Why do diode's need a threshold voltage?

doing some revision and looking at the I against V graph, please also explain why the graph is shape like this.

i know that the current increases a fair amount


http://upload.wikimedia.org/wikibooks/en/e/eb/V-I_diode.png

they don't need, they have   lol

There is actually no real "threshold".  The forward current i of a diode increases exponentially with voltage and decreases with temperature, typically i = i0[1 + exp(ev/kT)] where i0 is the so-called 'leakage' current (the current for a highly reverse-biased diode); e is the electronic charge, v is the forward diode voltage, T is Kelvin temperature, and k is Bolzmann's constant.  kT/e is about 26 mV at room temperature.  Any exponential builds up slowly at first and then faster.  The slower region is typically called the threshold voltage.  So e.g. a silicon diode will operate around a forward drop of about 700 mV because below that voltage there is very little current, as the formula shows.  Then i = i0exp(ev/kT) is obviously entirely adequate since exp(700/26) = 4.9e11 >> 1.

If i0 = 2.04e-12 then we'd have 1 ampere at 700 mV.  But look at what happens for 500 mV: 2.2e8*2.04e-12 = only 0.45 mA.  So people tend to call 700 mV as the 'threshold' voltage but it's a matter of semantics only.