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 16.32P

(a)

To determine

The transition points for the p-channel and the n-channel MOSFET.

To sketch: The voltage transfer characteristics.

To find: The value of the input voltage for the different values of the input output voltages.

(a)

Expert Solution
Check Mark

Answer to Problem 16.32P

The value of the transition voltages are VIt is 1.65V , VOPt for 2.05V , VONt is 1.25V . The plot for the voltage transfer characteristics is shown in Figure 2. The value of the input voltage for the output voltage of 0.25V is 2.032V and for the output voltage of 3.05V is 1.268V .

Explanation of Solution

Calculation:

The given diagram is shown in Figure 1

  Microelectronics: Circuit Analysis and Design, Chapter 16, Problem 16.32P , additional homework tip  1

The expression to determine the trans-conductance parameter for NMOS is given by,

  Kn=( k n2)(WL)n

Substitute 2 for (WL)n and 100μA/V2 for kn in the above equation.

  Kn=( 100 μA/ V 2 2)2=100μA/V2

The expression to determine the trans-conductance parameter for PMOS is given by,

  Kp=( k p2)(WL)p

Substitute 5 for (WL)p and 40μA/V2 for kp in the above equation.

  Kp=( 40 μA/ V 2 2)5=100μA/V2

The expression to determine the transition points VIt is given by,

  VIt=VDD+VTP+ k n k p VTN1+ k n k p VTN

Substitute 3.3V for VDD , 0.4V for VTP , 100μA/V2 for kn and 100μA/V2 for kp and 0.4V for VTN in the above equation.

  VIt=3.3V+( 0.4V)+ 100 μA/ V 2 100 μA/ V 2 ( 0.4V)1+ 100 μA/ V 2 100 μA/ V 2 ( 0.4V)=1.65V

The expression to determine the value of the voltage VOPt is given by,

  VOPt=VItVTP

Substitute 1.65V for VIt and 0.4V for VTP in the above equation.

  VOPt=1.65V(0.4V)=2.05V

The expression to determine the value of the voltage VONt is given by,

  VONt=VItVTP

Substitute 1.65V for VIt and 0.4V for VTN in the above equation.

  VONt=1.65V0.4V=1.25V

The voltage transfer characteristics is shown below.

The required diagram is shown in Figure 2

  Microelectronics: Circuit Analysis and Design, Chapter 16, Problem 16.32P , additional homework tip  2

The expression to determine the value of the input voltage is given by,

  Kn[2(vIVTN)vOvO2]=Kp(V DDvI+V TP)2

Substitute 100μA/V2 for Kp , 100μA/V2 for Kn , 0.25V for vO , 0.4V for VTN , 3.3V for VDD and 0.4V for VTP in the above equation.

  100μA/V2[2( v I0.4V)(0.25V)( 0.25V)2]=100μA/V2(3.3V v I0.4V)2vI26.3vI+8.6725=0vI=2± 2 2 4( 1 )( 0.49 )2vI=4.268V,2.032V

The value of the voltage must be less than the supply voltage, thus the input voltage is 2.032V .

The expression to determine the value of the input voltage when the output voltage is more than 1.25V is given by,

  Kn(vIV TN)2=Kp[2(VDDvI+VTP)(VDDvO)( V DD v O)2]

Substitute 100μA/V2 for Kp , 100μA/V2 for Kn , 3.05V for vO , 0.4V for VTN , 3.3V for VDD and 3.3V for VDD in equation (1).

  100μA/V2( v I0.4V)2=100μA/V2[2(3.3V v I0.4V)(3.3V3.05V)( 3.3V v O )2]vI20.3vI1.2275=0vI=0.3± ( 0.3 ) 2 4( 1.2275 )( 1 )2vI=1.268V

Conclusion:

Therefore, the value of the transition voltages are VIt is 1.65V , VOPt for 2.05V , VONt is 1.25V . The plot for the voltage transfer characteristics is shown in Figure 2. The value of the input voltage for the output voltage of 0.25V is 2.032V and for the output voltage of 3.05V is 1.268V .

