Microelectronics: Circuit Analysis and Design
Microelectronics: Circuit Analysis and Design
4th Edition
ISBN: 9780073380643
Author: Donald A. Neamen
Publisher: McGraw-Hill Companies, The
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Chapter 7, Problem 7.69P

For the PMOS common−source circuit shown in Figure P769, the transistor parameters are: V T P = 2 V , K P =1mA/V 2 , λ = 0 , C g s = 15 pF , and C g d = 3 pF . (a) Determine the upper 3dB frequency. (b) What is the equivalent Miller capacitance? State any assumptions or approximations that you make. (c) Find the midband voltage gain.

Chapter 7, Problem 7.69P, For the PMOS commonsource circuit shown in Figure P769, the transistor parameters are: VTP=2V ,
Figure P7.69

(a)

Expert Solution
Check Mark
To determine

The upper 3dB frequency

Answer to Problem 7.69P

The upper 3dB frequency is 10.4MHz

Explanation of Solution

Given:

The given circuit is shown below.

  Microelectronics: Circuit Analysis and Design, Chapter 7, Problem 7.69P , additional homework tip  1

Calculation:

Calculate the gate voltage VG

  VG=(R2R1+R2)[10(10)]VDD=(R2R1+R2)[20]VDD

Substitute 22×103 for R2,8×103 for R1, and 10 for VDD

  VG=(2222+8)2010

  VG=4.67V

The expression for drain current ID is,

  ID=VCCVSGVGRS

Substitute 10 for VCC,0.5×103 for RS and 4.67 for VG

  ID=10VSG4.670.5×103...........(1)

Consider the another expression for drain current ID

  ID=103(VSG2)2(2)

Equate equations (1) and (2)

  10VSG4.670.5×103=103(VSG2)2

  10VSG4.67=0.5(VSG24VSG+4)

  202VSG9.34=VSG24VSG+4

  VSG22VSG6.66=0

Calculate source-to-gate voltage VSG

  VSG=2±44(1)(6.66)2(1)

  =3.77V

Calculate transconductance gm

  gm=2Kp(VSG+VTP)

Substitute 103 for KP,3.77 for VSC and -2 for VTP

  gm=2(103)(3.772)

  gm=3.54mA/V

Calculate the Miller Capacitance.

  CM=CSd[1+g=(RDRL)]

Substitute 3×1012 for CPD,3.54×103 for gm,2×103 for RD, and 5×103 for RL

  CM=3×1012[1+3.54×103(2×1035×103)]=3×1012[1+3.54×103(2×5×1032+5)]=18.2pF

Calculate the time constant τ

  τ=Req(Cgs+CM)............(3)

Where

  Req=RiR1R2

  Req=0.5822

  Req=0.461

Recall equation (3).

  τ=Req(Cgs+CM)

Substitute 0.461×103 for Req,15×1012 for Cgs, and 18.2×1012 for CM

  τ=(0.461×103)(15+18.2)×1012

  τ=15.30ns

Calculate the upper 3 dB frequency fB

  fB=12πτSubstitute15.3×109forτfB=12π(15.30×109)=10.4MHz

Therefore, the upper 3dB frequency is 10.4MHz

(b)

Expert Solution
Check Mark
To determine

The value of the Miller capacitance

Answer to Problem 7.69P

The value of the Miller capacitance is 18.2pF

Explanation of Solution

Given:

The given circuit is shown below.

  Microelectronics: Circuit Analysis and Design, Chapter 7, Problem 7.69P , additional homework tip  2

Calculation:

Calculate the gate voltage VG

  VG=(R2R1+R2)[10(10)]VDD=(R2R1+R2)[20]VDD

Substitute 22×103 for R2,8×103 for R1, and 10 for VDD

  VG=(2222+8)2010

  VG=4.67V

The expression for drain current ID is,

  ID=VCCVSGVGRS

Substitute 10 for VCC,0.5×103 for RS and 4.67 for VG

  ID=10VSG4.670.5×103...........(1)

Consider the another expression for drain current ID

  ID=KP(VSG+VTP)2

Substitute 103 for KP and 2 for VTP

  ID=103(VSG2)2(2)

Equate equations (1) and (2)

  10VSG4.670.5×103=103(VSG2)2

  10VSG4.67=0.5(VSG24VSG+4)

  202VSG9.34=VSG24VSG+4

  VSG22VSG6.66=0

Calculate source-to-gate voltage VSG

  VSG=2±44(1)(6.66)2(1)

  =3.77V

Calculate transconductance gm

  gm=2Kp(VSG+VTP)

Substitute 103 for KP,3.77 for VSC and -2 for VTP

  gm=2(103)(3.772)

  gm=3.54mA/V

Calculate the Miller Capacitance.

  CM=Cgd[1+gm(RDRL)]

Substitute 3×1012 for CPD,3.54×103 for gm,2×103 for RD, and 5×103 for RL

  CM=3×1012[1+3.54×103(2×1035×103)]=3×1012[1+3.54×103(2×5×1032+5)]=18.2pF

The value of the Miller capacitance is 18.2pF

(c)

Expert Solution
Check Mark
To determine

The mid-band voltage gain.

Answer to Problem 7.69P

The mid-band voltage gain is 4.66

Explanation of Solution

Given:

The given circuit is shown below.

  Microelectronics: Circuit Analysis and Design, Chapter 7, Problem 7.69P , additional homework tip  3

Calculation:

Draw the small-signal model of the circuit shown in figure

  Microelectronics: Circuit Analysis and Design, Chapter 7, Problem 7.69P , additional homework tip  4

Determine the expression for output voltage vo

Apply the voltage division principle.

  vo=gmVsg(RDRL)(4)

Determine source-to-gate voltage Vgs

  Vsg=[(R1R2)(R1R2)+Ri]vi

Substitute 8×103 for R1,22×103 for R2, and 0.5×103 for Ri

  Vsg=[(8×10322×103)(8×10322×103)+0.5×103]vi=[(5.8666×103)(5.8666×103)+0.5×103]vi=0.9215vi

Substitute 0.9215vi for Vsg in equation (4)

  vo=gm(0.9215vi)(RDRL)

  vovi=gm(0.9215)(RDRL)

Therefore, the expression for the mid-band voltage gain is,

  Av=gm(0.9215)(RDRL)

Substitute 3.54×103 for 3.54×103 for gm,2×103 for RD, and 5×103 for RL

  Av=3.54×103(0.9215)(2×1035×103)=4.66

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Chapter 7 Solutions

Microelectronics: Circuit Analysis and Design

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