The cut−in voltage of each diode in the circuit shown in Figure P2.52 is V γ = 0.7 V . Determine I D 1 , I D 2 , I D 3 , and V A for (a) R 3 = 14 k Ω , R 4 = 24 k Ω ; (b) R 3 = 3.3 k Ω , R 4 = 5.2 k Ω ; and (c) R 3 = 3.3 k Ω , R 4 = 1.32 k Ω . Figure P2.52

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Answer 1

Textbook Question

Chapter 2, Problem 2.52P

The cut−in voltage of each diode in the circuit shown in Figure P2.52 is $V γ = 0.7 V$ . Determine $I D 1 , I D 2 , I D 3$ , and $V A$ for (a) $R 3 = 14 k Ω , R 4 = 24 k Ω$ ; (b) $R 3 = 3.3 k Ω , R 4 = 5.2 k Ω$ ; and (c) $R 3 = 3.3 k Ω , R 4 = 1.32 k Ω$ .

Chapter 2, Problem 2.52P, The cutin voltage of each diode in the circuit shown in Figure P2.52 is V=0.7V . Determine
Figure P2.52

(a)

Expert Solution

To determine

The values of $ID1,ID2,ID3, and VA$

Answer to Problem 2.52P

The current flowing through diodes is

$ID3=0.5 mA$

$ID2=0.5 mA$ , $ID1=1 mA$ and $VA=2.7 V$

Explanation of Solution

Given:

Microelectronics: Circuit Analysis and Design, Chapter 2, Problem 2.52P , additional homework tip 1

$R3=14kΩ R4=24kΩ$

Calculation:

The cut-in voltage for each diode is, $Vγ=0.7V$

Assume all diodes are ON.

Draw the circuit diagram with node voltages and cut-in voltages.

Microelectronics: Circuit Analysis and Design, Chapter 2, Problem 2.52P , additional homework tip 2

Apply Kirchhoff’s current law node A.

$VA−156.15k+VA−0.72k+VA−0.7−(−5)R3+VA−0.7−(−10)R4=0VA−156.15k+VA−0.72k+VA+4.3R3+VA+9.3R4=0$

Substitute $14 kΩ$ for $R3$ and $24 kΩ$ for $R4$ .

$VA−156.15k+VA−0.72k+VA+4.314k+VA+9.324k=0VA−156.15+VA−0.72+VA+4.314+VA+9.324=0(16.15+12+114+124)VA−156.15−0.72+4.314+9.324=00.7756VA−2.094=0VA=2.7 V$

Therefore, the voltage at node $A, VA$ is $2.7 V$

Calculate the current flowing through diode, $D1$ .

$ID1=VA−0.72k$

Substitute $2.7V$ for $VA$ .

$ID1=2.7−0.72k=22k=1 mA$

Therefore, the current flowing through diode $D1,$ is $ID1=1 mA$ .

Calculate the current flowing through diode, $D2$ .

$ID2=VA−0.7−(−5)R3$

Substitute $2.7V$ for $VA$ and $14 kΩ$ for $R3$ .

$ID2=2.7−0.7+514k=714k=0.5 mA$

Therefore, the current flowing through diode $D2,$ is $ID2=0.5 mA$ .

Calculate the current flowing through diode, $D3$ .

$ID3=VA−0.7−(−10)R4$

Substitute $2.7V$ for $VA$ and $24 kΩ$ for $R4$ .

$ID3=2.7−0.7+1024k=1224k=0.5 mA$

Therefore, the current flowing through diodes is $ID3=0.5 mA$

Conclusion:

Therefore, the current flowing through diodes is

$ID3=0.5 mA$

$ID2=0.5 mA$ , $ID1=1 mA$ and $VA=2.7 V$

(b)

Expert Solution

To determine

The values of $ID1,ID2,ID3, and VA$

Answer to Problem 2.52P

The current flowing through diodes is $VA=−0.991V$ , $ID1= 0 A$ , $ID2=1.003 mA$ and $ID3=1.6 mA$

Explanation of Solution

Given:

$R3=3.3kΩ R4=5.2kΩ$

Calculation:

Assume diode $D1$ is OFF and remaining diodes are ON.

Draw the circuit diagram with node voltages and cut-in voltages.

Microelectronics: Circuit Analysis and Design, Chapter 2, Problem 2.52P , additional homework tip 3

Apply Kirchhoff’s current law node A.

$VA−156.15k+VA−0.7−(−5)R3+VA−0.7−(−10)R4=0VA−156.15k+VA+4.3R3+VA+9.3R4=0$

Substitute $3.3 kΩ$ for $R3$ and $5.2 kΩ$ for $R4$ .

$VA−156.15k+VA+4.33.3k+VA+9.35.2k=0VA−156.15+VA+4.33.3+VA+9.35.2=0(16.15+13.3+15.2)VA−156.15+4.33.3+9.35.2=00.6579VA+0.6525=0VA=−0.65250.6579=−0.991V$

Therefore, the voltage at node $A, VA$ is $−0.991V$

The diode $D1$ is reverse biased. Therefore, the current flowing through diode $D1$ is zero.

That is, $ID1=0 A$ .

Therefore, the current flowing through diode $D1,$

$ID1= 0 A$

Calculate the current flowing through diode, $D2$ .

$ID2=VA−0.7−(−5)R3$

Substitute $−0.991V$ for $VA$ and $3.3 kΩ$ for $R3$ .

$ID2=−0.991−0.7+53.3k=3.3093.3k=1.003 mA$

Therefore, the current flowing through diode $D2,$ is $ID2=1.003 mA$ .

Calculate the current flowing through diode, $D3$ .

$ID3=VA−0.7−(−10)R4$

Substitute $−0.991V$ for $VA$ and $5.2 kΩ$ for $R4$ .

$ID3=−0.991−0.7+105.2k=8.3095.2k=1.6 mA$

Therefore, the current flowing through diode $D3,$ is $ID3=1.6 mA$ .

Conclusion:

Therefore, the current flowing through diodes is $VA=−0.991V$ , $ID1= 0 A$ , $ID2=1.003 mA$ and $ID3=1.6 mA$

(c)

Expert Solution

To determine

The values of $ID1,ID2,ID3, and VA$

Answer to Problem 2.52P

The current flowing through diodes is $VA=−5V$ , $ID1=0 A$ , $ID2=0 A$

$ID3=3.25 mA$

Explanation of Solution

Given:

$R3=3.3kΩR4=1.32kΩ$

Calculation:

Assume diode $D1$ and $D2$ are OFF and diode $D3$ is ON.

Draw the circuit diagram with node voltages and cut-in voltages.

