Is your transformer secondary 12 turns total or 12 turns on each side of the center tap? If it's the former, that's why you can only get 30 V under load. I calculate it this way:
Input voltage is 220 VRMS full-wave rectified to about 310 VDC.
This means that your half-bridge is driving the transformer with a voltage whose peak is half of this, or 155 V.
The 33:12 transformer is going to turn this into a peak voltage of about 56 V.
If the secondary is center-tapped, then you're only hitting the rectifiers with a peak of 28 V.
As for the excessive rise at low loads — well, that's why lots of SMPS specify a minimum load. It's actually quite difficult to design an efficient one that also has a huge dynamic range. One problem might be excessive leakage inductance (i.e., less than perfect coupling) in your transformer.
EDIT: Since I can't put this drawing in the comments, I'll add it here. Your transformer drive waveform always needs to be symmetric. At 50% duty cycle, it should look like a square wave, with a small amount of "crossover distortion" created by the dead time:
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But at lower duty cycles, it still needs to be symmetric, with longer "off" periods between the alternating pulses. It should look like this:
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- ------ ------ ------ --
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This is the sort of waveform that the drivers on the SG3525 are designed to produce.
It takes almost 2 mA just to charge and discharge the gate of your MOSFET. You're also wasting about 5 mA in R1, since it is grounded through pin 7 about half the time. Your voltage feedback divider is drawing about 1 mA from the high-voltage rail, which translates to more than 20 mA at the input.
There's a problem with using a 555 to drive a large MOSFET: The limited output current of the 555 means that the MOSFET can't switch quickly from full-off to full-on and back again. It spends a lot of time (relatively speaking) in a transition region, in which it dissipates a significant amount of your input power instead of delivering that power to the output. The MOSFET has a total gate charge of 63 nC, and the 555 has a maximum output current of about 200 mA, which means it takes a minimum of 63 nC / 200 mA = 315 ns to charge or discharge the gate. If you're using a CMOS 555, the output current is much less and the switching time is correspondingly longer.
If you add a gate driver chip between the 555 and the MOSFET (one that's capable of peak currents of 1-2A), you'll see a marked increase in overall efficiency. A real boost controller chip will often have such drivers built in.
If you're serious about developing switchmode power converters, you definitely need to get an oscilloscope so that you can see these effects for yourself.
That regulator design is also rather crappy for another reason. The power through a boost mode converter is regulated by varying the duty cycle of the switching element. In this circuit, the feedback is created by using a transistor to pull down on the control voltage node of the 555, which reduces the upper switching threshold. However, because of the way the 555 is constructed, this also reduces the lower switching threshold by a proportional amount. This means that the change in duty cycle as the ouptut voltage rises is much less than you might otherwise think. It has a bigger effect on the frequency of the output pulses, but this isn't relevant. Again, switching to a proper boost controller chip would solve this problem.
By the way, the "regulator" part of the circuit is NOT using the input voltage as its reference, it's using the forward voltage of Q1's B-E junction as its reference.
As Spehro points out, a 100 µH inductor at a switching frequency of 30 kHz — nominal on time = 16 µs — with a 9V source is going to reach a peak current of 1.44 A. This is really abusing the hell out of a 9V battery, not to mention the I2R losses in both the inductor and the MOSFET. This is also uncomfortably close to the saturation current of the inductor, which only exacerbates the losses.
Best Answer
With 240 W out, the input power will be 240 W plus whatever the supply uses/wastes internally. You say the supply is 89% efficient at full load. How efficient is it at the 40% load of 240 W?
To be pessimistic, let's assume 80% efficiency at 240 W. Then do the math. (240 W)/80% = 300 W. To get the current, divide that by the voltage: (300 W)/(230 V) = 1.3 A.
Cable size is a function of the maximum current it must be able to sustain. Get a wire chart and look it up.