Peavey TNT-100 Repair Problems

[jrem]

Guys:

I have a Peavey TNT100 bass amp on my bench. The schematic and layout is at jrem.webhop.org/tnt_100_schematic.pdf.

I narrowed the problem (no output) down to the output section (the signal follows through the collector of Q3).

My voltages showed that Q7 was blown, so I changed it out. When I fired the amp up (with a load on the output) I smoked R42.

So now I'm at an impass . . . and could used some help. After drewling over the schematic for a few stale afternoons at work I'm confused and need some direction . . . is the current through R47 coming from the feedback circuit (Q4& Q5)?

Could someone please take a peek and give me a clue? I don't need the answer, but I know I'm so close that it will kill me to pay someone to fix this just to find out it's something stupid I overlooked.

Thanks in advance,

John.

[Steve Conner]

Was it R42 or R47 you smoked? You mentioned both numbers, and they suggest different failures. Whatever the cause, a shotgun replacement of all the silicooties (except maybe Q1 Q2 and Q3 which are very unlikely to be damaged) would probably fix it. Usually when a SS amp dies, it's because an output transistor blows out and takes a bunch of other stuff with it.

[jrem]

Was it R42 or R47 you smoked? You mentioned both numbers, and they suggest different failures. Whatever the cause, a shotgun replacement of all the silicooties (except maybe Q1 Q2 and Q3 which are very unlikely to be damaged) would probably fix it. Usually when a SS amp dies, it's because an output transistor blows out and takes a bunch of other stuff with it.

Sorry . . . R42, right after Q3 at the input section of the output stage.

Shotgun . . . you thinking replacing Q8 & Q9 too?

Thanks . . .

[Steve Conner]

Hmm, the most likely path for the fault current that smoked R42 would be through Q6 and Q7 if they both failed short. You already tried replacing Q7, so try changing out Q6 (and replace Q7 too in case you just blew the new one you put in)

Due to the low price of transistors, it's probably as well to replace all of them except Q1 Q2 Q3, that's what repair meister Enzo always seems to recommend. I know the output devices are more expensive, so you could maybe get away with leaving the old ones in there if they test good. But failures of the driver transistors could weaken the output devices too.

BTW, check to make sure the diode assembly CR9 isn't shorted to anything, it ought to be fixed to the heatsink somehow.

[Amp Kat]

Well there's no doubt that Enzo's the SS master as he taught me all I know which still isn't sometimes enough to crack these things. The main thing I do is check the positive and negative rail voltages and make sure your getting your +15/- 15 for the low voltage op-amps or transistors in the preamp then check your +40/-40 or whatever they use for the power transistors. If a transistor device is bad it will most likely take that voltage to near 0 or very low and for some reason it's usually the negative side. If you pull the transistors out of the circuit one by one starting with the drivers when you get the faulty one out the voltage will come back up saying the voltage regulators or zeners are good. I've seen drivers that checked good under diode test and were still bad but the output trannys are usually pretty acurate.

[jrem]

cool, thanks . . . I did check my voltages, I know the power supply is good, and that's how I detailed it out to be that one driver transistor.

I also checked the output trannies on a test fixture, they looked okay. So I'm thinking it's one of the drivers or the feedback circuit.

Thanks for the input, though, I will take both of the drivers out and check my voltages.

Oh, and one more thing, what does the 'F.P.' stand for on the resistor designations? I don't think I've seen that before . . . film? They appear to be ceramic encapsulated which would lead me to believe precision wire, but maybe it's something else?

[tim]

F.P. stands for Flameproof....supposedly when a F.P. resistor scorches, it won't burst into flames, but I've seen a fiberglass circuitboard burst into flames under one. If it calls for one, I'd use it, just because....

[Enzo]

Flame proof - basically metal film with a ceramic coating. Yes, a good idea to use the right part.

This is a solid state amp, so it needs no load. In fact it is preferable to use NO load until you are sure no DC is on the output. If the output slaps over to one of the rails, it doesn't hurt it without a load, but with a load, then the load will draw lots of current and burn more stuff up. Once it is DC free, then connect a load and se how it sounds.

You found a shot Q7 and a burnt R42. That is an odd resistor to burn alright. R49,50 burn all the time along with their drivers.

First, verify your resistors. Are the ballasts OK - 0.33 ohm R51,52,53? Are both 47 ohm OK - R49,50? Mainly looking for opens, but any charcoal should be replaced. If you pulled the outputs and tested them, I'll trust that.

An open ballast means more stress to all those 100 ohm resistors around Q4,5. SO if a ballast is open, pay more attention to the 100s.

