Range of output voltage for the given specifications.

Question

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Answer 1

Question

Chapter 14, Problem 14.51P

a.

To determine

Range of output voltage for the given specifications.

a.

Expert Solution

Answer to Problem 14.51P

The range of output voltage is,

4 mV≤vO≤76 mV

Explanation of Solution

Given:

The given circuit is shown below.

Microelectronics: Circuit Analysis and Design, Chapter 14, Problem 14.51P , additional homework tip 1

Input offset voltage is VOS=3 mV

Average input bias current is IB=0.4 μA

Offset bias current is IOS=0.06 μA

vI=0 and R=0

Calculation:

As,

IB=I B1+I B22IB1+IB2=2IBIB1+IB2=2×0.4×10−6IB1+IB2=0.8×10−6 ...... (1)

And

IOS=|IB1−IB2|IB1−IB2=±IOSIB1−IB2=±0.06×10−6 ..... (2)

Adding (1) and (2)

IB1+IB2+IB1−IB2=0.8×10−6±0.06×10−62IB1=0.8×10−6±0.06×10−6IB1=0.8× 10 −6±0.06× 10 −62=0.8× 10 −6+0.06× 10 −62 or 0.8× 10 −6−0.06× 10 −62⇒ IB1=0.43 μA or 0.37 μA

From (1)

IB1=0.8×10−6−IB2

For IB1=0.43 μA

IB1=0.8×10−6−IB2IB1=0.8×10−6−0.43×10−6⇒ IB1=0.37 μA

For IB1=0.37 μA

IB1=0.8×10−6−IB2IB1=0.8×10−6−0.37×10−6⇒ IB1=0.43 μA∴ IB1=0.37 μA or IB1=0.43 μA

Now, the modified circuit with offset voltage and bias current is shown below.

Microelectronics: Circuit Analysis and Design, Chapter 14, Problem 14.51P , additional homework tip 2

KCL at non-inverting terminal,

VY−( v I + V OS )R+IB2=0VY−( v I + V OS )R=−IB2VY−(vI+V OS)=−IB2RVY=vI+VOS−IB2R

Now, for ideal op-amp VX=VY

VX=vI+VOS−IB2R ..... (3)

Now, KCL at inverting terminal,

VX10 k+VX−vO100 k+IB1=010VX+VX−vO100 k+IB1=010VX+VX−vO=−IB1×100 k11VX−vO=−IB1×100 kvO=11VX+(I B1×100 k)

Putting VX from (3)

vO=11(vI+VOS−IB2R)+(IB1×100 k) ..... (4)

Now,

For vI=0 and R=0

vO=11(0+V OS−I B2(0))+(I B1×100 k) vO=11VOS+(I B1×100 k)

For IB1=0.43 μA and VOS=3 mV

vO(max)=11(3 m)+(0.43 μ×100 k) vO(max)=33×10−3+43×10−3⇒ vO(max)=76 mV

For IB1=0.37 μA and VOS=-3 mV

vO(min)=11(−3 m)+(0.37 μ×100 k) vO(min)=(−33× 10 −3)+(37× 10 −3)⇒ vO(min)=4 mV

0So, the range of output voltage is,

vO(min)≤vO≤vO(max)⇒ 4 mV≤vO≤76 mV

b.

To determine

Range of output voltage for the given specifications.

b.

Expert Solution

Answer to Problem 14.51P

The range of output voltage is,

−39 mV≤vO≤39 mV

Explanation of Solution

Given:

The given circuit is shown below.

Microelectronics: Circuit Analysis and Design, Chapter 14, Problem 14.51P , additional homework tip 3

Input offset voltage is VOS=3 mV

Average input bias current is IB=0.4 μA

Offset bias current is IOS=0.06 μA

vI=0 and R=9.09 kΩ

Calculation:

From equation (4)

vO=11(vI+VOS−IB2R)+(IB1×100 k)

For vI=0 and R=9.09 kΩ

vO=11(0+V OS−I B2( 9.09 k))+(I B1×100 k) vO=11VOS−99.99 k×IB2+(I B1×100 k)

Putting the values,

For IB1=0.43 μA, IB2=0.37 μA and VOS=3 mV

vO(max)=11(3 m)−(99.99 k×0.37 μ)+(0.43 μ×100 k) vO(max)=33×10−3−36.99×10−3+43×10−3⇒ vO(max)=39 mV

For IB1=0.37 μA, IB2=0.43 μA and VOS=−3 mV

vO(max)=11(−3 m)−(99.99 k×0.43 μ)+(0.37 μ×100 k) vO(max)=(−33× 10 −3)−(42.99× 10 −3)+(37× 10 −3)⇒ vO(max)=−39 mV

So, the range of output voltage is,

vO(min)≤vO≤vO(max)⇒ −39 mV≤vO≤39 mV

c.

To determine

Range of output voltage.

c.

Expert Solution

Answer to Problem 14.51P

The range of output voltage is,

2.161 V≤vO≤2.239 V

Explanation of Solution

Given:

The given circuit is shown below.

