Understanding constant current circuit
Zach, this circuit is pretty easy to understand if you understand the BJT first. (You will understand diodes, if you understand the BJT, so that's a given.) Everyone struggles with these things at some point, so it's fine you don't apprehend this well right now. Take it one step at a time.

There is plenty of information on diodes here (and elsewhere.) You are awash in information about them. I won't try and replicate any of that. It's enough for this circuit that you accept two things about diodes:


A forward-biased diode has a fixed voltage across it. For regular silicon diodes, this value is \$700\:\text{mV}\$. (For LEDs, which are also diodes, it varies with the color and type and you have to look at the datasheet for that.)
Everything I just said in point #1 is actually wrong. But for these purposes, you don't need to worry about that fact.


Now to the BJT. It also has a diode from base to emitter. So the rules above apply. But we add the following about the BJT:


When the BJT's base-emitter diode is forward-biased, the collector current is the same as the emitter current.
What I just said in point #3 is also wrong. But #3 is close enough for these purposes to not matter.


So. Now we can describe the circuit.


The \$20\:\text{k}\$ resistor forward biases the two diodes by providing a path for the current to go to ground.
The total voltage across the two diodes is therefore \$1.4\:\text{V}\$, with the rest left over for the resistor. Therefore, the base voltage for the BJT is \$10\:\text{V}-1.4\:\text{V}=8.6\:\text{V}\$.
Therefore also the resistor current is \$\frac{10\:\text{V}-1.4\:\text{V}}{20\:\text{k}\Omega}\approx 430\:\mu\text{A}\$.
The BJT's emitter is forward biased and therefore the emitter will be \$700\:\text{mV}\$ above the base or \$8.6\:\text{V}+700\:\text{mV}\approx 9.3\:\text{V}\$.
So the voltage across the \$500\:\Omega\$ resistor is \$10\:\text{V}-9.3\:\text{V}=700\:\text{mV}\$ (one diode drop -- which if you look closely you should see why this will be the case in this circuit.) From this, we can compute that the current in that resistor is \$\frac{700\:\text{mV}}{500\:\Omega}\approx 1.4\:\text{mA}\$.
Since by rule #3 above, the emitter current and collector currents are the same, it follows that the collector current is also \$1.4\:\text{mA}\$.


The collector current is always the same as the emitter current (within a reasonable approximation.) So, it doesn't matter what resistor you place between the collector and ground.

Except,


The above conclusion isn't right if the collector current we just worked out causes a voltage drop across the collector resistor that exceeds the base voltage. So this means that the resistor cannot be larger than \$R=\frac{8.6\:\text{V}}{1.4\:\text{mA}}\approx 6100 \:\Omega\$. So it has limits.