How much voltage does the first lm338 drop ?
How does it actually track the voltage, I thought there should be 1.25 V across adj and output?
I am not able to understand the logic of using the pre regulator, does it equally distribute power dissipation?
Can anybody explain this with an example of input voltages, current draw and output voltages on the two regulators?
I have searched hard and was not able to find an adequate answer.
If you look at the LM338 data sheet http://www.ti.com/lit/ds/symlink/lm138.pdf, you'll find the standard circuit on page 7.
simulate this circuit – Schematic created using CircuitLab
The circuit works by changing the output so that the voltage aross R1 stays fixed at 1.25 volts. Then (assuming no significant diversion of current by the Vadj pin), the voltage out is (1 + R2/R1) x 1.25 volts.
So, let's start by assuming that the right-hand regulator has some output output VO. Then , looking at the resistors R1 and R2, the left-hand regulator (the preregulator) will have its output, call it V2, set so that (V2 - VO) x (R1 / R1 + R2) = 1.25 volts. It's just like the standard circuit except that the preregulator is referenced to the primary regulator rather than ground.
For R1 and R2 in the schematic, 1 + R2 / R1 = 1 + 720 / 240 = 1 + 3 = 4. So the preregulator will maintain V2 at VO + 5 volts.
At this point, you are probably deeply suspicious. How, after all, can you be certain that the main regulator is giving the correct output? The whole thing forms a vicious circle, right? If the main regulator is off, the preregulator is off, and it makes your head hurt, right? The secret is the presence of C1. This makes the preregulator just a little slower than the main regulator. So the preregulator always provides 5 volts more than the output, but doesn't respond quite as fast. This allows the main regulator to do its' thing without causing the whole circuit to start oscillating.