Microelectronics: Circuit Analysis and Design, Chapter 14, Problem 14.51P , additional homework tip 4

Input offset voltage is VOS=3 mV

Average input bias current is IB=0.4 μA

Offset bias current is IOS=0.06 μA

vI=0.2 V and R=9.09 kΩ

Calculation:

From equation (4)

vO=11(vI+VOS−IB2R)+(IB1×100 k)

For vI=0.2 V and R=9.09 kΩ

vO=11(0.2+V OS−I B2( 9.09 k))+(I B1×100 k) vO=2.2+11VOS−99.99 k×IB2+(I B1×100 k)

Putting the values,

For IB1=0.43 μA, IB2=0.37 μA and VOS=3 mV

vO(max)=2.2+11(3 m)−(99.99 k×0.37 μ)+(0.43 μ×100 k) vO(max)=2.2+33×10−3−36.99×10−3+43×10−3⇒ vO(max)=2.239 V

For IB1=0.37 μA, IB2=0.43 μA and VOS=−3 mV

vO(max)=2.2+11(−3 m)−(99.99 k×0.43 μ)+(0.37 μ×100 k) vO(max)=2.2+(−33× 10 −3)−(42.99× 10 −3)+(37× 10 −3)vO(max)=2.2−39×10−3 V⇒ vO(max)=2.161 V

So, the range of output voltage is,

vO(min)≤vO≤vO(max)⇒ 2.161 V≤vO≤2.239 V

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Answer 2

Question

Chapter 14, Problem 14.51P

a.

To determine

Range of output voltage for the given specifications.

a.

Expert Solution

Answer to Problem 14.51P

The range of output voltage is,

4 mV≤vO≤76 mV

Explanation of Solution

Given:

The given circuit is shown below.

Microelectronics: Circuit Analysis and Design, Chapter 14, Problem 14.51P , additional homework tip 1

Input offset voltage is VOS=3 mV

Average input bias current is IB=0.4 μA

Offset bias current is IOS=0.06 μA

vI=0 and R=0

Calculation:

As,

IB=I B1+I B22IB1+IB2=2IBIB1+IB2=2×0.4×10−6IB1+IB2=0.8×10−6 ...... (1)

And

IOS=|IB1−IB2|IB1−IB2=±IOSIB1−IB2=±0.06×10−6 ..... (2)

Adding (1) and (2)

IB1+IB2+IB1−IB2=0.8×10−6±0.06×10−62IB1=0.8×10−6±0.06×10−6IB1=0.8× 10 −6±0.06× 10 −62=0.8× 10 −6+0.06× 10 −62 or 0.8× 10 −6−0.06× 10 −62⇒ IB1=0.43 μA or 0.37 μA

From (1)

IB1=0.8×10−6−IB2

For IB1=0.43 μA

IB1=0.8×10−6−IB2IB1=0.8×10−6−0.43×10−6⇒ IB1=0.37 μA

For IB1=0.37 μA

IB1=0.8×10−6−IB2IB1=0.8×10−6−0.37×10−6⇒ IB1=0.43 μA∴ IB1=0.37 μA or IB1=0.43 μA

Now, the modified circuit with offset voltage and bias current is shown below.

Microelectronics: Circuit Analysis and Design, Chapter 14, Problem 14.51P , additional homework tip 2

KCL at non-inverting terminal,

VY−( v I + V OS )R+IB2=0VY−( v I + V OS )R=−IB2VY−(vI+V OS)=−IB2RVY=vI+VOS−IB2R

Now, for ideal op-amp VX=VY

VX=vI+VOS−IB2R ..... (3)

Now, KCL at inverting terminal,

VX10 k+VX−vO100 k+IB1=010VX+VX−vO100 k+IB1=010VX+VX−vO=−IB1×100 k11VX−vO=−IB1×100 kvO=11VX+(I B1×100 k)

Putting VX from (3)

vO=11(vI+VOS−IB2R)+(IB1×100 k) ..... (4)

Now,

For vI=0 and R=0

vO=11(0+V OS−I B2(0))+(I B1×100 k) vO=11VOS+(I B1×100 k)

For IB1=0.43 μA and VOS=3 mV

vO(max)=11(3 m)+(0.43 μ×100 k) vO(max)=33×10−3+43×10−3⇒ vO(max)=76 mV

For IB1=0.37 μA and VOS=-3 mV

vO(min)=11(−3 m)+(0.37 μ×100 k) vO(min)=(−33× 10 −3)+(37× 10 −3)⇒ vO(min)=4 mV

0So, the range of output voltage is,

vO(min)≤vO≤vO(max)⇒ 4 mV≤vO≤76 mV

b.

To determine

Range of output voltage for the given specifications.

b.

Expert Solution

Answer to Problem 14.51P

The range of output voltage is,

−39 mV≤vO≤39 mV

Explanation of Solution

Given:

The given circuit is shown below.

Microelectronics: Circuit Analysis and Design, Chapter 14, Problem 14.51P , additional homework tip 3

Input offset voltage is VOS=3 mV

Average input bias current is IB=0.4 μA

Offset bias current is IOS=0.06 μA

vI=0 and R=9.09 kΩ

Calculation:

From equation (4)

vO=11(vI+VOS−IB2R)+(IB1×100 k)

For vI=0 and R=9.09 kΩ

vO=11(0+V OS−I B2( 9.09 k))+(I B1×100 k) vO=11VOS−99.99 k×IB2+(I B1×100 k)

Putting the values,

For IB1=0.43 μA, IB2=0.37 μA and VOS=3 mV

vO(max)=11(3 m)−(99.99 k×0.37 μ)+(0.43 μ×100 k) vO(max)=33×10−3−36.99×10−3+43×10−3⇒ vO(max)=39 mV

For IB1=0.37 μA, IB2=0.43 μA and VOS=−3 mV

vO(max)=11(−3 m)−(99.99 k×0.43 μ)+(0.37 μ×100 k) vO(max)=(−33× 10 −3)−(42.99× 10 −3)+(37× 10 −3)⇒ vO(max)=−39 mV

So, the range of output voltage is,

vO(min)≤vO≤vO(max)⇒ −39 mV≤vO≤39 mV

c.

To determine

Range of output voltage.

c.

Expert Solution

Answer to Problem 14.51P

The range of output voltage is,

2.161 V≤vO≤2.239 V

Explanation of Solution

Given:

The given circuit is shown below.

