They have a polar nature to them because biasing them one way helps the current to flow, but biasing them the other way (reverse bias) forms a boundary to charge flow, so very little current flows when the diode is in reverse bias.
I understand the concept of the biasing but why do you have these regions of exponential change in the reverse and forward biased directions. From what I have read,when the diode is reverse biased the current which it trying to pass through it is exciting the electrons but the potential drop is too much to overcome for a time (besides the small leakage current). However, once the reverse breakdown voltage is achieved the excited electrons rush out in an exponential inverse voltage. But why is this exponential region apparent in the forward biased direction since the diode is meant to allow current to flow in that direction. For example why must silicon reach a voltage on .65 V before the current takes off so to speak? Does the diode inhibit current in the forward biased direction as well?
I misspoke when I said the "current" is exciting the electrons, the current is electrons. duh
The p and n regions are at different intrinsic potentials, with an energy (voltage) gap preventing current flow. The charge carriers can't flow "uphill." When you apply a reverse bias, the gap gets bigger and again no current flows (except for a little leakage). When you apply a forward bias voltage, it narrows the gap. At room temperature in Silicon, 0.6 to 0.7V sufficiently flattens the gap that carriers can flood across.
The exponential behavior comes from the thermal Boltzmann distribution of carrier energies. At room temperature some carriers have thermal (kinetic) energy, but energetic ones are exponentially less numerous than slow or stationary ones. Carriers with more energy than the gap can flow across (jump the hill), carrying a current. As you increase the forward bias voltage, exponentially more carriers have enough energy. That's why the I-V curve is exponential.