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Problem with Vintage Pioneer receiver SX-1000TW Output

Have a vintage Pioneer SX-1000TX receiver that the right channel doesn't seem to be working 100%. The output is slightly lower than the left channel and "thin" as best I can describe it (e.g. lacking any body to the sound). I have the schematics and have done some basic trouble shooting but at a loss where to go from here as I'm still learning about electronic/circuits etc. Here's what I've determined so far.

1) Using a old audio signal injector, I believe I narrowed it down to the Main amp unit or beyond. Initially, I thought I thought the problem was in the control amp unit where the tone controls/preamp circuitry all lie, Since it seemed like the lack of gain and lack of "body" could be related to the bass circuitry but injecting the audio signal right at Pins 2 In on the Main Amp unit (post Control Amp Unit shown in the block diagram below) still shows a disparaging difference between the channels.I was afraid to put the injector any further down the circuit as it would getting more severe of a "pop" as I moved the injector through the circuit. It's an old tube EICO Model 324 frequency generator and I'm not sure if it has the proper protection necessary or not, so I didn't want to chance it any further.Given it was designed for old tube radio diagnostics and they have much higher voltages, it probably is, but would like a confirmation from anyone that knows how far down a transistor based circuit it's safe to attempt signal injection.

2) I believe I esr checked many of caps in the main amp section (2nd schematic below), at least those greater than 10uf. Also esr'd the big 1000uf caps that follow the main amp. They all seem fine though it was done with an in circuit measure, so I realize under some cases those readings won't be accurate.

3) I measured the voltages around the transistors in the main amp circuit. I color coded those voltage readings in Red on the 2nd schematic below. I find a few interesting things from these measurements which I can't explain:
a) the first transistor Q801/Q802 in the circuit doesn't seems to react the same between the left and right channels. The right channel seems "abnormal" to me that all the voltage all around 10 volts. On the left channel, the emitter has a much higher output from the base/collector.
b) The right channels has higher voltages in general across the circuit. For example Q812 has 26.3 volts at the base, yet driven from the same 40 volts that the left channel's Q811 gets, but it's pulled much lower to 22.1 volts. I want to understand what in the circuit of the right channel could be causing that large differential. The Cap C808 failing (even though esr'd ok), One of the resistors R814,816etc out of tolerance? Diode D802 failing? Or something else in the circuit affecting it?
c) The output of this main amp unit feeds two power transistors from pins 7 & 11. On the right channel, Pin 7 (which feed q2 on the block diagram) is a couple of volts higher on the right than the left channel, yet Pin 4 (that feeds Q4) a 10th of a volt lower. I don't understand the relationship between these two power transistors, but it seems the overall voltage output for the amp is higher on the right channel, yet the signal is weaker/thinner.

4) One last item of note - The power resistor R4 that follows Q2/Q4 on the block diagram seems pretty heavily out of tolerance. Should be 330K and measures about 220K. What is the effect of that lower resistance on the circuit?

Appreciate any specific response to my questions above, as well as anything else to check for, measure etc. Obviously I want to fix this but in truth I want to learn "why" things are doing what they are doing, and what the circuit purpose is. I'm really at a beginner level in a lot of this stuff, but looking to learn and improve my diagnostic and repair ability.


Pioneer_main_amp_block.jpg



Pioneer_main_amp.jpg
 
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KrisBlueNZ

Sadly passed away in 2015
Hi there and welcome to Electronics Point :)

I would be suspecting C8 for loss of capacitance causing loss of bass, but possibly also C802, C804, C810, C814, C818.

If you replace them, you might as well replace the ones in the left channel as well - when one component fails, other similar ones are likely to fail as well.

You've made a good start by checking the DC conditions. They look OK but the base voltages on the bottom output transistors are pretty low. Can you measure the voltages across the two bottom emitter resistors, R29 and R30 please.

Old equipment like this tends to have multiple problems. Electrolytics are always a potential problem, especially ones in hot areas, but resistors also drift, and transistors often fail, and often in strange ways - they go intermittent, leaky, or noisy. Just a heads up for you!
 
Hi there and welcome to Electronics Point :)

I would be suspecting C8 for loss of capacitance causing loss of bass, but possibly also C802, C804, C810, C814, C818.

If you replace them, you might as well replace the ones in the left channel as well - when one component fails, other similar ones are likely to fail as well.

You've made a good start by checking the DC conditions. They look OK but the base voltages on the bottom output transistors are pretty low. Can you measure the voltages across the two bottom emitter resistors, R29 and R30 please.

Old equipment like this tends to have multiple problems. Electrolytics are always a potential problem, especially ones in hot areas, but resistors also drift, and transistors often fail, and often in strange ways - they go intermittent, leaky, or noisy. Just a heads up for you!
Chris, I will check that out but note that I inadvertently posted before I was finished editing. I just saved after your post so there is much more information there (and questions I have). Thanks!
 

KrisBlueNZ

Sadly passed away in 2015
Re Q801/802, yes there is a significant difference between left and right channels. But I think you have the collector and emitter voltage markings exchanged on the left channel. The 13.6V measurement will be the collector voltage.

The voltages you've marked on the right channel might be right but I suspect you have the collector and base swapped. In either case, the transistor is conducting too hard. This could be caused by R804 resistance being low or by R806 resistance being high, but you would need to lift one end to measure them otherwise the other components in the circuit would affect your measurement. Heating the resistor may affect its value if it's old. You could just replace both of them on suspicion, then re-measure the voltages.

If the voltages don't improve, I would suspect Q802. You could exchange Q801 and Q802 and see if the problem follows the transistor, or you could just replace them on suspicion. According to posts at http://www.audiokarma.org/forums/showthread.php?t=189009 the 2SC458 is a known reliability issue and should be replaced with 2SC1222 or 2SC2240. Both of those are obsolete. Another source recommends 2SC2675 but these aren't available through Digi-Key or Mouser either.

There's nothing special about the 2SC458 and it could be replaced with a BC549B if you rearrange the pins.

Yes, Q811/812 base and emitter voltages are noticeably different too. This could be due to a faulty transistor but first check both base resistors on both transistors (R813, R814, R815, R816). Again you need to lift one end out of circuit. If you find they are more than 10% from their nominal value, you may find it's worthwhile simply replacing all the resistors. (Assuming you want to put in the effort to preserve the thing.)

Regarding how far you can go injecting signals, I wouldn't go past Q803/804 collector. Beyond that point, the signal gets signal splits into two signals, one for the top output transistor and one for the bottom one. Injecting the signal beyond there could concievably cause excess current flow through the output transistors, and in any case will probably not work very well.

The output stage is a type called "quasi-complementary". The DC voltages are set by Q803/804 in conjunction with (for the left channel) R837, VR801, R839 and R841. You can adjust VR801/802 to get the "CENTER" terminal voltage closer to half of the "+82V" rail voltage if you like. Ideally it should be about 1V less than half the rail voltage. But the relatively small variations you're seeing won't cause a problem by themselves.

