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.
It is hard to determine from the datasheet alone and may need testing to be sure.
I have seen supplies similar to this one do one of two things when reaching the limit:
- Gradually reduce voltage to keep load current at the limit (matching your conclusions)
- Pulse the load. i.e. when the current limit is crossed, voltage goes to zero then tries again a moment later, only to reset again immediately.
I see nothing in the linked datasheet that clues me into which of these two situations would apply.
Best Answer
There is no theoretical limit other than the laws of thermodynamics. Efficiency cannot be 100%.
Typically switching supplies designed to a reasonable cost have an efficiency of >80% over a range of outputs and inputs. Typically efficiency falls as the output load is reduced because there is some overhead in the supply, and it may fall a bit at the highest rated current as I^2R and other losses mount.
Transformers can be very low loss (especially if they are designed for low losses) but that is not necessarily true - in fact relatively recent State of California legislation basically outlaws the sale of many cheap transformer AC adapters because their losses are higher than switchmode types, especially at low loads. Some run warm to the touch even with no load. As an Engineer we have some control over this- by specifying the better (and more expensive) laminations the transformer can have dramatically lower core losses, but this is unsuitable for a product that must sell for a very low price.
In your earlier question you were asking about linear supplies, which generally (but not always) have a low efficiency. A perfect linear regulator supplying 1A at 1V from a 12V source has an efficiency of 1W/12W = 8.3%, which is much lower than 80%. On the other hand, a perfect LDO (linear regulator) supplying clean 5.0V from a noisy 5.6V source has an 89% efficiency, possibly better than a switching supply.
Current in itself is not a problem- supply currents in modern PC motherboards are probably in the 100A range (power of a CPU alone might be 120W+), and the supplies themselves do not require heat sinks.