Microelectronics: Circuit Analysis and Design, Chapter 14, Problem 14.51P , additional homework tip 4

Input offset voltage is VOS=3 mV

Average input bias current is IB=0.4 μA

Offset bias current is IOS=0.06 μA

vI=0.2 V and R=9.09 kΩ

Calculation:

From equation (4)

vO=11(vI+VOS−IB2R)+(IB1×100 k)

For vI=0.2 V and R=9.09 kΩ

vO=11(0.2+V OS−I B2( 9.09 k))+(I B1×100 k) vO=2.2+11VOS−99.99 k×IB2+(I B1×100 k)

Putting the values,

For IB1=0.43 μA, IB2=0.37 μA and VOS=3 mV

vO(max)=2.2+11(3 m)−(99.99 k×0.37 μ)+(0.43 μ×100 k) vO(max)=2.2+33×10−3−36.99×10−3+43×10−3⇒ vO(max)=2.239 V

For IB1=0.37 μA, IB2=0.43 μA and VOS=−3 mV

vO(max)=2.2+11(−3 m)−(99.99 k×0.43 μ)+(0.37 μ×100 k) vO(max)=2.2+(−33× 10 −3)−(42.99× 10 −3)+(37× 10 −3)vO(max)=2.2−39×10−3 V⇒ vO(max)=2.161 V

So, the range of output voltage is,

vO(min)≤vO≤vO(max)⇒ 2.161 V≤vO≤2.239 V

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Answer 3

Question

Chapter 14, Problem 14.51P

a.

To determine

Range of output voltage for the given specifications.

a.

Expert Solution

Answer to Problem 14.51P

The range of output voltage is,

4 mV≤vO≤76 mV

Explanation of Solution

Given:

The given circuit is shown below.

Microelectronics: Circuit Analysis and Design, Chapter 14, Problem 14.51P , additional homework tip 1

Input offset voltage is VOS=3 mV

Average input bias current is IB=0.4 μA

Offset bias current is IOS=0.06 μA

vI=0 and R=0

Calculation:

As,

IB=I B1+I B22IB1+IB2=2IBIB1+IB2=2×0.4×10−6IB1+IB2=0.8×10−6 ...... (1)

And

IOS=|IB1−IB2|IB1−IB2=±IOSIB1−IB2=±0.06×10−6 ..... (2)

Adding (1) and (2)

IB1+IB2+IB1−IB2=0.8×10−6±0.06×10−62IB1=0.8×10−6±0.06×10−6IB1=0.8× 10 −6±0.06× 10 −62=0.8× 10 −6+0.06× 10 −62 or 0.8× 10 −6−0.06× 10 −62⇒ IB1=0.43 μA or 0.37 μA

From (1)

IB1=0.8×10−6−IB2

For IB1=0.43 μA

IB1=0.8×10−6−IB2IB1=0.8×10−6−0.43×10−6⇒ IB1=0.37 μA

For IB1=0.37 μA

IB1=0.8×10−6−IB2IB1=0.8×10−6−0.37×10−6⇒ IB1=0.43 μA∴ IB1=0.37 μA or IB1=0.43 μA

Now, the modified circuit with offset voltage and bias current is shown below.

Microelectronics: Circuit Analysis and Design, Chapter 14, Problem 14.51P , additional homework tip 2

KCL at non-inverting terminal,

VY−( v I + V OS )R+IB2=0VY−( v I + V OS )R=−IB2VY−(vI+V OS)=−IB2RVY=vI+VOS−IB2R

Now, for ideal op-amp VX=VY

VX=vI+VOS−IB2R ..... (3)

Now, KCL at inverting terminal,

VX10 k+VX−vO100 k+IB1=010VX+VX−vO100 k+IB1=010VX+VX−vO=−IB1×100 k11VX−vO=−IB1×100 kvO=11VX+(I B1×100 k)

Putting VX from (3)

vO=11(vI+VOS−IB2R)+(IB1×100 k) ..... (4)

Now,

For vI=0 and R=0

vO=11(0+V OS−I B2(0))+(I B1×100 k) vO=11VOS+(I B1×100 k)

For IB1=0.43 μA and VOS=3 mV

vO(max)=11(3 m)+(0.43 μ×100 k) vO(max)=33×10−3+43×10−3⇒ vO(max)=76 mV

For IB1=0.37 μA and VOS=-3 mV

vO(min)=11(−3 m)+(0.37 μ×100 k) vO(min)=(−33× 10 −3)+(37× 10 −3)⇒ vO(min)=4 mV

0So, the range of output voltage is,

vO(min)≤vO≤vO(max)⇒ 4 mV≤vO≤76 mV

b.

To determine

Range of output voltage for the given specifications.

b.

Expert Solution

Answer to Problem 14.51P

The range of output voltage is,

−39 mV≤vO≤39 mV

Explanation of Solution

Given:

The given circuit is shown below.

Microelectronics: Circuit Analysis and Design, Chapter 14, Problem 14.51P , additional homework tip 3

Input offset voltage is VOS=3 mV

Average input bias current is IB=0.4 μA

Offset bias current is IOS=0.06 μA

vI=0 and R=9.09 kΩ

Calculation:

From equation (4)

vO=11(vI+VOS−IB2R)+(IB1×100 k)

For vI=0 and R=9.09 kΩ

vO=11(0+V OS−I B2( 9.09 k))+(I B1×100 k) vO=11VOS−99.99 k×IB2+(I B1×100 k)

Putting the values,

For IB1=0.43 μA, IB2=0.37 μA and VOS=3 mV

vO(max)=11(3 m)−(99.99 k×0.37 μ)+(0.43 μ×100 k) vO(max)=33×10−3−36.99×10−3+43×10−3⇒ vO(max)=39 mV

For IB1=0.37 μA, IB2=0.43 μA and VOS=−3 mV

vO(max)=11(−3 m)−(99.99 k×0.43 μ)+(0.37 μ×100 k) vO(max)=(−33× 10 −3)−(42.99× 10 −3)+(37× 10 −3)⇒ vO(max)=−39 mV

So, the range of output voltage is,

vO(min)≤vO≤vO(max)⇒ −39 mV≤vO≤39 mV

c.

