Electronic – Can someone explain how this oscillator circuit works

The base-emitter junction is like a diode. When the voltage across it (Vbe) exceeds approximately 0.65V (can be as low as 0.55V and as high as 0.9V, check the datasheet for your transistor) it begins conducting.

The current (not voltage!) through the base emitter junction is amplified by the gain of the transistor, which is known as HFE. Ic(collector current) = Ib(base current) * HFE. Remember HFE is not constant for transistors, it varies from transistor to transistor and depending on temperature, previous usage, etc., so don't rely on it for controlled amplification. For the 2N2222 it is around 160, plus or minus 30.

By applying a base-emitter voltage exceeding 0.65V to the transistor you can use it as a switch.

^{simulate this circuit – Schematic created using CircuitLab}

(It is an NPN transistor you want. 2N3904 or 2N2222 will do.)

If you want to use an LED which is not blue or white then use a 47 ohm resistor in series with it.

When you press the switch the LED will come on.

So as you mentioned it says that the transistor is essentially two diodes.

You should, but may not, know that the typical voltage drop required over a diode to make it conduct is ~0.7V but can vary depending on the diode of course. So if you just 'stick' a voltage across the terminals as when you increase the voltage over the diode current flows:

As the resistance across a diode is very low when this voltage is applied we can work out that the current would be extremely high: I = V/R, simple to see that the lower R is the higher the current, and this can be very damaging to the base terminal, I believe a datasheet of the particular transistor will give you more information on what size current it can take.

What this is saying is you need to have a current limiting resistor in front of the base terminal on the transistor which does exactly what its name describes, limits the current. As the voltage drop across the transistor will remain at 0.6-0.8V we can work out the size resistor we would need quite easily. R = (Vin - Vdrop)/I, 'I' being the base current that it can take, Vdrop being the voltage drop from the base to the emitter and Vin being the supply that is going into base, you also need to look at the hfe of the transistor so see if it will be able to give you the amount of current you need, which coincidentally can be limited, or 'tailored' with a resistor at the emitter pin so the transistor is less reliant on the hfe, but I am sure you will get on to that in the future!