Short: Add a 1 ohm resistor in series with the transformer :-).
Longer:
A "perfect" transformer and 'perfect" capacitor will have infinite current spikes, as I know you realise.
While real world results will vary with transformer maker's 'ethos and philosophy', the real world experience is that you wil usually get superior results by adding a small "conduction angle spreading resistor" in series with the transformer winding feed to the capacitors. This is counter intuitive to what you may expect from an efficiency point of view and is often not done in practice. Theoretical calculation of the effect of such a resistor is surprisingly annoying but simulation will show the effects instantly.
Given that the mean DC level under load is 0.7071 ( = sqrt(2) ) of V peak, you have quite a lot of headroom to work with and can afford a modest amount of drop in the series resistance. There are several scondary effects which may be useful depending on environment. Spreading the conduction angle improves the power factor of the otherwise very peaked load - but probably not enough to make a difference in meeting or failing formal power factor requirements. Sometimes more importantly, spreading the conduction angle greatly reduces peak loads on the diodes and reduces EMC issues (ie less radiated electromagnetic noise) - probably not an intuitive effect of adding a few ohms of series resistance.
Lets have a play with some figures:
You have 15 VAC secondary voltage and are aiming at 12VDC at 2A.
Assume for now that about 15VDC minimum on the filter caps is acceptable 9giving the regulator 3V headroom minimum).
Vpeak is 15 x 1.414 = 21.2 V
Load power is VI = 12 x 2 = 24 Watts.
If you managed to filter this well enough to achieve say about 20VDC on the cap you would dissipate Vdrop x I = (20-12)x 2 = 16 Watts in the regulator and "as a bonus" achieve massive ripple CURRENT in the caps but little ripple VOLTAGE. This does not seem like a marvellous idea :-).
If you can manage to spread conduction over 25% of the voltage cycle you will get mean current during conduction down to 4 x Iavg = 8A.
Assuming 21V peak, 25% conduction occurs at about 19V transformer output, and a very useful 50% conduction happens at just under 15V. See graph below.
This suggests that inserting even one ohm series resistance is going to have a substantial effect. If the 8A mean that is required for 25% conduction is dropped across 1 ohm the 8 volt voltage drop is going to ensure that the 8A does not happen (as 21-8 = 13V which is lower than the 15V DC target this was based on).
If 50% conduction occurs then mean current during this period will be 4A and mean drop across 1 ohm would be 4V so this may be "about right" as if the filter cap was at about 15V you'd get (21-15)/1 = 6A peak at waveform peak - and as the cap will have "rippled up" in voltage by then you'll get less than 6A). And so on.
Yes, you can analytically work out what happens. But, just put 1 ohm in the simulator and see what happens.
This has the effect of putting MORE ripple voltage on the capacitor(s), LESS ripple current, less regulator losses and less transformer losses, less diode EMI.
The series resistnce could be in the transformer but then addes to heat generatoion inside a relatively costly component where you'd rather be trying to optimise power transfer rather than heat loss. A 5 Watt 1 ohm resistor will probably work OK here. 10W would be safer due to peaks. eg 4A at 50% = I^2R x 50% = 15=6W x 0.4 = 8W BUT waveform is complex so actual heating needs to be calculated.
Note that in many cases the ripple current rating of two capacitors is superior to that of a single capacitor of equal total capacitance.
Use 105C (or better) caps as a matter of course in this sort of application. 2000 hours+ a good idea. Cap life ~~~ 2^((Trated-tactual)/10) x Rated_life
From the article:
As it appears, Haswell's C6/C7 states require a minimum load of 0.05A on the 12V2 rail, and many desktop power supply units (PSUs) just cannot provide that low current, reports The Tech Report web-site. Meanwhile, numerous older PSUs, which comply with ATX12V v2.3 design guidelines only called for a minimum load of 0.5A on the CPU power rail, hence a less sophisticated internal feedback loop/protection could be used, reports VR-Zone web-site. As a result, unless C6/C7 power states are disabled in the BIOS, PCs with older/cheap PSUs may become unstable when processors enter these states.
A minimum load specification signifies the smallest load that can be drawn from the power supply while meeting all of the other requirements in the specification (regulation, transient response, etc.)
The power supply may or may not be able to deliver less current than what is specified as its minimum. It may deliver but drift out of voltage regulation; it may become unstable and oscillate; it may hiccup on and off; it may even go into overvoltage protection and latch off. Because the load is outside the specification, "anything goes".
The article's statement "just cannot provide that low current" is (to me) a gross simplification of the matter, and is a bit misleading. Current power supplies were never designed to meet this specific condition, so behaviour at this condition is undefined.
Best Answer
Holy cow, talk about an undersized heat sink.
Your 24 VAC transformer will produce about 32 volts at the main filter caps. Producing a 5 volt output from the LM317 means that you are dropping about 27 volts across it. A current of .14 amps says the LM317 is dissipating about 3.8 watts. The data sheet for your heatsink says that the temperature rise AT THE HEATSINK will be about 80 C, for a heatsink temperature of ~105 C. This is consistent with your measurement. If you look up the datasheet for the LM317, it specifies a thermal resistance (chip to case) of about 5 degrees per watt, so 4 watts on the chip will give it a 20 C rise over the case. Since the case is tied to the heatsink, your chip temperature is right about 125 C, which is the absolute maximum allowed. Dave Tweed is right about the regulator shutting down in order to keep from self-destructing.
If you could pull 1 amp, you would be dissipating 25 watts, and your heatsink temperature would be (roughly) 500 C. Plus another 125 C rise to the chip itself, and there is simply no way that's going to happen.