Re "R4" you mentioned, do you mean R40/41? They are supposed to be 330Ω not 330k. The value won't make much difference. It is just there to pull the final output (after C7/C8) to 0V.
 
The effective voltage across R29 (left channel) and R30 (Right channel) is zero. Not sure I'm surprised given a .5 volt or less at each of those connected transistor bases.

Unrelated - I had an part # error in my original text regarding item #4 and that 330K resistor - it was R41 not R4.
 

KrisBlueNZ

Sadly passed away in 2015
No voltage at all across R29? Nor across R30? That means there's no DC current flowing in the output stages. You may be able to get some current flowing by adjusting VR5 and VR6. Adjust VR5 for 0.014V across R29 and adjust VR6 for 0.014V across R30.
 
Meant 330 ohm (typo). I will check everything else you list tomorrow
and get back here. Thanks!
No voltage at all across R29? Nor across R30? That means there's no DC current flowing in the output stages. You may be able to get some current flowing by adjusting VR5 and VR6. Adjust VR5 for 0.014V across R29 and adjust VR6 for 0.014V across R30.

I'll have to check that resistor voltage again tomorrow as well . I'm pretty sure one was definitely zero. The other might have been very very small so I need to remeasure and se how close it was to your .014V target (where did you get that from by the way?). As long as I'm adjusting things, should I adjust the Center rail voltage first, prior to that Vr5/Vr6 adjustment or visa versa? Also what's weird is the the block diagram shows the powersupply as 78v, but as it goes to pin 6 on the Main amp schematic it says 82V (yet is actually reading 86.9v). Is that a typo on their schematics or are the power filter caps (C12/C13) somehow boosting the voltage? I get the powersupply board could have it's own issues not supplying the correct voltage (and maybe I should have started my diagnosis there, schematic for that below), but the question was really more how we go from a schematic 78v to schematic 82v without any obvious voltage circuit boost (at least not obvious to me) And if I adjust the center to 1V less then half the rail - should it be half of the actual ((86.9/2) - 1) or intended spec ((82/2) -1)? I assume it wants to be half (minus 1) of actual voltage, whatever is there.



Pioneer_power_supply.jpg
 

KrisBlueNZ

Sadly passed away in 2015
I'll have to check that resistor voltage again tomorrow as well . I'm pretty sure one was definitely zero. The other might have been very very small so I need to remeasure and se how close it was to your .014V target (where did you get that from by the way?).
The first diagram shows the current into the collector of Q1 as 20 mA. Almost all of the DC current in Q1 flows straight down, through the two output transistors. It will cause a voltage drop across the emitter resistors according to Ohm's Law: V = I × R where I is the current, 0.02 amps, and R is the resistance, which is shown as 0.7Ω. So V = 0.02 × 0.7 which is 0.014 volts. That's the voltage you should expect across each emitter resistor if 20 mA is flowing in the output stage.
As long as I'm adjusting things, should I adjust the Center rail voltage first, prior to that Vr5/Vr6 adjustment or visa versa?
Adjust the current (VR5/6) first, then the centre voltage. The voltage adjustment won't affect the current significantly, but with no current flowing, the voltage adjustment may not work properly.
Also what's weird is the the block diagram shows the powersupply as 78v, but as it goes to pin 6 on the Main amp schematic it says 82V (yet is actually reading 86.9v). Is that a typo on their schematics or are the power filter caps (C12/C13) somehow boosting the voltage?
That output (78V, 82V, whatever) is not regulated and the stated voltages are just nominal, typical values. The actual voltage will vary depending on the AC mains voltage and the amount of load on the power supply. If the load current is lower than it should be - for example, because the output stages aren't drawing their 20 mA each, as they're supposed to - then the voltage will be higher than it should be. Re-measure the power supply voltage once you've adjusted the output stage current. It should be a bit lower.
I get the powersupply board could have it's own issues not supplying the correct voltage (and maybe I should have started my diagnosis there, schematic for that below), but the question was really more how we go from a schematic 78v to schematic 82v without any obvious voltage circuit boost (at least not obvious to me) And if I adjust the center to 1V less then half the rail - should it be half of the actual ((86.9/2) - 1) or intended spec ((82/2) -1)? I assume it wants to be half (minus 1) of actual voltage, whatever is there.
That's not easy to answer. I guess it should be 1V less than half the power supply voltage when the output stage is running near maximum power, actually, for best performance. But it's not worth worrying about until all the other problems are fixed. So yeah, in the meantime just go for 1V less than half of what you measure.

Are you aware that getting this unit working well could take quite a lot of time and could involve replacing a LOT of parts? Are you prepared to do that?
 
I expect that it will, especially after what happened today (more on that following) but I am PT and about to retire in two months so I have plenty of time on my hands.

I verified you were correct about the mismarked collector/emitter values. just got them mixed when writing them.
I adjusted VR5/6 first to .014v. They were definitely lower initially. Didn't resolve anything but didn't seem to cause any harm either, at least initially.
I did check the resistors R804/806 for high/low. Both right on the money value wise.
I decided to swap the 801/802 transistors to see if the problem moved. Did that and was trying to verify if it shifted the problem to the left channel or not and check measurements and that's when the fuse blew. I had sound just prior and I was in the process of comparing left vs right sound to determine if it got better or not when it went.

Pulled the dead fuse out. Noted that it was a 4amp fuse yet the block diagram in the original post shows 1.5amp. I had a 1 and a 2 so I went with the 2amp. Blew again. Disconnected both the 40V and 78v/64V outputs from the power supply board and unit stayed on. Measured those outputs and got 40 & 64 Receptively (the 64V threw me, because I had been referring to a 78 and 82 V source from the schematic pins but I also realized that the schematic has 64V in parathesis, so I'm guessing it's really ok. Also measured the other Power supply outputs and they all are ok as well. I tried just connecting the 40 and then just the 64 but blew the fuse each time. At that point I realized (and should have about 5 fuses earlier) I needed to build a bulb limiter, which I did. Put both 40/64v connections back and bulb is glowing bright and hear a hum out the speakers as well.
Re-ESR'd both the power filter caps and the output caps and they check out ok. Tried to measure voltages but with the limiter circuit I'm assuming that radically affects all the reading due to the current change, as I only saw a volt or two on the 40 and 64 lines....

When I swapped the transistors, I'm sure they were in correctly. I also had a heatsink on the pins both when removing and soldering back though I did a real dummy move and left the heatsink across all three pins on Q801 when powering on. It was a brief moment of stupidity and I realized what I had done the second I turned it on - but the fuse didn't blow then. I turned off removed the heatsink and back on again, and it was shortly after that the fuse blew when I was comparing channels.

