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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Textbook Question
Chapter 2, Problem 2.59P

Each diode cut−in voltage in the circuit in Figure P2.59 is 0.7 V. Determine I D 1 , I D 2 , I D 3 , and υ O for (a) υ I = 0.5 V , (b) υ I = 1.5 V , (c) υ I = 3.0 V , and (d) υ I = 5.0 V .

Chapter 2, Problem 2.59P, Each diode cutin voltage in the circuit in Figure P2.59 is 0.7 V. Determine ID1,ID2,ID3 , and O for
Figure P2.59

(a).

Expert Solution
Check Mark
To determine

The values of ID1,ID2,ID3,vo .

Answer to Problem 2.59P

  ID1=ID2=ID3=0Avo=0.5V

Explanation of Solution

Given Information:

The given circuit is shown below.

  Microelectronics: Circuit Analysis and Design, Chapter 2, Problem 2.59P , additional homework tip  1

  vI=0.5Vvγ=0.7V

Calculation:

For vI=0.5V , diode D1 is in cut off region and acts as open circuit because the voltage difference between positive and negative terminal is less than 0.7 V. The diode D2,D3 are in reverse bias mode and act like open circuit. Hence, the currents through the diodes are zero.

  ID1=ID2=ID3=0A

The input voltage is dropped across the output because the current in the circuit is zero.

  vo=0.5V

(b).

Expert Solution
Check Mark
To determine

The values of ID1,ID2,ID3,vo .

Answer to Problem 2.59P

  ID1=0.0667mAID2=ID3=0Avo=1.234V

Explanation of Solution

Given Information:

The given circuit is shown below.

  Microelectronics: Circuit Analysis and Design, Chapter 2, Problem 2.59P , additional homework tip  2

  vI=1.5Vvγ=0.7V

Calculation:

For vI=1.5V , the diode D1 is forward biased and in active mode. The diode D2 is in cut off mode and acts as open circuit because the voltage difference between positive and negative terminal is less than 0.7 V. The diode D3 is in reverse bias region and acts like open circuit. Hence, the currents through the diodes D2 and D3 are zero.

  ID2=ID3=0A

The value of diode current ID1 is determined as follows:

The modified circuit is:

  Microelectronics: Circuit Analysis and Design, Chapter 2, Problem 2.59P , additional homework tip  3

Applying Kirchhoff’s voltage law:

  vI+ID1R1+ID1R2+0.7=0ID1=vI0.7R1+R2ID1=1.50.7(4+8)×103ID1=0.0667mA

The value of output voltage vo is:

  vo=ID1R2+0.7vo=(0.0667×103)(8×103)+0.7vo=1.234V

(c).

Expert Solution
Check Mark
To determine

The values of ID1,ID2,ID3,vo .

Answer to Problem 2.59P

  ID1=0.171mA,ID2=0.0615mAID3=0mA,vo=2.069V

Explanation of Solution

Given Information:

The given circuit is shown below.

  Microelectronics: Circuit Analysis and Design, Chapter 2, Problem 2.59P , additional homework tip  4

  vI=3Vvγ=0.7V

Calculation:

Assuming the diodes are in forward bias and in active mode.

Applying Kirchhoff’s current law at output node:

  vo34+vo0.78+vo1.76+vo2.74=0vo(14+18+16+14)=5.74+0.78+1.76vo(12+18+16)=34.2+2.1+6.824vo(12+3+424)=43.124vo=43.119vo=2.268V

From above calculation diodes D1,D2 are in forward bias active mode but the diode D3 is in cut off mode because the voltage difference between positive and negative terminal is less than 0.7 V. The current through diode D3 is zero.

  ID3=0A

Hence, the assumption is incorrect.

The modified figure is:

  Microelectronics: Circuit Analysis and Design, Chapter 2, Problem 2.59P , additional homework tip  5

Applying Kirchhoff’s current law at output node:

  vo34+vo0.78+vo1.76=0vo(14+18+16)=34+0.78+1.76vo(14+18+16)=18+2.1+6.824vo(6+3+424)=26.924vo=26.913vo=2.069V

The diode currents ID1,ID2 are determined as follows:

  ID1=vo0.78ID1=2.0690.78×103ID1=0.171mAID2=vo1.76ID2=2.0691.76×103ID2=0.0615mA

(d).

Expert Solution
Check Mark
To determine

The values of ID1,ID2,ID3,vo .

Answer to Problem 2.59P

  ID1=0.275mA,ID2=0.2mAID3=0.05mA,vo=2.9V

Explanation of Solution

Given Information:

The given circuit is shown below.

  Microelectronics: Circuit Analysis and Design, Chapter 2, Problem 2.59P , additional homework tip  6

  vI=5Vvγ=0.7V

Calculation:

For vI=5V, all diodes are forward biased and in active mode.

Applying Kirchhoff’s current law at output node:

  vo54+vo0.78+vo1.76+vo2.74=0vo(14+18+16+14)=7.74+0.78+1.76vo(12+18+16)=46.2+2.1+6.824vo(12+3+424)=55.124vo=43.119vo=2.9V

The values of diode currents are:

  ID1=vo0.78ID1=2.90.78×103ID1=0.275mAID2=vo1.76ID2=2.91.76×103ID2=0.2mAID3=vo2.74ID3=2.92.74×103ID3=0.05mA

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Chapter 2 Solutions

Microelectronics: Circuit Analysis and Design

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