But Q7 shorted C-B and took R42. That current went somewhere, so test the dual diode CR9 for shorts or opens. it tests like a regular diode but has twice the voltage drop across it. ANd the tip about looking to see if it touches ground is a great one. The fault current had to go somewhere after the diode, and Q6, R49 is the likely path.

Q4,5 are limiters. Check each with a meter on diode for any shorted legs. But they are not needed for the amp to function, so lifting CR10,11 essentially eliminates them from the game. Make sure to have them working before returning the amp to service. They detect too much voltage across R51,52, which means too much current through them. At that pioint they turn on and clamp the base drive of the outputs.

This is a simple circuit, so set the meter to diode test, and just check every diode on the board for short or open. Ten of them maybe? If yuo have +/-15, then the zeners are OK, and the main rectifiers are too or you'd have no rails. But who knows all the little ones? Test them. Since we tested all the others, check Q1,2,3 for shorts. SInce R42 burnt, it is possible Q3/R39 got involved at some point. Check them.

It's a dumb thing, but check C29 lytic next to the dual diode on the drawing.

If the two outputs are not shorted, I'd be inclined to let them live as far as the burn-in bench to see if they work under load. But with Q7 and R42 burning, I'd have to replace Q6 unless the dual diode is shorted to frame.

I'd wager you have an open R49 or 50.

[jrem]

wow, thanks Enzo, that's exactly the advice I was looking for. I'll go through it this weekend when I get the chance. I'm also planning or ordering new trannies for the output section.

Any advice where to get the dual diode? The Peavey part is SZ-138869, the Peavey cross reference is MZ-2361, nothing like that at Mouser . . .

[jrem]

Looks like R51 & R52 are open (13m and 2m, respectfully). R53 reads .5, so I guess it's okay, but three new ones will be in the mix.

I think the dual diode CR9 is shot. I lifted R42 and get 440k one way, 750k ohms the other (I don't have a diode checker function on my DMM). Also with the 'limiter' trannies out (I figured it was some sort of feedback circuit, but I never thought of them keeping the outputs from running away) I put the ohm meter on CR10 and 11, and they don't appear to do what diodes are supposed to do.

I'm pretty sure Q3 is okay cause I had a signal generator on this amp and scoped it to that point (what I would classify as the input section of the output stage) and it looked good.

Guess I need to invest in a cap checker and a diode checker. They work in circuit? Or do you need to lift one end every time? That gets tedious.

Also, I can say that someone has been in here before (someone gave me the amp in it's current state, and I've had it apart for some time but only have just gotten serious about fixing it my self, it's sort of a learning project) as R47 has been replaced (they left the smoke on R52).

Maybe I should just go with the shotgun approach and rebuild the entire output section. Looks like I'll be buying all the parts anyways!

So 'no DC on the output' means no bias, and that the output should be zero volts with no signal, and it swings based on the input signal. Just reiterating that as that is how I'm reading that statement.

Thanks again for everyone's help.

[Enzo]

No DC on output means exactly what it says - the amp sits there powered up and the output where the speaker would connect does not have any DC voltage present. If it is less than a quarter volt or something, ignore it. What we do not want is 2 volts, 20 volts, 40 volts sitting there. So no "offset", I use the term "bias" for idle current through the transistors. We want it to sit there without offsetting the output. If the signal moves it, that's a bonus, at this point all we want is for it not to blow up.

You need a diode test function. Buy it or make it. Even inexpensive meters these days have a diode test function, usually also labeled continuity. If I had to invent one, I would take a 9v battery and a 10,000 ohm resistor in series. Apply the voltage from that to a test part. In this case a diode. SO what we have is a 9v battery, in series with a 10k resistor and the diode. The resistor limits current through the circuit - and thus the diode - to a little under 1 milliamp. If the diode is facing the right way, that 1ma flows through it, and the diode will have a typical voltage drop across itself of maybe a half volt. Reverse the diode, and a good diode will stop the current - hey, it's a diode after all.

So with this tester - a resistor and battery - we apply it to a diode, and watch the voltage across the diode as we do. An open diode or one that is connectd in reverse will show no voltage drop across itself. That does not mean zero volts, that means the full 9v will appear on the meter. Zero volts across the diode would mean either it was shorted or not connected or the battery died. If I get a chance I will make one after dinner and see if it works right. ANyone see any holes in this invention?

SInce you don't have a diode test, you can't really test the power transistors other than for shorts. If a power transistor should show a base to emitter junction voltage of .4 volts and the test shows it as .2 volts, that means the transistor is very leaky. SOmething your ohm meter won't show you.

That dual diode really needs enough voltage across it to turn it on for a real test. Not shorted is a good sign actually.

