Astable multivibrator not producing a square output wave, using two BJTs

First all below concerns only CMOS gates with switching point at 1/2 of Vcc. So, not HCT series. HC series are OK and 4xxx too.

At first, R1 does not affect the frequency at all. It is placed there in order to make the input current of the inverter (through the protection diodes) to not affect the work of the schematic. That is why it should be much bigger than R2.

The frequency of the schematic is \$F=\frac{1}{2.2.R_2.C_1}\$

How it is derived?

^{simulate this circuit – Schematic created using CircuitLab}

At first notice that the voltage in points 2 and 3 can be only 0 or Vcc.

The schematic turns in the other stage when V1 is equal to the half of the power voltage.

When the second gate output flips from 1 to 0, the capacitor is charged to -0.5Vcc (the left plate is negative), so V1 becomes -0.5Vcc and starts to increase because R2 is connected to Vcc: $$ \tau = R_2.C_1 $$ $$ V_1 = V_{cc}.(1-e^{-\frac{t}{\tau}})-\frac{V_{cc}}{2}.e^{-\frac{t}{\tau}} = V_{cc} - \frac{3.V_{cc}}{2}.e^{-\frac{t}{\tau}} $$ The switching of the schematic will happen when V1 becomes equal to Vcc/2, so: $$ \frac{V_{cc}}{2} = V_{cc} - \frac{3.V_{cc}}{2}.e^{-\frac{t}{\tau}} $$ Or: $$ \frac{V_{cc}}{2} = \frac{3.V_{cc}}{2}.e^{-\frac{t}{\tau}} $$ $$ \frac{1}{3} = e^{-\frac{t}{\tau}} => 3 = e^{\frac{t}{\tau}} $$ $$ ln 3 = \frac{t}{\tau} $$ $$ t = \tau.ln 3 = R_2.C_1.ln 3 = 1.098612289.R_2.C_1 $$

This is the half of the period (because the schematic switches exactly on the half of the Vcc), so the period: $$ T = 2.t = 2.197224577.R_2.C_1 $$

BTW: This oscillator has very high frequency stability, both, by the temperature and by the Vcc. This way its use have to be encouraged for all schematics where quartz stability is not needed.

First of all, try following a schematic rather than a youtube tutorial showing a breadboard. There is a good chance you may have wired something incorrectly.

A 555 timer in an astable configuration would look like this:

^{simulate this circuit – Schematic created using CircuitLab}

Using this, you should be able to place your components on a breadboard, following the schematic closely. I assume you know the internal connections of a breadboard, but just in case, here is a diagram:

If you change the values of R1/R2 or C1 then you will be able to vary the frequency. Or stick a potentiometer in the place of R2.

This link HERE will be able to help you calculate which values to use for your desired frequency.

If you wanted to work it out yourself, you could use the forumula:

f= 1.44/(R1+2R2)C

Although I have found that to be a tad off sometimes! But it gives you a ballpark idea as real world electronics will never be exact!

So, basically, take it apart and rebuild it, this time following a schematic!

P.S make sure you have a working LED!