What kind circuit damage would that Q801 pin short do? That's certainly one possibility.
Another is some heat damage to either transistor, or some else close on the trace (maybe the connected collector diode?)
Another is the VR5/6 voltage adjustment pushed a partially faulty/failing output transistor over the edge. I guess I could pull them off and do a diode check on them to see if they are shorted or open?

Other thoughts to check what might be causing the fuse to blow?
 
Update - With unit on under bulb limiter, it's only the right channel that has a loud hum. Left is quiet (at least with volume set to zero.) Checked all 4 output transistors q1-q4. Right channel Q2 has a short between collector and emitter. Which certainly would contribute to the loud hum and I assume the blown fuse as well. Interesting that it was the right channel that I had the weak signal all along. Perhaps the Voltage adjustment to .014v just took that transistor over the edge? Or that in combo with a better Q802 transistor driving the amp? My heatsink short mistake would have no bearing, since that was on the left channel.

So next step obviously to get a replacement Q2, but anything else in the circuit that should (or even can be) checked with Q2 out before I risk a new one?
 

KrisBlueNZ

Sadly passed away in 2015
You're right. Having Q801 completely shorted out wouldn't cause the fuse to blow. The DC conditions at that part of the circuit cannot affect the output stage. It would have just stopped any signal from getting through.

You're also right that the most likely cause of fuse blowing is failed output transistors, and if Q2 has failed short, it's wise to replace all four of them. I will look up the 2SC793 and see what I can find as a suitable replacement.

I take it you found a service manual? Could you post a URL so I can get a copy of it? If it's not already online, could you upload it somewhere and post a URL?

Yes, it's possible that adjusting the output stage current brought on the failure if there were other problems with the components.

It sounds like the resistors may all be OK, although I think it would be wise to measure a selection of resistors (lift one end out of circuit first) to get an idea of what proportion of them are out of specification. Measure some low-value ones, but mostly measure the high-value ones - 100 kΩ and higher - because higher-value resistors normally fail earlier than lower-value resistors. If you don't find any out-of-spec, it might be safe to leave all the original resistors in the board.

You've already replaced the electrolytics, and small-value capacitors will probably be fine. It might be wise to upload a photo of the board though, as I (or someone else here) may recognise certain components that have known reliability problems.

So that leaves the semiconductors - transistors and diodes. Personally I would replace them all without hesitation. If there are intermittent problems, and you want to identify the exact transistor that is failing, you would be wasting a huge amount of time. Most of the transistors will be pretty cheap, and the diodes are very cheap.

Using a bulb limiter is a great idea. The 64V coming from the power supply is a worry. It's supposed to be around 80V. If it's only 64V even when the amplifier section is disconnected, there might be something wrong with the regulator on the power supply board that produces the 40V rail. But most likely the fault would be the transistor and/or the zener diode, and I hope you'll agree to replace them.

Edit: What's your location? There is a field for this in your profile. It helps us to know which retailers to look at.
 

KrisBlueNZ

Sadly passed away in 2015
OK, here are the specs for the 2SC793 output transistor.
Silicon NPN in TO-3 package.
VCBO = 100V; VCEO = 80V.
IC max = 7A; PD max = 60W
fT = 9 MHz
hFE 30~200 at IC = 1A and VCE = 5V

Digikey have the ON Semiconductor 2N3442G: 140V 10A USD 2.85 http://www.digikey.com/product-detail/en/2N3442G/2N3442GOS-ND/1475100 but the data sheet at http://www.onsemi.com/pub_link/Collateral/2N3442-D.PDF says that the collector cutoff current, ICEO, is a maximum of 200 mA! Admittedly, that's specified at 140V, but that figure is crazy high. I have to assume they're serious, so that transistor is not suitable.

Digikey have the Central Semiconductor 2N3716: 80V 10A USD 6.05 http://www.digikey.com/product-detail/en/2N3716/2N3716-ND/4807603. This looks like a good option, though it's more than twice the price of the 2N3442G.

You might be able to get original 2SC793 transistors through a buying house but they might be old, and they might even be counterfeit. I think you would be best to go for the 2N3716 at USD 6.05 each.

I'll look into suitable replacements for the other transistors in the circuit. I may not be able to do this for a few days though.
 
I have the original user manual which I just scanned and posted here http://home.comcast.net/~jimkarl/Pioneer_SX-1000tw_User_Manual.pdf

I'm in Maryland. Not really anyone local to my area, short of a radio shack though. I tend to buy most of my parts on ebay, unless it's a harder to find item. Since I buy stuff very onesy/twosey, it generally hasn't made sense to buy from Mouser/digi etc because the shipping usually costs as much or more than the part itself. But there have been some hard to find bulbs, for example, that I needed for a different receiver that I had to bite the bullet and pay some heavy shipping for.

I could get a NOS direct replacement (same part #) for Q2 off ebay, a pair actually for about $13.50 - one for the fix, one as a spare. It will take a while to get here but I'm not really in any specific hurry. I find that if the "alternative" replacement has some slightly different performance characteristics, and they often do,, then I need to match all the others as well (and the cost starts to rise) or things might be unbalanced/uneven. .As I'm just starting repairing things (this being about my 6th or 7th item) I'm starting to build up a cache of parts but unfortunately transistors aren't in there (yet). Have a few different diodes, but a wider kit range still on order. In truth, my goal for this is not a complete restoration on this particular item. It's really more to just get rid of any major operational problems (in this case, the weaker right channel), as these items are for resale and I will disclose what I have done to the unit and what I haven't or any known issues/concerns. So I have to be mindful about parts costs because they can quickly exceed the product value. I found that especially to be the case when I had leaky vintage power caps, especially on things like vintage radios. Those parts are harder to find and get pricey quick! Having said that, I've found myself "investing" in parts as I go. I generally try to buy a spare or in some cases a whole kit (e.g. I needed a 50V cap for this same receiver's power supply board last week (failed ESR and capacitance was way out of spec) and since I didn't have ANY 50Vcaps, regardless of uf value, I figured go ahead and buy a wide range kit of them, as somewhere down the line I'll need another, and likely a different value too. Anyway, the point being that if the replacement transistor fixed the weak channel problem, I would probably consider that a stopping point, barring no obvious other problem like noise etc.. If it was still weak or channel was noisy, I would be back to the control amp board and further part diagnostics/replacement. I appreciate you pointing out the troublesome 2SC458 transistor as well. That BC549B is certainly cheap enough that I can buy a small lot of them for the next time I might need them and once I can compare my channels again may find one channel still needs some work over the other but at the moment, it sounds like getting a replacement going is my top priority.

I will check out that transistor on the power supply board as well. I think I did the diode checks last week but I'll do them again just to be certain.
 

KrisBlueNZ

Sadly passed away in 2015
In no circumstances would I replace only one of the output transistors - even if the unit was brand new. The other one is likely to have been stressed by the high current flow. And there's no point replacing the output transistors until you've fixed the problem that caused them to fail. Replacing the driver transistors would be a good start.