Microelectronics: Circuit Analysis and Design, Chapter 2, Problem 2.52P , additional homework tip 4

Apply Kirchhoff’s current law node A.

$VA−156.15k+VA−0.7−(−10)R4=0VA−156.15k+VA+9.3R4=0$

Substitute $1.32 kΩ$ for $R4$ .

$VA−156.15k+VA+9.31.32k=0VA−156.15+VA+9.31.32=0(16.15+11.32)VA−156.15+9.31.32=00.92VA+4.606=0VA=−5V$

Therefore, the voltage at node $A, VA$ is $−5V$ .

The Diode $D1$ and $D2$ are reverse biased. Therefore, the current flowing through diode $D1$ is zero.

That is, $ID1=0 A$ .

Therefore, the current flowing through diode $D1,ID1 is 0 A.$

The current flowing through the diode, $D2$ is zero.

That is,

$ID2=0 A.$

Therefore, the current flowing through diode $D2,ID2 is 0 A.$

Calculate the current flowing through diode, $D3$ .

$ID3=VA−0.7−(−10)R4$

Substitute $−5V$ for $VA$ and $1.32 kΩ$ for $R4$ .

$ID3=−5−0.7+101.32k=4.31.32k=3.25 mA$

Therefore, the current flowing through diode $D3,$ is $ID3=3.25 mA$ .

Conclusion:

Therefore, the current flowing through diodes is $VA=−5V$ , $ID1=0 A$ , $ID2=0 A$

$ID3=3.25 mA$

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Answer 2

Textbook Question

Chapter 2, Problem 2.52P

The cut−in voltage of each diode in the circuit shown in Figure P2.52 is $V γ = 0.7 V$ . Determine $I D 1 , I D 2 , I D 3$ , and $V A$ for (a) $R 3 = 14 k Ω , R 4 = 24 k Ω$ ; (b) $R 3 = 3.3 k Ω , R 4 = 5.2 k Ω$ ; and (c) $R 3 = 3.3 k Ω , R 4 = 1.32 k Ω$ .

Chapter 2, Problem 2.52P, The cutin voltage of each diode in the circuit shown in Figure P2.52 is V=0.7V . Determine
Figure P2.52

(a)

Expert Solution

To determine

The values of $ID1,ID2,ID3, and VA$

Answer to Problem 2.52P

The current flowing through diodes is

$ID3=0.5 mA$

$ID2=0.5 mA$ , $ID1=1 mA$ and $VA=2.7 V$

Explanation of Solution

Given:

Microelectronics: Circuit Analysis and Design, Chapter 2, Problem 2.52P , additional homework tip 1

$R3=14kΩ R4=24kΩ$

Calculation:

The cut-in voltage for each diode is, $Vγ=0.7V$

Assume all diodes are ON.

Draw the circuit diagram with node voltages and cut-in voltages.

Microelectronics: Circuit Analysis and Design, Chapter 2, Problem 2.52P , additional homework tip 2

Apply Kirchhoff’s current law node A.

$VA−156.15k+VA−0.72k+VA−0.7−(−5)R3+VA−0.7−(−10)R4=0VA−156.15k+VA−0.72k+VA+4.3R3+VA+9.3R4=0$

Substitute $14 kΩ$ for $R3$ and $24 kΩ$ for $R4$ .

$VA−156.15k+VA−0.72k+VA+4.314k+VA+9.324k=0VA−156.15+VA−0.72+VA+4.314+VA+9.324=0(16.15+12+114+124)VA−156.15−0.72+4.314+9.324=00.7756VA−2.094=0VA=2.7 V$

Therefore, the voltage at node $A, VA$ is $2.7 V$

Calculate the current flowing through diode, $D1$ .

$ID1=VA−0.72k$

Substitute $2.7V$ for $VA$ .

$ID1=2.7−0.72k=22k=1 mA$

Therefore, the current flowing through diode $D1,$ is $ID1=1 mA$ .

Calculate the current flowing through diode, $D2$ .

$ID2=VA−0.7−(−5)R3$

Substitute $2.7V$ for $VA$ and $14 kΩ$ for $R3$ .

$ID2=2.7−0.7+514k=714k=0.5 mA$

Therefore, the current flowing through diode $D2,$ is $ID2=0.5 mA$ .

Calculate the current flowing through diode, $D3$ .

$ID3=VA−0.7−(−10)R4$

Substitute $2.7V$ for $VA$ and $24 kΩ$ for $R4$ .

$ID3=2.7−0.7+1024k=1224k=0.5 mA$

Therefore, the current flowing through diodes is $ID3=0.5 mA$

Conclusion:

Therefore, the current flowing through diodes is

$ID3=0.5 mA$

$ID2=0.5 mA$ , $ID1=1 mA$ and $VA=2.7 V$

(b)

Expert Solution

To determine

The values of $ID1,ID2,ID3, and VA$

Answer to Problem 2.52P

The current flowing through diodes is $VA=−0.991V$ , $ID1= 0 A$ , $ID2=1.003 mA$ and $ID3=1.6 mA$

Explanation of Solution

Given:

$R3=3.3kΩ R4=5.2kΩ$

Calculation:

Assume diode $D1$ is OFF and remaining diodes are ON.

Draw the circuit diagram with node voltages and cut-in voltages.

Microelectronics: Circuit Analysis and Design, Chapter 2, Problem 2.52P , additional homework tip 3

Apply Kirchhoff’s current law node A.

$VA−156.15k+VA−0.7−(−5)R3+VA−0.7−(−10)R4=0VA−156.15k+VA+4.3R3+VA+9.3R4=0$

Substitute $3.3 kΩ$ for $R3$ and $5.2 kΩ$ for $R4$ .

$VA−156.15k+VA+4.33.3k+VA+9.35.2k=0VA−156.15+VA+4.33.3+VA+9.35.2=0(16.15+13.3+15.2)VA−156.15+4.33.3+9.35.2=00.6579VA+0.6525=0VA=−0.65250.6579=−0.991V$

Therefore, the voltage at node $A, VA$ is $−0.991V$

The diode $D1$ is reverse biased. Therefore, the current flowing through diode $D1$ is zero.

That is, $ID1=0 A$ .

Therefore, the current flowing through diode $D1,$

$ID1= 0 A$

Calculate the current flowing through diode, $D2$ .

$ID2=VA−0.7−(−5)R3$

Substitute $−0.991V$ for $VA$ and $3.3 kΩ$ for $R3$ .

$ID2=−0.991−0.7+53.3k=3.3093.3k=1.003 mA$

Therefore, the current flowing through diode $D2,$ is $ID2=1.003 mA$ .

Calculate the current flowing through diode, $D3$ .

$ID3=VA−0.7−(−10)R4$

Substitute $−0.991V$ for $VA$ and $5.2 kΩ$ for $R4$ .

