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 17, Problem 17.11EP

(a)

To determine

The value of the current iD,iB and iC .

(a)

Expert Solution
Check Mark

Answer to Problem 17.11EP

The value of the current iD is 1.747mA , iB is 0.253mA and iC is 3.791mA .

Explanation of Solution

Calculation:

The given diagram is shown in Figure 1

  Microelectronics: Circuit Analysis and Design, Chapter 17, Problem 17.11EP

Apply KVL in the above circuit.

  5V=iC(2.25kΩ)Vγ(SD)+VBE(on)

Substitute 0.3V for Vγ(SD) and 0.7V for VBE(on) in the above equation.

  5V=iC(2.25kΩ)0.3V+0.7ViC=2.0444mA

The expression for the value of the current iC is given by,

  iC=iB+iC1+(1β)

Substitute 2mA for iB , 2.0444mA for iC and 15 for β in the above equation.

  iC=2mA+2.0444mA1+(115)=3.791mA

The expression for the value of the base current iB is given by,

  iB=iCβ

Substitute 3.791mA for iC and 15 for β in the above equation.

  iB=3.791mA15=0.253mA

The expression for the value of the drain current is given by,

  iD=iBiB

Substitute 2mA for iB and 0.253mA for iB in the above equation.

  iD=2mA0.253mA=1.747mA

(b)

To determine

The value of the current iD , iB and iC .

(b)

Expert Solution
Check Mark

Answer to Problem 17.11EP

The value of the current iD is 1.122mA , iB is 0.878mA and iC is 13.166mA .

Explanation of Solution

Calculation:

Apply KVL in the above circuit.

  5V=iC(2.25kΩ)Vγ(SD)+VBE(on)

Substitute 0.3V for Vγ(SD) and 0.7V for VBE(on) in the above equation.

  5V=iC(2.25kΩ)0.3V+0.7ViC=2.0444mA

The expression for the value of the current iC is given by,

  iC=iD+iC+iL ……. (1)

The expression for the value of the current iB is given by,

  iB=iB+iD ……. (2)

The expression for the value of the current iC is given by,

  iC=βiBiB=iCβ ……. (3)

From equation (2) and equation (3), the value of the current iD is given by,

  iD=iBiB

Substitute iCβ for iB in the above equation.

  iD=iBiCβ

Substitute iBiCβ for iD in the above equation

  iC=iBiCβ+iC+iLiC=iB+iC+iL1(+1β)

Substitute 10mA for iL , 2mA for iD , 2.0444mA for iC and 15 for β in the above equation.

  iC=2mA+2.0444mA+10mA1+(115)=13.166mA

Substitute 15 for β and 13.166mA for iC in equation (3).

  iB=13.166mA15=0.878mA

The expression for the value of the drain current is given by,

  iD=iBiB

Substitute 2mA for iB and 0.878mA for iB in the above equation.

  iD=2mA0.878mA=1.122mA

Conclusion:

Therefore, the value of the current iD is 1.122mA , iB is 0.878mA and iC is 13.166mA .

(c)

To determine

The value of the maximum load current.

(c)

Expert Solution
Check Mark

Answer to Problem 17.11EP

The maximum value of the load current is 28mA .

Explanation of Solution

Calculation:

The expression for the value of the VCE is given by,

  VCE(sat)=VBE(on)Vγ(SD)

Substitute 0.7V for VBE(on) and 0.3V for Vγ(SD) in the above equation.

  VCE(sat)=0.7V0.3V=0.4V

The expression for condition of the edge saturation is given by,

  VCC(iCiDiL)RCVCE(sat)

Substitute 0.4V for VCE(sat) in the above equation.

  VCC(iCiDiL)RC0.4V

Substitute 0 for iD , 5V for VCC , 2.25kΩ for RC in the above equation.

  5V(iCiL)(2.25kΩ)0.4V4.6V2.25kΩiCiL(iCiL)2.0444mA

The expression for the value of the load current is given by,

  iB+iC+iL1+(1β)iL2.044mAiB+iC+iLβ2.044mA(1+1β)

Substitute 15 for β , 2.044mA for iC , 2mA for iB in the above equation.

  2mA+2.044mA+iL152.044mA(1+115)iL28mA

Conclusion:

Therefore, the maximum value of the load current is 28mA .

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Chapter 17 Solutions

Microelectronics: Circuit Analysis and Design

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