
R2 (1 โ x) Rx R ร Rx
โ 1 = 0.0198
R2 + (1 โ x) Rx R1 + xRx
(0.5)(1 โ x)(50) (0.5)(50) x
โ = 0.0198
(0.5) + (1 โ x)(50) (0.5) + x(50)
25(1 โ x) 25 x
โ = 0.0198
50.5 โ 50 x 0.5 + 50 x
(0.5 + 50 x)(25 โ 25 x) โ (25 x)(50.5 โ 50 x)
= 0.0198
(50.5 โ 50 x )(0.5 + 50 x)
25 {0.5 โ 0.5 x + 50 x โ 50 x 2 โ 50.5 x + 50 x 2 } = 0.0198 {25.25 + 2525 x โ 25 x โ 2500 x 2 }
25 {0.5 โ x} = 0.0198 {25.25 + 2500 x โ 2500 x 2 }
0.5 โ x = 0.019998 + 1.98 x โ 1.98 x 2
1.98 x 2 โ 2.98 x + 0.48 = 0
2.98 ยฑ (2.98) 2 โ 4(1.98)(0.48)
x=
2(1.98)
So
x = 0.183
and
1 โ x = 0.817
14.32
R1โฒ = R1 ||15 = 0.5 ||15 = 0.4839 kฮฉ
โฒ
R2 = R2 || 35 = 0.5 || 35 = 0.4930 kฮฉ
From Equation (14.75),
โi โ โi โ
(0.026) ln โ C1 โ + iC1 R1โฒ = (0.026) ln โ C 2 โ + iC 2 R2
โฒ
โ IS3 โ โ IS 4 โ
โi โ
(0.026) ln โ C1 โ = iC 2 R2 โ iC1 R1โฒ
โฒ
โ iC 2 โ
โi โ โก i Rโฒ โค
โฒ
(0.026) ln โ C1 โ = iC 2 R2 โข1 โ C1 โ 1 โฅ
โ iC 2 โ โฃ โฒ
iC 2 R2 โฆ
โi โ โก โ i โโค
(0.026) ln โ C1 โ = iC 2 (0.4930) โข1 โ (0.9815) โ C1 โ โฅ
โ iC 2 โ โฃ โ iC 2 โ โฆ
By trial and error:
19. iC1 = 252 ฮผ A and iC 2 = 248 ฮผ A
or
iC1
= 1.0155
iC 2
14.33
From Eq. (14.79), we have
โ R โ
vo = I B1 R2 โ I B 2 R3 โ 1 + 2 โ
โ R1 โ
I B1 = 1 ฮผ A I B 2 = 2 ฮผ A
Setting vo = 0, we have
โ 200 โ
0 = (10โ6 )( 200 ร 103 ) โ ( 2 ร 10โ6 ) R3 โ 1 + โ
โ 20 โ
200 ร 10โ3
R3 = โ R = 9.09 K
( 2 ร10โ6 ) (11) 3
14.34
R2
1+ = 80
R1
R1 = 6.329 K
V f = vI = 5sin ฯ t ( mV )
I1 + I B = I 2
(a) VI = 0 โ VX = 0 I 2 = I B
VO = (10โ6 )(500 ร 103 ) โ vo = 0.50 V
VX v โ VX
(b) + IB = o
R1 R2
โ 1 1 โ v โ 1 1 โ
VX โ + โ + I B = 0 โ vo = R2 โ + โ vI + I B R2
โ R1 R2 โ R1 โ R1 R2 โ
vo = 80 โก5sin ฯ t ( mV ) โค + (10โ6 )( 500 ร 103 )
โฃ โฆ
vo = [ 0.5 + 0.4sin ฯ t ] ( v )
14.35
a.
20. For I B 2 = 1 ฮผ A, then v0 = โ (10โ6 )(104 )
or v0 = โ0.010 V
b. If a 10 kฮฉ resistor is included in the feedback loop
Now v0 = โ I B 2 (10) + I B1 (10) = 0
Circuit is compensated if I B1 = I B 2 .
14.36
From Equation (14.83), we have
v0 = R2 I 0S
where R2 = 40 kฮฉ and I 0 S = 3 ฮผ A.