$ID3=−0.991−0.7+105.2k=8.3095.2k=1.6 mA$

Therefore, the current flowing through diode $D3,$ is $ID3=1.6 mA$ .

Conclusion:

Therefore, the current flowing through diodes is $VA=−0.991V$ , $ID1= 0 A$ , $ID2=1.003 mA$ and $ID3=1.6 mA$

(c)

Expert Solution

To determine

The values of $ID1,ID2,ID3, and VA$

Answer to Problem 2.52P

The current flowing through diodes is $VA=−5V$ , $ID1=0 A$ , $ID2=0 A$

$ID3=3.25 mA$

Explanation of Solution

Given:

$R3=3.3kΩR4=1.32kΩ$

Calculation:

Assume diode $D1$ and $D2$ are OFF and diode $D3$ is ON.

Draw the circuit diagram with node voltages and cut-in voltages.

Microelectronics: Circuit Analysis and Design, Chapter 2, Problem 2.52P , additional homework tip 4

Apply Kirchhoff’s current law node A.

$VA−156.15k+VA−0.7−(−10)R4=0VA−156.15k+VA+9.3R4=0$

Substitute $1.32 kΩ$ for $R4$ .

$VA−156.15k+VA+9.31.32k=0VA−156.15+VA+9.31.32=0(16.15+11.32)VA−156.15+9.31.32=00.92VA+4.606=0VA=−5V$

Therefore, the voltage at node $A, VA$ is $−5V$ .

The Diode $D1$ and $D2$ are reverse biased. Therefore, the current flowing through diode $D1$ is zero.

That is, $ID1=0 A$ .

Therefore, the current flowing through diode $D1,ID1 is 0 A.$

The current flowing through the diode, $D2$ is zero.

That is,

$ID2=0 A.$

Therefore, the current flowing through diode $D2,ID2 is 0 A.$

Calculate the current flowing through diode, $D3$ .

$ID3=VA−0.7−(−10)R4$

Substitute $−5V$ for $VA$ and $1.32 kΩ$ for $R4$ .

$ID3=−5−0.7+101.32k=4.31.32k=3.25 mA$

Therefore, the current flowing through diode $D3,$ is $ID3=3.25 mA$ .

Conclusion:

Therefore, the current flowing through diodes is $VA=−5V$ , $ID1=0 A$ , $ID2=0 A$

$ID3=3.25 mA$

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Answer 3

Textbook Question

Chapter 2, Problem 2.52P

The cut−in voltage of each diode in the circuit shown in Figure P2.52 is $V γ = 0.7 V$ . Determine $I D 1 , I D 2 , I D 3$ , and $V A$ for (a) $R 3 = 14 k Ω , R 4 = 24 k Ω$ ; (b) $R 3 = 3.3 k Ω , R 4 = 5.2 k Ω$ ; and (c) $R 3 = 3.3 k Ω , R 4 = 1.32 k Ω$ .

Chapter 2, Problem 2.52P, The cutin voltage of each diode in the circuit shown in Figure P2.52 is V=0.7V . Determine
Figure P2.52

(a)

Expert Solution

To determine

The values of $ID1,ID2,ID3, and VA$

Answer to Problem 2.52P

The current flowing through diodes is

$ID3=0.5 mA$

$ID2=0.5 mA$ , $ID1=1 mA$ and $VA=2.7 V$

Explanation of Solution

Given:

Microelectronics: Circuit Analysis and Design, Chapter 2, Problem 2.52P , additional homework tip 1

$R3=14kΩ R4=24kΩ$

Calculation:

The cut-in voltage for each diode is, $Vγ=0.7V$

Assume all diodes are ON.

Draw the circuit diagram with node voltages and cut-in voltages.

Microelectronics: Circuit Analysis and Design, Chapter 2, Problem 2.52P , additional homework tip 2

Apply Kirchhoff’s current law node A.

$VA−156.15k+VA−0.72k+VA−0.7−(−5)R3+VA−0.7−(−10)R4=0VA−156.15k+VA−0.72k+VA+4.3R3+VA+9.3R4=0$

Substitute $14 kΩ$ for $R3$ and $24 kΩ$ for $R4$ .

$VA−156.15k+VA−0.72k+VA+4.314k+VA+9.324k=0VA−156.15+VA−0.72+VA+4.314+VA+9.324=0(16.15+12+114+124)VA−156.15−0.72+4.314+9.324=00.7756VA−2.094=0VA=2.7 V$

Therefore, the voltage at node $A, VA$ is $2.7 V$

Calculate the current flowing through diode, $D1$ .

$ID1=VA−0.72k$

Substitute $2.7V$ for $VA$ .

$ID1=2.7−0.72k=22k=1 mA$

Therefore, the current flowing through diode $D1,$ is $ID1=1 mA$ .

Calculate the current flowing through diode, $D2$ .

$ID2=VA−0.7−(−5)R3$

Substitute $2.7V$ for $VA$ and $14 kΩ$ for $R3$ .

$ID2=2.7−0.7+514k=714k=0.5 mA$

Therefore, the current flowing through diode $D2,$ is $ID2=0.5 mA$ .

Calculate the current flowing through diode, $D3$ .

$ID3=VA−0.7−(−10)R4$

Substitute $2.7V$ for $VA$ and $24 kΩ$ for $R4$ .

$ID3=2.7−0.7+1024k=1224k=0.5 mA$

Therefore, the current flowing through diodes is $ID3=0.5 mA$

Conclusion:

Therefore, the current flowing through diodes is

$ID3=0.5 mA$

$ID2=0.5 mA$ , $ID1=1 mA$ and $VA=2.7 V$

(b)

Expert Solution

To determine

The values of $ID1,ID2,ID3, and VA$

Answer to Problem 2.52P

The current flowing through diodes is $VA=−0.991V$ , $ID1= 0 A$ , $ID2=1.003 mA$ and $ID3=1.6 mA$

Explanation of Solution

Given:

$R3=3.3kΩ R4=5.2kΩ$

Calculation:

Assume diode $D1$ is OFF and remaining diodes are ON.

Draw the circuit diagram with node voltages and cut-in voltages.

Microelectronics: Circuit Analysis and Design, Chapter 2, Problem 2.52P , additional homework tip 3

Apply Kirchhoff’s current law node A.

$VA−156.15k+VA−0.7−(−5)R3+VA−0.7−(−10)R4=0VA−156.15k+VA+4.3R3+VA+9.3R4=0$

Substitute $3.3 kΩ$ for $R3$ and $5.2 kΩ$ for $R4$ .

