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Peavey Renown

Started by Hawk, June 26, 2015, 09:12:28 AM

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Enzo

I need to know.  How does the removal of one parallel transistor cause the rest to fail?  Remember we already stipulated the amp would not be producing full power this way, but will work sufficiently to service.

g1

Quote from: LateDev on July 03, 2015, 12:45:24 PM
QuoteA shorted Q13 could be C-E, so it would not necessarily affect the bias.
Actually it would be a miracle if it did not effect the base as the emitter and collector is physically separated by the base on a bipolar transistor, unlike an FET which you may be getting confused with.
No I'm not getting it confused with anything  ;).  We often see C-E shorts in power transistors yet the base is shorted to neither C or E.
But this is aside from the point, you were originally speaking of it as a driver.  In which case a C-E short would have affected the bias.  So no need to argue a point you never originally intended, right?


Quote from: LateDev on July 03, 2015, 12:45:24 PMNever assume you can run the amp with one O/P transistor removed, as this could blow the rest. If you need to know why, just ask.
Ok, I'm asking.  No one said to run an amp with one O/P transistor removed.  What was said was that when there are multiple pairs, extra pairs can be removed, for test purposes at lower power levels.  We do this all the time in the repair world.  You saw the caution about power levels when doing this?

Hawk

CR28 reads .014v one way and the other, in circuit?? As LateDev suggest testing diodes out of circuit so will try that.

QuoteProtection circuit is CR29, Q4, Q8, CR30, R105, R106. These act as clamps on the bases of Q5 and Q9, which are the drivers, via R107 and R108 . All transistors and diodes should be checked out of circuit
. Okay, good.

Thanks for the post on parallel diode readings. :cheesy:


QuoteNo one said to run an amp with one O/P transistor removed.  What was said was that when there are multiple pairs, extra pairs can be removed, for test purposes at lower power levels.  We do this all the time in the repair world.  You saw the caution about power levels when doing this?
If, for example, you mean Q3, Q12 and Q7, Q11, and Q6, Q10 then I get it. Correct?

QuoteOperating with one pair of the output transistors missing is not one less gain stage, it is less current gain.  This translates to less power available to the load, and more strain on the other output transistors at higher power levels.
Understood!

LateDev

#33
Quote from: g1 on July 03, 2015, 01:17:43 PM
But this is aside from the point, you were originally speaking of it as a driver.  In which case a C-E short would have affected the bias.  So no need to argue a point you never originally intended, right?
Did you not read my post ? I did say the numbering was different duh, in other words on the diagram I have, Q13 is shown at the position of Q5 on Hawk's cct, which is a driver.

I can only go by how a bipolar transistor is physically made. For there to be a short between collector and emitter the base region of a transistor must also be shorted, this does not mean that you have to see a short between the base leg to either the collector or the emitter as the leg is connected to a small portion of the base. You should really read up on how a bipolar transistor works to understand why the base current is effected when there is a CE short. Pay particular attention to doping of the base emitter region, and the now defunct depletion layer. which is blown away when there is a short of this nature.

Quote from: Enzo on July 03, 2015, 01:06:47 PM
I need to know.  How does the removal of one parallel transistor cause the rest to fail?  Remember we already stipulated the amp would not be producing full power this way, but will work sufficiently to service.
In actual fact Hawk was more sensible than this, and mistakenly suggested he remove a pair of transistors.
What we have here is that no explanation was given when it was said it would be fine to do so, and as you should know that without checking other components within the output section of the power amp, you cannot know that powering up will be safe.
Transistors are current operating devices, which means that the bias current for the output transistors is calculated to be shared between each of the 3 transistors. See Kirchoffs current law(KCL) What you now have is a circuit designed for 3 transistors each side, being biased on harder because there is only 2 transistors each side and you still have not checked other components in a DC connected circuit.

This is in answer to Hawk's question
QuoteSo if everything else is working but I take out Q12 and Q13, the amp should work, although not as powerful as I've removed a gain stage?
No, but it could cause damage, so take your time. don't rush, and check everything. I do not mean the whole circuit as it is an output transistor that blew.
If Q3 is OK and everything after that is fine, then there should be no problems. The furthest back you really need to check is to Q1 and Q2 but these never really blow, or it would be farer to say I have never seen them blown.
Once you are happy that you cannot see any other problems, replace any blown components, connect it up with the light bulb on the mains, then take voltage readings at different points. Never inject a signal until you know all other conditions, in a DC sense are correct.