Then
v0 = ( 40 ร 103 )( 3 ร 10โ6 )
or
v0 = 0.12 V
14.37
a. Assume all bias currents are in the same direction and into each op-amp.
v01 = I B1 (100 kฮฉ ) = (10โ6 )(105 ) โ v01 = 0.1 V
Then
v02 = v01 ( โ5 ) + I B1 ( 50 kฮฉ )
= ( 0.1)( โ5 ) + (10โ6 )( 5 ร 104 )
= โ0.5 + 0.05
or
v02 = โ0.45 V
b. Connect R3 = 10 ||100 = 9.09 kฮฉ resistor to noninverting terminal of first op-amp, and
R3 = 10 || 50 = 8.33 kฮฉ resistor to noninverting terminal of second op-amp.
21. 14.38
a. For a constant current through a capacitor.
1 t
v0 = โซ I dt
C 0
0.1ร 10โ6
or v0 = โ t โ v0 = (0.1)t
10โ6
b. At t = 10 s, v0 = 1 V
c. Then
100 ร 10โ12
v0 = โ6
โ t โ v0 = (10โ4 )t
10
At t = 10 s, v0 = 1 mV
14.39
a. Assume all bias currents are into the op-amp.
v01 = I B1 ( 50 kฮฉ ) = (10 ร10โ6 )( 50 ร103 )
or
v01 = v02 = 0.5 V
v03 = ( โ1)( v01 ) + (10 ร 10โ6 )( 20 ร 103 )
or
v03 = โ0.3 V
b. RA = 10 || 50 โ RA = 8.33 kฮฉ
RB = 20 || 20 โ RB = 10 kฮฉ
c. Assume the worst case offset current, that is, I 0 S = I B1 โ I B 2 or I 0 S = I B 2 โ I B1 .
From Equation (14.83),
v01 = R2 I 0 S = ( 50 ร 103 )( 2 ร 10โ6 )
or
v01 = v02 = 0.1 V
v03 = ( โ1) v01 โ I 0 S R2
= ( โ1)( 0.1) โ ( 2 ร 10โ6 )( 20 ร 103 )
or
v03 = โ0.14 V
14.40
a. Using Equation (14.79),
Circuit (a),
โ 50 โ
v0 = ( 0.8 ร 10โ6 )( 50 ร 103 ) โ ( 0.8 ร10โ6 )( 25 ร 103 ) โ1 + โ
โ 50 โ
or
v0 = 0
Circuit (b),
โ 50 โ
v0 = ( 0.8 ร 10โ6 )( 50 ร 103 ) โ ( 0.8 ร 10 โ6 )(103 ) โ1 + โ
โ 50 โ
โ2
= 4 ร 10 โ 1.6
or
v0 = โ1.56 V
b. Assume I B1 = 0.7 ฮผ A and I B 2 = 0.9 ฮผ A, then using Equation (14.79):
Circuit (a),
22. โ 50 โ
v0 = ( 0.7 ร 10โ6 )( 50 ร 103 ) โ ( 0.9 ร 10โ6 )( 25 ร 103 ) โ 1 + โ
โ 50 โ
= 0.035 โ 0.045
or v0 = โ0.010 V
Circuit (b),
โ 50 โ
v0 = ( 0.7 ร 10โ6 )( 50 ร 103 ) โ ( 0.9 ร 10โ6 )(106 ) โ1 + โ
โ 50 โ
= 0.035 โ 1.8
or v0 = โ1.765 V
14.41
a. If R = 0,
โ 100 โ
v0,max = โ1 + โ V0 S + I B (100 kฮฉ)
โ 10 โ
= (11)(10 ร 10โ3 ) + (2 ร 10โ6 )(100 ร 103 )
v0,max = 0.110 + 0.20
โ v0,max = 0.310 V
b.
R = 10.1|| 100 = 9.17 kฮฉ = R
14.42
โ Ri โ
a. โ โ (15) = 0.010 V
โ Ri + R2 โ
15
= 0.0006667
15 + R2
15(1 โ 0.0006667) = 0.0006667 R2
Then
R2 = 22.48 Mฮฉ
b. R1 = Ri || RF = 15 || 10 โ R1 = 6 kฮฉ
14.43
a. Assume the offset voltage polarities are such as to produce the worst case values, but the bias
currents are in the same direction.
