Electronic – Current Transformers and Energy Harvesting

A metal core is advisable because it concentrates the magnetic field. This makes magnetic coupling between coils more effective, but you can do it without the core too, especially if both coils are tightly coupled (mechanically close to each other). If you don't have a core and the two coils are some distance apart, part of the generated field will pass outside of the other coil.
(In a transformer the outside metal "shortcuts" the field, increasing it. Without this outside metal the field would fill a large volume around the core.)

I have current carrying conductor with a current transformer around it. If the load is rated at 5V, 2A and only requires 2W of power, but the primary side conductor is carrying 100V 12A.

The first problem is that the CT primary current will vary from 0 at no load to a maximum depending on the load on the line. This makes it very difficult to extract even power from such a scheme and it is the reason that constant voltage is the preferred method of power distribution.

I understand that constant current power is used for runway lighting with CTs driving lights along the sides of the runway. The advantage is that there is no drop in lamp power at the far end of the runway as the current is constant throughout. Note that the current is constant and that changes everything!

Back to your question:

You are proposing to draw \$ 5 V \times 2 A = 10 W \$. The nearest standard type might be a 50:5 A unit such as an INSTRUMENT TRANSFORMER 15RL-500 Current Transformer, 50:5, 50 A available from Farnell / Newark. A quick look through the datasheet shows:

Figure 1. Note the maximum burden is 2.5 VA on this CT.

That's not enough power so we need to go bigger.

Figure 2. This one can handle a 12.5 VA burden.

Let's use this one. Note that you've now spent $50 to $100. (I can't find a price.)

To get 5 A out of the secondary we'll need 200 A in the primary but we only have 12 A in our circuit. Let's wind the primary through the core to increase the ampere-turns closer to 200. \$ \frac {200 Aturns}{12 A} = 16.66~turns \$. Since you can only have integer turns let's go for 16. Now 12 A on the primary will give us \$ \frac {5}{200} 12 \cdot 16 = 4.8~A \$ on the secondary.

The maximum secondary voltage we can get is given by \$ \frac {VA~rating}{max~current} = \frac {12.5}{4.8} = 2.6~V \$. Now we can see that this is getting awkward. 2.6 V RMS (it's AC, remember) isn't a great starting point for an efficient low-voltage PSU.

Add to this the problems of variable current and ensuring that, should the current increase for some reason, that you don't exceed the VA rating of the transformer and you can see that the exercise becomes far from trivial and far from cheap.

By the way, your secondary impedance, \$ \frac {2.6V}{4.8A} = 0.54~\Omega \$ will be reflected back to the primary as \$ R_P = R_S (\frac {N_P}{N_S})^2 = 0.54 (\frac {16}{40})^2 = 0.086~\Omega \$ and this will cause a \$ V = IR = 12 \cdot 0.086 = 1.04~V \$ drop in voltage to your mains load.

^{simulate this circuit – Schematic created using CircuitLab}

Background: I am designing an energy harvesting unit to power my Raspberry Pi.

No you're not. Harvesting suggests that you're getting energy that would otherwise be wasted. This is a complex way of connecting a load to a mains supply. As you can see, power is drawn from the mains and you (or someone else, if you're stealing it) will have to pay for it.