I think if you try to replace only the components that are definitely faulty, (a) it will take a lot of effort to find out which components those are, (b) you may find that many or all of the semiconductors are failing, to various degrees, and (c) the unit may fail shortly after you sell it, when the next semiconductor reaches the point of complete failure.

When you're working with old equipment where semiconductors are starting to fail, it doesn't make sense to take that approach. Wholesale replacement of semiconductors is the way to go. If this means you would blow your budget, then you're probably better off binning the thing and avoiding the hassle.
 
I agree with you. I'll replace both output transistors on that channel. I was also planning on getting those BC549B replacements for Q801/802 since you pointed out they are high failure parts and they are so inexpensive and I get get a bulk of them for future repairs.
I uploaded a pic of the amp board in case there are others you see and recommend as well. I have to be a little realistic that this isn't a restoration project and I can't replace every part that is old (otherwise the whole device should be replaced LOL). Where I can identify failed parts or significantly out of tolerance I need to replace them. Where they have known failures "histories" parts (but are currently still working and in tolerance) and the parts are so inexpensive and can get a bulk of them and will high a high reuse of the leftovers over time, I want to do those too. but I have to draw the line someplace so I don't want to replace parts that aren't bad yet because that isn't the end goal, at least for this particular unit.

The whole purpose of coming to this forum is to actually learn how I can tell a bad part from a good one, or a marginal one that should be replaced. Prior to this unit, I learned that I really needed an esr meter to check caps. Caps I thought were good from a capacitance reading, but really weren't and only the ESR meter could tell me that. I learned how to check diodes, and transistors the same way. That's how I found the short on this Q2 output one. I suspected a problem with Q802 originally, not because I knew what the expected B/E/C voltages should be but from a comparison of the other channel's transistor relative values. Though ideally I would be able to learn and understand what the expected values should be without a comparison part in circuit.
I also learned about the light bulb limiter (though built it a few fuses too late LOL).


In this thread - I've learned a number of things so far from you and all that information is much appreciated. What the target center voltage needs to be (though I'm not sure I got an explanation for the -1V below center, does that have something to do with a circuit voltage increase downstream that will get it ultimately closer to center?) How to check and set the output transistor current/voltage. I saw those variable pots on a number of other receivers I had and wish I had known then what I know now. Though I knew ohm's law, I learned a real practical application for it here - determine what the measure voltage should be when not stated on the schematic - buy using the target amperage value. When you told me that, it was like a lightbulb going off for me. Now I get how I can apply that to solve and debug next time.

There's still so much I don't know yet or understand. I'm going to check out that power supply right after I hit send here, but the moment, let's assume the 64V output with the power supply out of circuit is correct (it does after all match one of the two voltages listed on the schematic. My assumption is they listed the unloaded voltage as 64V and the loaded one as 78V). When the unit was "working" I was getting 86v, certainly well over both targets. Without anything obvious in the circuit to boost that voltage (power runs from the power supply, through the filter caps and to the amp board and also to the output transistors) the applied Ohm's law would teach me that either the resistance of the circuit changed, or the current did. Certainly connecting the amp board and those output transistors to the circuit adds some resistance. I guess it's possible some transistor in the circuit (amp board or possibly that then failing Q2 output one) is/was leaking current onto that path, which would drive the voltage higher. Of course, won't know for sure until I can get the Q2 parts and get it back up and running again. I'm hoping to see that 86V come back down to a more nominal 78v as the schematic specifies. That 86V was a red flag for me and like the Q802 transistor voltages, while I don't' understand the "why" or the cause, I know (or at least feel) something isn't right.
Another example is you mentioned the base voltages (.5v and .4v) looked low on the output transistor. How do you know what expected base voltages are when not printed on the circuit schematic?
 

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KrisBlueNZ

Sadly passed away in 2015
I'm happy to help where I can. You're obviously intelligent and you have figured out a lot by yourself without any hand-holding. This is a pleasant change from a certain category of users who ask vague questions and expect a perfect solution to be provided.

I agree with you. I'll replace both output transistors on that channel.
[...]
I have to draw the line someplace so I don't want to replace parts that aren't bad yet because that isn't the end goal, at least for this particular unit
Then you don't agree with me. Isn't your end goal to get the unit to a point where you can sell it and not expect it to come back within a few weeks or months with another failed semiconductor? That would be my end goal. There are only 12 transistors on the amp board. Replacing only the ones that are definitely faulty now is false economy.
In this thread - I've learned a number of things so far from you and all that information is much appreciated. What the target center voltage needs to be (though I'm not sure I got an explanation for the -1V below center, does that have something to do with a circuit voltage increase downstream that will get it ultimately closer to center?)
No, it's because the output can swing closer to the 0V rail than it can to the positive rail, because Q1/Q3 is operated as an emitter follower, which has a minimum voltage drop of around 0.7V, while Q2/Q4 is operated in common emitter mode, which has a much lower minimum voltage drop. So if the output was set half way between the positive supply and 0V, it would clip at the top before it started clipping at the bottom.
How to check and set the output transistor current/voltage. I saw those variable pots on a number of other receivers I had and wish I had known then what I know now. Though I knew ohm's law, I learned a real practical application for it here - determine what the measure voltage should be when not stated on the schematic - buy using the target amperage value. When you told me that, it was like a lightbulb going off for me. Now I get how I can apply that to solve and debug next time.
Great!
There's still so much I don't know yet or understand. I'm going to check out that power supply right after I hit send here, but the moment, let's assume the 64V output with the power supply out of circuit is correct (it does after all match one of the two voltages listed on the schematic. My assumption is they listed the unloaded voltage as 64V and the loaded one as 78V).
It would be the other way round. An unregulated supply always has a higher voltage when there's no load. The load current pulls the voltage down.
When the unit was "working" I was getting 86v, certainly well over both targets.
That could be explained by the low operating current in the output stages. That could easily cause an increase from 82V to 86V. A drop to 64V is mysterious though.
Without anything obvious in the circuit to boost that voltage (power runs from the power supply, through the filter caps and to the amp board and also to the output transistors) the applied Ohm's law would teach me that either the resistance of the circuit changed, or the current did. Certainly connecting the amp board and those output transistors to the circuit adds some resistance. I guess it's possible some transistor in the circuit (amp board or possibly that then failing Q2 output one) is/was leaking current onto that path, which would drive the voltage higher.
You've got some wrong ideas there.

An unregulated power supply produces a voltage that will be highest when it is unloaded. Unloaded means that no current is being drawn from it. As you draw more current, the voltage will drop. But the behaviour of the power supply is not dictated by Ohm's Law, because Ohm's Law really only applies to resistors, or resistive or "ohmic" components, and the unregulated power supply is a complicated combination of an unregulated AC voltage source (which includes some series resistance), rectifier diodes, and smoothing capacitance.

