Q:

Voltage drop assumption on diodes (LEDs, etc.)

I've always wondered why is it valid to assume some voltage drop (different from the junction voltage drop) on diodes or LEDs from the power up of a circuit. It simplifies the analysis, but it's unclear to me why it is a valid approach.
For example, in the question Capacitors charging in a DC circuit, the assumption is a drop of \$2V\$ across the LED.
What is the explanation?

A:

Diodes have very steep output characteristics - there are almost no current below some voltage value, and after this voltage is reached the current increase is very steep:

This steep increase in current at \$V_d\$ essentially means that no matter what current you'll force through the diode, the voltage will not vary much from \$V_d\$. 
You can argue that for very large currents the voltage can increase by a significant amount, but in practice this will not happen. What will happen is that when the current increases over a nominal diode's current, the diode will be permanently damaged (short-circuit, open-circuit or some other bad effect).
Due to the above reasons, if the diode is present and is conducting - it will have a voltage drop of magnitude \$\approx V_d\$ across it.
Now, you may ask: "But how do we know a priori that the diode is conducting?". Good question. We really can't know this a priori, but we can ASSUME that this is the case and perform an analysis of the circuit. After the analysis is done we must ensure that all the voltages and the currents in our circuit are consistent. If they are - our assumption was correct. If, on the other hand, we find that the voltage drop (as follows from the analysis) on the diode is below its  \$V_d\$, then it is a contradiction and our assumption was wrong - the diode is in cut-off region of operation and behaves as an open circuit.

A:

It's a valid approach just for general cases simply because most LEDs happen to have a forward voltage drop near that value.  
Here is a table take from wikipedia on the voltage-drop of different LEDs:

You can see that for some reds, oranges, yellows, and some greens, 2V is close to the value.  I suppose if you were making a circuit with ultraviolet LEDs you could assume 3.5, or 4V.  Now if you actually know the forward voltage drop on the LEDs you are using, it would make a lot more sense to use that value.
Going off of what echad said, the constant voltage drop model is the simplest one, and speeds up analysis.  In reality, voltage drop on diodes have an exponential relationship.
Also, there are several different models for analyzing circuits that contain diodes. 
Taken from a textbook I use at school, Microelectronic Circuits 6th Ed, by Sedra and Smith:
Graphical Analysis of the Exponential Model, using a load line

Constant Voltage Drop Model

Now this is for plain silicon diodes, but the same math holds true for all diodes, just the parameters are slightly different and the drop for LEDs comes out different based on how they are manufactured.

