the diode I-V curve
If you plot a diode's current (vertical) against the voltage across it (horizontal), you do not get a straight line like a resistor — you get a hockey-stick. That curve, the I-V (current-voltage) characteristic, is the diode's whole personality on one graph: flat and near-zero for reverse and small forward voltages, then suddenly shooting up once you pass the turn-on knee.
The forward part follows the Shockley diode equation, I = Is times (e^(V/(n times VT)) - 1), where Is is a tiny leakage current, VT is the 'thermal voltage' (about 26 mV at room temperature), and n is a fudge factor near 1 to 2. The takeaway is the exponential: current multiplies for each fixed step of voltage. That steep wall near 0.7 V is why we can pretend the voltage is 'stuck' there. To the left, reverse bias gives almost-flat tiny leakage — until the curve plunges down at the breakdown voltage.
The slope of the curve at your operating point is what matters in real design. Because it is curved, the diode's effective resistance changes with current: steep where it conducts (low resistance), flat where it blocks (huge resistance). That changing slope, called dynamic or small-signal resistance (about 26 mV / I at current I), is how a diode shapes signals and why simple models exist to approximate the curve.
At 1 mA the dynamic resistance is about 26 mV / 1 mA = 26 ohm; at 10 mA it falls to ~2.6 ohm. So the more current it carries, the more wire-like the diode looks for small wiggles.
The I-V curve's slope sets the diode's small-signal resistance, about 26 mV / I.
The exponential is so steep that reading Vf to three decimals is meaningless without stating the current and temperature. Always quote a diode voltage with its test current, e.g. '0.7 V at 10 mA'.