Pioneer SX-1250 Repair Manual

Discussion in 'Solid State' started by t j, Mar 13, 2018.

  1. steveUK

    steveUK AK Subscriber Subscriber

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    Well, is he "so wrong"? I mean, from what's been said, neither 10mV or 100mV is optimum. Granted, on the face of it 10mV is 'more wrong' than 100mV, but it seems the latter is not a great choice either. Dunno. I think there's more mileage in this thread yet.

    So, any ideas why Pioneer specified 100mV if, as has been said, it causes needlessly higher running temperatures? Surely this 'high' figure of 100mV has been quoted for a reason? Or do you think it is simply a case of it being 'not quite right although it works' when in actual fact something like 70mV is technically a better value? What's going on?!
     
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  2. gslikker

    gslikker Super Member

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    Guessing possibilities only:

    - "lesser" behavior of the output stage for thermal feedback if not preheated (read: too much heat difference for good behavior between no ouput and cranked up)
    - "lesser" electrical behavior at very low output when not having much bias (distortion/stability)
    - "better" behavior for distortion figures over the full range of low to high power (performance at expense of power consumption and "reliability-when-old").

    IMO, as steveUK states, the tech can be wrong but he is not "so wrong".
    He is probably very concerned if he can give any warranty on his work when this amount of heating up a 40 year old receiver is wanted.
    Myself, I would not warrant it, but I do not repair audio for other people nor as work, so it is easy to conclude.

    My own SX1250 is on the pile to ever restore (I did not anything this winter due eye surgery and average is only 2 or 3 amps each winter), but if and when it happens I surely will try how it behaves at much lower bias.
     
    Last edited: Mar 14, 2018
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  3. t j

    t j Active Member

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    Well SteveUK, for me it's a simple matter of communication. I wanted to verify this discrepancy and he goes screwball on me. Now, after paying the dude, I'm left with nothing but questions. I asked him to confirm his findings and he says he has no time, but offers to "mentor" me for a fee. Just a bunch of BS as far as I'm concerned.

    If it's not a cut and dry 100mV, then how can the guy be so cocksure of his 10mV proclamation? Bad science, bad customer service, and he even had bad breath.
     
  4. tarior

    tarior Dirty pool, old man? Subscriber

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    When looking at the schematic, the diagram shows each channel receiving 200mA of current at a no signal condition. That would work out to 50mA per output transistor, minus a little taken up by the drivers.
    If it were me, I'd be shooting for an average of about 25mV measured across each emitter resistor (DO NOT USE BARE PROBES), and then see how that correlates with what is measured at the summed test points.
    This is a good time to note that a good output transistor (drivers, too) to heatsink interface is absolutely crucial. Fresh thermal grease on the outputs and drivers is mandatory.
     
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  5. gslikker

    gslikker Super Member

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    I see two pairs of output transistors only, each channel, in my schematic. so the 200mA divides over two pairs??? each transistor 100 milliamps. As measuring (for voltage) is basically over 0.5 ohms resistors in series, you would expect 50 mV each resistor, corresponding to 100mA each resistor / each emitter.
    Anyway, to me it looks measuring the normal way will give you no clue if one of the output resistors would fail open circuit, as the test points are summed through the 22 ohm resistors which would give the average between two emitters.
    I am still thinking looking at it, the right way.

    Tarior I agree with you to check over the emitter resistors themselves!
    An open-circuit failure of an output transistor as well as the resistor or pcb track/transistor wire will go unnoticed measuring according the test points only, as the voltage is just the summing of averages over the power resistors using the four 22 ohms resistors. Also suspect HFE/VBE differences will be detected doing your way.

    BTW I also do not like the bias pot and stv diodes setup, it looks to me if a diode or a bias pot gets open circuit (for example during setting bias), bias can rise to immediate destruction of the amp?
     
    Last edited: Mar 14, 2018
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  6. steveUK

    steveUK AK Subscriber Subscriber

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    I don't see the need to replace thermal grease given the way in which it works, on the contrary I think it is better left alone if all is well. To that end I just googled the issue and this is one of the 'responses' (full web page address follows):

    "There is no need to reapply it at all. Once its on that its.
    The only time you need to reapply it is when you remove and replace the heatsink for whatever reason.
    Just incase you don't know the Thermal Interface Material or 'TIM' is there only to fill the 'void' between the CPU heat spreader and the bottom of the heatsink to make better thermal contact. As neither are entirely smooth, its not visable to the naked eye however.
    It is not a lubrication issue of keeping it 'wet', infact quite the opposite a heatsink that is stuck down to the heat spreader is a good sign the TIM is doing its job."


    http://www.tomshardware.co.uk/forum/285023-28-long-thermal-paste-replaced
     
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  7. tarior

    tarior Dirty pool, old man? Subscriber

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    From looking at the engineering diagram, rather than the procedure put forth by the technical manual writer, 200mA for one channel comes out to each output transistor sinking 50mA, not 100mA. Probably even a little less, as each pair of drivers needs to be biased on as well, and there are other losses through the bias string and whatnot.

