Chapter 5, Problem 84P

A four-bit R-2R ladder DAC is presented in Fig. 5.103.

1. (a) Show that the output voltage is given by

   $-V_0=R_f\left(\frac{V_1}{2R}+\frac{V_2}{4R}+\frac{V_3}{8R}+\frac{V_4}{16R}\right)$

2. (b) If R_f = 12 kΩ and R = 10 kΩ, find |V₀| for [V₁V₂V₃V₄] = [1011] and [V₁V₂V₃V₄] = [0101].

To determine

Show that the output voltage of the four-bit $R-2R$ ladder DAC is $-V_o=R_f\left(\frac{V_1}{2R}+\frac{V_2}{4R}+\frac{V_3}{8R}+\frac{V_4}{16R}\right)$

Explanation of Solution

Calculation:

The given figure with the representation of node voltage is shown in Figure 1.

Apply Kirchhoff’s current law at node $V_x$.

$\begin{array}{l}\frac{V_x-V_4}{2R}+\frac{V_x-V_y}{R}+\frac{V_x-0}{2R}=0\\ \frac{V_x-V_4+2\left(V_x-V_y\right)+V_x}{2R}=0\\ V_x-V_4+2V_x-2V_y+V_x=0\end{array}$

$4V_x-2V_y=V_4$        (1)

Apply Kirchhoff’s current law at node $V_y$.

$\begin{array}{l}\frac{V_y-V_z}{R}+\frac{V_y-V_3}{2R}+\frac{V_y-V_x}{R}=0\\ \frac{2\left(V_y-V_z\right)+V_y-V_3+2\left(V_y-V_x\right)}{2R}=0\\ 2V_y-2V_z+V_y-V_3+2V_y-2V_x=0\end{array}$

$-2V_x+5V_y-2V_z=V_3$        (2)

Apply Kirchhoff’s current law at node $V_z$.

$\begin{array}{l}\frac{V_z-V_2}{2R}+\frac{V_z-V_y}{R}+\frac{V_z-V_n}{R}=0\\ \frac{V_z-V_2+2\left(V_z-V_y\right)+2\left(V_z-V_n\right)}{2R}=0\\ V_z-V_2+2V_z-2V_y+2V_z-2V_n=0\end{array}$

$-2V_y+5V_z-2V_n=V_2$        (3)

According to the properties of the ideal op amp, the voltage at non-inverting terminal of the op amp is same as the voltage at the inverting terminal of the op amp. The non-inverting terminal of the op amp is grounded. Therefore,

$V_n=0$

Substitute $0$ for $V_n$ in equation (3).

$-2V_y+5V_z-2\left(0\right)=V_2$

$-2V_y+5V_z=V_2$        (4)

Apply Kirchhoff’s current law at node $V_n$.

$\frac{V_n-V_1}{2R}+\frac{V_n-V_z}{R}+\frac{V_n-V_o}{R_f}=0$        (5)

Substitute $0$ for $V_n$ in equation (5).

$\begin{array}{l}\frac{0-V_1}{2R}+\frac{0-V_z}{R}+\frac{0-V_o}{R_f}=0\\ \frac{V_o}{R_f}=-\frac{V_1}{2R}-\frac{V_z}{R}\\ \frac{V_o}{R_f}=\frac{-V_1-2V_z}{2R}\end{array}$

$-V_o=\frac{R_f}{2R}\left(V_1+2V_z\right)$        (6)

Multiply equation (2) with 2.

$-4V_x+10V_y-4V_z=2V_3$        (7)

Add the equation (1) and (7).

$\begin{array}{l}4V_x-2V_y-4V_x+10V_y-4V_z=V_4+2V_3\\ 8V_y-4V_z=V_4+2V_3\\ 4\left(2V_y-V_z\right)=V_4+2V_3\end{array}$

$2V_y-V_z=\frac{V_4+2V_3}{4}$        (8)

Add the equation (4) and (8).

