Parallel current limiters, ballast resistors and transistor matching
The ballast resistors are not required to help deal with part variations. The current control transistor should be thermally isolated from the TIP142 and from the current-measuring resistor, though. Also, you probably should include a soft-start capacitor in the circuit (or at least be aware of the idea in case it may help avoid damage to BJTs.) And you should include an emitter degeneration resistor for the control transistor to help equilibriate the currents.



I'm going to completely ignore your choice to divide this up into three current supplies and just look at the datasheet, instead. What I gather is that the die temperature should not exceed \$150\:^\circ\text{C}\$. I'll choose \$125\:^\circ\text{C}\$ to keep a margin. I'll also assume an ambient of \$25\:^\circ\text{C}\$, so the difference allowed is \$100\:^\circ\text{C}\$. Acccepting your worst-case \$58\:\text{V}\$ across the TIP142 itself, this means your situation must be \$I\le \frac{100\:^\circ\text{C}}{58\:\text{V}\cdot\left(R_{\theta \text{JC}}+R_{\theta \text{CA}}\right)}\$ where we already know that \$R_{\theta \text{JC}}=1\:\frac{^\circ\text{C}}{\text{W}}\$. Counting in the vagaries of mounting and the availability of heat sinks for these packages, I'm guessing that you cannot expect better than \$R_{\theta \text{CA}}=4\:\frac{^\circ\text{C}}{\text{W}}\$. This means \$I\le \frac{100\:^\circ\text{C}}{58\:\text{V}\cdot\left(1\:\frac{^\circ\text{C}}{\text{W}}+4\:\frac{^\circ\text{C}}{\text{W}}\right)}=430\:\text{mA}\$.

Or about \$25\:\text{W}\$ per section.

So already I have a problem with your choices. I think you need to plan on at least 10 of these sections. 12 would actually be the specification. But you are allowed to make some balancing decisions here. So I'd say anything at or above 10 sections would be advisable, given your desire for \$5\:\text{A}\$ total.



Variations in the TIP142 don't really matter that much, since you have a control transistor observing the current and making individual adjustments for each TIP142.

The current-control resistors will heat up and may account for a small shift in resistance (assuming you don't scrimp on its power rating.) But all of these current-control resistors are going to heat up together and I don't expect a large variation of resistance between them. So the variations in current supplied by each section, relative to these resistors, should be within the specification for the resistors, themselves.

So what matters here will be the variations between the control transistors. And this can be on the order of variations as if they were up to \$\pm 15\:^\circ\text{C}\$ different from each other, worst case. Or on the order of \$V_\text{BE}\$ variations of about \$\pm 30\:\text{mV}\$. Given an expected \$V_\text{BE}\approx 700\:\text{mV}\$ this could mean another 10% maximum variation between each section. Whether or not this is terribly important to you isn't a question I can answer. You could just leave it alone and live with it. But if you want to do something about it, you can add a small degeneration resistor into the emitter leg to provide a reasonably known "added voltage drop" to reduce the effective percent-variation between the sections.

This is why I said that the ballast resistors are not needed. That's a horrible waste of power and not nearly as effective, either.



Here's an example circuit I'd consider putting more time into:







simulate this circuit &ndash; Schematic created using CircuitLab

Assuming that \$Q_4\$ supplies \$10\:\text{mA}\$ at its collector (adjust \$R_3\$ to achieve this depending on your \$V_\text{CC}\$ power supply rail voltage), and assuming also that the TIP142 won't require more than \$500\:\mu\text{A}\$ into its base, then this means we have a reliable \$9.5\:\text{mA}\$ into \$Q_2\$'s collector and out its emitter into \$R_1\$. This makes the voltage drop across \$R_1\$ predictable at \$33\:\Omega\cdot 9.5\:\text{mA}\approx 260\:\text{mV}\$. Rounding things out, this places a \$1\:\text{V}\$ drop across \$R_2\$.

\$C_1\$ adds a soft-start component to this. (You might also add a \$27\:\text{k}\Omega\$ bleed-resistor across it -- not shown.) The current source \$Q_4\$ will have to charge up \$C_1\$ before the circuit can fully operate. I've set the timing here to about \$100\:\text{ms}\$. You can adjust this differently... or entirely avoid \$C_1\$, if you want.

I've used TO-220 BJTs for \$Q_3\$ and \$Q_4\$. Use whatever pair you want, but since one of them will dissipate up to a watt of power, the TO-220 package would be desired here. No heat sink needed. The package's \$R_{\theta \text{JA}}\$ should be good enough.

I believe the above circuit will provide a reliable section without the need for those wasteful ballast resistors you've added. It does so at the expense of a little more circuitry. But there is less waste, better balancing of the currents, and you now know that you need more than three sections, too.



You'd probably be much better off with a switcher design, though. You might check out this page: Roman Black's constant current switching regulators.



Added Note

Per your question about the current mirror:







simulate this circuit

I think you can see the buffering/base current compensation BJT that's been added: \$Q_5\$. This will handle the added base currents.