Microelectronics: Circuit Analysis and Design
Microelectronics: Circuit Analysis and Design
4th Edition
ISBN: 9780073380643
Author: Donald A. Neamen
Publisher: McGraw-Hill Companies, The
bartleby

Videos

Textbook Question
Book Icon
Chapter 10, Problem 10.76P

For the circuit shown in Figure P10.76, the transistor parameters are
V T N = 0.5 V , V T P = 0.5 V , k n = 100 μ A / V 2 , k P = 60 μ A / V 2 , λ 1 = 0.02 V 1 ,and λ 2 = 0.03 V 1 .The quiescent drain current is I D Q = 200 μ A and the quiescent output voltage is V O = 1.25 V . (a) Determine ( W / L ) 1 such that the small-signal voltage gain is A V = 100 , (b) determine V I ,and (c) determine V G assuming K n 1 = K P 2 .

Chapter 10, Problem 10.76P, For the circuit shown in Figure P10.76, the transistor parameters are

(a)

Expert Solution
Check Mark
To determine

The (W/L)1 ratio.

Answer to Problem 10.76P

  (WL)1=25

Explanation of Solution

Given:

  K'n=100μA/V2K'p=60μA/V2λ1=0.02V1λ2=0.03V1VTN=0.5VVTP=0.5VVO=1.25VIDQ=200μA

  Av=100

Calculation:

The given circuit is,

  Microelectronics: Circuit Analysis and Design, Chapter 10, Problem 10.76P , additional homework tip  1

The small-signal voltage gain is given, the expression for gain is,

  Av=gm1(ro1ro2)

Substituting the given values,

  Av=gm1(ro1ro2)100=gm1(ro1ro2)100=gm1(ro1ro2)(1)

Substitute k'n(WL)1(VIVTN) for gm1 in equation (1)

  100=[k'n(WL)1(VIVTN)](ro1ro2)VI=VGS1

Now substitute the given values in above expression,

  100=[100×106(WL)1(VI0.5)](ro1ro2)(2)

It is known that,

  ro1=1λ1IDQ

  ro2=1λ2IDQ

On putting the given values,

  ro1=1(0.02)(200×106)ro1=250kΩ

  ro2=1(0.03)(200×106)ro2=166.67kΩ

Now substitute the value of ro1andro2 in equation (2)

  100=[100×106(WL)1(VI0.5)](250×103166.67×103)106=(WL)1(VI0.5)(250×103)×(166.67×103)250×103+166.67×103106=(WL)1(VI0.5)(100×103)10=(WL)1(VI0.5)(VI0.5)=10(WL)1(3)

The drain current IDQ for transistor M1 is expressed as,

   I DQ = K ' n 2 ( W L ) 1 ( V GS1 V TN ) 2 (1+ λ 1 V DS1 )

   = K ' n 2 ( W L ) 1 ( V G1 V S1 V TN ) 2 (1+ λ 1 ( V D1 V S1 ))

Hence VG1=VI, VD1=VO and VD1=VO substitute the values,

   I DQ = K ' n 2 ( W L ) 1 ( V GS1 V TN ) 2 (1+ λ 1 V DS1 ) 200× 10 6 = 100× 10 6 2 ( W L ) 1 ( V 1 00.5) 2 (1+0.02( V O 0))

   4= ( W L ) 1 ( V 1 0.5) 2 (1+0.02 V O )

Substitute 1.25V for VO ,

   4= ( W L ) 1 ( V 1 0.5) 2 (1+0.02(1.25))

   4= ( W L ) 1 ( V 1 0.5) 2 (1.025)

   3.902= ( W L ) 1 ( V 1 0.5) 2

Put the value of (V10.5) from equation (3).

  3.902=(WL)1(10(WL)1)2(WL)1=1003.902(WL)1=25.63

  (WL)1=25

Conclusion:

  (WL)1=25

(b)

Expert Solution
Check Mark
To determine

The voltage VI .

