Chapter 16, Problem 16.99P

(a)

To determine

The maximum allowed tolerance in the $R_1$

Answer to Problem 16.99P

The maximum allowed value of the tolerance is $5.88\%$ .

Explanation of Solution

Calculation:

The given diagram is shown in Figure 1

  

The expression to determine the analog output voltage $v_O$ when the resistor $R_1$ is connected to the reference voltage is given by,

  $v_O=\left(\frac{b_1}{2}+\frac{b_2}{4}+\frac{b_3}{8}+\frac{b_4}{16}\right)\left(-V_R\right)$

Substitute $-5\text{V}$ for $V_R$ , $1$ for $b_1$ , $0$ for $b_2$ , $0$ for $b_3$ and $0$ for $b_4$ in the above equation.

  $\begin{array}{c}{v}_O=\left(\frac{1}{2}+\frac{0}{4}+\frac{0}{8}+\frac{0}{16}\right)\left(-\left(-5\text{V}\right)\right)\\ =2.5\text{V}\end{array}$

The expression to determine the value of the least significant digit is given by,

  $\text{LSB}=\frac{V_R}{2^n}$

Substitute $-5\text{V}$ for $V_R$ and $4$ for $n$ in the above equation.

  $\begin{array}{c}\text{LSB}=\frac{-5\text{V}}{2^4}\\ =-0.3125\text{V}\end{array}$

The expression to determine the value of the output analog voltage with maximum error is given by,

  $v_{O\left(\max \right)}=v_O\pm \frac{1}{2}\text{LSB}$

Substitute $2.5\text{V}$ for $v_O$ and $-0.3125\text{V}$ for $\text{LSB}$ in the above equation.

  $\begin{array}{c}{v}_{O\left(\max \right)}=\left(2.5\text{V}\right)\pm \frac{1}{2}\left(-0.3125\text{V}\right)\\ =2.34375\text{V},2.65625\text{V}\end{array}$ \

The expression to determine the value of the output analog voltage with maximum error $\left(v_{O\left(\max \right)}\right)$ in terms of the maximum allowable tolerance in the resistance $R_1$ from the circuit is given by,

  $v_{O\left(\max \right)}=\left(\frac{R_F}{R_1+\Delta R_1}\right)\left(-V_R\right)$

Substitute $10\text{k}\Omega$ for $R_F$ , $20\text{k}\Omega$ for $R_1$ and $-5\text{V}$ for $V_R$ in the above equation.

  $\begin{array}{l}{v}_{O\left(\max \right)}=\left(\frac{10\text{k}\Omega }{20\text{k}\Omega +\Delta R_1}\right)\left(-\left(-5\text{V}\right)\right)\\ \Delta R_1=\left(\frac{10\text{k}\Omega }{v_{O\left(\max \right)}}\right)\left(5\text{V}\right)-20\text{k}\Omega \end{array}$   ...... (1)

Substitute $2.34375\text{V}$ for $v_{O\left(\max \right)}$ in the above equation.

  $\begin{array}{c}\Delta R_1=\left(\frac{10\text{k}\Omega }{2.34375\text{V}}\right)\left(5\text{V}\right)-20\text{k}\Omega \\ =1.333\text{k}\Omega \end{array}$

The expression to determine the percentage allowed tolerance in the value of resistance $R_1$ in terms of percentage $\%\Delta R_1$ is given by,

  $\%\Delta R_1=\left(\frac{\Delta R_1}{R_1}\right)100$   ...... (2)

Substitute $1.333\text{k}\Omega$ for $\Delta R_1$ and $20\text{k}\Omega$ for $R_1$ in the above equation.

  $\begin{array}{c}\%\Delta R_1=\left(\frac{1.333\text{k}\Omega }{20\text{k}\Omega }\right)100\\ =6.7\%\end{array}$

Substitute $2.65625\text{V}$ for $v_{O\left(\max \right)}$ in equation (1)

  $\begin{array}{c}\Delta R_1=\left(\frac{10\text{k}\Omega }{2.65625\text{V}}\right)\left(5\text{V}\right)-20\text{k}\Omega \\ =\left|-1.1765\text{k}\Omega \right|\\ =1.1765\text{k}\Omega \end{array}$

Substitute $1.1765\text{k}\Omega$ for $\Delta R_1$ and $20\text{k}\Omega$ for $R_1$ in equation (2)

  $\begin{array}{c}\%\Delta R_1=\left(\frac{1.1765\text{k}\Omega }{20\text{k}\Omega }\right)100\\ =5.88\%\end{array}$

Conclusion:

Therefore, the maximum allowed value of the tolerance is $5.88\%$ .

