I'm just gonna do a lone ranger thing, pop in and out and try to steer a bit to alleviate frustration
edit: CHECK THE ORIENTATION OF D8 AND D9, they may be in backwards. Both face the same way, with the band facing AWAY from the heat sink "wall".
Ok, the voltages to watch are pins 14 (+56) 10 (-56) and 8 (+13v), the others are derived from them.
secondly, what wattage is the DBT? 150 watts would be best - still protects yet won't affect the operating voltages all that much.
general advice specific to the SX-1010 AWR-054: there can be almost invisible breaks in traces found under and around the "wall" heat sink and it's mounting points, as well as where the power transistors are mounted.
If you could, would you list the following voltages found ON THE ACTUAL TRANSISTOR LEADS: (copy the text and edit in your numbers (s xxx.xV), so I can "process" the info faster)
Q1 C s/b +66.0v IS: s xxx.xV Raw DC to Darlington series pass element
Q1 B s/b +56.6v IS: s xxx.xV Base drive for Darlington series pass element
Q1 E s/b +56.0v IS: s xxx.xV Emitter output of Darlington series pass element
Q2 C s/b +66.0v IS: s xxx.xV Raw DC to Darlington gain element
Q2 B s/b +57.2v IS: s xxx.xV Base drive for Darlington gain element
Q2 E s/b +56.6v IS: s xxx.xV Emitter output of Darlington gain element (to Q1 B)
Q3 C s/b +57.2v IS: s xxx.xV control voltage from feedback element
Q3 B s/b +32.6v IS: s xxx.xV divider feedback voltage for feedback element
Q3 E s/b +32.0v IS: s xxx.xV reference voltage for feedback element ************************************** priority!!!! **************
and the mirrored negative regulator
Q4 C s/b -66.0v IS: s xxx.xV Raw DC to Darlington series pass element
Q4 B s/b -56.6v IS: s xxx.xV Base drive for Darlington series pass element
Q4 E s/b -56.0v IS: s xxx.xV Emitter output of Darlington series pass element
Q5 C s/b -66.0v IS: s xxx.xV Raw DC to Darlington gain element
Q5 B s/b -57.2v IS: s xxx.xV Base drive for Darlington gain element
Q5 E s/b -56.6v IS: s xxx.xV Emitter output of Darlington gain element (to Q1 B)
Q6 C s/b -57.2v IS: s xxx.xV control voltage from feedback element
Q6 B s/b -32.6v IS: s xxx.xV divider feedback voltage for feedback element
Q6 E s/b -32.0v IS: s xxx.xV reference voltage for feedback element ************************************** priority!!!! **************
until the +56 is working, we won't even address pin 8 (+13v) because due to R19 (10k) the regulator is starved for operating current for it's reference.
edit 2, this is added on later.
Darlington? What is a Darlington? It is a "super transistor" pair where one transistor does the heavy lifting, and the other transistor supplies a HUGE amount of gain, so that the base current needed to drive the collector to emitter current of the heavy lifter is minuscule. The trade off? A normal transistor "loses" 0.6v in that type of circuit, a darlington loses 1.2v because there are TWO transistors.
In this power supply circuit type, be it a single transistor for just Q1, or a darlington pair like Q1 & Q2, it uses a series pass element that tries to maintain it's emitter AT the base voltage minus either the 0.6v or 1.2v passing more or less current to get the voltage "just right". It is called a series pass element as either a single transistor or darlington pair - and the pair of course has a series pass element and a gain element..
The Q3 feedback system really isn't complicated, but it is ESSENTIAL to having a constant stable predictable voltage to control the power supply. So the whole shebang exists to service the zener diode and it's constant voltage (at a specified, stable current), to keep the zener diode happy, and then to manipulate the series pass element to KEEP the voltage stable as the load varies, while NOT letting through the MUCH higher voltage found at the collector of the series pass element.
Q3's emitter is kept at a constant voltage by the zener diode, and the feedback transistor tries to keep IT'S base 0.6v greater than the emitter by varying IT'S collector to emitter current, which results in a voltage at the collector - which is then used to control the separate series pass element.
The feedback system gets different voltages and can even vary them manually with a pot by using a voltage divider (R5 and R7) to feed the voltage to Q3's base instead of connecting the base directly to the output voltage.
The rest of the resistors in that Q3 furball supply the various voltages/currents that operate it.
extra credit:
Look at Q8, it is a "naked" series pass element. The base is hooked up to a voltage divider (R16 and R17) wired to the regulated voltage. The collector is hooked up to the regulated voltage (ignore R18 for a minute - we'll get back to it). Thus Q8's emitter will try to be 0.6v less than the base voltage that comes from the voltage divider. A handy way to generate a stable lower voltage.
Calculate the voltage divider and it will show 37.3 v out for 56v in.
We get 35v out of Q8, so HUH??? 37.3 - 35 = 2.3v ??
Welcome to the concept of base current - the base current of Q8 is affecting the voltage divider a bit, pulling it down like R17 is LESS than 30,000 ohms....
We could fool around a bit more with math, figure out the apparent resistance and use that to get the base current - but ..... maybe too deep...
OH, remember R18? We're back to it... some of the energy being bled off to get the lower voltage at whatever current being drawn from that regulator is instead left behind in R18 - keeping it's heat OUT of Q8, so Q8 runs cooler - the resistor is MADE to take the heat.