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 10, Problem 10.69P
To determine

The currents IREF , I1 , I2 , I3 and I4 .

Expert Solution & Answer
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Answer to Problem 10.69P

  IREF=59.09μA

  I1=11.82μA

  I2=73.87μA

  I3=47.27μA

  I4=236.4μA

Explanation of Solution

Given:

  VTN=0.8VVTP=0.8VK'n=100μA/V2K'P=60μA/V2λn=λp=0R=100kΩ

Calculation:

The given circuit is,

  Microelectronics: Circuit Analysis and Design, Chapter 10, Problem 10.69P

According to the circuit reference current IREF will be,

  IREF=(K'n2)(11)(VGSNVTN)2(1)

And

  IREF=(K'p2)(11)(VSGP+VTP)2(2)

Now substitute the given values in equation (1) and equation (2),

   I REF =( K ' n 2 )( 1 1 ) ( V GSN V TN ) 2 I REF =( 100μ 2 )( 1 1 ) ( V GSN 0.8) 2 ( 3 )

   I REF =( K ' p 2 )( 1 1 ) ( V SGP + V TP ) 2 I REF =( 60μ 2 )( 1 1 ) ( V SGP 0.8) 2 ( 4 )

On comparing equation (3) and equation (4),

   ( 100μ 2 )( 1 1 ) ( V GSN 0.8) 2 =( 60μ 2 )( 1 1 ) ( V SGP 0.8) 2 50 ( V GSN 0.8) 2 =30 ( V SGP 0.8) 2 50(VGSN0.8)=30(VSGP0.8)7.071(VGSN0.8)=5.477(VSGP0.8)7.0715.477(VGSN0.8)=(VSGP0.8)VSGP=7.0715.477(VGSN0.8)+0.8(5)

Now reference current expression is,

  IREF=V+VVSGPVGSNR

  IREF=5(5)VSGPVGSN100kΩ(6)

Now compare equation (3) and equation (6)

  IREF=(100μ2)(11)(VGSN0.8)2=IREF=5(5)VSGPVGSN100kΩ

   ( 100μ 2 )( 1 1 ) ( V GSN 0.8) 2 = 5(5) V SGP V GSN 100kΩ ( 100μ 2 )( 1 1 ) ( V GSN 0.8) 2 ×100× 10 3 =10 7.071 5.477 ( V GSN 0.8) V GSN 5 ( V GSN 0.8) 2 =9.21.291( V GSN 0.8) V GSN 5 ( V GSN 0.8) 2 =9.21.291 V GSN +1.032) V GSN

   5 ( V GSN 0.8) 2 =10.2322.291 V GSN ( V GSN 0.8) 2 = 24.2322.291 V GSN 5 VGSN2+0.641.6VGSN=2.0460.458VGSNVGSN21.142VGSN1.406=0

Solve the above expression by quadratic degree method,

  VGSN=1.887VVGSN=0.745V

Consider VGSN=1.887V and substitute in equation (5)

  VSGP=7.0715.477(VGSN0.8)+0.8

  VSGP=7.0715.477(1.8870.8)+0.8VSGP=1.291×1.887+0.8VSGP=2.2033V

Calculate reference current,

  IREF=5(5)VSGPVGSN100kΩ

  IREF=102.20331.887100×103

  IREF=59.09μA

Now calculate current I1,I2,I3andI4

  I1=(WL)1IREF

  I1=(0.21)×59.09μA

  I1=11.82μA

  I2=(WL)2IREF

  I2=1.251×59.09μA

  I2=73.87μA

  I3=(WL)3IREFI3=0.81×59.09μA

  I3=47.27μA

  I4=(WL)4IREFI4=41×59.09μA

  I4=236.4μA

Conclusion:

  IREF=59.09μA

  I1=11.82μA

  I2=73.87μA

  I3=47.27μA

  I4=236.4μA

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Chapter 10 Solutions

Microelectronics: Circuit Analysis and Design

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