Chapter 4, Problem 74E

A light-sensing circuit is in Fig. 4.90, including a resistor that changes value under illumination (photoresistor R_light) and a variable resistor (potentiometer R_pot). The circuit is in the Wheatstone bridge configuration such that a “balanced” condition results in V_out = 0 for a defined value of incident light and a corresponding value for R_light. (a) Derive an algebraic expression for V_out in terms of R_S, R₁, R₂, R_light, and R_pot. (b) Using the numerical values given in the circuit, calculate the value of R_pot required to balance the circuit at 500 lux, where R_light = 200 Ω. (c) If the resistance of the photoresistor decreases by 2% for a light increase to 600 lux (and assuming the resistance change with light is linear), what will the light level be if you measure V_out = 150 mV?

■ FIGURE 4.90

To determine

Write an algebraic expression for the output voltage $V_{\text{out}}$ in terms of $R_S$, $R_1$, $R_2$, $R_{\text{light}}$, and $R_{\text{pot}}$.

Explanation of Solution

Given data:

Refer to Figure 4.90 in the textbook.

Calculation:

The given circuit of Figure 4.90 is redrawn as shown in Figure 1.

Apply Kirchhoff’s voltage law for the loop current $i_1$ in Figure 1.

$\begin{array}{l}{i}_1{R}_S+R_1\left(i_1-i_2\right)+R_2\left(i_1-i_2\right)=V_S\\ i_1{R}_S+R_1{i}_1-R_1{i}_2+R_2{i}_1-R_2{i}_2=V_S\\ i_1{R}_S+R_1{i}_1+R_2{i}_1=V_S+R_1{i}_2+R_2{i}_2\end{array}$

$\left(R_S+R_1+R_2\right)i_1-\left(R_1+R_2\right)i_2=V_S$        (1)

Apply Kirchhoff’s voltage law for the loop current $i_1$ in Figure 1.

$\begin{array}{l}{R}_2\left(i_2-i_1\right)+R_1\left(i_2-i_1\right)+R_{light}{i}_2+R_{pot}{i}_2=0\\ R_2{i}_2-R_2{i}_1+R_1{i}_2-R_1{i}_1+R_{light}{i}_2+R_{pot}{i}_2=0\\ R_2{i}_2+R_1{i}_2+R_{light}{i}_2+R_{pot}{i}_2=R_2{i}_1+R_1{i}_1\\ \left(R_2+R_1+R_{light}+R_{pot}\right)i_2=\left(R_2+R_1\right)i_1\end{array}$

Rearrange the equation as follows,

$i_2=\left(\frac{R_2+R_1}{R_2+R_1+R_{light}+R_{pot}}\right)i_1$        (2)

Substitute equation (2) in equation (1).

$\begin{array}{l}\left(R_S+R_1+R_2\right)i_1-\left(R_1+R_2\right)\left(\frac{R_2+R_1}{R_2+R_1+R_{light}+R_{pot}}\right)i_1=V_S\\ \left(R_S+R_1+R_2\right)i_1-\left(\frac{{\left(R_1+R_2\right)}^2}{R_2+R_1+R_{light}+R_{pot}}\right)i_1=V_S\\ \left[\left(R_S+R_1+R_2\right)-\left(\frac{{\left(R_1+R_2\right)}^2}{R_2+R_1+R_{light}+R_{pot}}\right)\right]i_1=V_S\\ \left[\frac{\left(R_S+R_1+R_2\right)\left(R_2+R_1+R_{light}+R_{pot}\right)-{\left(R_1+R_2\right)}^2}{R_2+R_1+R_{light}+R_{pot}}\right]i_1=V_S\end{array}$

Rearrange the equation as follows,

$i_1=V_S\frac{R_2+R_1+R_{light}+R_{pot}}{\left(R_S+R_1+R_2\right)\left(R_2+R_1+R_{light}+R_{pot}\right)-{\left(R_1+R_2\right)}^2}$        (3)

Substitute equation (3) in equation (2).