$VA−156.15k+VA+4.33.3k+VA+9.35.2k=0VA−156.15+VA+4.33.3+VA+9.35.2=0(16.15+13.3+15.2)VA−156.15+4.33.3+9.35.2=00.6579VA+0.6525=0VA=−0.65250.6579=−0.991V$

Therefore, the voltage at node $A, VA$ is $−0.991V$

The diode $D1$ is reverse biased. Therefore, the current flowing through diode $D1$ is zero.

That is, $ID1=0 A$ .

Therefore, the current flowing through diode $D1,$

$ID1= 0 A$

Calculate the current flowing through diode, $D2$ .

$ID2=VA−0.7−(−5)R3$

Substitute $−0.991V$ for $VA$ and $3.3 kΩ$ for $R3$ .

$ID2=−0.991−0.7+53.3k=3.3093.3k=1.003 mA$

Therefore, the current flowing through diode $D2,$ is $ID2=1.003 mA$ .

Calculate the current flowing through diode, $D3$ .

$ID3=VA−0.7−(−10)R4$

Substitute $−0.991V$ for $VA$ and $5.2 kΩ$ for $R4$ .

$ID3=−0.991−0.7+105.2k=8.3095.2k=1.6 mA$

Therefore, the current flowing through diode $D3,$ is $ID3=1.6 mA$ .

Conclusion:

Therefore, the current flowing through diodes is $VA=−0.991V$ , $ID1= 0 A$ , $ID2=1.003 mA$ and $ID3=1.6 mA$

(c)

Expert Solution

To determine

The values of $ID1,ID2,ID3, and VA$

Answer to Problem 2.52P

The current flowing through diodes is $VA=−5V$ , $ID1=0 A$ , $ID2=0 A$

$ID3=3.25 mA$

Explanation of Solution

Given:

$R3=3.3kΩR4=1.32kΩ$

Calculation:

Assume diode $D1$ and $D2$ are OFF and diode $D3$ is ON.

Draw the circuit diagram with node voltages and cut-in voltages.

Microelectronics: Circuit Analysis and Design, Chapter 2, Problem 2.52P , additional homework tip 4

Apply Kirchhoff’s current law node A.

$VA−156.15k+VA−0.7−(−10)R4=0VA−156.15k+VA+9.3R4=0$

Substitute $1.32 kΩ$ for $R4$ .

$VA−156.15k+VA+9.31.32k=0VA−156.15+VA+9.31.32=0(16.15+11.32)VA−156.15+9.31.32=00.92VA+4.606=0VA=−5V$

Therefore, the voltage at node $A, VA$ is $−5V$ .

The Diode $D1$ and $D2$ are reverse biased. Therefore, the current flowing through diode $D1$ is zero.

That is, $ID1=0 A$ .

Therefore, the current flowing through diode $D1,ID1 is 0 A.$

The current flowing through the diode, $D2$ is zero.

That is,

$ID2=0 A.$

Therefore, the current flowing through diode $D2,ID2 is 0 A.$

Calculate the current flowing through diode, $D3$ .

$ID3=VA−0.7−(−10)R4$

Substitute $−5V$ for $VA$ and $1.32 kΩ$ for $R4$ .

$ID3=−5−0.7+101.32k=4.31.32k=3.25 mA$

Therefore, the current flowing through diode $D3,$ is $ID3=3.25 mA$ .

Conclusion:

Therefore, the current flowing through diodes is $VA=−5V$ , $ID1=0 A$ , $ID2=0 A$

$ID3=3.25 mA$

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Answer 4

Textbook Question

Chapter 2, Problem 2.52P

The cut−in voltage of each diode in the circuit shown in Figure P2.52 is $V γ = 0.7 V$ . Determine $I D 1 , I D 2 , I D 3$ , and $V A$ for (a) $R 3 = 14 k Ω , R 4 = 24 k Ω$ ; (b) $R 3 = 3.3 k Ω , R 4 = 5.2 k Ω$ ; and (c) $R 3 = 3.3 k Ω , R 4 = 1.32 k Ω$ .

Chapter 2, Problem 2.52P, The cutin voltage of each diode in the circuit shown in Figure P2.52 is V=0.7V . Determine
Figure P2.52

(a)

Expert Solution

To determine

The values of $ID1,ID2,ID3, and VA$

Answer to Problem 2.52P

The current flowing through diodes is

$ID3=0.5 mA$

$ID2=0.5 mA$ , $ID1=1 mA$ and $VA=2.7 V$

Explanation of Solution

Given:

Microelectronics: Circuit Analysis and Design, Chapter 2, Problem 2.52P , additional homework tip 1

$R3=14kΩ R4=24kΩ$

Calculation:

The cut-in voltage for each diode is, $Vγ=0.7V$

Assume all diodes are ON.

Draw the circuit diagram with node voltages and cut-in voltages.

Microelectronics: Circuit Analysis and Design, Chapter 2, Problem 2.52P , additional homework tip 2

Apply Kirchhoff’s current law node A.

$VA−156.15k+VA−0.72k+VA−0.7−(−5)R3+VA−0.7−(−10)R4=0VA−156.15k+VA−0.72k+VA+4.3R3+VA+9.3R4=0$

Substitute $14 kΩ$ for $R3$ and $24 kΩ$ for $R4$ .

$VA−156.15k+VA−0.72k+VA+4.314k+VA+9.324k=0VA−156.15+VA−0.72+VA+4.314+VA+9.324=0(16.15+12+114+124)VA−156.15−0.72+4.314+9.324=00.7756VA−2.094=0VA=2.7 V$

Therefore, the voltage at node $A, VA$ is $2.7 V$

Calculate the current flowing through diode, $D1$ .

$ID1=VA−0.72k$

Substitute $2.7V$ for $VA$ .

$ID1=2.7−0.72k=22k=1 mA$

Therefore, the current flowing through diode $D1,$ is $ID1=1 mA$ .

Calculate the current flowing through diode, $D2$ .

$ID2=VA−0.7−(−5)R3$

Substitute $2.7V$ for $VA$ and $14 kΩ$ for $R3$ .

$ID2=2.7−0.7+514k=714k=0.5 mA$

Therefore, the current flowing through diode $D2,$ is $ID2=0.5 mA$ .

Calculate the current flowing through diode, $D3$ .

$ID3=VA−0.7−(−10)R4$

Substitute $2.7V$ for $VA$ and $24 kΩ$ for $R4$ .

