Chapter 7, Problem 85P

A simple relaxation oscillator circuit is shown in Fig. 7.145. The neon lamp fires when its voltage reaches 75 V and turns off when its voltage drops to 30 V. Its resistance is 120 Ω when on and infinitely high when off.

1. (a) For how long is the lamp on each time the capacitor discharges?
2. (b) What is the time interval between light flashes?

To determine

Calculate the discharge time of the capacitance when the lamp is on in the given circuit of Figure 7.145.

Answer to Problem 85P

The discharge time of the capacitance is $659.7\mu \text{F}$ when the lamp is on.

Explanation of Solution

Given data:

The neon lamp is on when its voltage reaches 75 V and turn off when its voltage drops to 30 V.

The resistance is $120\Omega$ when neon lamp is on and infinitely high when neon lamp is off.

Refer to Figure 7.145 in the textbook.

The value of capacitance $\left(C\right)$ is $6\mu \text{F}$.

Formula used:

Write the general expression to find the complete voltage response for an RC circuit.

$v\left(t\right)=v\left(\infty \right)+\left[v\left(0\right)-v\left(\infty \right)\right]e^{-\frac{t}{\tau }}$ (1)

Here,

$\tau$ is the time constant for the RC circuit,

$v\left(0\right)$ is the initial capacitor voltage, and

$v\left(\infty \right)$ is the final capacitor voltage.

Write the expression to find the time constant for an RC circuit.

$\tau =R_{\text{Th}}C$ (2)

Here,

$R_{\text{Th}}$ is the Thevenin resistance, and

C is the capacitance of the capacitor.

Calculation:

The neon lamp is on when it reaches 75 V. Therefore, the initial capacitor voltage $v\left(0\right)$ is equal to 75 V. That is,

$v\left(0\right)=75\text{V}$

The neon lamp is off when it drops to 30 V. Therefore, the final capacitor voltage $v\left(\infty \right)$ is equal to 0 V. That is,

$v\left(\infty \right)=0\text{V}$

When the neon lamp on and off, during that time a $120\Omega$ resistor is parallel with a $6\mu \text{F}$ capacitor. Therefore, the Thevenin resistance $R_{\text{Th}}$ is equal to the $120\Omega$ resistor. That is,

$R_{\text{Th}}=120\Omega$

Substitute $120\Omega$ for $R_{\text{Th}}$ and $6\mu \text{F}$ for C in equation (2) to find the time constant $\tau$.

$\tau =\left(120\Omega \right)\left(6\mu \text{F}\right)$ (3)

Substitute the units $\frac{\text{V}}{\text{A}}$ for $\Omega$ and $\frac{\text{A}\cdot \text{s}}{\text{V}}$ for F in equation (3) to find the time constant $\tau$ in seconds.

$\begin{array}{c}\tau =\left(120\frac{\text{V}}{\text{A}}\right)\left(6\mu \frac{\text{A}\cdot \text{s}}{\text{V}}\right)\\ =\left(120\frac{\text{V}}{\text{A}}\right)\left(6\times {10}^{-6}\frac{\text{A}\cdot \text{s}}{\text{V}}\right)\left\{\because 1\mu ={10}^{-6}\right\}\\ =7.2\times {10}^{-4}\text{s}\end{array}$

Substitute $75\text{V}$ for $v\left(0\right)$, $0\text{V}$ for $v\left(\infty \right)$, and $7.2\times {10}^{-4}\text{s}$ for $\tau$ in equation (1) to find the output voltage across the capacitor $v\left(t\right)$ in volts.

$v\left(t\right)=0\text{V}+\left[75\text{V}-0\text{V}\right]e^{-\frac{t}{7.2\times {10}^{-4}\text{s}}}$

$v\left(t\right)=75e^{-\frac{t}{7.2\times {10}^{-4}\text{s}}}\text{V}$ (4)

When the neon lamp is off, the voltage drops to 30 V. That is, $v\left(t_1\right)=30\text{V}$.

At $t=t_1$, the equation (4) can be rewritten as,

$v\left(t_1\right)=75e^{-\frac{t_1}{7.2\times {10}^{-4}\text{s}}}\text{V}$

Substitute $30\text{V}$ for $v\left(t_1\right)$ to find the time $t_1$ in seconds.

$\begin{array}{l}30\text{V}=75e^{-\frac{t_1}{7.2\times {10}^{-4}\text{s}}}\text{V}\\ e^{-\frac{t_1}{7.2\times {10}^{-4}\text{s}}}=\frac{30}{75}\end{array}$

Taking ln on both sides of the equation.

$\begin{array}{l}\ln \left(e^{-\frac{t_1}{7.2\times {10}^{-4}\text{s}}}\right)=\ln \left(\frac{30}{75}\right)\\ -\frac{t_1}{7.2\times {10}^{-4}\text{s}}=-0.9163\\ \frac{t_1}{7.2\times {10}^{-4}\text{s}}=0.9163\end{array}$

Rearrange the equation as follows,

$\begin{array}{c}{t}_1=0.9163\times 7.2\times {10}^{-4}\text{s}\\ =6.597\times {10}^{-4}\text{s}\\ =6.597\times {10}^{-4}\times {10}^{-2}\times {10}^2\text{s}\\ =659.7\times {10}^{-6}\text{s}\end{array}$

Reduce the equation as follows,

$t_1=659.7\mu \text{s}\left\{\because 1\mu ={10}^{-6}\right\}$

Conclusion:

Thus, the discharge time of the capacitance is $659.7\mu \text{F}$ when the lamp is on.