    The pot/diode string is a simple and effective way to bias the outputs, if a little dangerous and not DIY friendly. One should really not mess with the bias trimmers if one doesn't know what he is doing.
    Also, every single unit that I ever work on that uses a biasing scheme like this one (it's VERY common) gets brand new Bourns trimmers for bias and DC offset. Always.
     
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  8. tarior

    tarior Dirty pool, old man? Subscriber

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    Yeah. OK. I'm sure that dried out powder is just as effective at preventing air voids and transmitting thermal energy as fresh "wet" grease.
     
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  9. gslikker

    gslikker Super Member

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    The 200 milliamps have to go somewhere.
    It goes 4 ways, one small predriver transistor, a bigger driver and the two collectors of assciated power transistors. It can go nowhere else.
    As it corresponds with the voltage markup in the manual I do not see any reason to think different...the manual markups are correct.
    Are you sure you are not just lured by the way the circuit is drawn?
    20180314_195507.jpg
     
    Last edited: Mar 14, 2018
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  10. steveUK

    steveUK AK Subscriber Subscriber

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    If it's gone to powder then yes I agree. Here in the UK I don't recall seeing many, if any examples of heat sink compound that has gone bone dry or as you say, turned to dust.
     
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  11. tarior

    tarior Dirty pool, old man? Subscriber

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    50mA X 4 = 200mA.

    The way the procedure is written gives about double that. 100mV across the test points (pin 7, 19) gives about 50mV across each (0.5ohm) emitter resistor for 100mA per output transistor for 400mA total (28 watts!), not including drivers and whatever else isn't being powered by the regulated P/S.
     
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  12. tarior

    tarior Dirty pool, old man? Subscriber

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    I've seen it a lot here. More often than not, in fact.
    I can see the point of view of the writer from Tom's Hardware. Which, near as I can tell, is to discourage non-qualified personnel from just taking stuff apart and putting it back together again with the risk of getting wrong and causing trouble. A valid POV, IMO.
     
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  13. gslikker

    gslikker Super Member

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    You can not add up the currents for each transistor in one channel, since the current running through the NPN transistors is the same current flowing through the PNP ones...
    So 100 milliamps each output transistor results in 200 milliamps total for each channel. I always see the NPN and PNP as a pair. 100 mA each transistor = 100 mA each pair.
    Just look at the drawing I attached previous post.
     
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  14. tarior

    tarior Dirty pool, old man? Subscriber

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    Oh yes, quite right. I should know better.:)

    So there we have it. 200 mA for each channel, roughly 100mV across the test points, best verified by checking at the emitter resistors.
     
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  15. gslikker

    gslikker Super Member

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    If the schematic looks like spaghetti, and it does, it happens :)
    Better not have this receiver in some cabinet, when it operates. 56W for bias, maybe assume 24 watt at least more losses for electronics and lamps, it may burn 80 watts or more at idle.
     
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  16. steveUK

    steveUK AK Subscriber Subscriber

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    So, maths aside, what's the conclusion then regarding that 100mV Pioneer stated figure? Is it correct? If not what, ideally, should it be?
     
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  17. gslikker

    gslikker Super Member

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    I have no reason to doubt the correctness of the manual, if a yes/no statement would be asked.

    ( in the end, what "ideal" means, will be debatable forever)
     
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  18. steveUK

    steveUK AK Subscriber Subscriber

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    Well not really if you take "ideal" to mean the value that gives optimum performance in terms of distortion and reliability. A lower figure runs cooler, but a higher figure improves the receiver's distortion figure at lower listening levels. There has to be a happy medium, an ideal figure, and that's what Pioneer should have (and maybe have) given in the service manual. I have never said it's right or wrong, others have suggested it may be. I'm asking them now, in conclusion, what do you think is the ideal value?
     
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  19. gslikker

    gslikker Super Member

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    I have no idea, I will try the very day if I ever start working on it....My own needs are not 100 kHz +/- 1dB at 160 watts.
     
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  20. drumbum

    drumbum Super Member

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    With bipolar outputs, if you observe on a distortion analyzer, you adjust the bias until the distortion falls away, usually around 10mv. The idea is to switch on both comp pairs so the switching dist. goes away.

    Some feel there is no benefit to more quiescent current if both transistors are on.

    Douglas Self writes that a range of about 43mv-52mv is enough for the commonly used emitter resistors, i.e. 22,33,47, 1ohm.

    But there is obviously differing opinions amongst amp designers.

    Hope this helps.
     
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