DC supply for control circuit of a DC-DC converter?
A common approach is to make a low power supply which will start the microcontroller and which is then either 


disabled when the main circuit starts. Or,
the low power circuit may be low enough power to be left running.


The arrangement below shows how this can be done using a very minor  modification  to @Pentium100 's circuit. The load in his circuit is R2. I've left R2 in place but it now serves as a very light nominal load when no power is being drawn by the microcontroller.
 If you supply the microcontroller via a diode from the emitter of Q, then, if you subsequently supply the load from a slightly higher voltage, the low power circuit will supply no power as the diode will be reversed bias.
 eg Say you microcontroller will operate with 3V &lt; Vdd &lt; 5V.
 If you arrange for the emitter of Q to be at 4V then the output end of the diode will be a diode voltage drop below Q1 emitter.
 A diode drop is typically about 0.6V so the microcontroller will see about 4V-0.6V = about 3.4V. So the microcontroller has Vdd =~3.4V Vdd to start up with.
 BUT if you then feed 5V to Vdd the diode will be reverse biased and supply no power.
 The only power then drawn by the low power circuit is the power fed to DZ1 by R1. This could be say 0.1 mA for a power dissipation of V x I = 72 V x 0.0001A = 7 mW. Even at 1 mA its only 72 mW. 

Super rough also borrowed diagram ...



Upper uode is part of the buck regulator feed - may or may not actually use a diode in the location shown but this gives the general idea. 

The above diagram is modified from one found here. An interesting and useful page but only somewhat related to this topic.