To determine

Range of output voltage.

c.

Expert Solution

Answer to Problem 14.51P

The range of output voltage is,

2.161 V≤vO≤2.239 V

Explanation of Solution

Given:

The given circuit is shown below.

Microelectronics: Circuit Analysis and Design, Chapter 14, Problem 14.51P , additional homework tip 4

Input offset voltage is VOS=3 mV

Average input bias current is IB=0.4 μA

Offset bias current is IOS=0.06 μA

vI=0.2 V and R=9.09 kΩ

Calculation:

From equation (4)

vO=11(vI+VOS−IB2R)+(IB1×100 k)

For vI=0.2 V and R=9.09 kΩ

vO=11(0.2+V OS−I B2( 9.09 k))+(I B1×100 k) vO=2.2+11VOS−99.99 k×IB2+(I B1×100 k)

Putting the values,

For IB1=0.43 μA, IB2=0.37 μA and VOS=3 mV

vO(max)=2.2+11(3 m)−(99.99 k×0.37 μ)+(0.43 μ×100 k) vO(max)=2.2+33×10−3−36.99×10−3+43×10−3⇒ vO(max)=2.239 V

For IB1=0.37 μA, IB2=0.43 μA and VOS=−3 mV

vO(max)=2.2+11(−3 m)−(99.99 k×0.43 μ)+(0.37 μ×100 k) vO(max)=2.2+(−33× 10 −3)−(42.99× 10 −3)+(37× 10 −3)vO(max)=2.2−39×10−3 V⇒ vO(max)=2.161 V

So, the range of output voltage is,

vO(min)≤vO≤vO(max)⇒ 2.161 V≤vO≤2.239 V

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Answer 4

Question

Chapter 14, Problem 14.51P

a.

To determine

Range of output voltage for the given specifications.

a.

Expert Solution

Answer to Problem 14.51P

The range of output voltage is,

4 mV≤vO≤76 mV

Explanation of Solution

Given:

The given circuit is shown below.

Microelectronics: Circuit Analysis and Design, Chapter 14, Problem 14.51P , additional homework tip 1

Input offset voltage is VOS=3 mV

Average input bias current is IB=0.4 μA

Offset bias current is IOS=0.06 μA

vI=0 and R=0

Calculation:

As,

IB=I B1+I B22IB1+IB2=2IBIB1+IB2=2×0.4×10−6IB1+IB2=0.8×10−6 ...... (1)

And

IOS=|IB1−IB2|IB1−IB2=±IOSIB1−IB2=±0.06×10−6 ..... (2)

Adding (1) and (2)

IB1+IB2+IB1−IB2=0.8×10−6±0.06×10−62IB1=0.8×10−6±0.06×10−6IB1=0.8× 10 −6±0.06× 10 −62=0.8× 10 −6+0.06× 10 −62 or 0.8× 10 −6−0.06× 10 −62⇒ IB1=0.43 μA or 0.37 μA

From (1)

IB1=0.8×10−6−IB2

For IB1=0.43 μA

IB1=0.8×10−6−IB2IB1=0.8×10−6−0.43×10−6⇒ IB1=0.37 μA

For IB1=0.37 μA

IB1=0.8×10−6−IB2IB1=0.8×10−6−0.37×10−6⇒ IB1=0.43 μA∴ IB1=0.37 μA or IB1=0.43 μA

Now, the modified circuit with offset voltage and bias current is shown below.

Microelectronics: Circuit Analysis and Design, Chapter 14, Problem 14.51P , additional homework tip 2

KCL at non-inverting terminal,

VY−( v I + V OS )R+IB2=0VY−( v I + V OS )R=−IB2VY−(vI+V OS)=−IB2RVY=vI+VOS−IB2R

Now, for ideal op-amp VX=VY

VX=vI+VOS−IB2R ..... (3)

Now, KCL at inverting terminal,

VX10 k+VX−vO100 k+IB1=010VX+VX−vO100 k+IB1=010VX+VX−vO=−IB1×100 k11VX−vO=−IB1×100 kvO=11VX+(I B1×100 k)

Putting VX from (3)

vO=11(vI+VOS−IB2R)+(IB1×100 k) ..... (4)

Now,

For vI=0 and R=0

vO=11(0+V OS−I B2(0))+(I B1×100 k) vO=11VOS+(I B1×100 k)

For IB1=0.43 μA and VOS=3 mV

vO(max)=11(3 m)+(0.43 μ×100 k) vO(max)=33×10−3+43×10−3⇒ vO(max)=76 mV

For IB1=0.37 μA and VOS=-3 mV

vO(min)=11(−3 m)+(0.37 μ×100 k) vO(min)=(−33× 10 −3)+(37× 10 −3)⇒ vO(min)=4 mV

0So, the range of output voltage is,

vO(min)≤vO≤vO(max)⇒ 4 mV≤vO≤76 mV

b.

To determine

Range of output voltage for the given specifications.

b.

Expert Solution

Answer to Problem 14.51P

The range of output voltage is,

−39 mV≤vO≤39 mV

Explanation of Solution

Given:

The given circuit is shown below.

