The circuit parameters for the two-transistor current source shown inFigure 10.2(b) are V + = 3 V , V − = − 3 V , and R 1 = 47 k Ω . The transistor parameters are β = 120 , V B E ( o n ) = 0.7 V , and V A = ∞ . Determine I R E F , I O , and I B 1 .(Ans. I R E F = 0.1128 m A , I O = 0.1109 m A , I B 1 = 0.9243 μ A )

Question

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Answer 1

Textbook Question

Chapter 10, Problem 10.1EP

The circuit parameters for the two-transistor current source shown inFigure 10.2(b) are $V + = 3 V$ , $V − = − 3 V$ , and $R 1 = 47 k Ω$ . The transistor parameters are $β = 120 , V B E ( o n ) = 0.7 V$ , and $V A = ∞$ . Determine $I R E F , I O$ , and $I B 1$ .(Ans. $I R E F = 0.1128 m A , I O = 0.1109 m A , I B 1 = 0.9243 μ A$ )

Expert Solution & Answer

To determine

The value of $IREF$ , $I0$ and $IB1$ .

Answer to Problem 10.1EP

The values are:

$IREF= 0.1128 mA$

$I0 =0.1109 mA$

$IB1= 0.9242 μA$

Explanation of Solution

Given:

The given circuit is shown below.

Microelectronics: Circuit Analysis and Design, Chapter 10, Problem 10.1EP

$V−=−3VV+= 3V$

$R1=47 kΩ$

$β=120$

$VBE(on)= 0.7 V$

Calculation:

$VA=∞$ (Ideal case)

Apply KVL in input Loop

$V+− IREFR1−VBE(on)−V−=0 (1)$

From equation,

$IREF=V+−V BE(on)−V−R1IREF=3−0.7−(−3)47×1000= 0.1128 mA$

From figure,

$I0=Ic= Iref1+2βI0= 0.1128 × 10 −31+2 120= 0.1109 mA$

Also, both transistors are identical ( $IB1=IB2$ ). So,

$2Icβ= IB1+IB2 2Icβ=2IB1$

So,

$IB1= Icβ$

$IB1= 0.1109× 10 −3120= 0.9245 μA≈0.923μA$

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Answer 2

Textbook Question

Chapter 10, Problem 10.1EP

The circuit parameters for the two-transistor current source shown inFigure 10.2(b) are $V + = 3 V$ , $V − = − 3 V$ , and $R 1 = 47 k Ω$ . The transistor parameters are $β = 120 , V B E ( o n ) = 0.7 V$ , and $V A = ∞$ . Determine $I R E F , I O$ , and $I B 1$ .(Ans. $I R E F = 0.1128 m A , I O = 0.1109 m A , I B 1 = 0.9243 μ A$ )

Expert Solution & Answer

To determine

The value of $IREF$ , $I0$ and $IB1$ .

Answer to Problem 10.1EP

The values are:

$IREF= 0.1128 mA$

$I0 =0.1109 mA$

$IB1= 0.9242 μA$

Explanation of Solution

Given:

The given circuit is shown below.

Microelectronics: Circuit Analysis and Design, Chapter 10, Problem 10.1EP

$V−=−3VV+= 3V$

$R1=47 kΩ$

$β=120$

$VBE(on)= 0.7 V$

Calculation:

$VA=∞$ (Ideal case)

Apply KVL in input Loop

$V+− IREFR1−VBE(on)−V−=0 (1)$

From equation,

$IREF=V+−V BE(on)−V−R1IREF=3−0.7−(−3)47×1000= 0.1128 mA$

From figure,

$I0=Ic= Iref1+2βI0= 0.1128 × 10 −31+2 120= 0.1109 mA$

Also, both transistors are identical ( $IB1=IB2$ ). So,

$2Icβ= IB1+IB2 2Icβ=2IB1$

So,

$IB1= Icβ$

$IB1= 0.1109× 10 −3120= 0.9245 μA≈0.923μA$

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Answer 3

Textbook Question

Chapter 10, Problem 10.1EP

The circuit parameters for the two-transistor current source shown inFigure 10.2(b) are $V + = 3 V$ , $V − = − 3 V$ , and $R 1 = 47 k Ω$ . The transistor parameters are $β = 120 , V B E ( o n ) = 0.7 V$ , and $V A = ∞$ . Determine $I R E F , I O$ , and $I B 1$ .(Ans. $I R E F = 0.1128 m A , I O = 0.1109 m A , I B 1 = 0.9243 μ A$ )

Expert Solution & Answer

To determine

The value of $IREF$ , $I0$ and $IB1$ .

Answer to Problem 10.1EP

The values are:

$IREF= 0.1128 mA$

$I0 =0.1109 mA$

$IB1= 0.9242 μA$

Explanation of Solution

Given:

The given circuit is shown below.