In very simple terms, an unregulated power supply can be modelled as a perfect voltage source with a resistor in series with it. This resistance is called the "internal resistance" of the supply, and voltage will be dropped across it according to the amount of current you draw from the power supply. The more current you draw, the more voltage will be dropped across the internal resistance, and the less voltage will be available from the output. But this is only a very rough approximation. An unregulated power supply's output voltage does not drop linearly in relation to the current drain.

Resistance is a quantity that sometimes seems to be defined in reverse to what you would expect. An open circuit is an infinitely high resistance, and a piece of wire has almost zero resistance. You know if you connect a piece of wire across a power supply, that will short it out. A short circuit across a power supply will draw a high current and cause the voltage to drop to zero. A high resistance, being an open circuit, represents the minimum possible load on the power supply.

Ohm's Law, I = V / R, shows this relationship. An open circuit has infinite resistance, and no current flows through it, regardless of the voltage. If R is infinite, then I (current) will be zero, regardless of V. Current (I) will only start to increase when R decreases.
Without anything obvious in the circuit to boost that voltage (power runs from the power supply, through the filter caps and to the amp board and also to the output transistors) the applied Ohm's law would teach me that either the resistance of the circuit changed, or the current did.
Voltage from the power supply is applied across the output stage of the amplifier. The design of the amplifier causes the transistors in the output stage to conduct electricity, i.e. instead of being like an infinite resistance, they are a finite resistance, so current flows through them. Since transistors are not ohmic (they are semiconductors), it is misleading to describe them as having a certain resistance. It is better to say that the output stage of the power amplifier conducts a certain amount of current through the output transistors. This current is adjustable via the trimpot.

This current affects the output of the power supply, because it is an unregulated power supply (at least, the main high voltage output is unregulated). This current is a source of load on the power supply, and it drags the output voltage down somewhat.
Certainly connecting the amp board and those output transistors to the circuit adds some resistance.
It adds some load on the power supply, i.e. it draws current from the power supply, which will cause its output voltage to drop. "Adds resistance" refers to increasing the resistance, which causes the current to drop. That's why it's not an appropriate thing to say in this context.
I guess it's possible some transistor in the circuit (amp board or possibly that then failing Q2 output one) is/was leaking current onto that path, which would drive the voltage higher.
The positive unregulated output from the power supply is the highest voltage in the circuit. There is no higher voltage that could "boost" the power supply's output. The output will only increase if the load (current) is reduced, or if the AC mains voltage increases.

Leakage is current flow through an unintended path. Due to a faulty component, or contamination, or whatever, a normally non-conductive path (infinite resistance, so no current can flow through it) can begin to conduct (non-infinite resistance, so current starts to flow). This current flow can cause two normally isolated parts of the circuit to affect each other. It's possible to get leakage current flowing from a high-voltage part of a circuit into a low-voltage part of a circuit, causing voltages to increase and possibly damaging components etc. But this can't be the cause of the increased power supply voltage here, because there is no higher voltage source that could be leaking current into the power supply and causing its voltage to increase.
Of course, won't know for sure until I can get the Q2 parts and get it back up and running again. I'm hoping to see that 86V come back down to a more nominal 78v as the schematic specifies. That 86V was a red flag for me and like the Q802 transistor voltages, while I don't' understand the "why" or the cause, I know (or at least feel) something isn't right.
I'm concerned at the 64V you measured. If it's real, it could indicate that something is drawing a lot of current from the power supply and pulling the voltage down significantly. Unintended high current flow is a concern, because it means that power is being dissipated somewhere (power is equal to voltage multiplied by current), and power dissipation corresponds to heat being generated. This is not normally a good thing.

There is no point replacing the output transistors without replacing the driver transistors as well. I strongly recommend replacing ALL of the transistors on the amp board. Here are recommended replacements. These are ALL in different packages from the originals, and you need to consult the data sheets to figure out how to rearrange the wires. Even worse than that, although all of the replacement transistors are in TO-92 packages, there are three different pinout variants! The BC549B and BC547B use the European variant (CBE from left to right looking at the flat surface with the pins pointing downwards), 2N5551 uses the American variant (EBC), and BC639 and BC640 use the Japanese variant (ECB). I suggest you draw up a translation diagram for each substitution, showing the transistors viewed from above, and post it here so I can check it before you switch the thing on!

2SC458 or 2SC871 (NPN, silicon, 30V low noise) --> BC549B (x2)

2SC538A (TO-18, NPN, silicon, 45V, 50 mA, 0.3W, 100 MHz, hFE 90(250)+) --> BC547B (x2)

2SC627 (TO-39, NPN, silicon, 200V, 100 mA, 0.7W, 10(20?) MHz, hFE -(360)+) --> 2N5551 (x4)

2SC485 (TO-39, NPN, silicon, 80V, 1.5A, 0.8W, 10 MHz, hFE 30+) --> BC639 (x2)

2SA537A (TO-5/TO-39, PNP, silicon, 80V, 0.7A, 0.75W, typ 200 MHz, hFE 35(80)200@50mA; 20(40)@400mA) --> BC640 (x2)

Another example is you mentioned the base voltages (.5v and .4v) looked low on the output transistor. How do you know what expected base voltages are when not printed on the circuit schematic?
When a transistor is operating in its linear region, i.e. partly conducting, it will have a base-emitter voltage of around 0.6~0.8V (for a silicon transistor).
 
I'm happy to help where I can. You're obviously intelligent and you have figured out a lot by yourself without any hand-holding. This is a pleasant change from a certain category of users who ask vague questions and expect a perfect solution to be provided.


Then you don't agree with me. Isn't your end goal to get the unit to a point where you can sell it and not expect it to come back within a few weeks or months with another failed semiconductor? That would be my end goal.


There are only 12 transistors on the amp board. Replacing only the ones that are definitely faulty now is false economy.

No, it's because the output can swing closer to the 0V rail than it can to the positive rail, because Q1/Q3 is operated as an emitter follower, which has a minimum voltage drop of around 0.7V, while Q2/Q4 is operated in common emitter mode, which has a much lower minimum voltage drop. So if the output was set half way between the positive supply and 0V, it would clip at the top before it started clipping at the bottom.

Great!

It would be the other way round. An unregulated supply always has a higher voltage when there's no load. The load current pulls the voltage down.

That could be explained by the low operating current in the output stages. That could easily cause an increase from 82V to 86V. A drop to 64V is mysterious though.

You've got some wrong ideas there.

An unregulated power supply produces a voltage that will be highest when it is unloaded. Unloaded means that no current is being drawn from it. As you draw more current, the voltage will drop. But the behaviour of the power supply is not dictated by Ohm's Law, because Ohm's Law really only applies to resistors, or resistive or "ohmic" components, and the unregulated power supply is a complicated combination of an unregulated AC voltage source (which includes some series resistance), rectifier diodes, and smoothing capacitance.

