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 16, Problem 16.62P

(a)

To determine

To construct: The table for the state of each transistor.

To find: The output voltage vO1 , vO2 and vO3 for the input logic state.

(a)

Expert Solution
Check Mark

Answer to Problem 16.62P

The value of the output voltages is obtained and is shown in Table 2

Explanation of Solution

Calculation:

The given diagram is shown in Figure 1

  Microelectronics: Circuit Analysis and Design, Chapter 16, Problem 16.62P

The given table is shown in Table 1

Table 1

    StateCLKvXvYvZ
    10000
    21111
    30000
    41011
    50000
    61101

For state 1

The input clock signal input X, Y and Z are at logic zero and the logic function of the output voltage vO1 is the compliment of the input voltage vX an is given by,

  vO1=v¯X   ....... (1).

Substitute 0 for vX in the above equation.

  vO1=0¯=1

From the circuit the NMOS device with the input vY is in series with the parallel combination of the NMOS device of input vZ and the compliment vo1 .

The expression for the logic function of the output voltage vo2 is given by,

  vO2=vY( v ¯ O1+vZ)¯   ...... (2)

Substitute 1 for vO1 , 0 for vY and 0 for vZ in the above equation.

  vO2=0( 1 ¯ +0)¯=1

The logic function of the output voltage vO3 is the compliment of the output voltage vO2 and is given by,

  vO3=vO2¯   ...... (3)

Substitute 1 for vO2 in the above equation.

  vO3=1¯=0

For state 2.

All the inputs are at logic 1.

Substitute 1 for vX in equation (1)

  vO1=1¯=0

Substitute 0 for vO1 , 1 for vY and 1 for vZ in equation (2)

  vO2=1( 0 ¯ +1)¯=1

Substitute 0 for vO2 in equation (3)

  vO3=0¯=1

For state 3.

All the inputs are at logic 0.

Substitute 0 for vX in equation (1)

  vO1=0¯=1

Substitute 1 for vO1 , 0 for vY and 0 for vZ in equation (2)

  vO2=0( 1 ¯ +0)¯=1

Substitute 1 for vO2 in equation (3)

  vO3=1¯=0

For state 4.

The input clock, vY and the input vZ are at logic zero and the input vX is at logic zero.

Substitute 0 for vX in equation (1)

  vO1=0¯=1

Substitute 1 for vO1 , 1 for vY and 1 for vZ in equation (2)

  vO2=1( 1 ¯ +1)¯=0

Substitute 0 for vO2 in equation (3)

  vO3=0¯=1

For state 5.

All the inputs are at logic zero.

Substitute 0 for vX in equation (1)

  vO1=0¯=1

Substitute 1 for vO1 , 0 for vY and 0 for vZ in equation (2)

  vO2=0( 1 ¯ +0)¯=1

Substitute 1 for vO2 in equation (3)

  vO3=1¯=0

For state 6.

The input clock, vX and the input vZ are at logic zero and the input vY is at logic zero.

Substitute 1 for vX in equation (1)

  vO1=1¯=0

Substitute 0 for vO1 , 0 for vY and 1 for vZ in equation (2)

  vO2=0( 0 ¯ +1)¯=1

Substitute 1 for vO2 in equation (3)

  vO3=1¯=0

The table for the state of the transistors and the output voltage is shown in table below.

The required table is shown in Table 2

Table 2

    StateCLKvXvYvZvO1vO2vO3
    100005V5V0V
    211110V0V5V
    300005V5V0V
    410115V5V
    500005V5V0V
    611010V5V0V

The table for the state of the transistor is shown below.

Conclusion:

Therefore, the value of the output voltages is obtained and is shown in Table 2

(b)

To determine

The logic function that the circuit implements.

(b)

Expert Solution
Check Mark

Answer to Problem 16.62P

The logic that the circuit performs is (vXORvZ)ANDvY .

Explanation of Solution

Calculation:

The expression for the output voltage vO3 is given by,

  vO3=v¯O3

Substitute vY( v ¯ O1+vZ)¯ for vO2 in the above equation.

  vO3= v Y ( v ¯ O1 + v Z )¯¯=(vX+vZ)vY=(vXORvZ)ANDvY

Conclusion:

Therefore, the logic that the circuit performs is (vXORvZ)ANDvY .

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Chapter 16 Solutions

Microelectronics: Circuit Analysis and Design

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