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 8, Problem 8.20P

(a)

To determine

The maximum and the minimum value of the output voltage and the minimum corresponding input voltages for the circuit to operate in the linear region for given value of RL .

(a)

Expert Solution
Check Mark

Answer to Problem 8.20P

The maximum value of the output voltage is 4.8V and the minimum value of the output voltage is 3.2V . The maximum value of the input voltage is 5.5V and the minimum value of the input voltage is 2.5V .

Explanation of Solution

Calculation:

The given diagram is shown in Figure 1

  Microelectronics: Circuit Analysis and Design, Chapter 8, Problem 8.20P

The expression for the maximum value of the output voltage is given by,

  vO(max)=V+VCE(sat)

Substitute 5V for VGS and 0.2V for VCE(sat) in the above equation.

  vO(max)=5V0.2V=4.8V

The expression for the drain to source saturation voltage is given by,

  VDS(sat)=VGSVTN

Substitute 0 for VGS and 1.8V for VTN in the above equation.

  VDS( sat)=0(1.8V)=1.8V

The expression for the minimum value of the output voltage is given by,

  vO(min)=VDS(sat)+V

Substitute 1.8V for VDS(sat) and 5V for VTN in the above equation.

  vO( min)=1.8V+(5V)=3.2V

The expression for the minimum value of the input voltage is given by,

  vI(min)=VBE(ON)+vO(min)

Substitute 0.7V for VBE(ON) and 3.2V for vO(min) in the above equation.

  vI( min)=0.7V3.2V=2.5V

The expression for the maximum value of the input voltage is given by,

  vI(max)=VBE(ON)+vO(max)

Substitute 0.7V for VBE(ON) and 4.8V for vO(max) in the above equation.

  vI( min)=0.7V+4.8V=5.5V

Conclusion:

Therefore, the maximum value of the output voltage is 4.8V and the minimum value of the output voltage is 3.2V . The maximum value of the input voltage is 5.5V and the minimum value of the input voltage is 2.5V .

(b)

To determine

The maximum and the minimum value of the output voltage and the minimum corresponding input voltages for the circuit to operate in the linear region for given value of RL .

(b)

Expert Solution
Check Mark

Answer to Problem 8.20P

The maximum value of the output voltage is 4.8V and the minimum value of the output voltage is 3.2V . The maximum value of the input voltage is 5.5V and the minimum value of the input voltage is 2.5V .

Explanation of Solution

Calculation:

The expression for the maximum value of the output voltage is given by,

  vO(max)=V+VCE(sat)

Substitute 5V for VGS and 0.2V for VCE(sat) in the above equation.

  vO(max)=5V0.2V=4.8V

The expression for the drain to source saturation voltage is given by,

  VDS(sat)=VGSVTN

Substitute 0 for VGS and 1.8V for VTN in the above equation.

  VDS( sat)=0(1.8V)=1.8V

The expression for the minimum value of the output voltage is given by,

  vO(min)=VDS(sat)+V

Substitute 1.8V for VDS(sat) and 5V for VTN in the above equation.

  vO( min)=1.8V+(5V)=3.2V

The expression for the minimum value of the input voltage is given by,

  vI(min)=VBE(ON)+vO(min)

Substitute 0.7V for VBE(ON) and 3.2V for vO(min) in the above equation.

  vI( min)=0.7V3.2V=2.5V

The expression for the maximum value of the input voltage is given by,

  vI(max)=VBE(ON)+vO(max)

Substitute 0.7V for VBE(ON) and 4.8V for vO(max) in the above equation.

  vI( min)=0.7V+4.8V=5.5V

Conclusion:

Therefore, the maximum value of the output voltage is 4.8V and the minimum value of the output voltage is 3.2V . The maximum value of the input voltage is 5.5V and the minimum value of the input voltage is 2.5V .

(c)

To determine

The smallest value of the RL possible and the conversion efficiency.

(c)

Expert Solution
Check Mark

Answer to Problem 8.20P

The minimum value of the load resistance is 5.144kΩ and the conversion efficiency is 10% .

Explanation of Solution

Calculation:

The expression for the current through the transistor Q1 when VO=2V is given by,

  ID=Kn(V GSv TN)2

Substitute 12mA/V2 for Kn , 0V for VGS and 1.8V for VTN in the above equation.

  ID=(12mA/ V 2)(0( 1.8V))2=38.89mA

The expression for the minimum value of the load resistance is RL(min) is given by,

  RL(min)=VOID

Substitute 2V for VO and 38.89mA for ID in the above equation.

  RL( min)=2V38.89mA=5.144kΩ

The expression for the power delivered to the load is given by,

  P¯L=VO22RL

Substitute 2V for VO and 5.144kΩ for RL in the above equation.

  P¯L= ( 2V )22( 5.144kΩ)=38.89mW

The expression for the average power supplied by the source is given by,

  P¯S=ID(V+V)

Substitute 38.88mA for ID , 5V for V+ and 5V for V in the above equation.

  P¯S=(38.88mA)(5V( 5V))=388.89mW

The expression for the value of power conversion efficiency is given by,

  %η=( P ¯L P ¯S)100

Substitute 388.89mW for PS and 38.89mW for P¯L in the above equation.

  %η=( 38.89mW 388.89mW)100=10%

Conclusion:

Therefore, the minimum value of the load resistance is 5.144kΩ and the conversion efficiency is 10% .

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Chapter 8 Solutions

Microelectronics: Circuit Analysis and Design

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