A few quick points:
0. I'm not sure where your original equations came from. They don't smell right. First, Vbe = 2V is way way beyond normal operation for a single BJT. Most folks use 0.6 to 0.7 when they're doing back-of-the-envelope calculations. In real life, 2V across the base-emitter terminals is an indication that not all is well.
1. As Connoiseur indicated, a key concept that is behind some of the troubles with this design is the "SAFE OPERATING AREA" In general, for a power transistor, this relates tempearature, power dissipation, and likelihood of smoke. Specifically, for a BJT it is a graph giving a bounding polygon where the transistor is in a "safe" operating condition if the voltage (Vce) and current (Ic) are within the polygon. Note that the safe area is normally specified assuming some duty cycle, and a case temperature. Knowing the case temperature requires a little math on its own.
2. What you've got in the drawing is an emitter follower. Unless there's some other criteria that you're considering, this is not often the best choice for a regulated power supply. See the pretty good discussion here:
https://www.eevblog.com/forum/beginners/variable-regulator-pass-transistor/But let's look at what's going to happen with the circuit in your drawing.
Assume for the moment that the input voltage to the LM138 is 40V. (Just for the sake of discussion). And now let's say that we want a 20V output at about 4 A. Working back from the output, we've got four transistors in parallel, each with its own emitter resistor.
Those resistors need to be small, or they'll have to be very high power widgets. At 4A, with four parallel pass devices, you'll see 1W per ohm in each of the Re . Push that current up to 10A and we're talking 6.25 W/ohm -- A few watts here, and a few watts there, pretty soon you're talking real power. Your original post suggested Re = 5 ohms.
If we go to the 5ohms that your initial figures suggested, this is what happens:
If we set the LM138 feedback when we have 0 load, the voltage at the 138 output will be something like 21 V. Let's take the load up to 4A. Then we'll be dropping 1A * 5 ohms or 5V through each of the series devices. That means your original 20V open circuit voltage dropped to 15V under load.
If the load draws 20A then you're realling in the soup, since each Re is passing 5A through 5ohms and the 25V drop across Re puts the emitter above the base. That's clearly nonsense, so what will really happen is Vout will drop to some voltage where the load current, emitter drop, Vbe, and Vb all conspire to please Kirchoff. But Vout won't be anywhere near 20V.
So, 5ohms for the emitter resistor is clearly not right. Normal values should put you on the order of 0.1 ohms or so...
Emitter followers don't make for very good regulation unless you sense the actual output. But there are interesting problems when you introduce that kind of feedback, since it is possible to design a combination DC supply and oscillator all in the same widget.
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A second consideration is the power dissipation. I'm assuming that you want this power supply to cover some range of voltages, perhaps 5 V to 20V? Let's now consider what happens in the transistors when you draw 4A at 5V. Vce will now be 35V (remember the input voltage was something like 40V). And each transistor is passing 1A. So each transistor has to dissipate 35 Watts of thermal energy somewhere. The data sheet shows a contour for "1s" duration. Frankly, I don't do power electronics much (I design loads, not supplies..
) but I suspect that continuous duty won't be quite as generous as the 1s contour.
The big deal here is that you're going to want to remove 35W of heat. The thermal resistance from the junction to the case is about 0.7 C/W. So, the junction is going to be about 20 C above the case temperature. If you get a heat sink of about 46mm x 46mm with some air blowing across it, you'll see a thermal resistance of about 1.5 C/W. Assuming a perfect bond between the case and the heat sink, the junction temperature is going to be about 70C above ambient. If you don't have a fan on the heat sink, the resistance rises to more than 6 C/W... Then the junction will be about 200 C above ambient. WHOOOSH! The smoke gets out. (And we were not even close to 10A load current...)
Heat sinks are important and easy to get wrong.
Take a look at the other thread, it had some good thinking points.