$i_2=\left(\frac{R_2+R_1}{R_2+R_1+R_{light}+R_{pot}}\right)\left(V_S\frac{R_2+R_1+R_{light}+R_{pot}}{\left(R_S+R_1+R_2\right)\left(R_2+R_1+R_{light}+R_{pot}\right)-{\left(R_1+R_2\right)}^2}\right)$

$i_2=V_S\frac{R_2+R_1}{\left(R_S+R_1+R_2\right)\left(R_2+R_1+R_{light}+R_{pot}\right)-{\left(R_1+R_2\right)}^2}$        (4)

In Figure 1, the output voltage $V_{\text{out}}$ is calculated as follows.

$\begin{array}{c}{V}_{out}=i_2{R}_{pot}+\left(i_2-i_1\right)R_2\\ =i_2{R}_{pot}+R_2{i}_2-R_2{i}_1\end{array}$

$V_{out}=-R_2{i}_1+\left(R_{pot}+R_2\right)i_2$        (5)

Substitute equation (3) and (4) in equation (5).

$V_{out}=\left[\begin{array}{l}-R_2\left(V_S\frac{R_2+R_1+R_{light}+R_{pot}}{\left(R_S+R_1+R_2\right)\left(R_2+R_1+R_{light}+R_{pot}\right)-{\left(R_1+R_2\right)}^2}\right)+\\ \left(R_{pot}+R_2\right)\left(V_S\frac{R_2+R_1}{\left(R_S+R_1+R_2\right)\left(R_2+R_1+R_{light}+R_{pot}\right)-{\left(R_1+R_2\right)}^2}\right)\end{array}\right]$

$V_{out}=\left[\begin{array}{l}-\frac{R_2{V}_S\left(R_2+R_1+R_{light}+R_{pot}\right)}{\left(R_S+R_1+R_2\right)\left(R_2+R_1+R_{light}+R_{pot}\right)-{\left(R_1+R_2\right)}^2}+\\ \frac{V_S\left(R_{pot}+R_2\right)\left(R_1+R_2\right)}{\left(R_S+R_1+R_2\right)\left(R_2+R_1+R_{light}+R_{pot}\right)-{\left(R_1+R_2\right)}^2}\end{array}\right]$        (6)

Conclusion:

Thus, the output voltage $V_{\text{out}}$ in terms of $R_S$, $R_1$, $R_2$, $R_{\text{light}}$, and $R_{\text{pot}}$ is expressed.

To determine

Calculate the value of variable resistor $R_{\text{pot}}$ required to balance the circuit at 500 lux.

Answer to Problem 74E

The value of variable resistor $R_{\text{pot}}$ required to balance the circuit at 500 lux is $206.06\Omega$.

Explanation of Solution

Given data:

Refer to Part (a).

The resistance $R_{\text{light}}$ is 200 ohms.

For balanced condition, $V_{\text{out}}=0$.

Calculation:

Refer to Part (a).

Substitute 0 for $V_{\text{out}}$ in equation (6) to find the variable resistor $R_{\text{pot}}$.

$\begin{array}{l}0=\left[\begin{array}{l}-\frac{R_2{V}_S\left(R_2+R_1+R_{light}+R_{pot}\right)}{\left(R_S+R_1+R_2\right)\left(R_2+R_1+R_{light}+R_{pot}\right)-{\left(R_1+R_2\right)}^2}+\\ \frac{V_S\left(R_{pot}+R_2\right)\left(R_1+R_2\right)}{\left(R_S+R_1+R_2\right)\left(R_2+R_1+R_{light}+R_{pot}\right)-{\left(R_1+R_2\right)}^2}\end{array}\right]\\ \left[\begin{array}{l}\frac{R_2{V}_S\left(R_2+R_1+R_{light}+R_{pot}\right)}{\left(R_S+R_1+R_2\right)\left(R_2+R_1+R_{light}+R_{pot}\right)-{\left(R_1+R_2\right)}^2}\\ =\frac{V_S\left(R_{pot}+R_2\right)\left(R_1+R_2\right)}{\left(R_S+R_1+R_2\right)\left(R_2+R_1+R_{light}+R_{pot}\right)-{\left(R_1+R_2\right)}^2}\end{array}\right]\\ R_2{V}_S\left(R_2+R_1+R_{light}+R_{pot}\right)=V_S\left(R_{pot}+R_2\right)\left(R_1+R_2\right)\\ R_2\left(R_2+R_1+R_{light}+R_{pot}\right)=\left(R_{pot}+R_2\right)\left(R_1+R_2\right)\end{array}$

Simplify the equation as follows,

$\begin{array}{l}{R}_2^2+R_1{R}_2+R_{light}{R}_2+R_{pot}{R}_2=R_{pot}{R}_1+R_1{R}_2+R_{pot}{R}_2+R_2^2\\ R_{light}{R}_2=R_{pot}{R}_1\end{array}$

$R_{pot}=\frac{R_{light}{R}_2}{R_1}$        (7)

Substitute $200\Omega$ for $R_{light}$, $198\Omega$ for $R_1$, and $204\Omega$ for $R_2$ in equation (7) to find $R_{\text{pot}}$ in ohms.

