The best DC power corresponds to a transformers secondary A.C. rating.
A transformer with 12v secondaries will make 16.92v D.C. after it is smoothed,
minus two .6v diode drops from the bridge rectifier; let's call it 15.72v D.C.
This is the D.C. peak voltage. The real power comes on lower, at 10.8v,
[.90 * secondary A.C.] This I have verified thru experiment. The most wattage
is developed at the transformers rated VA voltage minus the diodes & minus 10%.
[Average D.C. voltage].

My question: is this peak voltage good for anything? It seems to me it's just
an annoying useless fluff voltage, that doesn't have much power in it, & that
gets in the way of your design, preventing regulator I.C. from utilizing it's
maximum voltage capacity, [40v for LM317, LM723, etc.]

Should it, or can it be considered for a voltage regulator's overhead voltage of 2-3v?
Maybe I should just be rid of it, perhaps with a pre-regulator?

Anyway, when working with a transformer secondary of 12v, the problem is merely annoying.
If the secondary A.C. voltage is 100v, then the peak D.C. voltage is 141v, yet average
D.C. voltage is 90v. That's a 51v spread!

Again, this peak D.C. is just fluff, not much amperage in it, but often, more than enough
voltage to fry a regulator I.C. How should I think about it?

 

The peak voltage is what you use to rate your capacitors and diodes for maximum voltage requirements. All transformers load down. All rectifier/filter combinations have ripple that goes below the peak voltage. Dealing with that is the art and science of designing power supplies. You might call it, "fluff". I call it, "reality". Learn how to design for the reality available or fail. You would be amazed at the crap we had to work with 50 years ago. +/- 20% was, the usual design criteria for early TVs. Consider yourself lucky to be in the 21st century.

 

Consider this to 'peak' your interest in peak voltage...

• The peak value is a fact of life you have to deal with. It must not be ignored.
• The filter capacitor voltage rating needs to be at least the peak voltage value out of the transformer. Round up to the nearest standard voltage rating.
• From the peak voltage you can calculate the peak current into the capacitor and therefore the max ripple to expect under full load.
• Be sure to s.elect a capacitor that can handle that peak current
• The 3 volt head room is calculated from the lowest dip of the ripple. A series linear regulator also acts as a super efficient filter to remove virtually all the remaining ripple. This allows you to get away with a small filter cap.
• The voltage drop (51V) and the varying ripple you see when a load is applied or removed is also unavoidable. The drop in the input voltage to a 3 terminal regulator can be a good thing since it reduces the voltage drop across the regulator when current is the highest and therefore the power dissipation in the regulator.

Ifixit

 

Also consider that if you could somehow get rid of it, that your effective voltage would go down. It is making up for the times when the voltage output from the transformer is less than the effective voltage.

 

Alright, this is good, thank you.

I also saw this from TL783 datasheet:



A nifty way to deal with the peak voltage (sorry for calling it fluff).

 

How is a short-circuit protection circuit "dealing" with the peak voltage?

At first glance, I'm having a hard time seeing how this provides short-circuit protection in the first place. If the output is shorted, then the zener will hold the base of the Darlington at about 120 V which will place the emitter at about 118 V to 119 V. The regulator will still try to get 1.25 V across R1 by pumping out as much current as it can. Since the output is shorted, for every mA it delivers it will be dissipating 118 mW. The only thing I can think that this circuit does is claim the Vin-Vout voltage to a low enough value to guard against exceeding the spec on that while waiting for the internal thermal protection circuitry to shut it down.

Though, looking back at your OP, I can see that it might be useful for your application. Be sure you consider how much power is going to be dumped by the Darlington and heat sink it accordingly.

 

Also consider that if you could somehow get rid of it, that your effective voltage would go down. It is making up for the times when the voltage output from the transformer is less than the effective voltage.

I can appreciate this. Also, a simple battery charger could be constructed where the float voltage is equil to the
very top of the d.c. voltage output of the transformer. Very little amperage would flow, while a greatly discharged
lead acid battery would receive a bulk charge.

 

a simple battery charger could be constructed where the float voltage is equil to the
very top of the d.c. voltage output of the transformer.

Not good enough. You can't trust the power line voltage to be the same every where or every day. That's why we use voltage regulators. Remember, "Vout considered for all conditions of line and load".