Pioneer sx1010 troubles

D8 and d9 are in right that I know I checked them twice and just looked at them again the bulb is 100 watts should I can put two 60 watt bulbs or try to find 150 watt( hard to find here in Canada for some reason ) and those other numbers I will get after work tomorrow
 
keep the 100w in, it'll do - not a lot of droop.

the diode voltages are critical.
 
Well Holly crap I was up this morning reading over your post about pin 10,14 and the diode voltages being critical.started staring at the schematic and I actually understood what you were saying. The d8 need to switch on so q1,2,3 can get the right voltage...same with d9 but for Q4,5,6 I would have no clue why you wanted me to check those before this 1010. I've learned so much from this unit.
 
Ok, if you can absorb that, I ADDED some additional explanations to that post. </Yoda voice on> Open your eyes it will... </Yoda voice off>

Got me thinking about a power supply tutorial - when one's eyes are open, they will notice the similarities all the way up through and including the SX-1980!!
 
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.[/QU
so one of those three transistors is either in wrong ( which I'm sure is what happened when d8 blew) blown,or bad trace, I've already know there in right, there brand new so I'm pretty sure there fine, but out of curiosity which way should the front be on q8 facing the heat sink or away from the heatsink? I will get those voltages for you been busy kids have to much stuff to do at Christmas..and I actually understood that I've been trying to get someone to explain the different circuits in a way I understand you should have been a teacher
 
Q8, ksc2690 ? is a to-126 transistor with an ECB layout. I expect (without checking one of mine) the lettering to more face the heat sink.
it's specifically aimed at the inside 90 degree CORNER (or R15) of the heat sink. Draw a line between the e and b leads, and extend a perpendicular line in the direction of the collector.

Q1 (mje15032) is hard to mis install, it's a to-220 that is clamped to the heat sink.It has a BCE layout.

Q2 and Q3 are both ksc2383, a to-92L case (enlarged, taller to-92 epoxy teardrop) and they are ECB.
I'd expect both to have their faces (lettering) pointing towards R20 (and the heat sink, but with an "L" shaped heat sink, the definition of to or away from it gets stretched pretty thin).
That's because the pads for the collectors for these two are behind the lettering face, while q8's collector pad is in front by the lettering.

gotta find and quote a series of posts on installing transistors....

this ought to hold you for now:

hint, when trying to orient the transistor symbol to the transistor itself, rotate the board so that the emitter is to the left and the base is to the right for a transistor that is ECB, then the Collector will be in the middle, either forward of a line between the e and b or behind that line. Bend the center lead to meet the hole...

in other words, don't twist up your mind trying to rotate it, rotate the board itself, so that everything after that is as easy as falling off a log!!
ksa992.jpg
 
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When I restored my SX-950, I installed a bad (I think it was a 2690 but not sure) new OnSemi transistor, and had a runaway voltage out that would go to the flat amp IIRC. Since that experience, I test every new transistor before installing.
 
Q8, ksc2690 ? is a to-126 transistor with an ECB layout. I expect (without checking one of mine) the lettering to more face the heat sink.
it's specifically aimed at the inside 90 degree CORNER (or R15) of the heat sink. Draw a line between the e and b leads, and extend a perpendicular line in the direction of the collector.

Q1 (mje15032) is hard to mis install, it's a to-220 that is clamped to the heat sink.It has a BCE layout.

