Rectify AC and keep the voltage below V_max for DC step down converter
The following is not a managed design. You only have a few parts laying about and the only thing, apparently, capable of actually dissipating much power is the IRF520. My only goal here is to move some of the dissipation onto the FET and away from resistors you have. You will still have to dissipate something more than "tiny" in at least one of the resistors. But I'll keep it to a minimum.

You have a \$24\:\text{V}_\text{AC}\$ transformer. This means the peak DC voltage, ignoring bridge rectifier diode drops for now, would be \$\sqrt{2}\cdot 24\:\text{V}_\text{AC}\approx 34\:\text{V}_\text{DC}\$. Your output requirements for power are \$3.5\:\text{V}\cdot 500\:\text{mA}=1.75\:\text{W}\$. Given the 75% DC-DC converter efficiency you mentioned, this means \$\frac{1.75\:\text{W}}{.75}=2.\overline{3} \:\text{W}\$. So let's just call it \$2.5\:\text{W}\$ in round figures. This means that if you can arrange for about \$25\:\text{V}_\text{DC}\$ at the unregulated DC supply output, you'll need to support an average of \$100\:\text{mA}\$ from this unregulated DC supply.

I don't know what you have out of which to make a bridge rectifier. But just for a moment I'm going consider the idea of 1N4148 diodes. (These are very common.) A question is whether or not they are sufficient. They can, theoretically, handle a peak current of about \$2\:\text{A}\$ (for about \$1\:\mu\text{s}\$.) In your case, expect peak currents about 5-10 times more than your average current. This is under the absolute maximum specification for peak current, but the peak period is likely to be a lot more than \$1\:\mu\text{s}\$ -- at least two orders of magnitude larger. So I don't think it would be a good idea. Power dissipation is likely to be about their voltage drop, near peak, times the average current -- which is in very rough numbers likely to be about \$100\:\text{mW}\$. With the package thermal resistance to air from the datasheet, this suggests about \$35^\circ\text{C}\$ above ambient. If that holds, it should be okay. So if 1N4148 is all you have, you could at least attempt to use them. The worst case is that you destroy some. It would be better if you have some from the 1N400x series. 

Regardless, your bridge rectifier diodes are likely to drop more than volt per diode, or probably on the order of something above \$2\:\text{V}\$ and less than \$2.5\:\text{V}\$. So, without anything else your unregulated DC supply output is likely to be somewhere between a peak of \$31.5\:\text{V}\$ and a peak of \$32\:\text{V}\$. (This is assuming a load! Unloaded, it should be back towards \$34\:\text{V}\$.)

A capacitor is another consideration. Let's say you want to accept something on the order of about \$\pm 1.5\:\text{V}\$ variation during each cycle. Since the capacitor will be carrying the load for most of the cycle (and let's assume 60 Hz), then this works out to about \$270\:\mu\text{F}\$ for the filter capacitance. (You can ask me how I got that, if needed.) More is better. But that's a value I think you could live with.

Now, the \$\pm 1.5\:\text{V}\$ variation also means (because of the above peak values under load) that you should target about \$30\:\text{V}_\text{DC}\$. This is assuming there isn't a resistor to further drop the voltage. And now we are there. You really do need something to drop the voltage.

This can either just be your IRF520 or else you could use a series resistor and add a known load to the output to force a known voltage drop. Given your circumstances, I'd probably just try letting the FET do all the work. Something like this:







simulate this circuit &ndash; Schematic created using CircuitLab

You may need to adjust the resistor divider to target the voltage output at the capacitor that is safer for your DC-DC converter. The threshold voltage for the FET is as much as \$4\:\text{V}\$ but it can be as little as \$2\:\text{V}\$. So you really do need to be aware that the resistor divider may need adjustment. But I said it's unmanaged and that's about what I'm thinking you may be stuck using, without getting too complicated about it. Just be aware of the FET heating when you make adjustments. But the FET is just dropping some voltage while also passing along the peak bridge rectifier current pulses that charge up the filter capacitor. The average power should be close to \$500\:\text{mW}\$, I think. Given the thermal resistance for the device, it will definitely heat up but probably will be okay with or without a heat sink. (I'd advise using a heat sink, though.)

Another option would be to use a series resistance between the bridge rectifier and the capacitor, plus using the FET to act as a programmed load (in parallel to the DC-DC load) that would guarantee a certain average voltage drop across the series resistance. But this would require a series resistance that can dissipate and you've already said you have tiny resistors. So you'd have to fabricate something out of them. So I'll hold short talking about that kind of thing, for now.

I've added a resistor, \$R_3\$, to discharge the capacitance and to provide a tiny load. It really would be better if it dissipated more, but I set it to work as a "tiny" resistor. Make sure you do provide this load on the capacitor, though. Don't avoid it. I think it might also be a good thing to provide something like it (lower valued, though) on the other side of the DC-DC converter, as well. Yes, you may have a load attached to it. But, at times, you may not and will be glad of the resistance there, as well.

Before you connect up the DC-DC convert to the above circuit, measure the DC voltage at the capacitor and adjust \$R_1\$ and/or \$R_2\$ in order to achieve a voltage that is safe for your DC-DC converter. Also, feel free to add as many \$R_3\$ resistors as you are willing to add to the output. More is better. I'd recommend at the very least two of them. Very least. Once you are comfortable with the output voltage, go ahead and add the DC-DC converter. But also be sure that the DC-DC converter already has a load on its output, say \$1\:\text{k}\Omega\$, before hooking it up.