To determine

Calculate the time interval between the light flashes.

Answer to Problem 85P

The time interval between the light flashes is $16.636\text{s}$.

Explanation of Solution

Figure 1 shows the Thevenin resistance at the capacitance terminal.

In Figure 1, the Thevenin resistance $R_{\text{Th}}$ is equal to $4\text{M}\Omega$

Substitute $4\text{M}\Omega$ for $R_{\text{Th}}$ and $6\mu \text{F}$ for C in equation (2) to find the time constant $\tau$.

$\tau =\left(4\text{M}\Omega \right)\left(6\mu \text{F}\right)$ (5)

Substitute the units $\frac{\text{V}}{\text{A}}$ for $\Omega$ and $\frac{\text{A}\cdot \text{s}}{\text{V}}$ for F in equation (5) to find the time constant $\tau$ in seconds.

$\begin{array}{c}\tau =\left(4\text{M}\frac{\text{V}}{\text{A}}\right)\left(6\mu \frac{\text{A}\cdot \text{s}}{\text{V}}\right)\\ =\left(4\times {10}^6\frac{\text{V}}{\text{A}}\right)\left(6\times {10}^{-6}\frac{\text{A}\cdot \text{s}}{\text{V}}\right)\left\{\because 1\mu ={10}^{-6}\right\}\\ =24\text{s}\end{array}$

At $t=t_1$, the equation (1) can be rewritten as,

$v\left(t_1\right)=v\left(\infty \right)+\left[v\left(0\right)-v\left(\infty \right)\right]e^{-\frac{t_1}{\tau }}$

$v\left(t_1\right)-v\left(\infty \right)=\left[v\left(0\right)-v\left(\infty \right)\right]e^{-\frac{t_1}{\tau }}$ (6)

At $t=t_2$, the equation (1) can be rewritten as,

$v\left(t_2\right)=v\left(\infty \right)+\left[v\left(0\right)-v\left(\infty \right)\right]e^{-\frac{t_2}{\tau }}$

$v\left(t_2\right)-v\left(\infty \right)=\left[v\left(0\right)-v\left(\infty \right)\right]e^{-\frac{t_2}{\tau }}$ (7)

Dividing the equation (6) by (7).

$\begin{array}{c}\frac{v\left(t_1\right)-v\left(\infty \right)}{v\left(t_2\right)-v\left(\infty \right)}=\frac{\left[v\left(0\right)-v\left(\infty \right)\right]e^{-\frac{t_1}{\tau }}}{\left[v\left(0\right)-v\left(\infty \right)\right]e^{-\frac{t_2}{\tau }}}\\ =\frac{e^{-\frac{t_1}{\tau }}}{e^{-\frac{t_2}{\tau }}}\\ =e^{-\frac{t_1}{\tau }}{e}^{\frac{t_2}{\tau }}\end{array}$

$\frac{v\left(t_1\right)-v\left(\infty \right)}{v\left(t_2\right)-v\left(\infty \right)}=e^{\frac{t_2-t_1}{\tau }}$ (8)

Consider the time interval is,

$t_0=t_2-t_1$

Substitute $t_0$ for $t_2-t_1$ in equation (8).

$\frac{v\left(t_1\right)-v\left(\infty \right)}{v\left(t_2\right)-v\left(\infty \right)}=e^{\frac{t_0}{\tau }}$ (9)

Consider the neon lamp is on when its voltage $v\left(t_1\right)$ reaches 75 V and turn off when its voltage $v\left(t_2\right)$ drops to 30 V. The final capacitor voltage $v\left(\infty \right)$ is equal to the supply voltage 120 V.

Substitute 75 V for $v\left(t_1\right)$, 30 V for $v\left(t_2\right)$, $24\text{s}$ for $\tau$, and 120 V for $v\left(\infty \right)$ in equation (9) to find the time interval $t_0$ in seconds.

$\begin{array}{l}\frac{75\text{V}-120\text{V}}{30\text{V}-120\text{V}}=e^{\frac{t_0}{24\text{s}}}\\ 0.5=e^{\frac{t_0}{24\text{s}}}\end{array}$

Taking ln on both sides of the equation.

$\begin{array}{l}\ln \left(0.5\right)=\ln \left(e^{\frac{t_0}{24\text{s}}}\right)\\ -0.693147=\frac{t_0}{24\text{s}}\\ t_0=\left(-0.693147\right)\left(24\text{s}\right)\\ t_0=-16.636\text{s}\end{array}$

Since, the time interval must be taken as positive value. Therefore, $t_0=16.636\text{s}$.

Conclusion:

Thus, the time interval between the light flashes is $16.636\text{s}$.