Enzo

I don't know what to make of this.  If the C-E short exists, you say the base must also be "shorted".  Well, I will believe the base region may well be damaged.  But then you add that it doesn't mean there has to be a short from the base leg to the other terminals.  We do not tear transistors apart, so all we CAN do is test the relationship from one leg to another.  You might make some philosophical point about what goes on inside, but from a reasonable practical view, such as during troubleshooting, if we measure no short from the base leg, we will simply say ther is no short.  The transistor is already failed C-E, so anything else is moot.

QuoteIn actual fact Hawk was more sensible than this, and mistakenly suggested he remove a pair of transistors.
What we have here is that no explanation was given when it was said it would be fine to do so, and as you should know that without checking other components within the output section of the power amp, you cannot know that powering up will be safe.

Removing a pair is fine to do, more below.

What I suggested was that the amp does not need that one transistor to operate during test, I did not imply nothing else could be wrong.  I suggested that service could continue with that one part removed - or with two removed as a pair - and in fact I did explain that he should do so in order to find pout whatever else might still be wrong with the amp, so as to get all the parts into one order.  That specifically implies something additional may well be wrong.  So powering up would be done in the same service context as before.  I am sorry you did not understand that. And that is where current limiting bulbs or variacs with current meters would be exactly the tools to use.


As to removing Q12 and Q13, there will be no detriment to the amp.  In fact, MANY solid state amplifies are built on boards intended for more than one model, and the difference between a higher and lower power version will be the addition or deletion of a pair of output transistors on that board.  The bias of the solid state output is determined by the voltage difference at opposing bases.  removing a pair doesn't change that.

Yes, if we remove 1/3 of the output devices AND TRIED TO GET FULL OUTPUT from the amp, then each transistor would have to try to conduct the extra current.  But we already specified th amp would not produce full power under those conditions and should not be asked to.  SO again, your point is moot because it does not apply here.

It will not damage the amp in any way to operate it with Q12 and Q13 removed during test.

LateDev

Quote from: Enzo on July 03, 2015, 08:46:24 PM
I don't know what to make of this.  If the C-E short exists, you say the base must also be "shorted".
and where did I state that ?
Quote from: Enzo on July 03, 2015, 08:46:24 PM
Well, I will believe the base region may well be damaged.  But then you add that it doesn't mean there has to be a short from the base leg to the other terminals.  We do not tear transistors apart, so all we CAN do is test the relationship from one leg to another.  You might make some philosophical point about what goes on inside, but from a reasonable practical view, such as during troubleshooting, if we measure no short from the base leg, we will simply say ther is no short.  The transistor is already failed C-E, so anything else is moot.
Of course there does not have to be a short, if you care to reread the post I made, it was g1 that made that comment, and that was in respect to the different numbering of the schematic I was looking at, having different numbers to Hawks, again I explained all this.
Quoteg1 stated : A shorted Q13 could be C-E, so it would not necessarily affect the bias.
I merely pointed out that it would effect the bias. If you cannot understand this, then fine.

Same goes for the biasing of the power amp with a pair of transistors removed, if you note Hawk had decided to inject a signal, which should never be done at this stage. the four transistors are actually biased on more, than they normally would be, so the test is useless.

J M Fahey

Quote from: LateDev

Quote from: Enzo
I don't know what to make of this.  If the C-E short exists, you say the base must also be "shorted".
Quote from: LateDevand where did I state that ?

Quote from: LateDevFor there to be a short between collector and emitter the base region of a transistor must also be shorted

Enzo

QuoteSame goes for the biasing of the power amp with a pair of transistors removed, if you note Hawk had decided to inject a signal, which should never be done at this stage. the four transistors are actually biased on more, than they normally would be, so the test is useless.

Sorry, flat out wrong.  Removing one of the three pairs of outputs does not change the bias relationship between the bases.

Hawk

Thanks for everyone's input. I've learned a lot as usual. Will print off this thread and attach to schematic for future reading for future solid state repair work.

Will order parts and power up with my current meter, variac, and limiter and let everyone know my results.  Thanks :dbtu:

Hawk

Best techniques to save transistor legs from breaking!!!

