fallback = backup plan or alternative when primary plan fails. if you are on a cruise ship that is on fire, primary plan is to get to safety boats. but if that does not work, jumping into water and swimming away from a burning wreck is a fallback option.

current foldback = feature that in case of overload limits output current to a value less than rated.

if you have textbook covering this, it should first explain it based on first principle, and then use practical example such as 723 regulator. you should be able to use first principles and (by adding more components) accomplish the same on any regulator.

I think I found the whole problem. Just out of curiosity I removed the foldback current protection with Rsc, R5, R6 and left only Rsc for default current protection. And now the schematic works wonderfully with input voltages as low as 63V.

I don't have a clue why this would break the schematic. I guess that's why in the datasheet it is only shown in default current protection.

 

Foldback can be tricky to get right. Too much foldback and the first overcurrent can shut it down and prevent it from restarting.
And the overcurrent can simply be charging the output capacitor.

 

I removed the foldback current protection

I looked at that, and it seemed wrong, but I have not used it in that mode so did not say anything.

I though back to my LM723 supply. It is pretty high voltage and much higher in current. One problem is at high current and low voltage the transistors get very hot. The entire back side of the box is covered in large heatsinks. My transformer is probably 40Vct. (center papped) I added a switch in the diodes area. In one mode the diodes are across the entire secondary. In the low voltage mode, the diodes use only half the secondary thus reducing the voltage and reducing the voltage.
I used a 10-turn pot to set the voltage. It is hard to go from 0 to 40V on a 1-turn pot. I don't remember how, but I found a way to make the pot linear.

 

One cheap trick, which I have not seen, would be to use a lower performance rectifier scheme, such as a simple half wave rectifier??
OR is that too simple for the project assignment requirements.?? Really!! use a half wave single diode rectifier and then a two stage , large capacitors filter, and the supply voltage can be less.
IF you are already using an isolation/step-down transformer,, re-connect it as an autotransformer to get one third of the mains voltage instead of one half.

 

One cheap trick, which I have not seen, would be to use a lower performance rectifier scheme, such as a simple half wave rectifier??

And how does that reduce the voltage???

 

Foldback can be tricky to get right. Too much foldback and the first overcurrent can shut it down and prevent it from restarting.
And the overcurrent can simply be charging the output capacitor.

Exactly!
Sometimes being a positive feedback, foldback limiting prevents the supply to startup.
I would encourage you NOT TO GIVE UP, but experiment with a lower voltage fraction.
You may want to replace the 88 ohm resistor with a 100 ohm variable resistor

 

The short answer to how using a half wave rectifier reduces the voltage is that it cuts the delivered power in half.. It does not reduce the peak voltage at the filter input very much, BUT it does lead to some reduction at the output of the filter string, especially if it is a two resistor three capacitor filter string. If that reduction is enough is not obvious nor assured. BUT it should be worth a few minutes of investigation.

 

The short answer to how using a half wave rectifier reduces the voltage is that it cuts the delivered power in half..

no

 

no

If only half of the peaks get past the diodes I count that as half power. No, it does not deliver half the voltage, It certainly works in resistance heating applications.

 

If only half of the peaks get past the diodes I count that as half power. No, it does not deliver half the voltage, It certainly works in resistance heating applications.

Which, of course, doesn't apply in this application.......