You have the right idea for a basic unregulated supply. A transformer, four diodes, and as large a cap as you can manage will serve well enough for a lot of purposes, but isn't appropriate for all.
There are two main problems with such a unregulated supply. First, the voltage is not known well. Even with ideal components, so that the AC coming out of the transformer is a fixed fraction of the AC going in, you still have variations in that AC input. Wall power can vary by around 10%, and that's without considering unusual situations like brownouts. Then you have the impedance of the transformer. As you draw current, the output voltage of the transformer will drop.
Second, there will be ripple, possibly quite significant ripple. That cap is charged twice per line cycle, or every 8.3 ms. In between the line peaks, the cap is supplying the output current. This decreases the voltage on the cap. The only way to decrease this ripple in this type of design is to use a bigger cap or draw less current.
And don't even think about power factor. The power factor a full wave bridge presents to the AC line is "not nice". The transformer will smooth that out a little, but you will still have a crappy power factor regardless of what the load does. Fortunately, power factor is of little concern for something like a bench supply. Your refrigerator probably treats the power line worse than your bench supply ever will. Don't worry about it.
Some things you can't do with this supply is run a anything that has a tight voltage tolerance. For example, many digital devices will want 5.0 V or 3.3 V ± 10%. You're supply won't be able to do that. What you should probably do is aim for 7.5 V lowest possible output under load, with the lowest valid line voltage in, and at the bottom of the ripples. If you can guarantee that, you can use a 7805 regulator to make a nice and clean 5 V suitable for digital circuits.
Note that after you account for all the reasons the supply voltage might drop, that the nominal output voltage may well be several volts higher. If so, keep the dissipation of the regulator in mind. For example, if the nominal supply output is 9 V, then the regulator will drop 4 V. That 4 V times the current is the power that will heat the regulator. For example, if this is powering a digital circuit that draws 200 mA, then the dissipation in the regulator will be 4V x 200mA = 800mW. That's will get a 7805 in free air quite hot, but it will probably still be OK. Fortunately, 7805 regulators contain a thermal shutdown circuit, so they will just shut off the output for a while instead of allowing themselves to get cooked.
The printer draws large amounts (20+ Amps) of current at 12V and these supply exactly that - nothing more, nothing less!
The power supply will certainly deliver less current if the load demands less. If the 3D printer draws 1A, the supply will supply 1A. Your statement that the power supply delivers a certain power level - no more, no less - is incorrect.
I'm looking for a more stable power supply for my 3d printer.
You haven't mentioned stability in your posting. You did say that at high loads, the output voltage sags. The stiffness of a power supply is related to how well the voltage regulates at high load. The stability of a power supply is related to how well the voltage regulates when the output is subjected to a rapidly-changing dynamic load.
There could be two things going on causing the voltage sag:
1) The voltage sensing point is close to the power supply; it is regulating the voltage at that point and the loss you see at the load is due to resistive losses between the sense point and your measurement point
2) The power supply is entering a protection state and is limiting the voltage to limit the output power and keep the power supply thermally safe.
I just have a few concerns about the safety of using one of these devices.
A 'safe' power supply, in industry parlance, means it has been evaluated by a regulatory authority and found to comply with certain national / international safety standards for the application in which it was intended to be used. A single abnormal should not cause a safety hazard (shock / fire / shrapnel). The unit should bear one or more well-recognized safety marks (UL, CSA, TUV, etc.)
My main concern is that I noticed there is nowhere to simply plug a mains cable as an input to the power supply.
As Olin pointed out, this is a unit meant to be permanently installed into some other piece of equipment, not something that a user would be expected to swap-in or swap-out often.
However, isn't there a fair likelihood that someone may just pick it up by putting their fingers and short the live and neutral together? The exposed screw contacts look awfully prone to accidental contact with not just fingers, but nearby metallic objects. A built-in fuse won't exactly help here would it?
Notice in the photo that there's a clear insulating shield over the terminal block. That shield is part of the inherent safety of the unit and should prevent against mains shock from casual contact with the unit. If someone wants to hot-screw a powered mains cord onto this UUT, well, they deserve what they get. Not trying to be facetious, but these sorts of power supplies are meant to be installed by 'qualified' personnel who have some basic knowledge.
It's quite common for power supplies like this (meant for use inside other equipment) to get fed from a feed that has a fuse or breaker in it; there's the possibility that the internal wiring may make contact with the equipment itself. That doesn't mean there isn't a fuse in the power supply though (there should be!)
The other concern I have is whether it will be extremely dangerous if I accidentally reverse the live and neutral wires?
Yes, but only if there's a fault. The power supply will 'work' with reversed L and N. However, the power supply internal fuse is in series with the terminal marked L (line). Blowing the fuse means the neutral is now floating, which is a big no-no (netural must never be interrupted) and a big risk that your chassis can become a shock hazard.
At the moment, I am using a typical 320W PSU for a desktop computer to handle the power workload.
PC power supplies have minimum load requirements on various rails and are rarely a good choice for industrial applications like a 3D printer. Spend the money and get a single-output 12V supply that can deliver the power plus has remote sensing capabilities so that you get the best regulation possible.
Best Answer
If you are making a bench supply for conducting EE design work then you need to have "extra" protection because you may well be touching 0 volts or the "hot" DC output for significant periods of time. Fusing each leg of the DC output and using a zener diode to ground is something I would consider to be safe. When I say ground I mean a proper safety ground. Normally the zeners would not be called on to do anything but, if a transformer fault occured that generated a potential over-voltage situation, the fuses should blow. Zeners and fuses need to be rated accordingly.
For higher power protection, consider using a zener with a crow bar circuit like this: -
Here, the zener passes current to the SCR's gate when the output voltage of the regulator is exceeded. It turns on the SCR and it conducts current until the fuse blows. This is just one example of a crow bar protection circuit and please bear in mind that a lot of SMPS modules may have this feature already.
Other ideas along the same line