(b)

To determine

The maximum allowed tolerance in the value of $R_4$

Answer to Problem 16.99P

The maximum allowed value of the tolerance is $33.3\%$ .

Explanation of Solution

Calculation:

The expression to determine the analog output voltage $v_O$ when the resistor $R_4$ is connected to the reference voltage is given by,

  $v_O=\left(\frac{b_1}{2}+\frac{b_2}{4}+\frac{b_3}{8}+\frac{b_4}{16}\right)\left(-V_R\right)$

Substitute $-5\text{V}$ for $V_R$ , $0$ for $b_1$ , $0$ for $b_2$ , $0$ for $b_3$ and $1$ for $b_4$ in the above equation.

  $\begin{array}{c}{v}_O=\left(\frac{0}{2}+\frac{0}{4}+\frac{0}{8}+\frac{1}{16}\right)\left(-\left(-5\text{V}\right)\right)\\ =0.3125\text{V}\end{array}$

The expression to determine the value of the output analog voltage with maximum error is given by,

  $v_{O\left(\max \right)}=v_O\pm \frac{1}{2}\text{LSB}$

Substitute $0.3125\text{V}$ for $v_O$ and $-0.3125\text{V}$ for $\text{LSB}$ in the above equation.

  $\begin{array}{c}{v}_{O\left(\max \right)}=\left(0.3125\text{V}\right)\pm \frac{1}{2}\left(-0.3125\text{V}\right)\\ =0.15625\text{V},0.46875\text{V}\end{array}$

The expression to determine the value of the output analog voltage with maximum error $\left(v_{O\left(\max \right)}\right)$ in terms of the maximum allowable tolerance in the resistance $R_4$ from the circuit is given by,

  $v_{O\left(\max \right)}=\left(\frac{R_F}{R_4+\Delta R_4}\right)\left(-V_R\right)$

Substitute $10\text{k}\Omega$ for $R_F$ , $160\text{k}\Omega$ for $R_4$ and $-5\text{V}$ for $V_R$ in the above equation.

  $\begin{array}{l}{v}_{O\left(\max \right)}=\left(\frac{10\text{k}\Omega }{160\text{k}\Omega +\Delta R_4}\right)\left(-\left(-5\text{V}\right)\right)\\ \Delta R_4=\left(\frac{10\text{k}\Omega }{v_{O\left(\max \right)}}\right)\left(5\text{V}\right)-160\text{k}\Omega \end{array}$   ...... (3)

Substitute $0.15625\text{V}$ for $v_{O\left(\max \right)}$ in the above equation.

  $\begin{array}{c}\Delta R_4=\left(\frac{10\text{k}\Omega }{0.15625\text{V}}\right)\left(5\text{V}\right)-160\text{k}\Omega \\ =160\text{k}\Omega \end{array}$

The expression to determine the percentage allowed tolerance in the value of resistance $R_4$ in terms of percentage $\%\Delta R_4$ is given by,

  $\%\Delta R_4=\left(\frac{\Delta R_4}{R_4}\right)100$   ...... (4)

Substitute $160\text{k}\Omega$ for $\Delta R_4$ and $160\text{k}\Omega$ for $R_4$ in the above equation. $\Delta R_4$

  $\begin{array}{c}\%\Delta R_4=\left(\frac{160\text{k}\Omega }{160\text{k}\Omega }\right)100\\ =100\%\end{array}$

Substitute $0.46875\text{V}$ for $v_{O\left(\max \right)}$ in equation (1)

  $\begin{array}{c}\Delta R_4=\left(\frac{10\text{k}\Omega }{0.46875\text{V}}\right)\left(5\text{V}\right)-160\text{k}\Omega \\ =\left|-53.3334\text{k}\Omega \right|\\ =53.3334\text{k}\Omega \end{array}$

Substitute $53.3334\text{k}\Omega$ for $\Delta R_4$ and $160\text{k}\Omega$ for $R_4$ in equation (2)

  $\begin{array}{c}\%\Delta R_4=\left(\frac{53.3334\text{k}\Omega }{160\text{k}\Omega }\right)100\\ =33.3\%\end{array}$

Conclusion:

Therefore, the maximum allowed value of the tolerance is $33.3\%$ .