$ID3=2.7−0.7+1024k=1224k=0.5 mA$

Therefore, the current flowing through diodes is $ID3=0.5 mA$

Conclusion:

Therefore, the current flowing through diodes is

$ID3=0.5 mA$

$ID2=0.5 mA$ , $ID1=1 mA$ and $VA=2.7 V$

(b)

Expert Solution

To determine

The values of $ID1,ID2,ID3, and VA$

Answer to Problem 2.52P

The current flowing through diodes is $VA=−0.991V$ , $ID1= 0 A$ , $ID2=1.003 mA$ and $ID3=1.6 mA$

Explanation of Solution

Given:

$R3=3.3kΩ R4=5.2kΩ$

Calculation:

Assume diode $D1$ is OFF and remaining diodes are ON.

Draw the circuit diagram with node voltages and cut-in voltages.

Microelectronics: Circuit Analysis and Design, Chapter 2, Problem 2.52P , additional homework tip 3

Apply Kirchhoff’s current law node A.

$VA−156.15k+VA−0.7−(−5)R3+VA−0.7−(−10)R4=0VA−156.15k+VA+4.3R3+VA+9.3R4=0$

Substitute $3.3 kΩ$ for $R3$ and $5.2 kΩ$ for $R4$ .

$VA−156.15k+VA+4.33.3k+VA+9.35.2k=0VA−156.15+VA+4.33.3+VA+9.35.2=0(16.15+13.3+15.2)VA−156.15+4.33.3+9.35.2=00.6579VA+0.6525=0VA=−0.65250.6579=−0.991V$

Therefore, the voltage at node $A, VA$ is $−0.991V$

The diode $D1$ is reverse biased. Therefore, the current flowing through diode $D1$ is zero.

That is, $ID1=0 A$ .

Therefore, the current flowing through diode $D1,$

$ID1= 0 A$

Calculate the current flowing through diode, $D2$ .

$ID2=VA−0.7−(−5)R3$

Substitute $−0.991V$ for $VA$ and $3.3 kΩ$ for $R3$ .

$ID2=−0.991−0.7+53.3k=3.3093.3k=1.003 mA$

Therefore, the current flowing through diode $D2,$ is $ID2=1.003 mA$ .

Calculate the current flowing through diode, $D3$ .

$ID3=VA−0.7−(−10)R4$

Substitute $−0.991V$ for $VA$ and $5.2 kΩ$ for $R4$ .

$ID3=−0.991−0.7+105.2k=8.3095.2k=1.6 mA$

Therefore, the current flowing through diode $D3,$ is $ID3=1.6 mA$ .

Conclusion:

Therefore, the current flowing through diodes is $VA=−0.991V$ , $ID1= 0 A$ , $ID2=1.003 mA$ and $ID3=1.6 mA$

(c)

Expert Solution

To determine

The values of $ID1,ID2,ID3, and VA$

Answer to Problem 2.52P

The current flowing through diodes is $VA=−5V$ , $ID1=0 A$ , $ID2=0 A$

$ID3=3.25 mA$

Explanation of Solution

Given:

$R3=3.3kΩR4=1.32kΩ$

Calculation:

Assume diode $D1$ and $D2$ are OFF and diode $D3$ is ON.

Draw the circuit diagram with node voltages and cut-in voltages.

Microelectronics: Circuit Analysis and Design, Chapter 2, Problem 2.52P , additional homework tip 4

Apply Kirchhoff’s current law node A.

$VA−156.15k+VA−0.7−(−10)R4=0VA−156.15k+VA+9.3R4=0$

Substitute $1.32 kΩ$ for $R4$ .

$VA−156.15k+VA+9.31.32k=0VA−156.15+VA+9.31.32=0(16.15+11.32)VA−156.15+9.31.32=00.92VA+4.606=0VA=−5V$

Therefore, the voltage at node $A, VA$ is $−5V$ .

The Diode $D1$ and $D2$ are reverse biased. Therefore, the current flowing through diode $D1$ is zero.

That is, $ID1=0 A$ .

Therefore, the current flowing through diode $D1,ID1 is 0 A.$

The current flowing through the diode, $D2$ is zero.

That is,

$ID2=0 A.$

Therefore, the current flowing through diode $D2,ID2 is 0 A.$

Calculate the current flowing through diode, $D3$ .

$ID3=VA−0.7−(−10)R4$

Substitute $−5V$ for $VA$ and $1.32 kΩ$ for $R4$ .

$ID3=−5−0.7+101.32k=4.31.32k=3.25 mA$

Therefore, the current flowing through diode $D3,$ is $ID3=3.25 mA$ .

Conclusion:

Therefore, the current flowing through diodes is $VA=−5V$ , $ID1=0 A$ , $ID2=0 A$

$ID3=3.25 mA$

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Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

(a)

Expert Solution

To determine

The values of $ID1,ID2,ID3, and VA$

Answer to Problem 2.52P

The current flowing through diodes is

$ID3=0.5 mA$

$ID2=0.5 mA$ , $ID1=1 mA$ and $VA=2.7 V$

Explanation of Solution

Given:

Microelectronics: Circuit Analysis and Design, Chapter 2, Problem 2.52P , additional homework tip 1

$R3=14kΩ R4=24kΩ$

Calculation:

The cut-in voltage for each diode is, $Vγ=0.7V$

Assume all diodes are ON.

Draw the circuit diagram with node voltages and cut-in voltages.

Microelectronics: Circuit Analysis and Design, Chapter 2, Problem 2.52P , additional homework tip 2

Apply Kirchhoff’s current law node A.

$VA−156.15k+VA−0.72k+VA−0.7−(−5)R3+VA−0.7−(−10)R4=0VA−156.15k+VA−0.72k+VA+4.3R3+VA+9.3R4=0$

Substitute $14 kΩ$ for $R3$ and $24 kΩ$ for $R4$ .

$VA−156.15k+VA−0.72k+VA+4.314k+VA+9.324k=0VA−156.15+VA−0.72+VA+4.314+VA+9.324=0(16.15+12+114+124)VA−156.15−0.72+4.314+9.324=00.7756VA−2.094=0VA=2.7 V$

Therefore, the voltage at node $A, VA$ is $2.7 V$

Calculate the current flowing through diode, $D1$ .

$ID1=VA−0.72k$

Substitute $2.7V$ for $VA$ .

$ID1=2.7−0.72k=22k=1 mA$

Therefore, the current flowing through diode $D1,$ is $ID1=1 mA$ .

Calculate the current flowing through diode, $D2$ .

$ID2=VA−0.7−(−5)R3$

Substitute $2.7V$ for $VA$ and $14 kΩ$ for $R3$ .

$ID2=2.7−0.7+514k=714k=0.5 mA$

Therefore, the current flowing through diode $D2,$ is $ID2=0.5 mA$ .

Calculate the current flowing through diode, $D3$ .

$ID3=VA−0.7−(−10)R4$

Substitute $2.7V$ for $VA$ and $24 kΩ$ for $R4$ .

