Question

An operational amplifier has output voltage limits of$+13\text{V}$ and $-14\text{V}$when used with$a\pm 15\text{V}$power supply. If the amplifier is used as a comparator, by what amount does $V_{+}$have to exceed $V_{-}$and $v_{-}$have to exceed $v_{+}$for the amplifier to be at limit if the open-loop gain A is

(a) 100,000

(b) 600,000

(c)$2.6\times {10}^6$

Short answer

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

Step-by-step solution

Step 1. Introduction

Write the expression for output of amplifier.

$v_{\text{o}}=A{v}_{\text{s}}=A\left(v_{+}-v_{-}\right)$

Rearrange for input difference voltage.

role="math" localid="1646896721915" $\left(v_{+}-v_{-}\right)=\frac{v_0}{A}\dots \dots \text{(1)}$

Here, A is open-loop voltage gain, and is$v_0$ output of amplifier.

Part (a)

If the open loop-gain $A=100000$.

Calculate the positive limit of amplifier.

Substitute $+13\text{V}$for $v_{\text{o}}$, and 100000 for A in equation (1).

$\left(v_{+}-v_{-}\right)=\frac{+13\text{V}}{100000}=\left(130\times {10}^{-6}\text{V}\right)\left(\frac{1\mu \text{V}}{{10}^{-6}\text{V}}\right)=130\mu \text{V}$

Substitute $-14\text{V}$for $v_0$, and 100000 for A in equation (1).

$\left(v_{+}-v_{-}\right)=\frac{-14\text{V}}{100000}=-\left(140\times {10}^{-6}\text{V}\right)\left(\frac{1\mu \text{V}}{{10}^{-6}\text{V}}\right)=-140\mu \text{V}$

Part (b)

If the open-loop gain .

Calculate the positive limit of amplifier.

Substitute $+13\text{V}$for $v_{\text{o}}$, and 600000 for A in equation (1).

$\left(v_{+}-v_{-}\right)=\frac{+13\text{V}}{600000}=\left(21.7\times {10}^{-6}\text{V}\right)\left(\frac{1\mu \text{V}}{{10}^{-6}\text{V}}\right)=21.7\mu \text{V}$

Calculate the negative limit of amplifier.

Substitute $-14\text{V}$for $v_{\text{o}}$, and 600000 for A in equation (1).

$\left(v_{+}-v_{-}\right)=\frac{-14\text{V}}{600000}=-\left(23.3\times {10}^{-6}\text{V}\right)\left(\frac{1\mu \text{V}}{{10}^{-6}\text{V}}\right)=-23.3\mu \text{V}$

Part(c)

If the open-loop gain $A=2.6\times {10}^6$.

Calculate the positive limit of amplifier.

Substitute +13V for $v_{\text{o}}$, and $2.6\times {10}^6$for A in equation (1).

$\left(v_{+}-v_{-}\right)=\frac{+13\text{V}}{2.6\times {10}^6}=\left(5.0\times {10}^{-6}\text{V}\right)\left(\frac{1\mu \text{V}}{{10}^{-6}\text{V}}\right)=5.0\mu \text{V}$

Calculate the negative limit of amplifier.

Substitute $-14\text{V}$for $v_0$, $2.6\times {10}^6$and A for in equation (1).

$\left(v_{+}-v_{-}\right)=\frac{-14\text{V}}{2.6\times {10}^6}=-\left(5.4\times {10}^{-6}\text{V}\right)\left(\frac{1\mu \text{V}}{{10}^{-6}\text{V}}\right)=-5.4\mu \text{V}$

Therefore, the positive limit must exceed by $5.0\mu \text{V}$, and the negative limit must exceed by $5.4\mu \text{V}$.