Microelectronics: Circuit Analysis and Design, Chapter 14, Problem 14.51P , additional homework tip 3

Input offset voltage is VOS=3 mV

Average input bias current is IB=0.4 μA

Offset bias current is IOS=0.06 μA

vI=0 and R=9.09 kΩ

Calculation:

From equation (4)

vO=11(vI+VOS−IB2R)+(IB1×100 k)

For vI=0 and R=9.09 kΩ

vO=11(0+V OS−I B2( 9.09 k))+(I B1×100 k) vO=11VOS−99.99 k×IB2+(I B1×100 k)

Putting the values,

For IB1=0.43 μA, IB2=0.37 μA and VOS=3 mV

vO(max)=11(3 m)−(99.99 k×0.37 μ)+(0.43 μ×100 k) vO(max)=33×10−3−36.99×10−3+43×10−3⇒ vO(max)=39 mV

For IB1=0.37 μA, IB2=0.43 μA and VOS=−3 mV

vO(max)=11(−3 m)−(99.99 k×0.43 μ)+(0.37 μ×100 k) vO(max)=(−33× 10 −3)−(42.99× 10 −3)+(37× 10 −3)⇒ vO(max)=−39 mV

So, the range of output voltage is,

vO(min)≤vO≤vO(max)⇒ −39 mV≤vO≤39 mV

c.

To determine

Range of output voltage.

c.

Expert Solution

Answer to Problem 14.51P

The range of output voltage is,

2.161 V≤vO≤2.239 V

Explanation of Solution

Given:

The given circuit is shown below.

Microelectronics: Circuit Analysis and Design, Chapter 14, Problem 14.51P , additional homework tip 4

Input offset voltage is VOS=3 mV

Average input bias current is IB=0.4 μA

Offset bias current is IOS=0.06 μA

vI=0.2 V and R=9.09 kΩ

Calculation:

From equation (4)

vO=11(vI+VOS−IB2R)+(IB1×100 k)

For vI=0.2 V and R=9.09 kΩ

vO=11(0.2+V OS−I B2( 9.09 k))+(I B1×100 k) vO=2.2+11VOS−99.99 k×IB2+(I B1×100 k)

Putting the values,

For IB1=0.43 μA, IB2=0.37 μA and VOS=3 mV

vO(max)=2.2+11(3 m)−(99.99 k×0.37 μ)+(0.43 μ×100 k) vO(max)=2.2+33×10−3−36.99×10−3+43×10−3⇒ vO(max)=2.239 V

For IB1=0.37 μA, IB2=0.43 μA and VOS=−3 mV

vO(max)=2.2+11(−3 m)−(99.99 k×0.43 μ)+(0.37 μ×100 k) vO(max)=2.2+(−33× 10 −3)−(42.99× 10 −3)+(37× 10 −3)vO(max)=2.2−39×10−3 V⇒ vO(max)=2.161 V

So, the range of output voltage is,

vO(min)≤vO≤vO(max)⇒ 2.161 V≤vO≤2.239 V

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Answer 5

Chapter 14, Problem 14.51P

a.

To determine

Range of output voltage for the given specifications.

a.

Expert Solution

Answer to Problem 14.51P

The range of output voltage is,

4 mV≤vO≤76 mV

Explanation of Solution

Given:

The given circuit is shown below.

Microelectronics: Circuit Analysis and Design, Chapter 14, Problem 14.51P , additional homework tip 1

Input offset voltage is VOS=3 mV

Average input bias current is IB=0.4 μA

Offset bias current is IOS=0.06 μA

vI=0 and R=0

Calculation:

As,

IB=I B1+I B22IB1+IB2=2IBIB1+IB2=2×0.4×10−6IB1+IB2=0.8×10−6 ...... (1)

And

IOS=|IB1−IB2|IB1−IB2=±IOSIB1−IB2=±0.06×10−6 ..... (2)

Adding (1) and (2)

IB1+IB2+IB1−IB2=0.8×10−6±0.06×10−62IB1=0.8×10−6±0.06×10−6IB1=0.8× 10 −6±0.06× 10 −62=0.8× 10 −6+0.06× 10 −62 or 0.8× 10 −6−0.06× 10 −62⇒ IB1=0.43 μA or 0.37 μA

From (1)

IB1=0.8×10−6−IB2

For IB1=0.43 μA

IB1=0.8×10−6−IB2IB1=0.8×10−6−0.43×10−6⇒ IB1=0.37 μA

For IB1=0.37 μA

IB1=0.8×10−6−IB2IB1=0.8×10−6−0.37×10−6⇒ IB1=0.43 μA∴ IB1=0.37 μA or IB1=0.43 μA

Now, the modified circuit with offset voltage and bias current is shown below.

Microelectronics: Circuit Analysis and Design, Chapter 14, Problem 14.51P , additional homework tip 2

KCL at non-inverting terminal,

VY−( v I + V OS )R+IB2=0VY−( v I + V OS )R=−IB2VY−(vI+V OS)=−IB2RVY=vI+VOS−IB2R

Now, for ideal op-amp VX=VY

VX=vI+VOS−IB2R ..... (3)

Now, KCL at inverting terminal,

VX10 k+VX−vO100 k+IB1=010VX+VX−vO100 k+IB1=010VX+VX−vO=−IB1×100 k11VX−vO=−IB1×100 kvO=11VX+(I B1×100 k)

Putting VX from (3)

vO=11(vI+VOS−IB2R)+(IB1×100 k) ..... (4)

Now,

For vI=0 and R=0

vO=11(0+V OS−I B2(0))+(I B1×100 k) vO=11VOS+(I B1×100 k)

For IB1=0.43 μA and VOS=3 mV

vO(max)=11(3 m)+(0.43 μ×100 k) vO(max)=33×10−3+43×10−3⇒ vO(max)=76 mV

For IB1=0.37 μA and VOS=-3 mV

vO(min)=11(−3 m)+(0.37 μ×100 k) vO(min)=(−33× 10 −3)+(37× 10 −3)⇒ vO(min)=4 mV

0So, the range of output voltage is,

vO(min)≤vO≤vO(max)⇒ 4 mV≤vO≤76 mV

b.