$\begin{array}{c}{R}_{pot}=\frac{\left(200\Omega \right)\left(204\Omega \right)}{198\Omega }\\ =206.06\Omega \end{array}$

Conclusion:

Thus, the value of variable resistor $R_{\text{pot}}$ required to balance the circuit at 500 lux is $206.06\Omega$.

To determine

Find the light level if the resistance of the photoresistor decreases by 2% for a light increase to 600 lux with the measure of output voltage $V_{\text{out}}=150\text{mV}$.

Answer to Problem 74E

The light level is 760 lux if the resistance of the photoresistor decreases by 2% for a light increase to 600 lux with the measure of output voltage $V_{\text{out}}=150\text{mV}$.

Explanation of Solution

Given data:

Refer to Part (b).

The variable resistor $R_{\text{pot}}$ is 206.06 ohms.

The output voltage $V_{\text{out}}$ is $150\text{mV}$ for a light level 600 lux.

Calculation:

When the photoresistor resistance decreases by 2% with a light increase to 600 lux, then $\begin{array}{c}{R}_{\text{light}}=\left(\left(100-2\right)\%\right)\left(200\Omega \right)\\ =\left(98\%\right)\left(200\Omega \right)\\ =\left(\frac{98}{100}\right)\left(200\Omega \right)\\ =196\Omega \end{array}$

Refer to Part (a), reduce the equation (6) as follows,

$V_{out}=\left[\frac{-R_2{V}_S\left(R_2+R_1+R_{light}+R_{pot}\right)+V_S\left(R_{pot}+R_2\right)\left(R_1+R_2\right)}{\left(R_S+R_1+R_2\right)\left(R_2+R_1+R_{light}+R_{pot}\right)-{\left(R_1+R_2\right)}^2}\right]$

Substitute 12 V for $V_S$, $196\Omega$ for $R_{light}$, $206.06\Omega$ for $R_{\text{pot}}$, $10\Omega$ for $R_S$, $198\Omega$ for $R_1$, and $204\Omega$ for $R_2$ in equation (7) to find $V_{\text{out}}$ in volts.

$\begin{array}{c}{V}_{out}=\frac{\begin{array}{l}-\left(204\Omega \right)\left(12\text{V}\right)\left(204\Omega +198\Omega +196\Omega +206.06\Omega \right)\\ +\left(12\text{V}\right)\left(206.06\Omega +204\Omega \right)\left(198\Omega +204\Omega \right)\end{array}}{\left(10\Omega +198\Omega +204\Omega \right)\left(204\Omega +198\Omega +196\Omega +206.06\Omega \right)-{\left(198\Omega +204\Omega \right)}^2}\\ =12\text{V}\left[\frac{-\left(204\Omega \right)\left(804.06\Omega \right)+\left(410.06\Omega \right)\left(402\Omega \right)}{\left(412\Omega \right)\left(804.06\Omega \right)-{\left(402\Omega \right)}^2}\right]\\ =0.057704\text{V}\\ =0.057704\times {10}^3\times {10}^{-3}\text{V}\end{array}$

Reduce the equation as follows,

$V_{out}=57.704\text{mV}\left\{\because 1\text{m}={10}^{-3}\right\}$

For a linear change, the light level is calculated as follows.

$\begin{array}{c}\text{Light}=500\text{lux}+\left(\frac{150\text{mV}}{57.704\text{mV}}\right)100\text{lux}\\ =500\text{lux}+259.95\text{lux}\\ =759.95\text{lux}\\ \cong 760\text{lux}\end{array}$

Conclusion:

Thus, the light level is 760 lux if the resistance of the photoresistor decreases by 2% for a light increase to 600 lux with the measure of output voltage $V_{\text{out}}=150\text{mV}$.