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 $-14\text{V}$ on the negative side. With open loop gain as A

We have,

$v_{0+}=A(v_{+}-v_{-})=+13\text{V}$ and $v_{0-}=A(v_{+}-v_{-})=-14\text{V}$

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

Answer to Problem 3.1QAP

$v_{+}$ has to exceed $v_{-}$ by $\text{130}\mu \text{V}$ for the output to be at limit $+13\text{V}$ and $v_{-}$ has to exceed $v_{+}$ by $\text{140}\mu \text{V}$ for output to be at limit $-14\text{V}$

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 $-14\text{V}$ 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 $v_{0+}$ +13 V

$v_{0+}=A(v_{+}-v_{-})=+13\text{V}$

Here, $A=100,000$

Hence,

$\begin{array}{l}13=100,000(v_{+}-v_{-})\\ \text{or}(v_{+}-v_{-})=\frac{13}{100,000}\text{=130}\mu \text{V }\end{array}$

This implies $v_{+}$ has to exceed $v_{-}$ by $\text{130}\mu \text{V}$ for the output to be at limit

Case 2: For $v_{-}$ has to exceed $v_{+}$ corresponding to an output of $v_{0+}$

$-14\text{ V}$

$v_{0+}=A(v_{+}-v_{-})=-14\text{V}$

Here, $A=100,000$

Hence,

$\begin{array}{l}-14=100,000(v_{+}-v_{-})\\ \text{or}(v_{+}-v_{-})=\frac{-14}{100,000}\text{=}-\text{140}\mu \text{V }\end{array}$

This implies $v_{-}$ has to exceed $v_{+}$ by $\text{140}\mu \text{V}$ 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 $-14\text{V}$ on the negative side. With open loop gain as A

We have,

$v_{0+}=A(v_{+}-v_{-})=+13\text{V}$ and $v_{0-}=A(v_{+}-v_{-})=-14\text{V}$

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

Answer to Problem 3.1QAP

$v_{+}$ has to exceed $v_{-}$ by $21.67\mu \text{V}$ for the output to be at limit +13 V and $v_{-}$ has to exceed $v_{+}$ by $23.33\mu \text{V}$ for output to be at limit $-14\text{V}$

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 $-14\text{V}$ 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 $v_{0+}$ +13 V

$v_{0+}=A(v_{+}-v_{-})$ =+13 V

Here, $A=600,000$

Hence,

$\begin{array}{l}13=600,000(v_{+}-v_{-})\\ \text{or}(v_{+}-v_{-})=\frac{13}{600,000}\text{=21}\text{.67}\mu \text{V }\end{array}$

This implies $v_{+}$ has to exceed $v_{-}$ by $\text{21}\text{.67}\mu \text{V}$ for the output to be at limit

Case 2: For $v_{-}$ has to exceed $v_{+}$ corresponding to an output of $v_{0+}$

$-14\text{V}$

$v_{0+}=A(v_{+}-v_{-})=-14\text{V}$

Here, $A=600,000$

Hence,

$\begin{array}{l}-14=600,000(v_{+}-v_{-})\\ \text{or}(v_{+}-v_{-})=\frac{-14}{600,000}\text{=}-23.33\mu \text{V }\end{array}$

This implies $v_{-}$ has to exceed $v_{+}$ by $23.33\mu \text{V}$ 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 $-14\text{V}$ on the negative side. With open loop gain as A

We have,

$v_{0+}=A(v_{+}-v_{-})=+13\text{V}$ and $v_{0-}=A(v_{+}-v_{-})=-14\text{V}$

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

Answer to Problem 3.1QAP

$v_{+}$ has to exceed $v_{-}$ by $\text{5 }\mu \text{V}$ for the output to be at limit +13 V and $v_{-}$ has to exceed $v_{+}$ by $5.384\mu \text{V}$ for output to be at limit $-14\text{ V}$

Explanation of Solution

Given the open loop gain A is $2.6\times {10}^6$.

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 $-14\text{ V}$ 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 $v_{0+}$ +13 V

$v_{0+}=A(v_{+}-v_{-})$ =+13 V

Here, $A=2.6\times {10}^6$

Hence,

$\begin{array}{l}13=2.6\times {10}^6(v_{+}-v_{-})\\ or(v_{+}-v_{-})=\frac{13}{2.6\times {10}^6}\text{=5}\mu \text{V }\end{array}$

This implies $v_{+}$ has to exceed $v_{-}$ by $5\mu \text{V}$ for the output to be at limit

Case 2 : For $v_{-}$ has to exceed $v_{+}$ - corresponding to an output of $v_{0+}$

$-14\text{ V}$

$v_{0+}=A(v_{+}-v_{-})=-14\text{V}$

Here, $A=2.6\times {10}^6$

Hence,

$\begin{array}{l}-14=2.6\times {10}^6(v_{+}-v_{-})\\ \text{or}(v_{+}-v_{-})=\frac{-14}{2.6\times {10}^6}\text{=}-\text{5}\text{.384}\mu \text{V }\end{array}$

This implies $v_{-}$ has to exceed $v_{+}$ by $5.384\mu \text{V}$ for output to be at limit