(b)

To determine

The transition points for the p-channel and the n-channel MOSFET.

To sketch: The voltage transfer characteristics.

To find: The value of the input voltage for the different values of the input output voltages.

(b)

Expert Solution
Check Mark

Answer to Problem 16.32P

The value of the transition voltages are VIt is 1.436V , VOPt for 1.836V , VONt is 1.036V . The plot for the voltage transfer characteristics is shown in Figure 3. The value of the input voltage for the output voltage of 0.25V is 1.78V and for the output voltage of 3.05V is 1.056V .

Explanation of Solution

Calculation:

The expression to determine the trans-conductance parameter for NMOS is given by,

  Kn=( k n2)(WL)n

Substitute 4 for (WL)n and 100μA/V2 for kn in the above equation.

  Kn=( 100 μA/ V 2 2)4=200μA/V2

The expression to determine the trans-conductance parameter for PMOS is given by,

  Kp=( k p2)(WL)p

Substitute 5 for (WL)p and 40μA/V2 for kp in the above equation.

  Kp=( 40 μA/ V 2 2)5=100μA/V2

The expression to determine the transition points VIt is given by,

  VIt=VDD+VTP+ k n k p VTN1+ k n k p VTN

Substitute 3.3V for VDD , 0.4V for VTP , 200μA/V2 for kn and 100μA/V2 for kp and 0.4V for VTN in the above equation.

  VIt=3.3V+( 0.4V)+ 200 μA/ V 2 100 μA/ V 2 ( 0.4V)1+ 200 μA/ V 2 100 μA/ V 2 ( 0.4V)=1.436V

The expression to determine the value of the voltage VOPt is given by,

  VOPt=VItVTP

Substitute 1.436V for VIt and 0.4V for VTP in the above equation.

  VOPt=1.436V(0.4V)=1.836V

The expression to determine the value of the voltage VONt is given by,

  VONt=VItVTP

Substitute 1.436V for VIt and 0.4V for VTN in the above equation.

  VONt=1.436V0.4V=1.036V

The voltage transfer characteristics is shown below.

The required diagram is shown in Figure 3

  Microelectronics: Circuit Analysis and Design, Chapter 16, Problem 16.32P , additional homework tip  3

The expression to determine the value of the input voltage is given by,

  Kn[2(vIVTN)vOvO2]=Kp(V DDvI+V TP)2

Substitute 100μA/V2 for Kp , 200μA/V2 for Kn , 0.25V for vO , 0.4V for VTN , 3.3V for VDD and 0.4V for VTP in the above equation.

  200μA/V2[2( v I0.4V)(0.25V)( 0.25V)2]=100μA/V2(3.3V v I0.4V)2vI26.8vI+8.935=0vI=6.8± ( 6.8 )4( 8.935 )( 1 )2vI=5.02V,1.78V

The value of the voltage must be less than the supply voltage, thus the input voltage is 1.78V .

The expression to determine the value of the input voltage when the output voltage is more than 1.25V is given by,

  Kn(vIV TN)2=Kp[2(VDDvIVTP)(VDDvO)( V DD v O)2]

Substitute 100μA/V2 for Kp , 200μA/V2 for Kn , 3.05V for vO , 0.4V for VTN , 3.3V for VDD and 3.3V for VDD in equation (1).

  200μA/V2( v I0.4V)2=100μA/V2[2(3.3V v I0.4V)(3.3V3.05V)( 3.3V3.05)2]2vI21.1vI1.0675=0vI=1.1± ( 1.1 ) 2 4( 1.0675 )( 2 )2(2)vI=1.056V

Conclusion:

Therefore, the value of the transition voltages are VIt is 1.436V , VOPt for 1.836V , VONt is 1.036V . The plot for the voltage transfer characteristics is shown in Figure 3. The value of the input voltage for the output voltage of 0.25V is 1.78V and for the output voltage of 3.05V is 1.056V .

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Chapter 16 Solutions

Microelectronics: Circuit Analysis and Design

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