Chapter 4, Problem 94CP

An attenuator is an interface circuit that reduces the voltage level without changing the output resistance.

1. (a) By specifying R_s and R_p of the interface circuit in Fig. 4.150, design an attenuator that will meet the following requirements:

   $\frac{V_o}{V_g}=0.125,R_{\text{eq}}=R_{\text{Th}}=R_g=100\Omega$

2. (b) Using the interface designed in part (a), calculate the current through a load of R_L = 50 Ω when V_g = 12 V.

Figure 4.150

To determine

Find the value of the resistance $R_s$ and $R_p$ of the circuit shown in Figure 4.150.

Answer to Problem 94CP

The value of the resistance $R_s$ is $700\Omega$ and $R_p$ is $114.29\Omega$ in the given circuit.

Explanation of Solution

Given data:

Refer to Figure 4.150 in the textbook.

The voltage,

$\frac{V_o}{V_g}=0.125$

The resistance,

$R_{\text{eq}}=R_{\text{Th}}=R_g=100\Omega$

Calculation:

In the given circuit, find the voltage $V_o$ by removing the load resistor $R_L$ and the modified circuit is shown in Figure 1.

In Figure 1, the voltage $V_o$ is calculated by using voltage division principle as follows.

$V_o=\left(V_g\right)\left(\frac{R_p}{R_g+R_s+R_p}\right)$

$\frac{V_o}{V_g}=\frac{R_p}{R_g+R_s+R_p}$        (1)

Let us consider $\alpha =\frac{V_o}{V_g}$.

Substitute $\alpha$ for $\frac{V_o}{V_g}$ in equation (1) as follows.

$\begin{array}{l}\alpha =\frac{R_p}{R_g+R_s+R_p}\\ \frac{R_p}{\alpha }=R_g+R_s+R_p\\ \frac{R_p}{\alpha }-R_p=R_g+R_s\\ R_p\left(\frac{1}{\alpha }-1\right)=R_g+R_s\end{array}$

Simplify the equation as follows,

$R_p\left(\frac{1-\alpha }{\alpha }\right)=R_g+R_s$        (2)

Refer to Figure 4.150 in the textbook.

In the given circuit, find the Thevenin resistance by turning off the voltage source $V_g$ (replace with a short circuit) and remove the load resistor $R_L$ and the modified circuit is shown in Figure 2.

In Figure 2, the Thevenin resistance is,

$\begin{array}{l}{R}_{\text{Th}}=\left(R_g+R_s\right)\left|\right|\left(R_p\right)\\ R_g=\frac{\left(R_g+R_s\right)\left(R_p\right)}{\left(R_g+R_s+R_p\right)}\left\{\because R_{\text{Th}}=R_g\right\}\\ R_p{R}_g+R_g{}^2+R_g{R}_s=R_p{R}_g+R_p{R}_s\\ R_p{R}_g+R_g{}^2+R_g{R}_s-R_p{R}_g=R_p{R}_s\end{array}$

Simplify the equation as follows,

$\begin{array}{l}{R}_g{}^2+R_g{R}_s=R_p{R}_s\\ R_p{R}_s=R_g\left(R_g+R_s\right)\end{array}$

$\frac{R_p{R}_s}{R_g}=R_g+R_s$        (3)

On comparing equation (2) and (3),

$R_p\left(\frac{1-\alpha }{\alpha }\right)=\frac{R_p{R}_s}{R_g}$

$R_s=R_g\left(\frac{1-\alpha }{\alpha }\right)$        (4)

Substitute 100 for $R_g$ and $0.125$ for $\alpha$ in equation (4) to find the value of resistance $R_s$ in ohms.

$\begin{array}{c}{R}_s=100\left(\frac{1-0.125}{0.125}\right)\left\{\because \alpha =\frac{V_o}{V_g}=0.125\right\}\\ =700\Omega \end{array}$

Substitute 100 for $R_g$, 700 for $R_s$, and $0.125$ for $\alpha$ in equation (2) to find the value of the resistance $R_p$ in ohms.

$\begin{array}{l}{R}_p\left(\frac{1-0.125}{0.125}\right)=100+700\\ R_p\left(\frac{1-0.125}{0.125}\right)=800\\ R_p\left(7\right)=800\\ R_p=114.29\Omega \end{array}$

Conclusion:

Thus, the value of the resistance $R_s$ is $700\Omega$ and $R_p$ is $114.29\Omega$ in the given circuit.

To determine

Find the current through the load resistor $R_L$ when $V_g=12\text{V}$.

Answer to Problem 94CP

The current through the load resistor $R_L$ is $10\text{mA}$.

Explanation of Solution

Given data:

Refer to Figure 4.150 in the textbook.

The voltage $V_g=12\text{V}$.

The load resistor $R_L=50\Omega$.

The resistance,

$R_{\text{eq}}=R_{\text{Th}}=R_g=100\Omega$

The voltage,

$\frac{V_o}{V_g}=0.125$        (5)

Calculation:

The Thevenin equivalent is shown in Figure 3.

Rearrange the equation (5) as follows,

$V_o=0.125V_g$

Substitute 12 for $V_g$ to find the voltage $V_o$ in volts.

$\begin{array}{c}{V}_o=0.125\times 12\\ =1.5\text{V}\end{array}$

In Figure 3, the voltage $V_o$ is the Thevenin voltage. Therefore,

$V_{\text{Th}}=V_o=1.5\text{V}$

In Figure 3, the current through the load resistor $R_L$ is calculated by using Ohm’s law as follows.

$I=\frac{V_{\text{Th}}}{R_{\text{Th}}+R_L}$

Substitute 1.5 for $V_{\text{Th}}$, 50 for $R_L$, and 100 for $R_{\text{Th}}$ to find the current through the load resistor $R_L$ in amperes.

$\begin{array}{c}I=\frac{1.5}{100+50}\\ =0.01\times {10}^3\times {10}^{-3}\text{A}\\ =\text{10}\text{mA}\left\{\because 1\text{m}={10}^{-3}\right\}\end{array}$

Conclusion:

Thus, the current through the load resistor $R_L$ is $10\text{mA}$.