Problem

What range of output voltage is developed in the circuit of Fig. 10.67?FIG. 10.67

What range of output voltage is developed in the circuit of Fig. 10.67?

FIG. 10.67

Step-by-Step Solution

Solution 1

Step #1 of 2

Refer FIG. $10.67$ in textbook.

The amplifier circuit is in non-inverting condition.

Write the expression for output voltage of non-inverting amplifier.

$$ V_{o}=\left(1+\frac{R_{f}}{R_{1}}\right) V_{1} $$

For maximum value of the output voltage, the input resistance is kept small. So the input resistance for the amplifier is $10 \mathrm{k} \Omega$.

$$ R_{1}=10 \mathrm{k} \Omega $$

Substitute $200 \mathrm{k} \Omega$ for $R_{f}, 0.5 \mathrm{~V}$ for $V_{1}$, and $10 \mathrm{k} \Omega$ for $R_{1}$ in the equation.

$$ \begin{aligned} V_{o} &=\left(1+\frac{200 \mathrm{k} \Omega}{10 \mathrm{k} \Omega}\right)(0.5 \mathrm{~V}) \\ &=(1+20)(0.5) \\ &=10.5 \mathrm{~V} \end{aligned} $$

Hence, the output voltage of non-inverting amplifier is $10.5 \mathrm{~V}$.

Step #2 of 2

Write the expression for output voltage of non-inverting amplifier.

$$ V_{o}=\left(1+\frac{R_{f}}{R_{1}^{\prime}}\right) V_{1} $$

For minimum value of the output voltage, the input resistance is kept small. The input resistance for the amplifier is,

$$ \begin{aligned} R_{1}^{\prime} &=10 \mathrm{k} \Omega+10 \mathrm{k} \Omega \\ &=20 \mathrm{k} \Omega \end{aligned} $$

Substitute $200 \mathrm{k} \Omega$ for $R_{f}, 0.5 \mathrm{~V}$ for $V_{1}$, and $20 \mathrm{k} \Omega$ for $R_{1}^{\prime}$ in the equation.

$$ \begin{aligned} V_{o} &=\left(1+\frac{200 \mathrm{k} \Omega}{20 \mathrm{k} \Omega}\right)(0.5 \mathrm{~V}) \\ &=(1+10)(0.5) \\ &=5.5 \mathrm{~V} \end{aligned} $$

Hence, the output voltage of non-inverting amplifier is $5.5 \mathrm{~V}$.

Hence, the output voltage range for the amplifier operation is $(5.5 \mathrm{~V}-10.5 \mathrm{~V}) .$

Solution 2