Since those ballast resistors burnt open, that means the power transistors had to conduct hard enough to do it. They are suspect to me. A pair of MJ15003 would replace them nicely.

I'm pretty sure Q3 is okay cause I had a signal generator on this amp and scoped it to that point (what I would classify as the input section of the output stage) and it looked good.

To it maybe, but through it is the key. I don't know how you can test it without having the following circuits working right.

By the way, for testing, a regular diode in place of the dual diode will work well enough to get the rest of the circuit going. It would reduce the current through the finals and add a little crossover distortion, but would work. But don't leave it that way since that dual diode also tracks the output thermally and compensates. We don't want thermal runaway.

[jrem]

No DC on output means exactly what it says

yip, got it.

You need a diode test function. Buy it or make it. Even inexpensive meters these days have a diode test function, usually also labeled continuity. If I had to invent one, I would take a 9v battery and a 10,000 ohm resistor in series.

okay, got it, I can do that. I'll just set it all up on a test board and use two test leads to connect to the diodes. Thanks for the tip.

That dual diode really needs enough voltage across it to turn it on for a real test. Not shorted is a good sign actually.

okay, thanks, I'll reference the data sheet for the voltages.

Since those ballast resistors burnt open, that means the power transistors had to conduct hard enough to do it. They are suspect to me. A pair of MJ15003 would replace them nicely.

yikes! they're ~$11 each! this thing is a Peavey, not a McIntosh! Are they worth it in an amp like this? The 2N3055's are a buck and a half each (http://jrem.webhop.org/x_ref_list.pdf is the Peavey cross reference sheet)

To it maybe, but through it is the key. I don't know how you can test it without having the following circuits working right.

hmm, impedance and reactance, eh?

By the way, for testing, a regular diode in place of the dual diode will work well enough to get the rest of the circuit going. <snip> We don't want thermal runaway.

yes, that's bad. thanks for the tips, I'll try and implement.

Best regards, John.

[Enzo]

$11??? Where are they charging that much?

Allied sells them for under $4. They are currently out of stock on the MJ15003, but they have plenty MJ15001 at $3.71 each.

2N3055 is marginal for this application. When the circuit was designed decaddes ago it was a common part and they selected them for voltage. With +/-40v rails, the outputs need to be at least 80v parts. The MJ15001 is a 140v part so there is good margin. it is the same 15 amps of the 2N3055, and has dissipation rating of almost twice the 2N3055. A far superior choice.

The MJ15003 is also 140v, but rated to 20 amps. The 15001 is enough.

If you had some MJ15022 or MJ15024 in stock they would work fine too, though a bit of overkill here.

MJ15015 would be a nice option, but they were out of stock on them too.

62792 is obsolete, the 2N3055 is an alternate type listed, but if you order a 62792 or 2N3055 from PV, you will get a 70484140. WHich is their house number on a...MJ15003.

You can also order parts from PV and they won't cost $11 each either.

Did you look up MJ15003 in Mouser and see $11 for the NTE "equivalent?" One more reason NEVER to use NTE.

MY home made diode tester was a thought experiment, so try it out with some junk parts first to get the hang of it.

hmm, impedance and reactance, eh?

Not really, I just think if you scope up to the part, without the following drivers and stuff working right, how can signal current flow through the part to them?

It would be like following the signal up to the grid of a 12AX7 that had no plate resustor connected. Wouldn't tell us much about the 12ax7.

[jrem]

<snip>SInce R42 burnt, it is possible Q3/R39 got involved at some point. Check them.

okay, I have R42, R49, and R50 lifted (got new FP resistors from Parts Express), and I have Q4 & Q5 out, and pulled Q7 (I had +/-22vdc on the rails, they're supposed to be +/-36, and noticed that Q7 is shorted C-B).

Now I get 36vdc on the collector of Q3, which isn't good, I believe. I put a signal on the input and scoped the preamp, nothing on the collector of Q3.

I do have a signal on the collector of Q1, so I'm guessing I'm good up to there, but it looks like Q3 is toast.

[Enzo]

These parts all work together. You can pull Q4,5 and the amp still works, because Q4,5 don't do anything until excess current flows through Q8 or Q9. But pulling Q7 leaves a hole in the circuit. As does opening those resistors.

36v at Q3-C doesn't sound good, but in the absence of R42, ther is no path for current through Q3, so who knows what voltage would appear there. But if suspicious, replace Q3 with a 2N5400, 2N5401, or other good high voltage signal transistor like the common MPSA93 or MPSA92.

Transistors are current devices, as contrasted to tubes which are voltage devices. There needs to be a path for current through the transistor or it won't work right.