(a) Interpretation: The amount by which v + has to exceed v − and v − has to exceed v + for the amplifier to work in limit is to be calculated. Concept introduction: Limiting condition of operational amplifier as a comparator implies that the comparator output voltage reaches the limiting value of output which is given here as +13 V on the positive side and − 14 V on the negative side. With open loop gain as A We have, v 0 + = A ( v + − v − ) = + 13 V and v 0 − = A ( v + − v − ) = − 14 V Corresponding to the positive saturation value and negative saturation value respectively. We need to find the difference in input voltage in each case.

BuyFind

Principles of Instrumental Analysis

7th Edition
Douglas A. Skoog + 2 others
Publisher: Cengage Learning
ISBN: 9781305577213
BuyFind

Principles of Instrumental Analysis

7th Edition
Douglas A. Skoog + 2 others
Publisher: Cengage Learning
ISBN: 9781305577213

Solutions

Chapter 3, Problem 3.1QAP
Interpretation Introduction

(a)

Interpretation:

The amount by which v+ has to exceed v and v has to exceed v+ for the amplifier to work in limit is to be calculated.

Concept introduction:

Limiting condition of operational amplifier as a comparator implies that the comparator output voltage reaches the limiting value of output which is given here as +13 V on the positive side and 14V on the negative side. With open loop gain as A

We have,

v0+=A(v+v)=+13V and v0=A(v+v)=14V

Corresponding to the positive saturation value and negative saturation value respectively. We need to find the difference in input voltage in each case.

Expert Solution

Answer to Problem 3.1QAP

v+ has to exceed v by 130μV for the output to be at limit +13V and v has to exceed v+ by 140μV for output to be at limit 14V

Explanation of Solution

Given the open loop gain A is 100,000

Operational amplifier as a comparator compares two voltages at its respective input terminals v+ and v and produces an output depending on the difference of the two input voltages. In the limiting condition, the output value reaches a saturation which is given here as +13 V on the positive side and 14V on the negative side. We have the open loop gain as A

We have to find the amount by which v+ has to exceed v and vice versa to reach the saturation condition or limiting condition

Case 1: For v+ has to exceed v corresponding to an output of v0+ +13 V

v0+=A(v+v)=+13V

Here, A=100,000

Hence,

13=100,000(v+v)or(v+v)=13100,000=130μ

This implies v+ has to exceed v by 130μV for the output to be at limit

Case 2: For v has to exceed v+ corresponding to an output of v0+

14 V

v0+=A(v+v)=14V

Here, A=100,000

Hence,

14=100,000(v+v)or(v+v)=14100,000=140μ

This implies v has to exceed v+ by 140μV for output to be at limit

Interpretation Introduction

(b)

Interpretation:

The amount by which v+ has to exceed v and v has to exceed v+ for the amplifier to work in limit is to be calculated.

Concept introduction:

Limiting condition of operational amplifier as a comparator implies that the comparator output voltage reaches the limiting value of output which is given here as +13 V on the positive side and 14V on the negative side. With open loop gain as A

We have,

v0+=A(v+v)=+13V and v0=A(v+v)=14V

Corresponding to the positive saturation value and negative saturation value respectively. We need to find the difference in input voltage in each case.

Expert Solution

Answer to Problem 3.1QAP

v+ has to exceed v by 21.67 μV for the output to be at limit +13 V and v has to exceed v+ by 23.33 μV for output to be at limit 14V

Explanation of Solution

Given the open loop gain A is 600,000.

Operational amplifier as a comparator compares two voltages at its respective input terminals v+ and v and produces an output depending on the difference of the two input voltages. In the limiting condition, the output value reaches a saturation which is given here as +13 V on the positive side and 14V on the negative side. We have the open loop gain as A

We have to find the amount by which v+ has to exceed v and vice versa to reach the saturation condition or limiting condition

Case 1: For v+ has to exceed v corresponding to an output of v0+ +13 V

v0+=A(v+v) =+13 V

Here, A=600,000

Hence,

13=600,000(v+v)or(v+v)=13600,000=21.67μ

This implies v+ has to exceed v by 21.67μV for the output to be at limit

Case 2: For v has to exceed v+ corresponding to an output of v0+

14V

v0+=A(v+v)=14V

Here, A=600,000

Hence,

14=600,000(v+v)or(v+v)=14600,000=23.33μ

This implies v has to exceed v+ by 23.33 μV for output to be at limit

Interpretation Introduction

(c)

Interpretation:

The amount by which v+ has to exceed v and v has to exceed v+ for the amplifier to work in limit is to be calculated.

Concept introduction:

Limiting condition of operational amplifier as a comparator implies that the comparator output voltage reaches the limiting value of output which is given here as +13 V on the positive side and 14V on the negative side. With open loop gain as A

We have,

v0+=A(v+v)=+13V and v0=A(v+v)=14V

Corresponding to the positive saturation value and negative saturation value respectively. We need to find the difference in input voltage in each case.

Expert Solution

Answer to Problem 3.1QAP

v+ has to exceed v by μV for the output to be at limit +13 V and v has to exceed v+ by 5.384 μV for output to be at limit 14 V

Explanation of Solution

Given the open loop gain A is 2.6×106.

Operational amplifier as a comparator compares two voltages at its respective input terminals v+ and v and produces an output depending on the difference of the two input voltages. In the limiting condition, the output value reaches a saturation which is given here as +13 V on the positive side and 14 V on the negative side. We have the open loop gain as A.

We have to find the amount by which v+ has to exceed v and vice versa to reach the saturation condition or limiting condition

Case 1: For v+ has to exceed v - corresponding to an output of v0+ +13 V

v0+=A(v+v) =+13 V

Here, A=2.6×106

Hence,

13=2.6×106(v+v)or(v+v)=132.6×106=5μ

This implies v+ has to exceed v by 5 μV for the output to be at limit

Case 2 : For v has to exceed v+ - corresponding to an output of v0+

14 V

v0+=A(v+v)=14V

Here, A=2.6×106

Hence,

14=2.6×106(v+v)or(v+v)=142.6×106=5.384μ

This implies v has to exceed v+ by 5.384 μV for output to be at limit

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