Q2 and Q3 are both ksc2383, a to-92L case (enlarged, taller to-92 epoxy teardrop) and they are ECB.
I'd expect both to have their faces (lettering) pointing towards R20 (and the heat sink, but with an "L" shaped heat sink, the definition of to or away from it gets stretched pretty thin).
That's because the pads for the collectors for these two are behind the lettering face, while q8's collector pad is in front by the lettering.

gotta find and quote a series of posts on installing transistors....

this ought to hold you for now:

hint, when trying to orient the transistor symbol to the transistor itself, rotate the board so that the emitter is to the left and the base is to the right for a transistor that is ECB, then the Collector will be in the middle, either forward of a line between the e and b or behind that line. Bend the center lead to meet the hole...

in other words, don't twist up your mind trying to rotate it, rotate the board itself, so that everything after that is as easy as falling off a log!!
View attachment 1069139
Before I get those voltages I will check the leg placements again to be sure
 
When I restored my SX-950, I installed a bad (I think it was a 2690 but not sure) new OnSemi transistor, and had a runaway voltage out that would go to the flat amp IIRC. Since that experience, I test every new transistor before installing.

So do I - it's easier to test before it goes in, than after it is in. And not just transistors (diodes, resistors caps etc..) I trust my suppliers - but I suspect that they are playing the odds too, so I don't. My luck can sometimes turn the laws of probability to silly putty. Six sigma is 3.4 in a million, but that means SOMEONE is gonna get them.
 
The whole blowout started with a wrong resitors sent. This thing has been a nightmare since. The best part is after mtf tutorial on a Darlington circuit I now know why it caused the blowout.( As mentioned earlier dijjikey did come good for all the blown parts replaced everything. Even though I didn't use the resistors till 3 months after I ordered them)
 
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Q1 C s/b +66.0v IS: s 34V Raw DC to Darlington series pass element
Q1 B s/b +56.6v IS: s .6.0V Base drive for Darlington series pass element
Q1 E s/b +56.0v IS: s 10V Emitter output of Darlington series pass element
Q2 C s/b +66.0v IS: s19.2 V Raw DC to Darlington gain element
Q2 B s/b +57.2v IS: s 7.5V Base drive for Darlington gain element
Q2 E s/b +56.6v IS: s5.9 V Emitter output of Darlington gain element (to Q1 B)
Q3 C s/b +57.2v IS: s 5.8V control voltage from feedback element
Q3 B s/b +32.6v IS: s 0.5V divider feedback voltage for feedback element
Q3 E s/b +32.0v IS: s 0V reference voltage for feedback element
Here is the first set of voltages if you need me to check any again let me know I see q3 is not getting very much of anything
 
Q4 C s/b -66.0v IS: s -18.1V Raw DC to Darlington series pass element
Q4 B s/b -56.6v IS: s- 18.7V Base drive for Darlington series pass element
Q4 E s/b -56.0v IS: s -19.6V Emitter output of Darlington series pass element
Q5 C s/b -66.0v IS: s-19.6 V Raw DC to Darlington gain element
Q5 B s/b -57.2v IS: s-19.4 V Base drive for Darlington gain element
Q5 E s/b -56.6v IS: s -18.9V Emitter output of Darlington gain element (to Q1 B)
Q6 C s/b -57.2v IS: s -19.4V control voltage from feedback element
Q6 B s/b -32.6v IS: s -10.9V divider feedback voltage for feedback element
Q6 E s/b -32.0v IS: s 0V reference voltage for feedback element
Here are the negative voltages
 
tell me about the AC power. those RAW voltages are way too low - not enough for the regulator to function, what is between it's AC plug and the wall socket?
 
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There a power bar and my dim bulb 100watt when I first turn it on all the voltages are higher a fair bit higher then the bulb goes really bright
 
I will upload a video of it later on today. Its hard to explain for about a minute there all in close to 25 and 30 volts
 
If you worked on the amplifiers, OR THE POWER SUPPLY, did you do the idle current set procedure?

You start out after any work on the unit with the idle current turned off or down to minimum.

That means setting the idle current pots on both amplifier channels to a measured zero ohms in circuit - that gets minimum idle current and also bypasses any clockwise or counter clockwise confusion.

From there it called "tickling the dragon's tail".
 
I tried to its at Zero volts won't move I test all the outputs all good if I get it set will it draw more power from the wall.could have the blowout taken my outputs out to? All the fuses went except the main and 2 transistors went to plastic flew everywhere.
 
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