In the images I attached (sorry one is blurry but you'll see that the middle leg broke off), I removed the transistors to test out of circuit and they both tested good. So I was feeling good about that. But when I tried inserting the 5331 the middle leg broke off. I had kept the transistor screwed down and had pried the legs out as I applied the iron from the other side of the board. After testing, I tried to slide the legs back into the holes using a small screwdriver--the middle leg wore out very quickly and broke.
Is it best to have the transistor standing at a 90 degree angle to the board when removing then simply pull upwards? When re-inserting should I insert legs through holes, solder, then bend the entire trans. down to the heatsink and screw it down??

On the smaller transistor, when I removed it, the legs soon became bent at odd angles that were very hard to straighten out and after trying several times to get all three legs in the hole the transistor looked like an arthritic octopus and the middle leg broke off. What is the best method for taking out and re-insertion with these small sensitive resistors??

I've ordered duplicates from Peavey and I don't want to screw up putting the new ones back in the board.

What have the experts been doing for years that works?  Thanks for  your input in advance!


Also, when I ordered a replacement transistor (Q13) from Peavey they told me it had to be a matched pair so I purchased two.  So, like output tubes, does this mean I have to replace Q12 with the other new transistor?

Enzo

First thing I do is remove the nut from the screw, so the transistor can be moved.  I then suck the solder from around the leads, then a worry the part free and lift it up.  I do not unkink the leads.  I rarely have to remove thse to determine if they are faulty.

J M Fahey

Agree and add: the main point is to remove all solder from the joint, any left bridging pad to leg will make you twist and pull, and we want as little mechanical effort as possible.
Sometimes there'sonly a few atoms left, so lightly pulling the leg from above or lightly pushing it from below with a screwdriver blade is all that's needed.

Sometimes it pays to have some solder wick .

Obsolete for many, but helps remove that last tiny bit of solder.

Hawk

Thanks guys.

QuoteI rarely have to remove these to determine if they are faulty.

Well, from what I've read most people say to get a proper reading the transistor must be removed from circuit. Enzo, I'd rather not for sure. So, how would you have tested this one in-circuit to save you the hassle of removing it? Would none of the associated circuit effect the reading?  Thanks!

I will certainly try your method to save future removals...

Enzo

Of course external things can affect the readings.  I admit to a great deal of experience, which may color my approach, but I keep in mind what I am looking for.  For example, I do not just take a zillion readings and hope they suggest something. I am looking for specific information.  If I am blowing fuses or the amp is going to DC, I start to look for shorted or open parts or circuit paths.  If the transistor is shorted, there is nothing in the world external to it that can make it look not shorted.  So if I am looking for shorts, then no need to remove anything.  On the other hand, when I do see a short, it may be the part itself or the things parallel to it.   So since the part had to come out to replace if shorted anyway, THE I remove it and retest.  If the part still is shorted out of circuit, then I know and replace it.  If it now seems OK, I know the short was, and still is, elsewhere in parallel.

The part may not be good, but at least my meter will tell me it is not shorted, or may be.

Sometimes two transistors are connected together, this is common, and so a couple of their junctions may be in parallel.  If they are such that one transistor conducts on direction and the other the other, I will see a diode indication it conducts both ways.   In one transistor that would mean a bad junction, but I am looking at the schematic, and will see the parallel reverse junctions and know it is an expected reading.

To get a proper reading it must be removed?  Well, it depends what you are doing.  If you want to measure gain and leakage and stuff, sure.  But if we are looking for evidence of failure, we can usually find that while in circuit.   None of this is hard and fast rules, just general tendencies.  We roll with the punches.

A lot of output stages feature low resistance resistors.  SO some PV solid state may have a 22 ohm resistance from base to emitter.  Your meter will see that as a short on diode test, but n resistance, voila, the 22 ohms the schematic calls for.  The part could still be bad somehow, but not shorted.

There is of course an endless tale of ways to do this stuff.

Hawk

Quoteyes in tube radio world it is known as a Dim Bulb Tester.  high current is dropped across the bulb protecting your equipment instead of high current across expensive delicate components such as transformers.  you can gage the bulb size by device power consumption.  100W power consumption = 100W bulb.  not to be confused with power output which is not close to power consumption

Replaced power output transistor. So, according to this statement above, I need to have a 450 watt bulb as the Renown Amp is rated at 120V/450 Watts. Not sure this statement agrees with some of the other info on the current limiter info discussed in this forum.  I read the info but not sure exactly how to proceed. I have a current limiter and a variac. I know that SS amps can operate at lower voltages so would it be okay to start with a 60W/100 W bulb? The amp  has a 5 Amp fuse.
I also have a current monitor.

Thanks.