$ID3=2.7−0.7+1024k=1224k=0.5 mA$

Therefore, the current flowing through diodes is $ID3=0.5 mA$

Conclusion:

Therefore, the current flowing through diodes is

$ID3=0.5 mA$

$ID2=0.5 mA$ , $ID1=1 mA$ and $VA=2.7 V$

(b)

Expert Solution

To determine

The values of $ID1,ID2,ID3, and VA$

Answer to Problem 2.52P

The current flowing through diodes is $VA=−0.991V$ , $ID1= 0 A$ , $ID2=1.003 mA$ and $ID3=1.6 mA$

Explanation of Solution

Given:

$R3=3.3kΩ R4=5.2kΩ$

Calculation:

Assume diode $D1$ is OFF and remaining diodes are ON.

Draw the circuit diagram with node voltages and cut-in voltages.

Microelectronics: Circuit Analysis and Design, Chapter 2, Problem 2.52P , additional homework tip 3

Apply Kirchhoff’s current law node A.

$VA−156.15k+VA−0.7−(−5)R3+VA−0.7−(−10)R4=0VA−156.15k+VA+4.3R3+VA+9.3R4=0$

Substitute $3.3 kΩ$ for $R3$ and $5.2 kΩ$ for $R4$ .

$VA−156.15k+VA+4.33.3k+VA+9.35.2k=0VA−156.15+VA+4.33.3+VA+9.35.2=0(16.15+13.3+15.2)VA−156.15+4.33.3+9.35.2=00.6579VA+0.6525=0VA=−0.65250.6579=−0.991V$

Therefore, the voltage at node $A, VA$ is $−0.991V$

The diode $D1$ is reverse biased. Therefore, the current flowing through diode $D1$ is zero.

That is, $ID1=0 A$ .

Therefore, the current flowing through diode $D1,$

$ID1= 0 A$

Calculate the current flowing through diode, $D2$ .

$ID2=VA−0.7−(−5)R3$

Substitute $−0.991V$ for $VA$ and $3.3 kΩ$ for $R3$ .

$ID2=−0.991−0.7+53.3k=3.3093.3k=1.003 mA$

Therefore, the current flowing through diode $D2,$ is $ID2=1.003 mA$ .

Calculate the current flowing through diode, $D3$ .

$ID3=VA−0.7−(−10)R4$

Substitute $−0.991V$ for $VA$ and $5.2 kΩ$ for $R4$ .

$ID3=−0.991−0.7+105.2k=8.3095.2k=1.6 mA$

Therefore, the current flowing through diode $D3,$ is $ID3=1.6 mA$ .

Conclusion:

Therefore, the current flowing through diodes is $VA=−0.991V$ , $ID1= 0 A$ , $ID2=1.003 mA$ and $ID3=1.6 mA$

(c)

Expert Solution

To determine

The values of $ID1,ID2,ID3, and VA$

Answer to Problem 2.52P

The current flowing through diodes is $VA=−5V$ , $ID1=0 A$ , $ID2=0 A$

$ID3=3.25 mA$

Explanation of Solution

Given:

$R3=3.3kΩR4=1.32kΩ$

Calculation:

Assume diode $D1$ and $D2$ are OFF and diode $D3$ is ON.

Draw the circuit diagram with node voltages and cut-in voltages.

Microelectronics: Circuit Analysis and Design, Chapter 2, Problem 2.52P , additional homework tip 4

Apply Kirchhoff’s current law node A.

$VA−156.15k+VA−0.7−(−10)R4=0VA−156.15k+VA+9.3R4=0$

Substitute $1.32 kΩ$ for $R4$ .

$VA−156.15k+VA+9.31.32k=0VA−156.15+VA+9.31.32=0(16.15+11.32)VA−156.15+9.31.32=00.92VA+4.606=0VA=−5V$

Therefore, the voltage at node $A, VA$ is $−5V$ .

The Diode $D1$ and $D2$ are reverse biased. Therefore, the current flowing through diode $D1$ is zero.

That is, $ID1=0 A$ .

Therefore, the current flowing through diode $D1,ID1 is 0 A.$

The current flowing through the diode, $D2$ is zero.

That is,

$ID2=0 A.$

Therefore, the current flowing through diode $D2,ID2 is 0 A.$

Calculate the current flowing through diode, $D3$ .

$ID3=VA−0.7−(−10)R4$

Substitute $−5V$ for $VA$ and $1.32 kΩ$ for $R4$ .

$ID3=−5−0.7+101.32k=4.31.32k=3.25 mA$

Therefore, the current flowing through diode $D3,$ is $ID3=3.25 mA$ .

Conclusion:

Therefore, the current flowing through diodes is $VA=−5V$ , $ID1=0 A$ , $ID2=0 A$

$ID3=3.25 mA$

Answer 8

(a)

Expert Solution

To determine

The values of $ID1,ID2,ID3, and VA$

Answer to Problem 2.52P

The current flowing through diodes is

$ID3=0.5 mA$

$ID2=0.5 mA$ , $ID1=1 mA$ and $VA=2.7 V$

Explanation of Solution

Given:

Microelectronics: Circuit Analysis and Design, Chapter 2, Problem 2.52P , additional homework tip 1

$R3=14kΩ R4=24kΩ$

Calculation:

The cut-in voltage for each diode is, $Vγ=0.7V$

Assume all diodes are ON.

Draw the circuit diagram with node voltages and cut-in voltages.

Microelectronics: Circuit Analysis and Design, Chapter 2, Problem 2.52P , additional homework tip 2

Apply Kirchhoff’s current law node A.

$VA−156.15k+VA−0.72k+VA−0.7−(−5)R3+VA−0.7−(−10)R4=0VA−156.15k+VA−0.72k+VA+4.3R3+VA+9.3R4=0$

Substitute $14 kΩ$ for $R3$ and $24 kΩ$ for $R4$ .

$VA−156.15k+VA−0.72k+VA+4.314k+VA+9.324k=0VA−156.15+VA−0.72+VA+4.314+VA+9.324=0(16.15+12+114+124)VA−156.15−0.72+4.314+9.324=00.7756VA−2.094=0VA=2.7 V$

Therefore, the voltage at node $A, VA$ is $2.7 V$

Calculate the current flowing through diode, $D1$ .

$ID1=VA−0.72k$

Substitute $2.7V$ for $VA$ .

$ID1=2.7−0.72k=22k=1 mA$

Therefore, the current flowing through diode $D1,$ is $ID1=1 mA$ .

Calculate the current flowing through diode, $D2$ .

$ID2=VA−0.7−(−5)R3$

Substitute $2.7V$ for $VA$ and $14 kΩ$ for $R3$ .