To determine

Range of output voltage for the given specifications.

b.

Expert Solution

Answer to Problem 14.51P

The range of output voltage is,

−39 mV≤vO≤39 mV

Explanation of Solution

Given:

The given circuit is shown below.

Microelectronics: Circuit Analysis and Design, Chapter 14, Problem 14.51P , additional homework tip 3

Input offset voltage is VOS=3 mV

Average input bias current is IB=0.4 μA

Offset bias current is IOS=0.06 μA

vI=0 and R=9.09 kΩ

Calculation:

From equation (4)

vO=11(vI+VOS−IB2R)+(IB1×100 k)

For vI=0 and R=9.09 kΩ

vO=11(0+V OS−I B2( 9.09 k))+(I B1×100 k) vO=11VOS−99.99 k×IB2+(I B1×100 k)

Putting the values,

For IB1=0.43 μA, IB2=0.37 μA and VOS=3 mV

vO(max)=11(3 m)−(99.99 k×0.37 μ)+(0.43 μ×100 k) vO(max)=33×10−3−36.99×10−3+43×10−3⇒ vO(max)=39 mV

For IB1=0.37 μA, IB2=0.43 μA and VOS=−3 mV

vO(max)=11(−3 m)−(99.99 k×0.43 μ)+(0.37 μ×100 k) vO(max)=(−33× 10 −3)−(42.99× 10 −3)+(37× 10 −3)⇒ vO(max)=−39 mV

So, the range of output voltage is,

vO(min)≤vO≤vO(max)⇒ −39 mV≤vO≤39 mV

c.

To determine

Range of output voltage.

c.

Expert Solution

Answer to Problem 14.51P

The range of output voltage is,

2.161 V≤vO≤2.239 V

Explanation of Solution

Given:

The given circuit is shown below.

Microelectronics: Circuit Analysis and Design, Chapter 14, Problem 14.51P , additional homework tip 4

Input offset voltage is VOS=3 mV

Average input bias current is IB=0.4 μA

Offset bias current is IOS=0.06 μA

vI=0.2 V and R=9.09 kΩ

Calculation:

From equation (4)

vO=11(vI+VOS−IB2R)+(IB1×100 k)

For vI=0.2 V and R=9.09 kΩ

vO=11(0.2+V OS−I B2( 9.09 k))+(I B1×100 k) vO=2.2+11VOS−99.99 k×IB2+(I B1×100 k)

Putting the values,

For IB1=0.43 μA, IB2=0.37 μA and VOS=3 mV

vO(max)=2.2+11(3 m)−(99.99 k×0.37 μ)+(0.43 μ×100 k) vO(max)=2.2+33×10−3−36.99×10−3+43×10−3⇒ vO(max)=2.239 V

For IB1=0.37 μA, IB2=0.43 μA and VOS=−3 mV

vO(max)=2.2+11(−3 m)−(99.99 k×0.43 μ)+(0.37 μ×100 k) vO(max)=2.2+(−33× 10 −3)−(42.99× 10 −3)+(37× 10 −3)vO(max)=2.2−39×10−3 V⇒ vO(max)=2.161 V

So, the range of output voltage is,

vO(min)≤vO≤vO(max)⇒ 2.161 V≤vO≤2.239 V

Answer 6

Chapter 14, Problem 14.51P

a.

To determine

Range of output voltage for the given specifications.

a.

Expert Solution

Answer to Problem 14.51P

The range of output voltage is,

4 mV≤vO≤76 mV

Explanation of Solution

Given:

The given circuit is shown below.

Microelectronics: Circuit Analysis and Design, Chapter 14, Problem 14.51P , additional homework tip 1

Input offset voltage is VOS=3 mV

Average input bias current is IB=0.4 μA

Offset bias current is IOS=0.06 μA

vI=0 and R=0

Calculation:

As,

IB=I B1+I B22IB1+IB2=2IBIB1+IB2=2×0.4×10−6IB1+IB2=0.8×10−6 ...... (1)

And

IOS=|IB1−IB2|IB1−IB2=±IOSIB1−IB2=±0.06×10−6 ..... (2)

Adding (1) and (2)

IB1+IB2+IB1−IB2=0.8×10−6±0.06×10−62IB1=0.8×10−6±0.06×10−6IB1=0.8× 10 −6±0.06× 10 −62=0.8× 10 −6+0.06× 10 −62 or 0.8× 10 −6−0.06× 10 −62⇒ IB1=0.43 μA or 0.37 μA

From (1)

IB1=0.8×10−6−IB2

For IB1=0.43 μA

IB1=0.8×10−6−IB2IB1=0.8×10−6−0.43×10−6⇒ IB1=0.37 μA

For IB1=0.37 μA

IB1=0.8×10−6−IB2IB1=0.8×10−6−0.37×10−6⇒ IB1=0.43 μA∴ IB1=0.37 μA or IB1=0.43 μA

Now, the modified circuit with offset voltage and bias current is shown below.