Use superposition:
23. Offset voltages
โ 100 โ
| v01 | = โ1 + โ (10) = 110 mV =| v01 |
โ 10 โ
โ 50 โ
| v02 | = (5)(110) + โ 1 + โ (10)
โ 10 โ
โ | v02 | = 610 mV
Bias Currents:
v01 = I B (100 kฮฉ) = (2 ร10โ6 )(100 ร 103 ) = 0.2 V
Then
v02 = (โ5)(0.2) + (2 ร 10โ6 )(50 ร 103 ) = โ0.9 V
Worst case: v01 is positive and v02 is negative, then
v01 = 0.31 V and v02 = โ1.51 V
b. Compensation network:
If we want
โ RB โ + +
โ โ V = 20 mV and V = 10 V
โ RB + RC โ
โ 8.33 โ
โ โ (10) = 0.020
โ 8.33 + RC โ
or RC โ 4.15 Mฮฉ
14.44
Assume bias currents are in same direction, but assume polarity of offset voltages are such as to
produce the worst case output.
a. Let I B1 = 5.5 ฮผ A, I B 2 = 4.5 ฮผ A
Bias Current Effects:
v01 = I B1 (50 kฮฉ) = 0.275 V โ v02 = 0.275 V
v03 = I B1 (20 kฮฉ) โ v01 โ v03 = โ0.165 V
Offset Voltage Effects:
โ 50 โ
v01 = (5) โ1 + โ = 30 mV โ v02 = 30 mV
โ 10 โ
โ 20 โ
v03 = โv01 โ 5 โ 1 + โ โ v03 = โ40 mV
โ 20 โ
Total Effect: v01 = 0.305 V and v02 = 0.305 V v03 = โ0.205 V
14.45
For circuit (a), effect of bias current:
v0 = (50 ร 103 )(100 ร 10โ9 ) โ 5 mV
Effect of offset voltage
24. โ 50 โ
v0 = (2) โ1 + โ = 4 mV
โ 50 โ
So net output voltage is v0 = 9 mV
For circuit (b), effect of bias current:
Let I B 2 = 550 nA, I B1 = 450 nA, then from Equation (14.79),
โ 50 โ
v0 = (450 ร 10โ9 )(50 ร 103 ) โ (550 ร 10โ9 )(106 ) โ1 + โ
โ 50 โ
โ2
= 2.25 ร 10 โ 1.1
or
v0 = โ1.0775 V
If the offset voltage is negative, then
v0 = (โ2)(2) = โ4 mV
So the net output voltage is
v0 = โ1.0815 V
14.46
a. At T = 25ยฐC, V0 S = 2 mV so the output voltage for each circuit is
v0 = 4 mV
b. For T = 50ยฐC, the offset voltage for is
V0 S = 2 mV + (0.0067)(25) = 2.1675 mV
so the output voltage for each circuit is
v0 = 4.335 mV
14.47
a. At T = 25ยฐC, V0 S = 1 mV, then
โ 50 โ
v01 = (1) โ 1 + โ โ v01 = 6 mV
โ 10 โ
and
โ 60 โ โ 60 โ
v02 = v01 โ 1 + โ + (1) โ 1 + โ
โ 20 โ โ 20 โ
= 6(4) + (1)(4) โ v02 = 28 mV
b. At T = 50ยฐC, V0 S = 1 + (0.0033)(25) = 1.0825 mV, then
v01 = (1.0825)(6) โ v01 = 6.495 mV
and
v02 = (6.495)(4) + (1.0825)(4)
or
v02 = 30.31 mV
14.48
25ยฐC; I B = 500 nA, I 0 S = 200 nA
50ยฐC, I B = 500 nA + (8 nA / ยฐC)(25ยฐC) = 700 nA
I 0 S = 200 nA + (2 nA / ยฐC)(25ยฐC) = 250 nA
a. Circuit (a): For I B , bias current cancellation, v0 = 0
Circuit (b): For I B , Equation (14.79),
โ 50 โ
v0 = (500 ร 10โ9 )(50 ร 103 ) โ (500 ร 10 โ9 )(106 ) โ 1 + โ
โ 50 โ
= 0.025 โ 1.00 โ v0 = โ0.975 V
b. Due to offset bias currents.