Answer to Problem 10.76P

  VI=0.9V

Explanation of Solution

Given:

  K'n=100μA/V2K'p=60μA/V2λ1=0.02V1λ2=0.03V1VTN=0.5VVTP=0.5VVO=1.25VIDQ=200μA

Calculation:

The given circuit is,

  Microelectronics: Circuit Analysis and Design, Chapter 10, Problem 10.76P , additional homework tip  2

The small-signal voltage gain is given, the expression for gain is,

  Av=gm1(ro1ro2)

Substituting the given values,

  Av=gm1(ro1ro2)100=gm1(ro1ro2)100=gm1(ro1ro2)(1)

Substitute k'n(WL)1(VIVTN) for gm1 in equation (1)

  100=[k'n(WL)1(VIVTN)](ro1ro2)VI=VGS1

Now substitute the given values in above expression,

  100=[100×106(WL)1(VI0.5)](ro1ro2)(2)

It is known that,

  ro1=1λ1IDQ

  ro2=1λ2IDQ

On putting the given values,

  ro1=1(0.02)(200×106)ro1=250kΩ

  ro2=1(0.03)(200×106)ro2=166.67kΩ

Now substitute the value of ro1andro2 in equation (2).

  100=[100×106(WL)1(VI0.5)](250×103166.67×103)106=(WL)1(VI0.5)(250×103)×(166.67×103)250×103+166.67×103106=(WL)1(VI0.5)(100×103)10=(WL)1(VI0.5)(VI0.5)=10(WL)1(3)

The drain current IDQ for transistor M1 is expressed as,

   I DQ = K ' n 2 ( W L ) 1 ( V GS1 V TN ) 2 (1+ λ 1 V DS1 ) = K ' n 2 ( W L ) 1 ( V G1 V S1 V TN ) 2 (1+ λ 1 ( V D1 V S1 ))

Hence VG1=VI,VD1=VO and VD1=VO substitute the values,

  IDQ=K'n2(WL)1(VGS1VTN)2(1+λ1VDS1)200×106=100×1062(WL)1(V100.5)2(1+0.02(VO0))4=(WL)1(V10.5)2(1+0.02VO)

Substitute 1.25v for VO ,

  4=(WL)1(V10.5)2(1+0.02(1.25))4=(WL)1(V10.5)2(1.025)3.902=(WL)1(V10.5)2

Put the value of (V10.5) from equation (3).

  3.902=(WL)1(10(WL)1)2(WL)1=1003.902(WL)1=25.63

  (WL)1=25

  (VI0.5)=10(WL)1

Substitute (WL)1=25 in above expression,

  (VI0.5)=1025(VI0.5)=0.4VI=0.4+0.5

  VI=0.9V

Conclusion:

  VI=0.9V

(c)

Expert Solution
Check Mark
To determine

The voltage VG for transistor M2

Answer to Problem 10.76P

  VG=1.607V

Explanation of Solution

Given:

  K'n=100μA/V2K'p=60μA/V2λ1=0.02V1λ2=0.03V1VTN=0.5VVTP=0.5VVO=1.25VIDQ=200μA

  Kn1=Kp2

Calculation:

The given circuit is,

  Microelectronics: Circuit Analysis and Design, Chapter 10, Problem 10.76P , additional homework tip  3

The small-signal voltage gain is given, the expression for gain is,

  Av=gm1(ro1ro2)

Substituting the given values,

  Av=gm1(ro1ro2)100=gm1(ro1ro2)100=gm1(ro1ro2)(1)

Substitute k'n(WL)1(VIVTN) for gm1 in equation (1)

  100=[k'n(WL)1(VIVTN)](ro1ro2)VI=VGS1

Now substitute the given values in above expression,

  100=[100×106(WL)1(VI0.5)](ro1ro2)(2)

It is known that,

  ro1=1λ1IDQ

  ro2=1λ2IDQ

On putting the given values,

  ro1=1(0.02)(200×106)ro1=250kΩ

  ro2=1(0.03)(200×106)ro2=166.67kΩ

Now substitute the value of ro1and ro2 in equation (2).

