#linear regulated

It might be the long week, and it may also be that I was mentally prepared to shelve the power supply project, since you know, I thought it was done!  But...

http://www.tymkrs.com/forum/viewtopic.php?id=27&p=2

So a comment from Youtube came:

"Your fuses are on the wrong side of the transformer. You'd want to protect the voltage regulator. What you usually would do is put a 1A fuse (if you want to protect it if it goes higher than 1A) after the capacitor. The current drawn from the 120 AC is much much lower than the current coming out of the voltage regulator."

Roy said: I think you want fuses where they are, but you might want an additional fuse where he says.  Every power supply I have been inside of has had a fuse between the wall and the transformer.  He's right that the current is higher on the output of the regulator than on the input to the transformer... but that's just because you have high voltage lower current on input and lower voltage and more current  on the output side

Brainwagon: Uh, the reason you have a fuse is so that if the transformer shorts, you don't start a fire.

Me: The transformer would limit it to 1.5A right?

Brainwagon: It doesn't limit anything.  That's what it's rated for.

And this throws my head into a tizzy.  I thought that what was coming out of the transformer was 18v 1.5A.  But apparently it's not?  It can draw more?

So I should have worked from this calculation:

18v * 1.5a = 115 @ x     and solving for x = 0.235A 

And gotten a fuse of 0.250A.

Essentially, the fuse that I have now would allow the transformer to draw up to 6.389A which is above its specified rating:

115V @ 1A = 18V @ 6.389A <--above rating of 1.5

Now to address the difference between the voltage regulator and I guess the transformer.

18V @ 1.5A = 28V @ 0.99A       (MAX Voltage Output) 18V @ 1.5A = 1.2V @ 22.5A      (MIN Voltage Output)

But if I have a circuit and it wants 1.2V, it won't necessarily pull 22.5A from this.  So as long as my circuit doesn't require more than 5-8A (max output of the LM338T chip), it should be okay?

But how do I account for that? The fuse protects the transformer from the wall, but doesn't protect the voltage regulator chip.  So what's the calculation to protect the voltage regulator chip?

Questionable equations:

1.2V @ 5A = 18V @ 0.333A

28V @ 5A = 18V @ 7.778A  (which is higher than the rating)

I'm genuinely confused.