Infinite Capacitor Oscillator

First, we can compute the total capacitance $C$ of the parallel capacitor array which, by observation, is given by the infinite sum: $\sum\limits_{k = 0}^\infty \left({\frac{1}{{2^k }}}\right) \mu F$

Therefore, by the infinite series formula: $C = 2 \mu F$

Next, we look at the states of the circuit which determine the period of $OUT$ .

Let the node of intersection of pin 2, pin 6, and the capacitor array be denoted as $\alpha$ . From the operation of the circuit, $OUT$ is at $V_{cc}$ when $V_{\alpha} \leq\frac {1}{3}V_{cc}$ . At this state, pin 7 has high impedance and the capacitor array can charge through resistors $R_{1}$ and $R_{2}$ , which have a total resistance of: $R_{1} + R_{2} = 1K\Omega + 1M\Omega$ $R_{1} + R_{2} = 1,001,000\Omega$

After some $t_{p}$ seconds, $V_{\alpha}$ reaches $\frac {2}{3}V_{cc}$ , changing the state of $OUT$ to $0V$ which also causes pin 7 to change to $0V$ . The capacitor array then discharges through $R_{2}$ until $V_{\alpha}$ reaches $\frac {1}{3}V_{cc}$ after some $t_{n}$ seconds, where the cycle repeats.

Let us assume, without loss of generality, that the period of $OUT$ begins when $OUT$ is at $V_{cc}$ , pin 7 has high impedance, and $V_{\alpha}=\frac {1}{3}V_{cc}$ .

The capacitor array $V_{\alpha}$ charges through resistors $R_{1}$ and $R_{2}$ with an initial voltage $E_{p}$ which is defined as the input voltage with respect to the voltage of the capacitor: $E_{p} = V_{cc} - \frac {1}{3}V_{cc} = \frac {2}{3}V_{cc}$

The $OUT = V_{cc}$ half of the period ends when the capacitor voltage $V_{\alpha}$ becomes $\frac {2}{3}V_{cc}$ after increasing by: $\Delta v_{p} = \frac {2}{3}V_{cc} - \frac {1}{3}V_{cc} = \frac {1}{3}V_{cc}$

We can compute the time $t_{p}$ taken to charge the capacitor array to $\frac {2}{3}V_{cc}$ by using the formula for capacitor charge rate: $t=RC\ln\left({\frac{1}{{1-{\frac{\Delta v}{{E}}}}}}\right)$

$t_{p}=(1,001,000\Omega)(2 \mu F)\ln\left({\frac{1}{{1-{\frac{\frac {1}{3}V_{cc}}{{\frac {2}{3}V_{cc}}}}}}}\right)$

$t_{p}=1.387680655 sec.$

Now, $OUT$ will become $0V$ and $V_{\alpha}$ will begin to discharge to $\frac {1}{3}V_{cc}$ (as previously mentioned) with a voltage $E_{n}$ . At this point, the voltage of the capacitor is $\frac {2}{3}V_{cc}$ and the input voltage to the capacitor is $0V$ due to pin 7, so $E_{n}$ is: $E_{n} = 0V - \frac {2}{3}V_{cc} = -\frac {2}{3}V_{cc}$

To reach $\frac {1}{3}V_{cc}$ , the voltage $V_{\alpha}$ must change by: $\Delta v_{n} = \frac {1}{3}V_{cc} - \frac {2}{3}V_{cc} = -\frac {1}{3}V_{cc}$

The capacitor now only discharges through $R_{2} = 1M\Omega$ . We compute for the time of $t_{n}$ in a similar way to $t_{p}$ . $t_{n}=(1M\Omega)(2 \mu F)\ln\left({\frac{1}{{1-{\frac{-\frac {1}{3}V_{cc}}{{-\frac {2}{3}V_{cc}}}}}}}\right)$

$t_{n}=1.386294361 sec.$

Therefore, the period is: $t_{p}+t_{n} = 1.387680655 + 1.386294361 sec.$ $t_{p}+t_{n} \approx \boxed{2.77 sec.}$

The circuit of the problem corresponds to the astable configuration of the LM555. As explained in the pages 10 and 11 of this PDF, the frecuency of oscillation is: $f = \frac{{1.44}}{{\left( {R_A + 2R_B } \right)C}}$ And the period is: $T = \frac{1}{f} = \frac{{\left( {R_A + 2R_B } \right)C}}{{1.44}}$ Where $R_A$ is the resistance between pins $7$ and $8$ , $R_B$ is the resistance between pins $7$ and $6$ , and $C$ is the capacitance between pin $6$ and ground. Therefore: $R_A = 1 k \Omega = 10^3 \Omega$ $R_B =1 M \Omega = 10^6 \Omega$ $C = \sum\limits_{k = 0}^\infty {\frac{1}{{2^k }}} \mu F = 2\mu F = 2 \cdot 10^{ - 6} F$ Replacing: $T = \frac{{\left( {10^3 + 2 \cdot 10^6 } \right) \times 2 \cdot 10^{ - 6} \Omega F}}{{1.44}} = \frac{{667}}{{240}} \approx \boxed{2.779}$