Circuit (a):
25. v0 = (200 ร 10โ9 )(50 ร 103 ) โ v0 = 0.010 V
Circuit (b):
Let I B 2 = 600 nA
I B1 = 400 nA
Then
โ 50 โ
v0 = (400 ร 10โ9 )(50 ร 103 ) โ (600 ร 10โ9 )(106 ) โ 1 + โ
โ 50 โ
= 0.020 โ 1.20 โ v0 = โ1.18 V
c. Circuit (a): Due to I B , v = 0
Circuit (b): Due to I B ,
โ 50 โ
v0 = (700 ร 10โ9 )(50 ร 103 ) โ (700 ร 10โ9 )(106 ) โ1 + โ
โ 50 โ
= 0.035 โ 1.40 โ v0 = โ1.365 V
Circuit (a): Due to I 0 S ,
v0 = (250 ร 10โ9 )(50 ร 103 ) โ v0 = 0.0125 V
Circuit (b): Due to I 0 S ,
Let I B 2 = 825 nA
I B1 = 575 nA
Then
โ 50 โ
v0 = (575 ร 10โ9 )(50 ร 103 ) โ (825 ร 10โ9 )(106 ) โ 1 + โ
โ 50 โ
= 0.02875 โ 1.65 โ v0 = โ1.62 V
14.49
25ยฐC; I B = 2 ฮผ A, I 0 S = 0.2 ฮผ A
50ยฐC, I B = 2 ฮผ A + (0.020 ฮผ A / ยฐC)(25ยฐC ) = 2.5 ฮผ A
I 0 S = 0.2 ฮผ A + (0.005 ฮผ A / ยฐC)(25ยฐC) = 0.325 ฮผ A
a. Due to I B : (Assume bias currents into op-amp).
v01 = I B (50 kฮฉ) = (2 ร 10โ6 )(50 ร 103 )
โ v01 = 0.10 V
โ 60 โ โ 60 โ
v02 = v01 โ 1 + โ + I B (60 kฮฉ) โ I B (50 kฮฉ) โ 1 + โ
โ 20 โ โ 20 โ
โ6 โ6
= (0.1)(4) + (2 ร 10 )(60 ร 10 ) โ (2 ร 10 )(60 ร 103 )4
3
or v02 = 0.12 V
b. Due to I 0 S :
1st op-amp. Let I B1 = 2.1 ฮผ A
2nd op-amp. Let I B1 = 2.1 ฮผ A
I B 2 = 1.9 ฮผ A
v01 = I B1 (50 kฮฉ) = (2.1ร 10โ6 )(50 ร 103 )
โ v01 = 0.105 V
โ 60 โ โ 60 โ
v02 = v01 โ 1 + โ + I B1 (60 kฮฉ) โ I B 2 (50 kฮฉ) โ1 + โ
โ 20 โ โ 20 โ
= (0.105)(4) + (2.1ร 10 )(60 ร 10 ) โ (1.9 ร 10 โ6 )(50 ร 103 )(4)
โ6 3
or
v02 = 0.166 V
c. Due to I B :
26. v01 = (2.5 ร 10 โ6 )(50 ร 103 ) โ v01 = 0.125 V
โ 60 โ โ 60 โ
v01 = v02 โ 1 + โ + I B (60 kฮฉ) โ I B (50 kฮฉ) โ1 + โ
โ 20 โ โ 20 โ
โ6
= (0.125)(4) + (2.5 ร 10 )(60 ร 10 ) โ (2.5 ร 106 )(50 ร 103 (4)
3
or v02 = 0.15 V
Due to I 0 S :
Let I B1 = 2.625 ฮผ A
I B 2 = 2.3375 ฮผ A
v01 = I B1 (50 kฮฉ) = (2.6625 ร 10โ6 )(50 ร 103 )
โ v01 = 1.133 V
โ 60 โ โ 60 โ
v02 = v01 โ 1 + โ + I B1 (60 kฮฉ) โ I B 2 (50 kฮฉ) โ 1 + โ
โ 20 โ โ 20 โ
= (0.133)(4) + (2.6625 ร 10โ6 )(60 ร 103 ) โ (2.3375 ร 10โ6 )(50 ร 103 )(4)
or
v02 = 0.224 V
14.50
โ R4 โ โ R2 โ
vB = โ โ vI 1 and v0 (vI 2 ) = vB โ 1 + โ
โ R3 + R4 โ โ R1 โ
or
โ R4 โ โ R2 โ
v0 (vI 2 ) = โ โ โ 1 + โ vI 2
โ R3 + R4 โ โ R1 โ
or vI 1 .
R2
v0 (vI 1 ) = โ โ vI
R1
Then
โ R4 โโ R2 โ R2
v0 = โ โโ 1 + โ vI 2 โ โ vI 1
โ R3 + R4 โ โ R1 โ R1
V V
we can write vI 2 = vcm + d and vI 1 = Vcm โ d โ Then
2 2
โ R4 โโ R2 โ โ Vd โ R2 โ Vd โ
v0 = โ โโ 1 + โ โ Vcm + โ โ โ โ Vcm โ โ
โ R3 + R4 โ โ R1 โ โ 2 โ R1 โ 2โ
Common-mode gain
v โ R4 โ โ R2 โ R2
Acm = 0 = โ โ โ1 + โ โ
Vcm โ R3 + R4 โ โ R1 โ R1