Pioneer sx1010 troubles

I wonder if not having the regulated supplies working the front of the amps, its causing the pulsing?
After having the blow outs on the power supply board a guy should replace everything. Which I think you did?
Snap an image of the PS so I can see the board components.
Then double check once again and fire it up on the DBT.
 
Here is the power supply photos are not the clearest but tried to get the transistors even with the fuses in I get the pulse
 

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I went through double checked everything d9 blew again. Trying to figure out how that diode keeps blowing. Bad solder joint maybe or solder splashed. Very strange
 
Do I see brown original-manufacture carbon film resistors on that power supply board? That might be your problem. Those old CFRs cannot be trusted, I'd strongly recommend you replace them with Metal Films, elevate them from the board for airflow/heat management. I attach the BOM from my SX-1010 power supply rebuild:

Power Supply AWR-054-0

R1 & R2: carbon film ¼ W 3.3k replace with metal film

R3 & R4: carbon film ¼ W 1k replace with metal film

R5 & R6: carbon film ¼ W 10k replace with metal film

R7 & R8: carbon film ¼ W 15k replace with metal film

R9 & R10: carbon film ¼ W 8.2k replace with metal film

R11 & R12: carbon film ¼ W 10 replace with metal film

R13: metal oxide resistor 2W 1.2k

R14: metal oxide resistor 2W 1.5k

R15: carbon film ¼ W 18k replace with metal film

R16: carbon film ¼ W 15k replace with metal film

R17: carbon film ¼ W 30k replace with metal film

R18: carbon film ¼ W 330 replace with metal film

R19: carbon film ¼ W 10k replace with metal film

R20: carbon film ¼ W 3.3 replace with metal film

R21: carbon film ¼ W 10 replace with metal film
 
Changed, q5,q6,q8,q2,q3,q7, q1 and Q4 tested ok I took some voltages and
15 18.3
14 28.2
13 17.4
12 0
11 9.8
10 0
9 2.0
8 2.2
There not right I'm at a loss here
 
I'm wondering why you have not replaced electrolytic capacitor C14 on the power supply board? In my opinion, it doesn't look too healthy. And I'd change all of the CFR's to MFR's, and while you're at it re-check all your pinouts and component orientations on the power supply board.

My transistor replacements were:
Q1: 2SD313: 863-MJE15032G
Q2: 2SC869: 512-KSC2383YBU
Q3: 2SC869: 512-KSC2383YBU
Q4: 2SB507: 863-MJE15033G
Q5: 2SA628A: 512-KSA1013YBU
Q6: 2SA628A: 512-KSA1013YBU
Q7: 2SD313: 863-MJE15032G
Q8: 2SC1384: 512-KSC2690AYS

The photo shows the component orientations for these replacements on my power supply board. Ignore the heat sink mods, sorry, they obscure some of the board.
SX1010_AWR_054B.jpg
 
Double check the 2383's for correct placement as I might see a discrepancy one one of them. Remember they may have the center collector leg forward.
 
There was a post last week where a guy had the main fuse blow.
I think the plan I had for the OP might work here. At least it might narrow the pulsing to a spot to look.
Remove the two fuses that supply the AC to the regulated. Fuses 5 and 6? Double check.
Turn it on using the DBT and see if it still pulses. If it goes bright shut it down quick.
I know I've seen this before. :dunno:
 
An invalid test on my part.
I believe the regulated supply needs to be running.
This 1010 is giving you the business.
Everything was fine until you rebuilt the PS?
I'm traveling today and will check in tonight.
 
Yes everything was fine till I changed the transistors it's been hell. Work was done on it before the traces are a mess. I've learned a lot from it that's for sure so not a complete waste of time I've fixed more units from what I've learned from this. Im trying to avoid soldering and resoldering on the board it's pretty toasty. I was thinking of just pulling everything (diodes,resitors,caps, ect) and replacing all new might be easier to clean the whole board.
 
Went through the the board fixed many broken traces and rechecked the voltages there better but still not right
15 4.6
14 10
13 5.9
12 0
11 - 8
10 -17
9 -4
8 4
These are on the dbt and the negative voltages are back
 
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.
 
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