Chapter 5, Problem 83P

Design a six-bit digital-to-analog converter.

1. (a) If |V_o| = 1.1875 V is desired, what should [V₁V₂V₃V₄V₅V₆] be?
2. (b) Calculate |V_o| if [V₁V₂V₃V₄V₅V₆| = [011011].
3. (c) What is the maximum value |V_o| can assume?

To determine

Find the required value of $\left[V_1{V}_2{V}_3{V}_4{V}_5{V}_6\right]$ when designing a six-bit digital-to-analog converter.

Answer to Problem 83P

The value of $\left[V_1{V}_2{V}_3{V}_4{V}_5{V}_6\right]$ is $\underset{\_}{\left[100110\right]}$.

Explanation of Solution

Given data:

The given value of $\left|V_o\right|$ is 1.1875 V.

Calculation:

To design a six-bit digital-to-analog converter, consider the following resistors.

$R_f=10\text{k}\Omega ,R_1=10\text{k}\Omega ,R_2=20\text{k}\Omega ,R_3=40\text{k}\Omega ,R_4=80\text{k}\Omega ,R_5=160\text{k}\Omega ,\text{and}{R}_6=320\text{k}\Omega$.

Consider the expression for the output voltage of 6 bit inverting summer that is a six-bit digital-to-analog converter.

$-V_o=\frac{R_f}{R_1}{V}_1+\frac{R_f}{R_2}{V}_2+\frac{R_f}{R_3}{V}_3+\frac{R_f}{R_4}{V}_4+\frac{R_f}{R_5}{V}_5+\frac{R_f}{R_6}{V}_6$

Substitute $10\text{k}\Omega$ for $R_f$, $10\text{k}\Omega$ for $R_1$, $20\text{k}\Omega$ for $R_2$, $40\text{k}\Omega$ for $R_3$, $80\text{k}\Omega$ for $R_4$, $160\text{k}\Omega$ for $R_5$, and $320\text{k}\Omega$ for $R_6$.

$-V_o=\frac{10\text{k}\Omega }{10\text{k}\Omega }{V}_1+\frac{10\text{k}\Omega }{20\text{k}\Omega }{V}_2+\frac{10\text{k}\Omega }{40\text{k}\Omega }{V}_3+\frac{10\text{k}\Omega }{80\text{k}\Omega }{V}_4+\frac{10\text{k}\Omega }{160\text{k}\Omega }{V}_5+\frac{10\text{k}\Omega }{320\text{k}\Omega }{V}_6$

$\left|V_o\right|=V_1+0.5V_2+0.25V_3+0.125V_4+0.0625V_5+0.03125V_6$ (1)

Here,

$\left|V_o\right|=-V_o$

In Equation (2), a digital input $\left[V_1{V}_2{V}_3{V}_4{V}_5{V}_6\right]=\left[100110\right]$ generates an analog output of

$\begin{array}{c}\left|V_o\right|=1+0.5\left(0\right)+0.25\left(0\right)+0.125\left(1\right)+0.0625\left(1\right)+0.03125\left(0\right)\\ =1+0+0+0.125+0.0625+0\end{array}$

$\left|V_o\right|=1+0+0+\left(\frac{1}{8}\right)+\left(\frac{1}{16}\right)+0$ (2)

Simplify the Equation (2) gives,

$\begin{array}{c}\left|V_o\right|=1+0+0+0.125+0.0625+0\\ =1.1875\text{V}\end{array}$.

Therefore, the required value of $\left[V_1{V}_2{V}_3{V}_4{V}_5{V}_6\right]$ is $\underset{\_}{\left[100110\right]}$.

Conclusion:

Thus, the value of $\left[V_1{V}_2{V}_3{V}_4{V}_5{V}_6\right]$ is $\underset{\_}{\left[100110\right]}$.

To determine

Find the value of $\left|V_o\right|$ when $\left[V_1{V}_2{V}_3{V}_4{V}_5{V}_6\right]=\left[011011\right]$.

Answer to Problem 83P

The value of $\left|V_o\right|$ is $\underset{\_}{843.75\text{mV}}$.

Explanation of Solution

Given data:

The given value of $\left[V_1{V}_2{V}_3{V}_4{V}_5{V}_6\right]=\left[011011\right]$.

Calculation:

For a digital input $\left[V_1{V}_2{V}_3{V}_4{V}_5{V}_6\right]=\left[011011\right]$, calculate the value of $\left|V_o\right|$.

Substitute 0 for $V_1$, 1 for $V_2$, 1 for $V_3$, 0 for $V_4$, 1 for $V_5$, and 1 for $V_1$ in Equation (1).

$\begin{array}{c}\left|V_o\right|=0+0.5\left(1\right)+0.25\left(1\right)+0.125\left(0\right)+0.0625\left(1\right)+0.03125\left(1\right)\\ =0+0.5+0.25+0+0.0625+0.03125\\ =0.84375\text{V}\\ =843.75\text{mV}\end{array}$

Conclusion:

Thus, the value of $\left|V_o\right|$ is $\underset{\_}{843.75\text{mV}}$.

To determine

Calculate the maximum value of $\left|V_o\right|$.

Answer to Problem 83P

The maximum value of $\left|V_o\right|$ is $\underset{\_}{1.96875\text{V}}$.

Explanation of Solution

Given data:

Consider that the digital input $\left[V_1{V}_2{V}_3{V}_4{V}_5{V}_6\right]=\left[111111\right]$ for the maximum value of $\left|V_o\right|$.

Calculation:

Substitute 1 for $V_1$, 1 for $V_2$, 1 for $V_3$, 1 for $V_4$, 1 for $V_5$, and 1 for $V_1$ in Equation (1).

$\begin{array}{c}\left|V_o\right|=1+0.5\left(1\right)+0.25\left(1\right)+0.125\left(1\right)+0.0625\left(1\right)+0.03125\left(1\right)\\ =1+0.5+0.25+0.125+0.0625+0.03125\\ =1.96875\text{V}\end{array}$

Conclusion:

Thus, the maximum value of $\left|V_o\right|$ is $\underset{\_}{1.96875\text{V}}$.