LED forward voltage - why does series resistor take excess voltage?
If you want a mathematical model for your LED it gets tough due to the non-linearity but we can make a simple model which may be what you require.



Figure 1. Forward current vs forward voltage for a Cree 503B LED.

Cree recommend operating this LED at 10 to 30 mA although it can be pulsed at &lt; 0.1 ms at 200 mA.

I've superimposed a tangent to the red LED curve at 20 mA and it conveniently gives a line between (1.9 V, 0 mA) and (2.7 V, 80 mA). The slope of the line is \$\frac {dI}{dV} = \frac {80 - 0}{2.7 - 1.9} = \frac {80}{0.8} = 100~mA/V \$ approx. over the range 0 to 60 mA. This slope (still ignoring the 1.9 V offset) is the same as a 10 Ω resistor. Therefore we could model an LED circuit as shown in Figure 2.







simulate this circuit &ndash; Schematic created using CircuitLab

Figure 2. LED model.

Now the LED forward voltage can be calculated as \$ 1.9 + 10 I \$ volts and for a given LED current, \$ I \$, and supply voltage, \$ V_S \$, the value of R1 is given by 

$$ I = \frac {V_S - V_{LED}}{R1 + R_{LED}} $$

$$ R1 = \frac {V_S - V_{LED}}{I} $$

$$ R1 = \frac {V_S - V_{LED}}{I} - R_{LED} $$

So if we want 20 mA through the LED on a 5 V supply we can calculate

$$ R1 = \frac {V_S - V_{LED}}{I} - R_{LED} = \frac {5 - 1.9}{0.02}-10 = 145~\Omega $$



Note that we could have avoided all this work by reading the forward voltage at 20 mA from the graph. It's 2.1 V. In our example that means that the voltage across R1 = 5 V - 2.1 V = 2.9 V. \$ R1 = \frac {V}{I} = \frac {2.9}{0.02} = 145 Ω \$. It's the same answer but with a simpler calculation.