$ID2=2.7−0.7+514k=714k=0.5 mA$

Therefore, the current flowing through diode $D2,$ is $ID2=0.5 mA$ .

Calculate the current flowing through diode, $D3$ .

$ID3=VA−0.7−(−10)R4$

Substitute $2.7V$ for $VA$ and $24 kΩ$ for $R4$ .

$ID3=2.7−0.7+1024k=1224k=0.5 mA$

Therefore, the current flowing through diodes is $ID3=0.5 mA$

Conclusion:

Therefore, the current flowing through diodes is

$ID3=0.5 mA$

$ID2=0.5 mA$ , $ID1=1 mA$ and $VA=2.7 V$

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

To determine

The values of $ID1,ID2,ID3, and VA$

Answer 13

Answer to Problem 2.52P

The current flowing through diodes is

$ID3=0.5 mA$

$ID2=0.5 mA$ , $ID1=1 mA$ and $VA=2.7 V$

Answer 14

Explanation of Solution

Given:

Microelectronics: Circuit Analysis and Design, Chapter 2, Problem 2.52P , additional homework tip 1

$R3=14kΩ R4=24kΩ$

Calculation:

The cut-in voltage for each diode is, $Vγ=0.7V$

Assume all diodes are ON.

Draw the circuit diagram with node voltages and cut-in voltages.

Microelectronics: Circuit Analysis and Design, Chapter 2, Problem 2.52P , additional homework tip 2

Apply Kirchhoff’s current law node A.

$VA−156.15k+VA−0.72k+VA−0.7−(−5)R3+VA−0.7−(−10)R4=0VA−156.15k+VA−0.72k+VA+4.3R3+VA+9.3R4=0$

Substitute $14 kΩ$ for $R3$ and $24 kΩ$ for $R4$ .

$VA−156.15k+VA−0.72k+VA+4.314k+VA+9.324k=0VA−156.15+VA−0.72+VA+4.314+VA+9.324=0(16.15+12+114+124)VA−156.15−0.72+4.314+9.324=00.7756VA−2.094=0VA=2.7 V$

Therefore, the voltage at node $A, VA$ is $2.7 V$

Calculate the current flowing through diode, $D1$ .

$ID1=VA−0.72k$

Substitute $2.7V$ for $VA$ .

$ID1=2.7−0.72k=22k=1 mA$

Therefore, the current flowing through diode $D1,$ is $ID1=1 mA$ .

Calculate the current flowing through diode, $D2$ .

$ID2=VA−0.7−(−5)R3$

Substitute $2.7V$ for $VA$ and $14 kΩ$ for $R3$ .

$ID2=2.7−0.7+514k=714k=0.5 mA$

Therefore, the current flowing through diode $D2,$ is $ID2=0.5 mA$ .

Calculate the current flowing through diode, $D3$ .

$ID3=VA−0.7−(−10)R4$

Substitute $2.7V$ for $VA$ and $24 kΩ$ for $R4$ .

$ID3=2.7−0.7+1024k=1224k=0.5 mA$

Therefore, the current flowing through diodes is $ID3=0.5 mA$

Conclusion:

Therefore, the current flowing through diodes is

$ID3=0.5 mA$

$ID2=0.5 mA$ , $ID1=1 mA$ and $VA=2.7 V$

Answer 15

(b)

Expert Solution

To determine

The values of $ID1,ID2,ID3, and VA$

Answer to Problem 2.52P

The current flowing through diodes is $VA=−0.991V$ , $ID1= 0 A$ , $ID2=1.003 mA$ and $ID3=1.6 mA$

Explanation of Solution

Given:

$R3=3.3kΩ R4=5.2kΩ$

Calculation:

Assume diode $D1$ is OFF and remaining diodes are ON.

Draw the circuit diagram with node voltages and cut-in voltages.

Microelectronics: Circuit Analysis and Design, Chapter 2, Problem 2.52P , additional homework tip 3

Apply Kirchhoff’s current law node A.

$VA−156.15k+VA−0.7−(−5)R3+VA−0.7−(−10)R4=0VA−156.15k+VA+4.3R3+VA+9.3R4=0$

Substitute $3.3 kΩ$ for $R3$ and $5.2 kΩ$ for $R4$ .

$VA−156.15k+VA+4.33.3k+VA+9.35.2k=0VA−156.15+VA+4.33.3+VA+9.35.2=0(16.15+13.3+15.2)VA−156.15+4.33.3+9.35.2=00.6579VA+0.6525=0VA=−0.65250.6579=−0.991V$

Therefore, the voltage at node $A, VA$ is $−0.991V$

The diode $D1$ is reverse biased. Therefore, the current flowing through diode $D1$ is zero.

That is, $ID1=0 A$ .

Therefore, the current flowing through diode $D1,$

$ID1= 0 A$

Calculate the current flowing through diode, $D2$ .

$ID2=VA−0.7−(−5)R3$

Substitute $−0.991V$ for $VA$ and $3.3 kΩ$ for $R3$ .

$ID2=−0.991−0.7+53.3k=3.3093.3k=1.003 mA$

Therefore, the current flowing through diode $D2,$ is $ID2=1.003 mA$ .

Calculate the current flowing through diode, $D3$ .

$ID3=VA−0.7−(−10)R4$

Substitute $−0.991V$ for $VA$ and $5.2 kΩ$ for $R4$ .

$ID3=−0.991−0.7+105.2k=8.3095.2k=1.6 mA$

Therefore, the current flowing through diode $D3,$ is $ID3=1.6 mA$ .

Conclusion:

Therefore, the current flowing through diodes is $VA=−0.991V$ , $ID1= 0 A$ , $ID2=1.003 mA$ and $ID3=1.6 mA$

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

To determine

The values of $ID1,ID2,ID3, and VA$

Answer 20

Answer to Problem 2.52P

The current flowing through diodes is $VA=−0.991V$ , $ID1= 0 A$ , $ID2=1.003 mA$ and $ID3=1.6 mA$

Answer 21

Explanation of Solution

Given:

$R3=3.3kΩ R4=5.2kΩ$

Calculation:

Assume diode $D1$ is OFF and remaining diodes are ON.

Draw the circuit diagram with node voltages and cut-in voltages.

Microelectronics: Circuit Analysis and Design, Chapter 2, Problem 2.52P , additional homework tip 3

Apply Kirchhoff’s current law node A.

$VA−156.15k+VA−0.7−(−5)R3+VA−0.7−(−10)R4=0VA−156.15k+VA+4.3R3+VA+9.3R4=0$

Substitute $3.3 kΩ$ for $R3$ and $5.2 kΩ$ for $R4$ .