In very simple terms, an unregulated power supply can be modelled as a perfect voltage source with a resistor in series with it. This resistance is called the "internal resistance" of the supply, and voltage will be dropped across it according to the amount of current you draw from the power supply. The more current you draw, the more voltage will be dropped across the internal resistance, and the less voltage will be available from the output. But this is only a very rough approximation. An unregulated power supply's output voltage does not drop linearly in relation to the current drain.

Resistance is a quantity that sometimes seems to be defined in reverse to what you would expect. An open circuit is an infinitely high resistance, and a piece of wire has almost zero resistance. You know if you connect a piece of wire across a power supply, that will short it out. A short circuit across a power supply will draw a high current and cause the voltage to drop to zero. A high resistance, being an open circuit, represents the minimum possible load on the power supply.

Ohm's Law, I = V / R, shows this relationship. An open circuit has infinite resistance, and no current flows through it, regardless of the voltage. If R is infinite, then I (current) will be zero, regardless of V. Current (I) will only start to increase when R decreases.

Voltage from the power supply is applied across the output stage of the amplifier. The design of the amplifier causes the transistors in the output stage to conduct electricity, i.e. instead of being like an infinite resistance, they are a finite resistance, so current flows through them. Since transistors are not ohmic (they are semiconductors), it is misleading to describe them as having a certain resistance. It is better to say that the output stage of the power amplifier conducts a certain amount of current through the output transistors. This current is adjustable via the trimpot.

This current affects the output of the power supply, because it is an unregulated power supply (at least, the main high voltage output is unregulated). This current is a source of load on the power supply, and it drags the output voltage down somewhat.

It adds some load on the power supply, i.e. it draws current from the power supply, which will cause its output voltage to drop. "Adds resistance" refers to increasing the resistance, which causes the current to drop. That's why it's not an appropriate thing to say in this context.

The positive unregulated output from the power supply is the highest voltage in the circuit. There is no higher voltage that could "boost" the power supply's output. The output will only increase if the load (current) is reduced, or if the AC mains voltage increases.

Leakage is current flow through an unintended path. Due to a faulty component, or contamination, or whatever, a normally non-conductive path (infinite resistance, so no current can flow through it) can begin to conduct (non-infinite resistance, so current starts to flow). This current flow can cause two normally isolated parts of the circuit to affect each other. It's possible to get leakage current flowing from a high-voltage part of a circuit into a low-voltage part of a circuit, causing voltages to increase and possibly damaging components etc. But this can't be the cause of the increased power supply voltage here, because there is no higher voltage source that could be leaking current into the power supply and causing its voltage to increase.

I'm concerned at the 64V you measured. If it's real, it could indicate that something is drawing a lot of current from the power supply and pulling the voltage down significantly. Unintended high current flow is a concern, because it means that power is being dissipated somewhere (power is equal to voltage multiplied by current), and power dissipation corresponds to heat being generated. This is not normally a good thing.

There is no point replacing the output transistors without replacing the driver transistors as well. I strongly recommend replacing ALL of the transistors on the amp board. Here are recommended replacements. These are ALL in different packages from the originals, and you need to consult the data sheets to figure out how to rearrange the wires. Even worse than that, although all of the replacement transistors are in TO-92 packages, there are three different pinout variants! The BC549B and BC547B use the European variant (CBE from left to right looking at the flat surface with the pins pointing downwards), 2N5551 uses the American variant (EBC), and BC639 and BC640 use the Japanese variant (ECB). I suggest you draw up a translation diagram for each substitution, showing the transistors viewed from above, and post it here so I can check it before you switch the thing on!

2SC458 or 2SC871 (NPN, silicon, 30V low noise) --> BC549B (x2)

2SC538A (TO-18, NPN, silicon, 45V, 50 mA, 0.3W, 100 MHz, hFE 90(250)+) --> BC547B (x2)

2SC627 (TO-39, NPN, silicon, 200V, 100 mA, 0.7W, 10(20?) MHz, hFE -(360)+) --> 2N5551 (x4)

2SC485 (TO-39, NPN, silicon, 80V, 1.5A, 0.8W, 10 MHz, hFE 30+) --> BC639 (x2)

2SA537A (TO-5/TO-39, PNP, silicon, 80V, 0.7A, 0.75W, typ 200 MHz, hFE 35(80)200@50mA; 20(40)@400mA) --> BC640 (x2)


When a transistor is operating in its linear region, i.e. partly conducting, it will have a base-emitter voltage of around 0.6~0.8V (for a silicon transistor).

It is a goal to make sure the device works well and doesn't come back. I did qualify "and the parts are so inexpensive and can get a bulk of them and will high a high reuse of the leftovers over time" which appears to be the case with the parts you listed. It looks like I can replace the entire board of transistors for less than $10 with lots of spares left over for next next time one comes along, and I see a lot of these older receivers, in fact this will be the third pioneer I've had/worked on. I was concerned this was going to end up being $30-$40 of transistors, with no spares for the future, in which case I would simply have to call it quits and sell it "as-is" because between that and what I paid for it I then wouldn't be able to get my money back out. But no worries here.

It would be the other way round. An unregulated supply always has a higher voltage when there's no load. The load current pulls the voltage down. So the schematic showing 78v (64v) out the power supply board is showing the unregulated (regulated) values? I also wanted to be clear when I saw 64V in measurement - it wasn't right off the power supply transformer - it was (one of the) outputs of the power supply board, with everything following (the amp boards and such) disconnected. I rechecked all of the diodes on that power supply board and they all seem ok. Re-esr'd the caps as well. The transistors on that board could have a problem I guess, and I could lift them out to do a diode check as well but I keep struggling with the fact the schematic says (64V) and that's what I'm seeing so why would there be concern?


When a transistor is operating in its linear region, i.e. partly conducting, it will have a base-emitter voltage of around 0.6~0.8V (for a silicon transistor).

Ahh - so that is target value for the circuit voltage there. So either via pot adjustment (if applicable) or part repair/replacement.

I've order all these transistors (all coming slow boat) but most had the data sheets attached to them so I've got those printed out and ready. A couple that didn't have the B/E/C printed on them but I can get the data sheets as well online. The challenge isn't the pin swapping but the upside down, back of board look and fix. That's how I got those Collector/Emitter voltages written down wrong initially :) I will definitely be plotting this out to be 100% sure all is right before proceeding.

Since all these parts will be coming in at different times, is there any safe way to test in stages? I had run a quick test last night under the bulb limiter and (with both channels in circuit but Right channel Q2 output tran removed since it's dead). Bulb went dim on power on and remain so for about 30+ seconds and then lit up brightly. Then right channel Transistor Q806 started smoking (or so I thought). Turn out it was just getting really (really) hot and burning the plastic sleeve on a part next to it. It had 75 or so volts coming in the collector, about 11 on the base and emitter. I did take that part out to test it, as I figured it was fried, but it actually still checked out ok. I'm assuming the lack of Q2 in circuit was casing it to heat up, or possibly some other circuit damage that happens as a result of Q2 dieing. My question is, if I disconnect Q4 will I get the same result (assuming it wasn't something else int he circuit that died)? Meaning would leaving Q2 in the circuit with no Q4 cause this effect? Do I need to remove the +75V from the board to test the amp board alone safely? In short, I'm looking for a way to put all these new opamp/driver transistors on the mainamp board and test first, without risking the more expensive output Q2/Q4 replacements, if there is a way to do that. If I make a mistake and burn out a some of the driver transistors, it's only pennies and I'll have lots of spares. That won't be the case for those output transistors though!
 