  100=[100×106(WL)1(VI0.5)](250×103166.67×103)106=(WL)1(VI0.5)(250×103)×(166.67×103)250×103+166.67×103106=(WL)1(VI0.5)(100×103)10=(WL)1(VI0.5)(VI0.5)=10(WL)1(3)

The drain current IDQ for transistor M1 is expressed as,

   I DQ = K ' n 2 ( W L ) 1 ( V GS1 V TN ) 2 (1+ λ 1 V DS1 )

   = K ' n 2 ( W L ) 1 ( V G1 V S1 V TN ) 2 (1+ λ 1 ( V D1 V S1 ))

Hence VG1=VI,

  VD1=VO and VD1=VO substitute the values,

  IDQ=K'n2(WL)1(VGS1VTN)2(1+λ1VDS1)200×106=100×1062(WL)1(V100.5)2(1+0.02(VO0))4=(WL)1(V10.5)2(1+0.02VO)

Substitute 1.25V for VO , I DQ = K ' n 2 ( W L ) 1 ( V GS1 V TN ) 2 (1+ λ 1 V DS1 )

   200× 10 6 = 100× 10 6 2 ( W L ) 1 ( V 1 00.5) 2 (1+0.02( V O 0))

   4= ( W L ) 1 ( V 1 0.5) 2 (1+0.02 V O )

Put the value of (V10.5) from equation (3).

  3.902=(WL)1(10(WL)1)2(WL)1=1003.902(WL)1=25.63

  (WL)1=25

Since Kn1=Kp2 so,

  K'n2(WL)1=K'p2(WL)2(WL)2=10060×25(WL)2=41.67

The drain current IDQ for transistor M2 is expressed as,

  IDQ=K'p2(WL)2(VSG2+VTP)2(1+λ2VSD2)=K'n2(WL)2(VS2VG2+VTP)2(1+λ2(VS2VD2))

On substituting the given values,

  200×106=60×1062(41.67)(VSG20.5)2(1+0.03(2.51.25))VSG2=0.893V

Now the gate voltage will be,

  VSVG=0.8932.5VG=0.893

  VG=1.607V

Conclusion:

  VG=1.607V

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!
Students have asked these similar questions
........ (Figure-1) R. RB= 380kN,Rc= 1kN B = 100, VBB = Vcc=12V RB ww Vec CC ......... I, V CE СЕ V ВЕ BB Q-1-b) Describe briefly the input / output characteristics and application of Common Emitter BJT Configuration
3. In the figure shown below, Vmax is measured as 5.9 V and V min measured as 1.2V. 18] In the figure shown below, is measured as 5.9 V an (a) Determine the value of V.. (b) Determine the value of Vm. (c) Determine the modulation index. (d) Suppose we can change the value of V. What is the maximum value that we could use for Vm without causing overmodulation?
Draw, Illustrate and label your schematic diagram before solving the problem. 3) Given an Emitter-Stabilize Biased transistor circuit with beta DC is 250,Base resistor is 150 ohms, collector resistor is 1.5k ohms ,emitter resistor is 500 ohms ,emitter voltage supply is -5v and  Voltage at common collector is +28V,Voltage at Base-emitter junction  is 0.7v,. Determine Base current, Collector current and Voltage at collector-emitter junction.