$\begin{array}{l}-2V_y+5V_z+2V_y-V_z=V_2+\frac{V_4+2V_3}{4}\\ 4V_z=V_2+\frac{2V_3+V_4}{4}\end{array}$

$2V_z=\frac{V_2}{2}+\frac{2V_3+V_4}{8}$        (9)

Substitute equation (9) in (6).

$\begin{array}{c}-V_o=\frac{R_f}{2R}\left(V_1+\frac{V_2}{2}+\frac{2V_3+V_4}{8}\right)\\ =\frac{R_f}{2R}\left(V_1+\frac{V_2}{2}+\frac{V_3}{4}+\frac{V_4}{8}\right)\end{array}$

$-V_o=\frac{R_f}{R}\left(\frac{V_1}{2}+\frac{V_2}{4}+\frac{V_3}{8}+\frac{V_4}{16}\right)$        (10)

$-V_o=R_f\left(\frac{V_1}{2R}+\frac{V_2}{4R}+\frac{V_3}{8R}+\frac{V_4}{16R}\right)$

Conclusion:

Thus, the output voltage of the four-bit $R-2R$ ladder DAC is proved as $-V_o=R_f\left(\frac{V_1}{2R}+\frac{V_2}{4R}+\frac{V_3}{8R}+\frac{V_4}{16R}\right)$

To determine

Find the value of $\left|V_o\right|$ when $\left[V_1{V}_2{V}_3{V}_4\right]=\left[1011\right]$ and $\left[V_1{V}_2{V}_3{V}_4\right]=\left[0101\right]$.

Answer to Problem 84P

The value of $\left|V_o\right|$ is $\underset{\_}{825\text{mV}}$ when $\left[V_1{V}_2{V}_3{V}_4\right]=\left[1011\right]$ and the value of $\left|V_o\right|$ is $\underset{\_}{375\text{mV}}$ when $\left[V_1{V}_2{V}_3{V}_4\right]=\left[0101\right]$.

Explanation of Solution

Given data:

The given value of $\left[V_1{V}_2{V}_3{V}_4\right]=\left[1011\right]$.

The value of $R$ and $R_f$ are $10\text{kΩ}$ and $12\text{kΩ}$ respectively.

Calculation:

For a digital input $\left[V_1{V}_2{V}_3{V}_4\right]=\left[1011\right]$, calculate the value of $\left|V_o\right|$.

Substitute $10\text{kΩ}$ for $R$, $12\text{kΩ}$ for $R_f$, 1 for $V_1$, 0 for $V_2$, 1 for $V_3$, and 1 for $V_4$ in equation (10).

$\begin{array}{c}\left|V_o\right|=\left(\frac{12\text{k}}{10\text{k}}\right)\left(\frac{1}{2}+\frac{0}{4}+\frac{1}{8}+\frac{1}{16}\right)\\ =\left(1.2\right)\left(\frac{8+2+1}{16}\right)\\ =0.825\text{V}\\ =\text{825}\text{mV}\end{array}$

For a digital input $\left[V_1{V}_2{V}_3{V}_4\right]=\left[0101\right]$, calculate the value of $\left|V_o\right|$.

Substitute $10\text{kΩ}$ for $R$, $12\text{kΩ}$ for $R_f$, 0 for $V_1$, 1 for $V_2$, 0 for $V_3$, and 1 for $V_4$ in equation (10).

$\begin{array}{c}\left|V_o\right|=\left(\frac{12\text{k}}{10\text{k}}\right)\left(\frac{0}{2}+\frac{1}{4}+\frac{0}{8}+\frac{1}{16}\right)\\ =\left(1.2\right)\left(\frac{4+1}{16}\right)\\ =0.375\text{V}\\ =37\text{5}\text{mV}\end{array}$

Conclusion:

Thus, the value of $\left|V_o\right|$ is $\underset{\_}{825\text{mV}}$ when $\left[V_1{V}_2{V}_3{V}_4\right]=\left[1011\right]$ and the value of $\left|V_o\right|$ is $\underset{\_}{375\text{mV}}$ when $\left[V_1{V}_2{V}_3{V}_4\right]=\left[0101\right]$.