# (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.

### Principles of Instrumental Analysis

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

### 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 AWe have,v0+=A(v+−v−)=+13 V and v0−=A(v+−v−)=−14 VCorresponding to the positive saturation value and negative saturation value respectively. We need to find the difference in input voltage in each case.

Expert Solution

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 AWe have,v0+=A(v+−v−)=+13 V and v0−=A(v+−v−)=−14 VCorresponding to the positive saturation value and negative saturation value respectively. We need to find the difference in input voltage in each case.

Expert Solution

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 AWe have,v0+=A(v+−v−)=+13 V and v0−=A(v+−v−)=−14 VCorresponding to the positive saturation value and negative saturation value respectively. We need to find the difference in input voltage in each case.

Expert Solution

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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