Chapter 10 Solutions

Microelectronics: Circuit Analysis and Design

Ch. 10 - Prob. 10.4TYUCh. 10 - Prob. 10.8EPCh. 10 - Prob. 10.9EPCh. 10 - Consider the JFET circuit in Figure 10.24. The...Ch. 10 - Consider Design Example 10.8. Assume transistor...Ch. 10 - The bias voltages of the MOSFET current source in...Ch. 10 - Prob. 10.7TYUCh. 10 - All transistors in the MOSFET modified Wilson...Ch. 10 - A simple BJT amplifier with active load is shown...Ch. 10 - Prob. 10.9TYUCh. 10 - Prob. 10.10TYUCh. 10 - Prob. 10.11TYUCh. 10 - Prob. 10.12EPCh. 10 - For the circuit in Figure 10.40(a), the transistor...Ch. 10 - Prob. 10.12TYUCh. 10 - Repeat Example 10.12 for the case where a resistor...Ch. 10 - Prob. 10.14TYUCh. 10 - Prob. 1RQCh. 10 - Explain the significance of the output resistance...Ch. 10 - Prob. 3RQCh. 10 - Prob. 4RQCh. 10 - What is the primary advantage of a BJT cascode...Ch. 10 - Prob. 6RQCh. 10 - Can a piecewise linear model of the transistor be...Ch. 10 - Prob. 8RQCh. 10 - Sketch the basic MOSFET two-transistor current...Ch. 10 - Discuss the effect of mismatched transistors on...Ch. 10 - Prob. 11RQCh. 10 - Sketch a MOSFET cascode current source circuit and...Ch. 10 - Discuss the operation of an active load.Ch. 10 - What is the primary advantage of using an active...Ch. 10 - Prob. 15RQCh. 10 - What is the impedance seen looking into a simple...Ch. 10 - What is the advantage of using a cascode active...Ch. 10 - Prob. 10.1PCh. 10 - The matched transistors Q1 and Q2 in Figure...Ch. 10 - Prob. 10.3PCh. 10 - Reconsider the circuit in Figure 10.2(a). Let...Ch. 10 - Prob. 10.5PCh. 10 - The transistor and circuit parameters for the...Ch. 10 - The bias voltages in the circuit shown in Figure...Ch. 10 - Consider the current source in Figure 10.2(b). The...Ch. 10 - Prob. 10.9PCh. 10 - Prob. 10.10PCh. 10 - Prob. D10.11PCh. 10 - In the circuit in Figure P10.11, the transistor...Ch. 10 - Prob. D10.13PCh. 10 - Consider the circuit shown in Figure P 10.14. The...Ch. 10 - Design a basic two-transistor current...Ch. 10 - The values of for the transistors in Figure P10.16...Ch. 10 - Consider the circuit in Figure P10.17. The...Ch. 10 - All transistors in the N output current mirror in...Ch. 10 - Design a pnp version of the basic three-transistor...Ch. 10 - Prob. D10.20PCh. 10 - Consider the Wilson current source in Figure...Ch. 10 - Consider the circuit in Figure P10.22. The...Ch. 10 - Consider the Wilson current-source circuit shown...Ch. 10 - Consider the Widlar current source shown in Figure...Ch. 10 - Prob. 10.25PCh. 10 - Consider the circuit in Figure P10.26. Neglect...Ch. 10 - (a) For the Widlar current source shown in Figure...Ch. 10 - Consider the Widlar current source in Problem...Ch. 10 - (a) Design the Widlar current source such that...Ch. 10 - Design a Widlar current source to provide a bias...Ch. 10 - Design the Widlar current source shown in Figure...Ch. 10 - The circuit parameters of the Widlar current...Ch. 10 - Consider the Widlar current source in Figure 10.9....Ch. 10 - Consider the circuit in Figure P10.34. The...Ch. 10 - The modified Widlar current-source circuit shown...Ch. 10 - Consider the circuit in Figure P10.36. Neglect...Ch. 10 - Consider the Widlar current-source circuit with...Ch. 10 - Assume that all transistors in the circuit in...Ch. 10 - In the circuit in Figure P10.39, the transistor...Ch. 10 - Consider the circuit in Figure P10.39, with...Ch. 10 - Consider the circuit shown in Figure P10.41....Ch. 10 - For the circuit shown in Figure P 10.42, assume...Ch. 10 - Consider the circuit in Figure P10.43. The...Ch. 10 - Consider