KrisBlueNZ

Sadly passed away in 2015
It is a goal to make sure the device works well and doesn't come back. I did qualify "and the parts are so inexpensive and can get a bulk of them and will high a high reuse of the leftovers over time" which appears to be the case with the parts you listed. It looks like I can replace the entire board of transistors for less than $10 with lots of spares left over for next next time one comes along, and I see a lot of these older receivers, in fact this will be the third pioneer I've had/worked on. I was concerned this was going to end up being $30-$40 of transistors, with no spares for the future, in which case I would simply have to call it quits and sell it "as-is" because between that and what I paid for it I then wouldn't be able to get my money back out. But no worries here.
Right, good. Yes, fair comment.
It would be the other way round. An unregulated supply always has a higher voltage when there's no load. The load current pulls the voltage down. So the schematic showing 78v (64v) out the power supply board is showing the unregulated (regulated) values?
There is no regulator in the power supply, so there's no such thing as "regulated" values. Based on the fact that the voltage is shown as +82V on the amplifier schematic, and you measured +86V at one time, you could say that the 86V measurement was the "unloaded" or "under-loaded" output voltage and the 82V measurement is the "normal load" voltage.

The unregulated rail also supplies the 40V regulator on the power supply board, so some of the current drawn by other parts of the unit is ultimately drawn from the unregulated rail, so 40 mA is probably less than half of the idle load on the power supply and I'm a bit surprised that losing that 40 mA load would cause a 5% increase in output voltage.

But that rail is unregulated so variations in the AC mains voltage could account for some or all of the voltage difference. But not a drop from 82V to 64V. That's 18V which is about 22%.
I also wanted to be clear when I saw 64V in measurement - it wasn't right off the power supply transformer - it was (one of the) outputs of the power supply board, with everything following (the amp boards and such) disconnected.
OK, let's pretend it didn't happen, unless you measure it again and can say which output it was.
I rechecked all of the diodes on that power supply board and they all seem ok. Re-esr'd the caps as well. The transistors on that board could have a problem I guess, and I could lift them out to do a diode check as well but I keep struggling with the fact the schematic says (64V) and that's what I'm seeing so why would there be concern?
Where does the schematic say 64V? I can't see it marked anywhere on the power supply diagram.

I think you should replace the transistors in the power supply as well. All three of them (Q901, Q902 and Q903) can be replaced with MPSA06. It's a bit over-rated for the Q902 and Q903 positions but it's a good transistor and pretty cheap. Back when that unit was designed, transistors were a lot more expensive so it was worthwhile using the cheapest one that would fit the requirements, but nowadays it's more important to use as few different parts as possible so you get the best price break.

If you want to get a stock of general purpose transistors for repair work, the MPSA06 is a good, versatile transistor. Here's a list for you to consider.

BC639 (NPN), BC640 (PNP): 1A 80V 1W
MPSA06 (NPN), MPSA56 (PNP): 0.8A 80V 0.625W fairly high gain
BC337-40 (NPN), BC327-40 (PNP): 0.8A 45V 0.625W high gain
BC547B (NPN), BC557B (PNP): 0.1A 45V 0.5W general purpose
BC549C (NPN), BC559C (PNP): 0.1A 30V low noise high gain

There's also the PN2222 (NPN), 0.6~1.0A depending on manufacturer and suffix, 40V, 0.625W, fairly high gain, if you want to avoid European part numbers for some political reason. (The BC... types are all European designators.)

When a transistor is operating in its linear region, i.e. partly conducting, it will have a base-emitter voltage of around 0.6~0.8V (for a silicon transistor).
Ahh - so that is target value for the circuit voltage there. So either via pot adjustment (if applicable) or part repair/replacement.
Right. Normally a component replacement, and often not the transistor that has the wrong voltage - normally something in the circuitry that is driving it.
I've order all these transistors (all coming slow boat) but most had the data sheets attached to them so I've got those printed out and ready. A couple that didn't have the B/E/C printed on them but I can get the data sheets as well online. The challenge isn't the pin swapping but the upside down, back of board look and fix. That's how I got those Collector/Emitter voltages written down wrong initially :) I will definitely be plotting this out to be 100% sure all is right before proceeding.
Yes, I remember having trouble with that too, but that was when I was a kid and my brain figured it out, so now it's second nature to me. So it will be easy to check your work.
Since all these parts will be coming in at different times, is there any safe way to test in stages? I had run a quick test last night under the bulb limiter and (with both channels in circuit but Right channel Q2 output tran removed since it's dead). Bulb went dim on power on and remain so for about 30+ seconds and then lit up brightly. Then right channel Transistor Q806 started smoking (or so I thought). Turn out it was just getting really (really) hot and burning the plastic sleeve on a part next to it. It had 75 or so volts coming in the collector, about 11 on the base and emitter. I did take that part out to test it, as I figured it was fried, but it actually still checked out ok. I'm assuming the lack of Q2 in circuit was casing it to heat up, or possibly some other circuit damage that happens as a result of Q2 dieing.
Right, with Q2 out of circuit, Q806 is forced to dissipate quite a lot of power.
My question is, if I disconnect Q4 will I get the same result (assuming it wasn't something else int he circuit that died)? Meaning would leaving Q2 in the circuit with no Q4 cause this effect? Do I need to remove the +75V from the board to test the amp board alone safely? In short, I'm looking for a way to put all these new opamp/driver transistors on the mainamp board and test first, without risking the more expensive output Q2/Q4 replacements, if there is a way to do that. If I make a mistake and burn out a some of the driver transistors, it's only pennies and I'll have lots of spares. That won't be the case for those output transistors though!
Right. Once you've installed the new driver transistors, you should test the amplifier with the collector of the top output transistors (Q1 and Q2) disconnected, and a 2k2 power resistor (e.g. http://www.digikey.com/product-detail/en/ERG-3SJ222/P2.2KW-3BK-ND/36811 or http://www.digikey.com/product-detail/en/AC05000002201JAC00/PPC5W2.2KCT-ND/596703) connected between the wire that used to go to the collector, and the emitter. This will provide about 20 mA of current through Q3/Q4 so the output stage can be adjusted, but if anything goes wrong, the current will be limited to 40 mA so the output transistors will be safe.

