This is OK, it's not a problem what you've done but note that the two individual bridge outputs are not isolated and this means they can't be connected together (N1 to N2) without the voltage levels changing. If you want a +V, 0V and -V supply use this: -
simulate this circuit – Schematic created using CircuitLab
When connected the circuit draws upwards of 10A
I'm not surprised. Consider the transformer primary as an inductor and forget about the secondary or any loading. The primary is still an inductor and you have to get the primary inductance high enough so that the current taken (with no secondary winding/load) is reasonably small.
Primary inductance (in fact it's called magnetization inductance) is proportional to turns squared so doubling the turns makes the inductance 4 times greater and, would reduce the intake of current by 4.
Eighteen turns is never going to be enough. A typical ferrite core (for example) might have an A\$_L\$ value of 10 micro henries per squared turn and this means 18 turns is going to be about 3 milli henries. At 60 Hz this has an impedance of just over an ohm and without a fuse you'd have a fire.
Usually AC transformers have round about 1000 turns on the primary and this of course means a primary inductance of 10 henries and a very much reduced magnetization current. Try looking at this site for some information on transformers.
Also note that transformers do not pass DC because the primary inductance becomes a short i.e. the lower the operating frequency you use the more turns you need to avoid excessive primary (only) current.
If you are still intent on rolling your own, here's the math: -
Also take a look at this medium power transformer: -
Note the number of primary turns. A quick estimation I did tells me there are about 12 layers of about 90 turns each. That's 1080 turns!
Best Answer
A conventional current transformer will do what you want.
As Jim Dearden says, you probably do not need or want a current step up transformer as voltage will be reduced AND the device will interfere with normal circuit action.
As shown below, if you use a conventional current transformer and feed the main circuit via a single turn "winding" on the core, then you will get an output voltage at lower current. Terminate the winding with a resistor and then use the voltage across the resistor as an energy source. The single turn usually consists of a conductor passed through the core laminations. This circuit produces voltage proportional to load current - so at low currents the voltage may be too low to be useful with a regulator. Replace the regulator with an "energy harvester" if you want to use a very wide range of voltages. If you remove R_CT_LOAD and place a minimum load on the regulator output it will work at low currents but Vin may rise to high values for heavy mains load. A dynamic load scheme could be 'easily enough' devised so that Vout was acceptable under most mains-load conditions.
simulate this circuit – Schematic created using CircuitLab
R_CT_Load OR a consistent minimum output load on the regulator is absolutely essential at all times. The resistor value is based on manufacturer's data for the CT used and is designed to give a target V when target current flows in the input single turn.
Note that with any current transformer, if you run it unloaded Vout can rise to very high values and may destroy the transformer or connected circuitry.