$VA−156.15k+VA+4.33.3k+VA+9.35.2k=0VA−156.15+VA+4.33.3+VA+9.35.2=0(16.15+13.3+15.2)VA−156.15+4.33.3+9.35.2=00.6579VA+0.6525=0VA=−0.65250.6579=−0.991V$

Therefore, the voltage at node $A, VA$ is $−0.991V$

The diode $D1$ is reverse biased. Therefore, the current flowing through diode $D1$ is zero.

That is, $ID1=0 A$ .

Therefore, the current flowing through diode $D1,$

$ID1= 0 A$

Calculate the current flowing through diode, $D2$ .

$ID2=VA−0.7−(−5)R3$

Substitute $−0.991V$ for $VA$ and $3.3 kΩ$ for $R3$ .

$ID2=−0.991−0.7+53.3k=3.3093.3k=1.003 mA$

Therefore, the current flowing through diode $D2,$ is $ID2=1.003 mA$ .

Calculate the current flowing through diode, $D3$ .

$ID3=VA−0.7−(−10)R4$

Substitute $−0.991V$ for $VA$ and $5.2 kΩ$ for $R4$ .

$ID3=−0.991−0.7+105.2k=8.3095.2k=1.6 mA$

Therefore, the current flowing through diode $D3,$ is $ID3=1.6 mA$ .

Conclusion:

Therefore, the current flowing through diodes is $VA=−0.991V$ , $ID1= 0 A$ , $ID2=1.003 mA$ and $ID3=1.6 mA$

Answer 22

(c)

Expert Solution

To determine

The values of $ID1,ID2,ID3, and VA$

Answer to Problem 2.52P

The current flowing through diodes is $VA=−5V$ , $ID1=0 A$ , $ID2=0 A$

$ID3=3.25 mA$

Explanation of Solution

Given:

$R3=3.3kΩR4=1.32kΩ$

Calculation:

Assume diode $D1$ and $D2$ are OFF and diode $D3$ is ON.

Draw the circuit diagram with node voltages and cut-in voltages.

Microelectronics: Circuit Analysis and Design, Chapter 2, Problem 2.52P , additional homework tip 4

Apply Kirchhoff’s current law node A.

$VA−156.15k+VA−0.7−(−10)R4=0VA−156.15k+VA+9.3R4=0$

Substitute $1.32 kΩ$ for $R4$ .

$VA−156.15k+VA+9.31.32k=0VA−156.15+VA+9.31.32=0(16.15+11.32)VA−156.15+9.31.32=00.92VA+4.606=0VA=−5V$

Therefore, the voltage at node $A, VA$ is $−5V$ .

The Diode $D1$ and $D2$ are reverse biased. Therefore, the current flowing through diode $D1$ is zero.

That is, $ID1=0 A$ .

Therefore, the current flowing through diode $D1,ID1 is 0 A.$

The current flowing through the diode, $D2$ is zero.

That is,

$ID2=0 A.$

Therefore, the current flowing through diode $D2,ID2 is 0 A.$

Calculate the current flowing through diode, $D3$ .

$ID3=VA−0.7−(−10)R4$

Substitute $−5V$ for $VA$ and $1.32 kΩ$ for $R4$ .

$ID3=−5−0.7+101.32k=4.31.32k=3.25 mA$

Therefore, the current flowing through diode $D3,$ is $ID3=3.25 mA$ .

Conclusion:

Therefore, the current flowing through diodes is $VA=−5V$ , $ID1=0 A$ , $ID2=0 A$

$ID3=3.25 mA$

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

To determine

The values of $ID1,ID2,ID3, and VA$

Answer 27

Answer to Problem 2.52P

The current flowing through diodes is $VA=−5V$ , $ID1=0 A$ , $ID2=0 A$

$ID3=3.25 mA$

Answer 28

Explanation of Solution

Given:

$R3=3.3kΩR4=1.32kΩ$

Calculation:

Assume diode $D1$ and $D2$ are OFF and diode $D3$ is ON.

Draw the circuit diagram with node voltages and cut-in voltages.

Microelectronics: Circuit Analysis and Design, Chapter 2, Problem 2.52P , additional homework tip 4

Apply Kirchhoff’s current law node A.

$VA−156.15k+VA−0.7−(−10)R4=0VA−156.15k+VA+9.3R4=0$

Substitute $1.32 kΩ$ for $R4$ .

$VA−156.15k+VA+9.31.32k=0VA−156.15+VA+9.31.32=0(16.15+11.32)VA−156.15+9.31.32=00.92VA+4.606=0VA=−5V$

Therefore, the voltage at node $A, VA$ is $−5V$ .

The Diode $D1$ and $D2$ are reverse biased. Therefore, the current flowing through diode $D1$ is zero.

That is, $ID1=0 A$ .

Therefore, the current flowing through diode $D1,ID1 is 0 A.$

The current flowing through the diode, $D2$ is zero.

That is,

$ID2=0 A.$

Therefore, the current flowing through diode $D2,ID2 is 0 A.$

Calculate the current flowing through diode, $D3$ .

$ID3=VA−0.7−(−10)R4$

Substitute $−5V$ for $VA$ and $1.32 kΩ$ for $R4$ .

$ID3=−5−0.7+101.32k=4.31.32k=3.25 mA$

Therefore, the current flowing through diode $D3,$ is $ID3=3.25 mA$ .

Conclusion:

Therefore, the current flowing through diodes is $VA=−5V$ , $ID1=0 A$ , $ID2=0 A$

$ID3=3.25 mA$

Answer 29

Ch. 2 - Repeat Example 2.1 if the input voltage is...Ch. 2 - Consider the bridge circuit shown in Figure 2.6(a)...Ch. 2 - Assume the input signal to a rectifier circuit has...Ch. 2 - The input voltage to the halfwave rectifier in...Ch. 2 - Consider the circuit in Figure 2.4. The input...Ch. 2 - The circuit in Figure 2.5(a) is used to rectify a...Ch. 2 - The secondary transformer voltage of the rectifier...Ch. 2 - Determine the fraction (percent) of the cycle that...Ch. 2 - The Zener diode regulator circuit shown in Figure...Ch. 2 - Repeat Example 2.6 for rz=4 . Assume all other...

The cut−in voltage of each diode in the circuit shown in Figure P2.52 is V γ = 0.7 V . Determine I D 1 , I D 2 , I D 3 , and V A for (a) R 3 = 14 k Ω , R 4 = 24 k Ω ; (b) R 3 = 3.3 k Ω , R 4 = 5.2 k Ω ; and (c) R 3 = 3.3 k Ω , R 4 = 1.32 k Ω . Figure P2.52

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Answer to Problem 2.52P

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Answer to Problem 2.52P

Explanation of Solution

Answer to Problem 2.52P

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