(For most amplifiers, the collectors of the top AND bottom output transistors should be disconnected for initial power-up, but this amplifier has a "quasi-complementary" output stage, which can't be adjusted unless the bottom output transistor is present.)
 
Right, good. Yes, fair comment.

There is no regulator in the power supply, so there's no such thing as "regulated" values. Based on the fact that the voltage is shown as +82V on the amplifier schematic, and you measured +86V at one time, you could say that the 86V measurement was the "unloaded" or "under-loaded" output voltage and the 82V measurement is the "normal load" voltage.

The unregulated rail also supplies the 40V regulator on the power supply board, so some of the current drawn by other parts of the unit is ultimately drawn from the unregulated rail, so 40 mA is probably less than half of the idle load on the power supply and I'm a bit surprised that losing that 40 mA load would cause a 5% increase in output voltage.

But that rail is unregulated so variations in the AC mains voltage could account for some or all of the voltage difference. But not a drop from 82V to 64V. That's 18V which is about 22%.

OK, let's pretend it didn't happen, unless you measure it again and can say which output it was.

Where does the schematic say 64V? I can't see it marked anywhere on the power supply diagram.

I think you should replace the transistors in the power supply as well. All three of them (Q901, Q902 and Q903) can be replaced with MPSA06. It's a bit over-rated for the Q902 and Q903 positions but it's a good transistor and pretty cheap. Back when that unit was designed, transistors were a lot more expensive so it was worthwhile using the cheapest one that would fit the requirements, but nowadays it's more important to use as few different parts as possible so you get the best price break.

If you want to get a stock of general purpose transistors for repair work, the MPSA06 is a good, versatile transistor. Here's a list for you to consider.

BC639 (NPN), BC640 (PNP): 1A 80V 1W
MPSA06 (NPN), MPSA56 (PNP): 0.8A 80V 0.625W fairly high gain
BC337-40 (NPN), BC327-40 (PNP): 0.8A 45V 0.625W high gain
BC547B (NPN), BC557B (PNP): 0.1A 45V 0.5W general purpose
BC549C (NPN), BC559C (PNP): 0.1A 30V low noise high gain

There's also the PN2222 (NPN), 0.6~1.0A depending on manufacturer and suffix, 40V, 0.625W, fairly high gain, if you want to avoid European part numbers for some political reason. (The BC... types are all European designators.)


Right. Normally a component replacement, and often not the transistor that has the wrong voltage - normally something in the circuitry that is driving it.

Yes, I remember having trouble with that too, but that was when I was a kid and my brain figured it out, so now it's second nature to me. So it will be easy to check your work.

Right, with Q2 out of circuit, Q806 is forced to dissipate quite a lot of power.

Right. Once you've installed the new driver transistors, you should test the amplifier with the collector of the top output transistors (Q1 and Q2) disconnected, and a 2k2 power resistor (e.g. http://www.digikey.com/product-detail/en/ERG-3SJ222/P2.2KW-3BK-ND/36811 or http://www.digikey.com/product-detail/en/AC05000002201JAC00/PPC5W2.2KCT-ND/596703) connected between the wire that used to go to the collector, and the emitter. This will provide about 20 mA of current through Q3/Q4 so the output stage can be adjusted, but if anything goes wrong, the current will be limited to 40 mA so the output transistors will be safe.

(For most amplifiers, the collectors of the top AND bottom output transistors should be disconnected for initial power-up, but this amplifier has a "quasi-complementary" output stage, which can't be adjusted unless the bottom output transistor is present.)

The 64V is printed on the first diagram in my post - At the bottom on the power supply block diagram right off pin 5 it says 78V and just the the right it shows 78V (64V). It is here that I initially measured 86V before the failure. After the failure and after disconnecting the main amp circuit from pins 5&7 and under bulb limiter connection, I measured 64V at pin 5. Reconnecting the main amp circuit as I tested last night but with Q2 out of circuit, it was measuring 75V. I could disconnect the main amp circuit again from pins 5 & 7 and double check, but I'm sure I it was 64V on pin 5 that I saw when everything was disconnected.

I'll get those power resistors but I was hoping for a more "baby step" test in which I could isolate, say, the first half of the circuit to see that those updates worked, without concerning about second half circuit center voltage adjustments, or needed Q2/4 in circuit etc. (but also not heating up Q806) so I didn't know if some of the pins 6-11 could be disconnected to allow the first part of the circuit to be tested in some fashion. Then move to the second half of the board and install and test those parts. Plus that Q802 was ordered first so should be here first and it gives me something to move on and verify until everything else arrives (which may be several weeks given where I bought it all from). That would also help me if something goes wrong since I would have some discrete changes between tests, rather than a flat out total replacement then test. But if that's the only way to do it, I'll just wait till everything gets here.
 

KrisBlueNZ

Sadly passed away in 2015
The 64V is printed on the first diagram in my post - At the bottom on the power supply block diagram right off pin 5 it says 78V and just the the right it shows 78V (64V).
So it does. There is no explanation in the text for what the value in parentheses means. My guess is that it's the voltage with the amplifier driving a heavy load - for example, operating at the maximum rated output power of 50W into 8Ω per channel.
It is here that I initially measured 86V before the failure. After the failure and after disconnecting the main amp circuit from pins 5&7 and under bulb limiter connection, I measured 64V at pin 5. Reconnecting the main amp circuit as I tested last night but with Q2 out of circuit, it was measuring 75V. I could disconnect the main amp circuit again from pins 5 & 7 and double check, but I'm sure I it was 64V on pin 5 that I saw when everything was disconnected
If you want, try to duplicate the reading. But be careful because something would have to be drawing some significant current out of the power supply to pull the voltage down that low, so watch out for components getting hot!
I'll get those power resistors but I was hoping for a more "baby step" test in which I could isolate, say, the first half of the circuit to see that those updates worked, without concerning about second half circuit center voltage adjustments, or needed Q2/4 in circuit etc. (but also not heating up Q806) so I didn't know if some of the pins 6-11 could be disconnected to allow the first part of the circuit to be tested in some fashion. Then move to the second half of the board and install and test those parts. Plus that Q802 was ordered first so should be here first and it gives me something to move on and verify until everything else arrives (which may be several weeks given where I bought it all from). That would also help me if something goes wrong since I would have some discrete changes between tests, rather than a flat out total replacement then test. But if that's the only way to do it, I'll just wait till everything gets here.
Well, the final part of the amplifier, involving the right-most three transistors in each channel on the circuit board as well as the output transistors, is a single block.

Once you've checked the power supply output voltages, you can test Q811/812 by checking for about 23V on their emitters, and if that voltage is right, you can check Q801/802 by confirming that the collector voltage is somewhere in the range 10~20V (I can't tell you an exact figure). You could also check that there is audio on the collector of Q801/802 but if the DC voltage is right, it's pretty likely that it will be amplifying properly.

The four 1N60 diodes are germanium types but this is not critical; they can be replaced with 1N914/1N4148 types.
 
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