Question

For a comparator with output voltage limits of$\pm 13\text{V}$ , what would the open-loop gain A need to be to keep the absolute value of the difference voltage$\left|vs\right|\le 2.0\mu \text{V}$ ?

Short answer

The required open loop gain of the amplifier is $6.5\times {10}^6$.

Step-by-step solution

Step 1. Introduction

The expression for gain of an operational amplifier.

$A=\frac{v_o}{v_{\text{s}}}$

Here,$v_o$ is output voltage change, and $v_s$is difference voltage voltage.

Step 2. Explanation

Calculate the open loop gain of an operational amplifier.

Substitute $13\text{V}$for $v_o$, and $2\mu \text{V}$for$v_s$in gain expression.

$\begin{array}{l}A=\frac{13\text{V}}{2\mu \text{V}}=\frac{13\text{V}}{\left(2\mu \text{V}\right)\left(\frac{{10}^{-6}\text{V}}{1\mu \text{V}}\right)}\\ =\frac{13\text{V}}{\left(2\times {10}^{-6}\text{V}\right)}\\ =6.5\times {10}^6\end{array}$

Therefore, the required open loop gain of the amplifier is$6.5\times {10}^6$ .