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Messages - Roly

#46
Quote from: EnzoProbably RCA 40409 and 40410 types.

I agree, but with one reservation.  These transistors also had downward lugs to attach the heatsink portion to the board, and I don't see matching holes in the board for these lugs, but it certainly looks like somebody intended the drivers to have heatsinks that aren't there.

I'd approach this one with considerable caution and suspicion because it looks like it has been badly messed about at some stage in its long life.  Until you get it going you are going to need to be in constant reality-check mode.

Here is an account of my own bruising encounter with a messed about Gibson amp.
#47
Quote from: Vitrolinwhat sets the limit for a power amps load

In a word - current.  The ability of the output stage to pass it, and for the power supply to deliver it, all without damage.

In simple terms what you propose, double up the output stage devices and beef up the supply, is exactly what manufacturers do.

In s.s. amps the power supply is actually an intimate part of the output stage, they really have to be considered together, and together they are designed for a certain current capacity determined by the lowest intended load.  So you seem to be on the right track, however it's not a case of either/or but the totality of the power supply and output stage.  The entire current loop from supply, through the OP stage, loudspeaker, and back to the power supply has to be up to the task.

HTH
#48
Hi cerealport, welcome.


The power board looks like it has had a couple of caps removed, and a fine blue wire kluged in there for some reason.  I would certainly investigate this before I applied power.

With an amp like this that is a total unknown quantity it's a good idea to initially power up via a limiting lamp.
#49
{here starteth the redesign...  8| }


Quote from: cbg Rickwhat is the proper way to get the current under control in the output transistors?

Okay, assuming that your output transistor are mounted on a sufficient heatsink (and by "sufficient" I really mean "excessive" - such as a couple of redundant CPU coolers o.n.o.), you are discovering why almost all amps are fitted with emitter resistors in the output pair.  These could be just about any value from 0.1 ohms to a couple of ohms, but typically 0.22/5W.

{the bigger the heatsinks the greater their thermal inertia and the more time you have to hit the "off" switch before your output transistors melt.}

What you are seeing is thermal runaway, what these resistors are there to prevent.

The reason for thermal runaway is that the VBE for each output transistor depends on temperature, falling with increasing temperature.  If the bias voltage is fixed then as the transistor gets warm and its VBE drops, it turns on more, and gets hotter.  This is a positive feedback loop that quickly leads (as you found) to thermal runaway, and if not quickly caught the destruction of the output pair.

With an emitter resistor between each emitter and the half-rail the result of more current is more voltage drop across the resistor, and this is in opposition to the drop of VBE.  If the resistor is large enough, and it doesn't need to be very large, the voltage across the emitter resistor will rise faster than VBE falls, reducing the thermal loop gain below unity and making the output stage thermally stable.

So the first thing to do is add a couple of emitter resistors.

Another thermal feedback loop exists between the transistor cases and the three bias diodes and ideally at least two of these should be tightly thermally couple to the output transistor cases (to increase the loop frequency response/reduce loop delay by greatly reducing the delay between the transistors getting hot and the diodes following).

Purely for physical reasons transistors in cases such as TO-220 are a whole lot easier to mount in intimate contact on the (TO-3) output transistor mounting bolts (thermally coupled but electrically isolated).  My personal favorite is to mount a couple of BD139's (or BD140's, doesn't matter, just get the polarity right) on or very near the output devices, and use the E-B junctions as two of the bias diodes.

In some stereo amps you will find a 3-way tagstrip mounted on one of the transistor bolts, carrying a normal diode bent over so it rests on the top of the transistor case and with a glob of thermal paste to improve coupling - a bit agricultural, but effective.


Quote from: cbg Ricklowering the values of R9 and R10

It might, but at the expense of increasing the dissipation in the drivers, and it doesn't address the basic problem of currently having a thermal loop gain above unity, and runaway - the output transistors get hotter faster than the thermal compensation diodes in the bias chain can pull back the base current injection.

A way to reduce the bias is to reduce the value of R6/100r, but that alone won't reduce the thermal loop gain, just start it off at a lower value.  This might make the amp thermally stable at idle but won't stop runaway once it is driven.

It may seem strange but this thermal loop between output transistors and their compensation diodes is subject to the same stability criteria as an electrical feedback loop.

If the loop gain remains well below unity, say x0.7, the output stage will finally settle to a steady state, but the closer it is to unity the longer it will take until it settles, and may significantly over- and under-shoot before it does.

As it comes to unity (e.g. x0.99) the whole amp may go into a slow thermal oscillation, continuously ramping up and down but never settling.  {in a marginal case you can even get an oscillation on an oscillation, an idle current wobble that dies away only to keep reappearing.  This can be over periods of tens of seconds to minutes, and can depend on the ambient temperature.}

As it comes above unity the output stage will be thermally unstable, what you have now, and may well self-destruct if not caught in time.