Microelectronics: Circuit Analysis and Design, Chapter 10, Problem 10.1EP

$V−=−3VV+= 3V$

$R1=47 kΩ$

$β=120$

$VBE(on)= 0.7 V$

Calculation:

$VA=∞$ (Ideal case)

Apply KVL in input Loop

$V+− IREFR1−VBE(on)−V−=0 (1)$

From equation,

$IREF=V+−V BE(on)−V−R1IREF=3−0.7−(−3)47×1000= 0.1128 mA$

From figure,

$I0=Ic= Iref1+2βI0= 0.1128 × 10 −31+2 120= 0.1109 mA$

Also, both transistors are identical ( $IB1=IB2$ ). So,

$2Icβ= IB1+IB2 2Icβ=2IB1$

So,

$IB1= Icβ$

$IB1= 0.1109× 10 −3120= 0.9245 μA≈0.923μA$

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Answer 4

Textbook Question

Chapter 10, Problem 10.1EP

The circuit parameters for the two-transistor current source shown inFigure 10.2(b) are $V + = 3 V$ , $V − = − 3 V$ , and $R 1 = 47 k Ω$ . The transistor parameters are $β = 120 , V B E ( o n ) = 0.7 V$ , and $V A = ∞$ . Determine $I R E F , I O$ , and $I B 1$ .(Ans. $I R E F = 0.1128 m A , I O = 0.1109 m A , I B 1 = 0.9243 μ A$ )

Expert Solution & Answer

To determine

The value of $IREF$ , $I0$ and $IB1$ .

Answer to Problem 10.1EP

The values are:

$IREF= 0.1128 mA$

$I0 =0.1109 mA$

$IB1= 0.9242 μA$

Explanation of Solution

Given:

The given circuit is shown below.

Microelectronics: Circuit Analysis and Design, Chapter 10, Problem 10.1EP

$V−=−3VV+= 3V$

$R1=47 kΩ$

$β=120$

$VBE(on)= 0.7 V$

Calculation:

$VA=∞$ (Ideal case)

Apply KVL in input Loop

$V+− IREFR1−VBE(on)−V−=0 (1)$

From equation,

$IREF=V+−V BE(on)−V−R1IREF=3−0.7−(−3)47×1000= 0.1128 mA$

From figure,

$I0=Ic= Iref1+2βI0= 0.1128 × 10 −31+2 120= 0.1109 mA$

Also, both transistors are identical ( $IB1=IB2$ ). So,

$2Icβ= IB1+IB2 2Icβ=2IB1$

So,

$IB1= Icβ$

$IB1= 0.1109× 10 −3120= 0.9245 μA≈0.923μA$

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Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

Expert Solution & Answer

To determine

The value of $IREF$ , $I0$ and $IB1$ .

Answer to Problem 10.1EP

The values are:

$IREF= 0.1128 mA$

$I0 =0.1109 mA$

$IB1= 0.9242 μA$

Explanation of Solution

Given:

The given circuit is shown below.

Microelectronics: Circuit Analysis and Design, Chapter 10, Problem 10.1EP

$V−=−3VV+= 3V$

$R1=47 kΩ$

$β=120$

$VBE(on)= 0.7 V$

Calculation:

$VA=∞$ (Ideal case)

Apply KVL in input Loop

$V+− IREFR1−VBE(on)−V−=0 (1)$

From equation,

$IREF=V+−V BE(on)−V−R1IREF=3−0.7−(−3)47×1000= 0.1128 mA$

From figure,

$I0=Ic= Iref1+2βI0= 0.1128 × 10 −31+2 120= 0.1109 mA$

Also, both transistors are identical ( $IB1=IB2$ ). So,

$2Icβ= IB1+IB2 2Icβ=2IB1$

So,

$IB1= Icβ$

$IB1= 0.1109× 10 −3120= 0.9245 μA≈0.923μA$

Answer 8

Expert Solution & Answer

To determine

The value of $IREF$ , $I0$ and $IB1$ .

Answer to Problem 10.1EP

The values are:

$IREF= 0.1128 mA$

$I0 =0.1109 mA$

$IB1= 0.9242 μA$

Explanation of Solution

Given:

The given circuit is shown below.

Microelectronics: Circuit Analysis and Design, Chapter 10, Problem 10.1EP

$V−=−3VV+= 3V$

$R1=47 kΩ$

$β=120$

$VBE(on)= 0.7 V$

Calculation:

$VA=∞$ (Ideal case)

Apply KVL in input Loop

$V+− IREFR1−VBE(on)−V−=0 (1)$

From equation,

$IREF=V+−V BE(on)−V−R1IREF=3−0.7−(−3)47×1000= 0.1128 mA$

From figure,

$I0=Ic= Iref1+2βI0= 0.1128 × 10 −31+2 120= 0.1109 mA$

Also, both transistors are identical ( $IB1=IB2$ ). So,

$2Icβ= IB1+IB2 2Icβ=2IB1$

So,

$IB1= Icβ$

$IB1= 0.1109× 10 −3120= 0.9245 μA≈0.923μA$

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

To determine

The value of $IREF$ , $I0$ and $IB1$ .

Answer 12

Answer to Problem 10.1EP

The values are:

$IREF= 0.1128 mA$

$I0 =0.1109 mA$

$IB1= 0.9242 μA$

Answer 13

Explanation of Solution

Given:

The given circuit is shown below.

Microelectronics: Circuit Analysis and Design, Chapter 10, Problem 10.1EP

$V−=−3VV+= 3V$

$R1=47 kΩ$

$β=120$

$VBE(on)= 0.7 V$

Calculation:

$VA=∞$ (Ideal case)

Apply KVL in input Loop

$V+− IREFR1−VBE(on)−V−=0 (1)$

From equation,

$IREF=V+−V BE(on)−V−R1IREF=3−0.7−(−3)47×1000= 0.1128 mA$

From figure,

$I0=Ic= Iref1+2βI0= 0.1128 × 10 −31+2 120= 0.1109 mA$

Also, both transistors are identical ( $IB1=IB2$ ). So,

$2Icβ= IB1+IB2 2Icβ=2IB1$

So,

$IB1= Icβ$

$IB1= 0.1109× 10 −3120= 0.9245 μA≈0.923μA$

Answer 14

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