Oscillation Frequencies...

Before we begin, we normalize all voltage, currents, time and elements by the values given below: $\begin{aligned} \text{voltages:} &&\SI{1}{V}, &&&& \text{currents:} &&\SI{1}{mA}, &&&& \text{time:} &&\SI{1}{ms}, &&&&\Rightarrow &&&&& R:&&\SI{1}{k\ohm}, &&&&C:&&\SI{1}{\micro F}, &&&& f:&&\SI{1}{kHz} \end{aligned}$ The network equations remain the same in the process, but now all variables represent their normalized, dimensionless counterparts.

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Short solution

A quick look into an (old) datasheet for the N*555* yields the formula between frequency and element values on page 11 (astable timer mode): $\begin{aligned} f &= \frac{1}{\ln(2)(R_1+2R_2)C}\approx\frac{1.44}{(R_1+2R_2)C} &&&\Rightarrow &&&& R_1 &= \frac{1}{\ln(2)fC}-2R_2 = \frac{1}{\ln(2)\cdot\num{37.7e-3}}-30 \approx\boxed{8.2}&&&(*) \end{aligned}$

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Longer solution

You think just using the frequency formula feels like cheating? Let's analyze how the circuit actually works - another look into the datasheet for the N*555* tells us which pins of the IC are relevant here. The pins for trigger- and threshold-voltage ( pin 2 and pin 6 ) are shorted, so the IC-logic simplifies to

$\textbf{Pin 2, 6:}\qquad \text{If }\begin{gathered} u_2(t)<3\\u_2(t)>6 \end{gathered},\text{ then the output pin 3 is set to }\begin{gathered} u_3(t)=9\\u_3(t)=0 \end{gathered}\text{ and the discharge pin 7 is set to }\begin{gathered} \text{not connected}\\u_7(t)=0 \end{gathered}$

All pin voltages $u_k(t)$ are measured from pin to ground.

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Let's assume the network was de-energized and we switch on the supply voltage $V_{cc}=9$ . Then the network will go through three distinct stages:

1. Initialization: $C$ will slowly charge through $R_1,\:R_2$ , while the output has $u_3(t)=9$ and the discharge is not connected. The network stays in this state until it reaches the voltage $u_2(t)=6$ $$
2. Low output: The IC connects discharge and output to ground ( $u_3(t)=u_7(t)=0$ ). Then $C$ will discharge through $R_2$ until it reaches the voltage $u_2(t)=3$ . Let's call the time it takes $T_0$ $$
3. High output: $C$ will slowly charge through $R_1,\:R_2$ , while the output has $u_3(t)=9$ and the discharge is not connected. The network stays in this state until it reaches the voltage $u_2(t)=6$ . Let's call the time it takes $T_1$ . Repeat stages 2., 3. $$

To calculate $T_0,\:T_1$ , we need to analyze two basic RC-circuits with different resistors and initial conditions. Using KVL and defining $i_C(t)$ as the current through $C$ from top to bottom, we get the following ODEs. We also shift time such that the initial conditions are set at $t=0$ in both cases: $\begin{aligned} T_0:&&&& 0 &= R_2i_C(t)+u_2(t)=R_2Cu_2^{(1)}(t)+u_2(t) &&&\Rightarrow &&&& u_2^{(1)}(t)&=-A_0u_2(t), & u_2(0)&=6\\[.5em] T_1:&&&& 0 &= -V_{cc}+(R_1+R_2)i_C(t) + u_2(t) = -V_{cc} + (R_1+R_2)C u_2^{(1)}(t) + u_2(t) &&& \Rightarrow &&&& u_2^{(1)}(t)&=-A_1u_2(t) +9A_1, &u_2(0)&=3 \end{aligned}$ The new short-hands are $A_0:=\frac{1}{R_2C},\: A_1:=\frac{1}{(R_1+R_2)C}$ . Solving both ODEs, we can calculate $T_0,\:T_1$ and finally the period $T=\frac{1}{f}$ of the output $u_3(t)$ : $\begin{aligned}\left.\begin{aligned} T_0:&&&& u_2(T_0) &= 6e^{-A_0T_0}\overset{!}{=}3&&&\Rightarrow &&&& T_0&=\ln(2) R_2C\\[.5em] T_1:&&&& u_2(T_1) &= 9-6e^{-A_1T_1}\overset{!}{=}6&&&\Rightarrow &&&& T_1&=\ln(2) (R_1+R_2)C \end{aligned}\right\}&&&&\Rightarrow &&&& T&=T_0+T_1=\ln(2)(R_1+2R_2)C\overset{!}{=}\frac{1}{f},&&& R_1&\underset{(*)}{\approx}\boxed{8.2} \end{aligned}$

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Rem.: The value $\SI{26.5}{ms}$ given for "pause time per wave" is actually the period $T=\frac{1}{f}$ and not $T_0$ , as I thought at first.

This question requires the use of the formula for the frequency of pulses which is :

$f = \frac { 1.44 }{ (R_{ 1 }+ 2{ R }_{ 2 }) \times C }$

In the frequency calculation, if the units k $\Omega$ and μF are used. The output frequency will be in kHz. The question states that:

• $R_{ 1 }$ is unknown
• $R_{ 2 }$ is 15k $\Omega$
• $C_{1}$ is 1μF
• The frequency $(f)$ is 37.7 Hz

To enable us to get a correct answer, we need to convert the frequency of 37.7Hz to kHz since we will be using k $\Omega$ and μF. If you need additional clarification look at the table on this page.

To convert Hz to kHz we $\div 1000$ . So...

• 37.7Hz = 0.0377 kHz

Lets re-arrange to get the value of $R_{ 1 }$ :

$R_{ 1 }=\frac { 1.44 }{ f\times C } -2{ R }_{ 2 }$

Substitute into the formula:

• $R_{ 1 }=\frac { 1.44 }{ 0.0377kHz \times 1μF } -2(15k\Omega)$

Which gives us 8.196 , rounded to 1 d.p which is $\boxed{8.2}k$ $\Omega$

Note: The time period is not used here. It is not required since we are only looking for the frequency which has a direct influence on the time delay between each pulse.