Diodes come in many shapes and sizes and are used for a huge range of applications like:
bridge rectifiers (AC to DC conversion):
Waveform clipping and clamping:
Circuit protection:
Voltage regulation (normally zener diodes):
Indicators (LED's):
And thats just to name a few!
Anyhow, diodes are basically made from a single block of semiconductor material which is then doped with certain impurities to leave one half with an excess amount of electrons (the cathode) and the other half a deficit of electrons (the anode). This now gives us a single PN junction better known as the diode.
Now that the diode is made, we have what is called the depletion region in the middle, separating the N and P type semiconductor material. This region is very high resistance and in order for us to get any current to flow through the diode, we need to overcome this high resistance depletion region. This voltage is better known as the 'barrier potential' or may also be referred to as the forward voltage drop of the diode.
So if we have a standard silicon diode, we actually need (roughly) .6 or .7 of a volt in order to overcome the barrier potential. So if we had a circuit with a battery (lets say 1.5volts) and then a resistor (lets say 1k ohm) and finally a silicon diode and they were all connected in series, then we would have about 0.6 volts dropped across the diode (because once it overcomes the barrier potential - it will only drop 0.6volts across it and thats it, even if we increase the power supply voltage, it will still only grab that 0.6 volts, this is it's forward voltage drop.
The resistor will get whatever is left, in this case 1.5 - 0.6 = 0.9volts. this means that the current flowing in the circuit is 0.9 / 1k = 900uA
If we increased the power supply to 10volts, then we would still have 0.6 volts across the diode (remember the forward voltage drop.) this means that the resistor will now get 10 volts - 0.6 volts = 9.4 volts. The current in the circuit is now, 9.4 / 1k = 9.4mA
This forward voltage drop is very important, for example you may have some LED's that you want to connect to your microcontroller. The microcontroller will only be able to handle a certain amount of current draw. On the same note, the LED can also only handle so much current. This is where you need to find out the forward voltage drop of your particular LED.
Let's say you have a red LED, and the forward voltage drop is 2volts. This means that the resistor you place in series with the LED will get 3 volts (assuming you are working with 5v VCC power) Therefor you can calculate the value of resistor that you need in order to keep the current within the specified limits.
And lastly for this introduction, if you connect a diode in the wrong way around (better known as reverse bias) the depletion region gets bigger and you get no current flow. So remember that in order to turn your diode on you need:
A power supply voltage at least as big as the forward voltage drop of the diode A resistor connected in series with the diode in order to drop the rest of the voltage and to limit current The positive side of your power supply connects to the anode, and the negative to the cathode in order to forward bias it (turn it on)