Microelectronics: Circuit Analysis and Design, Chapter 14, Problem 14.51P , additional homework tip 2

KCL at non-inverting terminal,

VY−( v I + V OS )R+IB2=0VY−( v I + V OS )R=−IB2VY−(vI+V OS)=−IB2RVY=vI+VOS−IB2R

Now, for ideal op-amp VX=VY

VX=vI+VOS−IB2R ..... (3)

Now, KCL at inverting terminal,

VX10 k+VX−vO100 k+IB1=010VX+VX−vO100 k+IB1=010VX+VX−vO=−IB1×100 k11VX−vO=−IB1×100 kvO=11VX+(I B1×100 k)

Putting VX from (3)

vO=11(vI+VOS−IB2R)+(IB1×100 k) ..... (4)

Now,

For vI=0 and R=0

vO=11(0+V OS−I B2(0))+(I B1×100 k) vO=11VOS+(I B1×100 k)

For IB1=0.43 μA and VOS=3 mV

vO(max)=11(3 m)+(0.43 μ×100 k) vO(max)=33×10−3+43×10−3⇒ vO(max)=76 mV

For IB1=0.37 μA and VOS=-3 mV

vO(min)=11(−3 m)+(0.37 μ×100 k) vO(min)=(−33× 10 −3)+(37× 10 −3)⇒ vO(min)=4 mV

0So, the range of output voltage is,

vO(min)≤vO≤vO(max)⇒ 4 mV≤vO≤76 mV

Answer 7

Chapter 14, Problem 14.51P

a.

To determine

Range of output voltage for the given specifications.

Answer 8

a.

Expert Solution

Answer to Problem 14.51P

The range of output voltage is,

4 mV≤vO≤76 mV

Answer 9

a.

Expert Solution

Answer 10

a.

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Given:

The given circuit is shown below.

Microelectronics: Circuit Analysis and Design, Chapter 14, Problem 14.51P , additional homework tip 1

Input offset voltage is VOS=3 mV

Average input bias current is IB=0.4 μA

Offset bias current is IOS=0.06 μA

vI=0 and R=0

Calculation:

As,

IB=I B1+I B22IB1+IB2=2IBIB1+IB2=2×0.4×10−6IB1+IB2=0.8×10−6 ...... (1)

And

IOS=|IB1−IB2|IB1−IB2=±IOSIB1−IB2=±0.06×10−6 ..... (2)

Adding (1) and (2)

IB1+IB2+IB1−IB2=0.8×10−6±0.06×10−62IB1=0.8×10−6±0.06×10−6IB1=0.8× 10 −6±0.06× 10 −62=0.8× 10 −6+0.06× 10 −62 or 0.8× 10 −6−0.06× 10 −62⇒ IB1=0.43 μA or 0.37 μA

From (1)

IB1=0.8×10−6−IB2

For IB1=0.43 μA

IB1=0.8×10−6−IB2IB1=0.8×10−6−0.43×10−6⇒ IB1=0.37 μA

For IB1=0.37 μA

IB1=0.8×10−6−IB2IB1=0.8×10−6−0.37×10−6⇒ IB1=0.43 μA∴ IB1=0.37 μA or IB1=0.43 μA

Now, the modified circuit with offset voltage and bias current is shown below.

Microelectronics: Circuit Analysis and Design, Chapter 14, Problem 14.51P , additional homework tip 2

KCL at non-inverting terminal,

VY−( v I + V OS )R+IB2=0VY−( v I + V OS )R=−IB2VY−(vI+V OS)=−IB2RVY=vI+VOS−IB2R

Now, for ideal op-amp VX=VY

VX=vI+VOS−IB2R ..... (3)

Now, KCL at inverting terminal,

VX10 k+VX−vO100 k+IB1=010VX+VX−vO100 k+IB1=010VX+VX−vO=−IB1×100 k11VX−vO=−IB1×100 kvO=11VX+(I B1×100 k)

Putting VX from (3)

vO=11(vI+VOS−IB2R)+(IB1×100 k) ..... (4)

Now,

For vI=0 and R=0

vO=11(0+V OS−I B2(0))+(I B1×100 k) vO=11VOS+(I B1×100 k)

For IB1=0.43 μA and VOS=3 mV

vO(max)=11(3 m)+(0.43 μ×100 k) vO(max)=33×10−3+43×10−3⇒ vO(max)=76 mV

For IB1=0.37 μA and VOS=-3 mV

vO(min)=11(−3 m)+(0.37 μ×100 k) vO(min)=(−33× 10 −3)+(37× 10 −3)⇒ vO(min)=4 mV

0So, the range of output voltage is,

vO(min)≤vO≤vO(max)⇒ 4 mV≤vO≤76 mV

Answer 13

b.

To determine

Range of output voltage for the given specifications.

b.

Expert Solution

Answer to Problem 14.51P

The range of output voltage is,

−39 mV≤vO≤39 mV

Explanation of Solution

Given:

The given circuit is shown below.

Microelectronics: Circuit Analysis and Design, Chapter 14, Problem 14.51P , additional homework tip 3

Input offset voltage is VOS=3 mV

Average input bias current is IB=0.4 μA

Offset bias current is IOS=0.06 μA

vI=0 and R=9.09 kΩ

Calculation:

From equation (4)

vO=11(vI+VOS−IB2R)+(IB1×100 k)

For vI=0 and R=9.09 kΩ

vO=11(0+V OS−I B2( 9.09 k))+(I B1×100 k) vO=11VOS−99.99 k×IB2+(I B1×100 k)

Putting the values,

For IB1=0.43 μA, IB2=0.37 μA and VOS=3 mV

vO(max)=11(3 m)−(99.99 k×0.37 μ)+(0.43 μ×100 k) vO(max)=33×10−3−36.99×10−3+43×10−3⇒ vO(max)=39 mV

For IB1=0.37 μA, IB2=0.43 μA and VOS=−3 mV

vO(max)=11(−3 m)−(99.99 k×0.43 μ)+(0.37 μ×100 k) vO(max)=(−33× 10 −3)−(42.99× 10 −3)+(37× 10 −3)⇒ vO(max)=−39 mV

So, the range of output voltage is,

vO(min)≤vO≤vO(max)⇒ −39 mV≤vO≤39 mV

Answer 14

b.

To determine

Range of output voltage for the given specifications.

Answer 15

b.