the MOSFET current-source circuit in...Ch. 10 - The MOSFET current-source circuit in Figure P10.44...Ch. 10 - Consider the basic two-transistor NMOS current...Ch. 10 - Prob. 10.47PCh. 10 - Consider the circuit shown in Figure P10.48. Let...Ch. 10 - Prob. 10.49PCh. 10 - The circuit parameters for the circuit shown in...Ch. 10 - Prob. 10.51PCh. 10 - Figure P10.52 is a PMOS version of the...Ch. 10 - The circuit shown in Figure P10.52 is biased at...Ch. 10 - The transistor circuit shown in Figure P10.54 is...Ch. 10 - Assume the circuit shown in Figure P10.54 is...Ch. 10 - The circuit in Figure P 10.56 is a PMOS version of...Ch. 10 - The transistors in Figure P10.56 have the same...Ch. 10 - Consider the NMOS cascode current source in Figure...Ch. 10 - Consider the NMOS current source in Figure P10.59....Ch. 10 - Prob. 10.60PCh. 10 - The transistors in the circuit shown in Figure...Ch. 10 - A Wilson current mirror is shown in Figure...Ch. 10 - Repeat Problem 10.62 for the modified Wilson...Ch. 10 - Prob. 10.64PCh. 10 - Prob. 10.65PCh. 10 - Prob. D10.66PCh. 10 - Prob. D10.67PCh. 10 - The parameters of the transistors in the circuit...Ch. 10 - Prob. 10.69PCh. 10 - Consider the circuit shown in Figure P10.70. The...Ch. 10 - Prob. 10.71PCh. 10 - Prob. D10.72PCh. 10 - Prob. 10.73PCh. 10 - Prob. D10.74PCh. 10 - Prob. 10.75PCh. 10 - For the circuit shown in Figure P10.76, the...Ch. 10 - Prob. 10.77PCh. 10 - Prob. 10.78PCh. 10 - The bias voltage of the MOSFET amplifier with...Ch. 10 - Prob. 10.80PCh. 10 - Prob. 10.81PCh. 10 - Prob. 10.82PCh. 10 - A BJT amplifier with active load is shown in...Ch. 10 - Prob. 10.84PCh. 10 - Prob. 10.85PCh. 10 - Prob. 10.86PCh. 10 - The parameters of the transistors in Figure P10.87...Ch. 10 - The parameters of the transistors in Figure P10.88...Ch. 10 - A BJT cascode amplifier with a cascode active load...Ch. 10 - Design a bipolar cascode amplifier with a cascode...Ch. 10 - Design a MOSFET cascode amplifier with a cascode...Ch. 10 - Design a generalized Widlar current source (Figure...Ch. 10 - The current source to be designed has the general...Ch. 10 - Designa PMOS version of the current source circuit...Ch. 10 - Consider Exercise TYU 10.10. Redesign the circuit...
Knowledge Booster
Background pattern image
Electrical Engineering
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, electrical-engineering and related others by exploring similar questions and additional content below.
Similar questions
SEE MORE QUESTIONS
Recommended textbooks for you
Text book image
Introductory Circuit Analysis (13th Edition)
Electrical Engineering
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:PEARSON
Text book image
Delmar's Standard Textbook Of Electricity
Electrical Engineering
ISBN:9781337900348
Author:Stephen L. Herman
Publisher:Cengage Learning
Text book image
Programmable Logic Controllers
Electrical Engineering
ISBN:9780073373843
Author:Frank D. Petruzella
Publisher:McGraw-Hill Education
Text book image
Fundamentals of Electric Circuits
Electrical Engineering
ISBN:9780078028229
Author:Charles K Alexander, Matthew Sadiku
Publisher:McGraw-Hill Education
Text book image
Electric Circuits. (11th Edition)
Electrical Engineering
ISBN:9780134746968
Author:James W. Nilsson, Susan Riedel
Publisher:PEARSON
Text book image
Engineering Electromagnetics
Electrical Engineering
ISBN:9780078028151
Author:Hayt, William H. (william Hart), Jr, BUCK, John A.
Publisher:Mcgraw-hill Education,
Diode Logic Gates - OR, NOR, AND, & NAND; Author: The Organic Chemistry Tutor;https://www.youtube.com/watch?v=9lqwSaIDm2g;License: Standard Youtube License