The acid test for thermal stability is a half hour run into a dummy load at about 2/3rds full power at a high ambient temperature (heat lamp), and you are looking for the heatsink temperature to settle under drive, then for the output pair current to ramp down to idle as soon as drive is removed.

HTH
#50
Quote from: EnzoThe nice thing about a circuit is that it is, well, a circuit.
:lmao:   :dbtu:



Hi Rynlander, welcome.


Big rack amps with humungus beer-can caps aside, I've never really found the need to discharge the power supply caps in s.s. amps.
#51
Quote from: ogeecheemanmy bad...junction of R50 and C29 is +32vdc. the other side of R50 is -17vdc.

"It's just a jump to the left.
And then a step to the right..."

New York to Chicago via Budapest, but we'll get there.
:lmao: :dbtu:

Quote from: ogeecheemanI have already replaced C29

In this rather rare situation I think that is a wise defensive move.


Quote from: ogeecheemanwaiting on MJE340 (VAS) and GI 752 diodes to arrive

The MJE340 should do it, but ... um ... why the GI752?  Did you find one shorted?
#52
Quote from: brodie1600Q11 from collector to chassis ground: 42 V

Quote from: Enzoas there is 84v across the C-E

These are in contradiction.  He can't have the 42V +ve rail on the collector of the VAS and have 84V across the VAS C-E.  I'm uncertain how the "84V" measurement was made, there was some confusion about "across", but I suspect it was VAS collector to -ve rail, which would be consistent with 42V to ground, and a shorted VAS.

Whatever, Q11 looks SNAFU.  The question now is what do we replace a 2SA1016K with? (150V/50mA/400mW)  An MJE350 should do.  {In Oz most retailers don't carry 2Sx series, you have to go to the equipment manufacturer for a "genuine spare part" which costs $40 for a $4 part, or substitute.}

I guess the mists will clear.
#53
Quote from: mladenuI do not like higher voltage and valve depreciation over time..  :-\

As you like, but I have to remark that over a hundred of these have been built by complete amateurs without a single injury, fatality, or dead cat.  The output valves are not run hard and will have a life in average duty of decades.




A circuit using a LM3886 with mixed voltage, R19, and current, R18, feedback.



Oh, and you're going to need a BIG heatsink;




#54
Quote from: brodie1600Q11 from collector to chassis ground: 42 V

Quote from: EnzoI'd be trying a new Q11 at this point.

Me too.

In the same way that the collector of Q9 is sitting about one diode drop above the half rail;

(400mV + VD28 - VBEQ18 - VBEQ20)

so the collector of Q11 should be sitting about one diode drop below the half rail;

(-400mV - VD29 + VBEQ18 + VBEQ20)

The fact that it isn't suggests that it has failed shorted C-E.
#55
Quote from: mladenui`m in search for suitable solid state design who will able to produce good side of characteristic valve sound... as much as possible...


I once had a conversation with a Bright Young Electronics Engineer that went something like this...

BYEE: "Why do you bother with all that old technology?  You could model the valves in a DSP and use a solid state output stage".

Now I've done a bit of programming in my time and I know that in one mouthful he was talking about a space project.  I also knew that to faithfully reproduce a 100 watt valve amp sound would require a s.s. power amp with considerable headroom.

"Yeah, I could do all that" I replied, "but here's a novel idea; if I want value sound, why don't I just use valves?"

e.g.
http://www.ozvalveamps.org/ava100/ava101lamington.htm

I subscribe to the school of thought that the (average quality) output transformer is a major contributor to valve sound (through progressive magnetic compression).


{I'll shut up now.  :-X }
#56
Quote from: ogeecheemanR50 readings strange. +15vdc on one side.. -24vdc on the other.

Indeed, one end of that goes to +ve rail via R66 47r and D12, should be close to the actual rail voltage.  If either of these have been damaged then it looks like it may have been on long enough to also damage C29 220uF.

The VAS isn't out of the woods yet, but we could be looking in the wrong direction and the actual fault is in the emitter circuit not the collector.  Whatever, the diff pair and VAS need sorting.
#57
{"Arrg dat's good" says the Irishman, "Den it'll be a beer for me, and a Cat-o-lik for me friend."}



Correction:  ::)  the seven volts across Q19 B-E is most likely due to the B-E junction zenering out in reverse mode because its emitter is being pulled so far negative, not being open B-E as I said.

I still don't like no apparent VBE drop on Q20.

Please test each of the transistors you replace to confirm which one(s) is/are dead.