Expert Solution

Answer to Problem 14.51P

The range of output voltage is,

−39 mV≤vO≤39 mV

Answer 16

b.

Expert Solution

Answer 17

b.

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Given:

The given circuit is shown below.

Microelectronics: Circuit Analysis and Design, Chapter 14, Problem 14.51P , additional homework tip 3

Input offset voltage is VOS=3 mV

Average input bias current is IB=0.4 μA

Offset bias current is IOS=0.06 μA

vI=0 and R=9.09 kΩ

Calculation:

From equation (4)

vO=11(vI+VOS−IB2R)+(IB1×100 k)

For vI=0 and R=9.09 kΩ

vO=11(0+V OS−I B2( 9.09 k))+(I B1×100 k) vO=11VOS−99.99 k×IB2+(I B1×100 k)

Putting the values,

For IB1=0.43 μA, IB2=0.37 μA and VOS=3 mV

vO(max)=11(3 m)−(99.99 k×0.37 μ)+(0.43 μ×100 k) vO(max)=33×10−3−36.99×10−3+43×10−3⇒ vO(max)=39 mV

For IB1=0.37 μA, IB2=0.43 μA and VOS=−3 mV

vO(max)=11(−3 m)−(99.99 k×0.43 μ)+(0.37 μ×100 k) vO(max)=(−33× 10 −3)−(42.99× 10 −3)+(37× 10 −3)⇒ vO(max)=−39 mV

So, the range of output voltage is,

vO(min)≤vO≤vO(max)⇒ −39 mV≤vO≤39 mV

Answer 20

c.

To determine

Range of output voltage.

c.

Expert Solution

Answer to Problem 14.51P

The range of output voltage is,

2.161 V≤vO≤2.239 V

Explanation of Solution

Given:

The given circuit is shown below.

Microelectronics: Circuit Analysis and Design, Chapter 14, Problem 14.51P , additional homework tip 4

Input offset voltage is VOS=3 mV

Average input bias current is IB=0.4 μA

Offset bias current is IOS=0.06 μA

vI=0.2 V and R=9.09 kΩ

Calculation:

From equation (4)

vO=11(vI+VOS−IB2R)+(IB1×100 k)

For vI=0.2 V and R=9.09 kΩ

vO=11(0.2+V OS−I B2( 9.09 k))+(I B1×100 k) vO=2.2+11VOS−99.99 k×IB2+(I B1×100 k)

Putting the values,

For IB1=0.43 μA, IB2=0.37 μA and VOS=3 mV

vO(max)=2.2+11(3 m)−(99.99 k×0.37 μ)+(0.43 μ×100 k) vO(max)=2.2+33×10−3−36.99×10−3+43×10−3⇒ vO(max)=2.239 V

For IB1=0.37 μA, IB2=0.43 μA and VOS=−3 mV

vO(max)=2.2+11(−3 m)−(99.99 k×0.43 μ)+(0.37 μ×100 k) vO(max)=2.2+(−33× 10 −3)−(42.99× 10 −3)+(37× 10 −3)vO(max)=2.2−39×10−3 V⇒ vO(max)=2.161 V

So, the range of output voltage is,

vO(min)≤vO≤vO(max)⇒ 2.161 V≤vO≤2.239 V

Answer 21

c.

To determine

Range of output voltage.

Answer 22

c.

Expert Solution

Answer to Problem 14.51P

The range of output voltage is,

2.161 V≤vO≤2.239 V

Answer 23

c.

Expert Solution

Answer 24

c.

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

Given:

The given circuit is shown below.

Microelectronics: Circuit Analysis and Design, Chapter 14, Problem 14.51P , additional homework tip 4

Input offset voltage is VOS=3 mV

Average input bias current is IB=0.4 μA

Offset bias current is IOS=0.06 μA

vI=0.2 V and R=9.09 kΩ

Calculation:

From equation (4)

vO=11(vI+VOS−IB2R)+(IB1×100 k)

For vI=0.2 V and R=9.09 kΩ

vO=11(0.2+V OS−I B2( 9.09 k))+(I B1×100 k) vO=2.2+11VOS−99.99 k×IB2+(I B1×100 k)

Putting the values,

For IB1=0.43 μA, IB2=0.37 μA and VOS=3 mV

vO(max)=2.2+11(3 m)−(99.99 k×0.37 μ)+(0.43 μ×100 k) vO(max)=2.2+33×10−3−36.99×10−3+43×10−3⇒ vO(max)=2.239 V

For IB1=0.37 μA, IB2=0.43 μA and VOS=−3 mV

vO(max)=2.2+11(−3 m)−(99.99 k×0.43 μ)+(0.37 μ×100 k) vO(max)=2.2+(−33× 10 −3)−(42.99× 10 −3)+(37× 10 −3)vO(max)=2.2−39×10−3 V⇒ vO(max)=2.161 V

So, the range of output voltage is,

vO(min)≤vO≤vO(max)⇒ 2.161 V≤vO≤2.239 V

Answer 27

Ch. 14 - Using the circuit and transistor parameters of...Ch. 14 - Prob. 14.2TYU Ch. 14 - Prob. 14.1EP Ch. 14 - Determine the closedloop input resistance at the...Ch. 14 - For a noninverting amplifier, the resistances are...Ch. 14 - An opamp with an openloop gain of AOL=105 is used...Ch. 14 - Prob. 14.3TYU Ch. 14 - An operational amplifier connected in a...Ch. 14 - Prob. 14.5TYU Ch. 14 - Prob. 14.6TYU

Range of output voltage for the given specifications.

Range of output voltage for the given specifications.

Answer to Problem 14.51P

Explanation of Solution

Range of output voltage for the given specifications.

Answer to Problem 14.51P

Explanation of Solution

Range of output voltage.

Answer to Problem 14.51P

Explanation of Solution

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