BJT Pair on Voltage Regulator Output
Pin 1 of the ribbon cable connector allows the target board to supply power to the serial adapter.
So it's probably better to &quot;see&quot; that section of the target board supply as the following:



simulate this circuit &ndash; Schematic created using CircuitLab
Note, above, that I've changed the viewpoint for \$Q_2\$ so that it is oriented correctly. The reason is that it's actually a BJT operated in reverse mode (active or saturated), which means \$\beta_{2_\text{SAT}}\$ will be rather low -- on the order of 1 or 2, maybe less. Meanwhile, \$\beta_{1_\text{SAT}}\$ will be much higher -- excessively convervative if estimated at 10, and much more likely operating okay nearer values of 40 or 50.
So there's quite a difference in the load currents that can be supported by that single resistor, \$R_2\$, when comparing the \$V_\text{D}\$ rail and the \$X_\text{PWR}\$ rail. And I think that's really the main point of this exercise. The connector pin on the target board is supposed to provide power to the serial adapter module. But it's an exposed pin and can be shorted or otherwise accidentally abused. If a short or an excessive load is placed on pin 1 (XPWR), then \$Q_2\$ just won't be able to saturate and the output voltage will drop to a safer level and so will its &quot;collector&quot; (reverse-emitter) current.
Meanwhile, \$V_\text{D}\$, which is safely and securely used by the target board itself (a known quantity and where fumble-fingers can't as easily screw things up) is being supplied via a higher-saturation-beta BJT, so it can handle a higher load current at its forward-collector.
However, \$R_2\$ does limit the total load that can be delivered either to \$V_\text{D}\$ or to the external pin to the adapter. So it's really a very nice way to handle things.
\$V_\text{BASE}\approx V_\text{CC}-700\:\text{mV}\$; available base drive current is about \$I_{R_2}\approx\frac{V_\text{CC}-700\:\text{mV}}{R_2=6.81\:\text{k}\Omega}\$.
This is one of those cases that reminds us why maintaining proficiency with a variety of discrete BJT design concepts still remains relevant today.
(A final note of smaller interest is that the voltage drop across the collector and emitter leads of a reverse-operated BJT, in saturation, is slightly less than it may be when forward-operated. So \$Q_1\$'s saturation voltage drop will likely be a little more than \$Q_2\$'s saturation voltage drop. Assuming sufficient base current is available for both BJTs, I'd expect the ratio of collector-emitter voltage drops between the two to be about a factor of 2 or 3. This has the nice effect that the voltage at \$X_\text{PWR}\$ will be closer to \$V_\text{D}\$ than \$V_\text{D}\$ is to \$V_\text{CC}\$. So those two rail voltages should be pretty close, so long as the load on \$X_\text{PWR}\$ is sufficiently light. I'm pretty sure that the designers of this circuit were aware of this, as well, and appreciated this side-effect, which came along for &quot;free.&quot;)