All of these s.s. power amps are DC coupled throughout and damage often propagates backwards from an output short, first the power transistors, then the drivers, then the VAS.  When I find any dead silicon on initial cold test I normally end up pulling all the transistors and testing them out of circuit just to be certain I've got everything first time.  An almost unique reverse polarity like this one?  Then it could be just about anything just about anywhere in the power amp.

Protection circuits aren't immune either.  They are intended to stop a device failing, but if a device does fail not all protection circuits are themselves protected against failure conditions should they arise due to a failure that the protection circuit couldn't prevent (like some nong reversing the bridge rectifier  :duh ).

We look forward to your early report of success and "loud noises off".    :dbtu:
#58
{"Yeah, of course we do" replies the bartender, nervously eying the big Croc by the Irishman's side.}



Okay, good, as long as we are on the same power amp circuit.

Quote from: ogeecheemanas I said previously the preamp stage is working perfectly

Not surprising because it would have been protected by the forward-biased zener diodes.

Quote from: ogeecheemancommon emitter voltage of Q19 & Q20 is  -17vdc
                                                                  Q19 BASE-10vdc                                               
                                                                  COLLECTOR= -37vdc
                                                                   Q20 BASE= -17vdc
                                                                   Q20 COLLECTOR= -32vdc

                                                                Q14 = COLLECTOR -31vdc
                                                                         BASE -31vdc         
                                                                         EMITTER -32vdc

Ah HA!  Lot's there that don't make sense.   :dbtu:

First transistor, Q19, SEVEN volts across it's B-E?  What does that say?  Must be open B-E, right?

Second transistor, Q20, you've got the same voltage on the emitter and base when it should be VBE = 0.7V different  :o  and moreover the base is being pulled down by 27k to the half rail at -21V.  Must be shorted B-E.

Your voltages around the VAS have a bit of a "spread"  ;) , but putting 31<->37 down to misreading the DVM or something, and it doesn't matter here anyway, because despite the -17V on the diff pair common emitters, neither of them appear to be drawing any collector currents worth talking about.  It looks like the common emitter voltage is only due to current passing Q20 (shorted) B-E towards the half rail.

With no collector currents (rather than differential collector currents as they should be), the reaction of the VAS stage by pulling the output south -21V isn't all that surprising (however it doesn't automatically mean that Q14, the VAS, is okay.  It may also be damaged, so remove and test, being careful to get it back the right way; or replace it defensively considering what else has gone down).


{attach: Crate gfx212 OPstage voltages 150509.jpg - power amp extract with voltages}
#59
That's kinda interesting, a bit Big Muff-ish, but can go right down to infinity attenuation in each band.
#60
{They walk up to the bar and the Irishman asks the bartender "Now would you be after serving Cat-o-liks in this bar me boyo?"}



Quote from: ogeecheemanit's in the power amp.there are no IC's there

You posted two circuits, <Crate GX-120, GX-212 Schematic(2).pdf> and <252SCH.pdf>.  Because of the thread title I have been using the former with the TL071/NE5532 op-amps, but because these two output stages are quite different we need to know which output stage circuit we are dealing with here, so we are all on the same page (literally).


Applying generally to both circuits; it seems that there is no distress, only that the OP stage is being driven to -21V out.  To me this implies that the power end is healthy and something is amiss in the differential input stage or VAS (IC5 and IC3, or Q19, Q20, and Q14)

{Attached is what appears to be the same output stage as in <252SCH.pdf> but in vastly more readable form. (Thanks to Jazz P Bass in http://music-electronics-forum.com/t16369/)}

You say "no IC's" so I'm assuming for the moment we are dealing with <252Xxc0.pdf>;

If you have -21V at the output then you should also have -21V at the base of Q20, the feedback side of the input diff pair, and it should be well saturated on.  This in turn should mean that the VAS, Q14, is turned right off, and that its collector should be close to the +ve supply rail.

I'd first check that you are actually getting something close to ground on the base of Q19 (to eliminate the possibility the amp is being driven correctly by a rogue voltage, which I doubt 'coz I can't see where it could come from).

My personal intuitive hit pick ATM would be the VAS Q14 gone leaky and the diff pair can no longer turn it off, but it could equally be either (or both) of the diff pair Q19 and Q20 themselves.

Please confirm that <252Xxc0.pdf> is the relevant circuit,
and measure and post the voltages at;

- the common emitters of the diff pair, Q19 and Q20

- the base and collector voltages of Q19 and Q20

- the collector voltage of Q14


attach: <252Xxc0.pdf>


Quote from: ogeecheemanoddly the bias voltage is right on the money. 7mv. and it responds when I turn the bias pot.

I = E/R

0.007/0.47 = 0.015 or 15mA idle (in each OP transistor).

This supports the idea that the power output end is actually working correctly.