Solid State Guitar Amp Forum | DIY Guitar Amplifiers

Solid State Amplifiers => Tubes and Hybrids => Topic started by: ilyaa on March 07, 2014, 03:43:05 AM

Title: blues deville got the blues
Post by: ilyaa on March 07, 2014, 03:43:05 AM
i like this amp a lot (the 2x12 combo) and it sounds good (usually)

this is a 93-ish deville (schematic attached)

lately ive noticed a sort of muffled/congested/sometimes even grainy quality to the drive channel. normally i have no complaints with this amp's tone, but this one is bugging me. it's sometimes crackly verging on staticy. why is it so hard to explain?! bottom line is, i keep wanting to turn it up but it just doesn't seem to break through the graininess. on top of that, sustained notes in particular will kind of crap out into a cheap sounding fuzz.

i got new power tubes. put them in. its fixed bias but i checked it out anyway and they seem fine. running about 37 mA at 474 on the plates so 17.5 watts. that's not too cold is it? (6L6GCs).

control grid voltage is only at -43 or so - should it be more like -50? im deliberating whether or not to put a bias pot in....

most voltages around the amp seem okay EXCEPT:

1) TP26 on the schematic (cathodes of the third pre-amp tube) should read 34 VDC but only reads 22 VDC
2) point Y, off of big filter caps, going to plates of third pre-amp tube and control grid of power tubes read 440 VDC should be 413
3) point X (going to plates of first two pre-amp tubes) reads over 400 VDC and should be only 364 VDC

now i know the high voltages have some leeway, but what about 1) ?? i cant help but feel like the problem is preamp related, seeing as new power tubes that seem to be running at the right bias didnt do much. if so, would a low voltage at preamp cathodes maybe cause the issue? ive swapped around preamp tubes with no luck yet (but i only have one known good one that im switching in and out)

any thoughts?

should i put in a bias pot and try to run the power tubes a bit hotter, see if that fixes it?
Title: Re: blues deville got the blues
Post by: DrGonz78 on March 07, 2014, 06:09:00 AM
As they always have told and I have read before, Divide and Conquer. Don't just guess that new tubes means the output side of the amp is not to blame, but rather put it to the test. Take a preamp out from another amp and plug it in to the power amp in. Test to see if the power amp side is the problem. Then if it is not the problem we have just removed 50% of where to look.

Crackly and static like sounds could be any number of problems related to that symptom. First declare positively that it is the preamp that is to blame.
Title: Re: blues deville got the blues
Post by: Enzo on March 07, 2014, 09:20:27 AM
You like this amp a lot, but lately it sounds wrong.  So NOW you think it needs a redesign?   You liked the way it was before.   Just fix what is wrong with it, don't redesign it.

This amp doesn't have a bias test point, and since it isn;t adjustable, it doesn;t need one, but in the very similar Hot Rod DeVille, the factory spec is 60ma, which is 30ma per tube.  B+ there is 485, not much different.  14 watts.

Besides, bias is not some super-critical thing, being "off" a volt or two will not harm anything or even usually be audible.

Quote
control grid voltage is only at -43 or so - should it be more like -50? im deliberating whether or not to put a bias pot in....

Well do things for the right reasons.   You are concerned that the amp is too cool already.  It isn't, but you can still prefer it hotter.  But now you are proposing to increase bias voltage to 50v from 43v.  Well, that is adjusting it cooler, the opposite of what you are wanting.

This amp is made so most any set of 6L6 tubes will work just plugging them in.

Gonz is correct, first determine where in the amp the problem lies.   Plug the guitar into the Power Amp In jack, still sound funny?   And run a cord from Preamp Out to some other amp and speaker.  SOund OK?  Or still has the issues?

TP26 is not a preamp tube, it is part of the power amp.  That is the cathode of the phase inverter.  Since your voltage there is low, it leads us to think one side of the phase inverter might not be running, thus reducing current through the cathodes.   So look at pins 1 and 6 of that tube and see if B+ is missing from one of them.  Suspecting an open plate resistor.
Title: Re: blues deville got the blues
Post by: Roly on March 07, 2014, 11:36:07 AM
ilyaa there is a logic to fault-finding amps, it's not magic or dumb luck.

"Divide and Conquer".  Is your problem before or after the main volume control (or Fx loop if fitted, which generally amounts to the same thing, splitting the whole amp half-and-half, into pre or main amp)?

The general approach is to first check that the power supplies are basically functional, that the voltages are roughly what they should be, then we proceed by binary division, first half or second half of the signal chain, pre or main, then splitting the faulty half in half, and so on until we come to the faulty stage, then the faulty component.

{assuming that the problem is in the main amp}

You find that the cathode voltage of the phase inverter V3a&b is low.  This means that the current in the cathode bias resistor is low, as Enzo says.

That leads him to think you may have an open anode resistor (R50, R51 or R52).  It leads me to think that the valve itself may be on the way out, possibly losing cathode emission.  It does not lead directly, or even indirectly, to replacing the output valves.  Either fault could lead to the amp sounding odd, but they obviously have quite different remedies.

You measure point Y at 440V instead of 413V.

Now stop for a minute and think what effect a high supply voltage would have on the phase inverter cathode voltage - would you expect it to be higher or lower than spec?

When you find an odd cathode voltage like this you then check the voltage(s) on the anode(s).  Low cathode and high anode means the valves isn't drawing as much current as intended (Ohm's Law x2), and you could perhaps try a new 12AX7; in this case one anode low or very low and Enzo is right, you have an open or high anode resistor.

One anode almost at the HT supply?  Are both sides lighting up?   No?  Then try rocking the valve in its socket to see if one of the heater connections is dodgy; pull the valve and measure the heater resistance of both sides to see if a heater is open, or is it at the socket?


You say the output stage grid voltage is -43V, but the grid voltage isn't specified, the bias voltage is specified at point -C, and they have a good reason for doing it that way.

Your meter has resistance, typically 1 megohm, so it draws current.

If the voltage at -C is -49.6V and you connect a one megohm resistor from one of the output grids to ground, what do you expect to read?

Well there is a 220k and a 1k5 resistor between point -C and each grid, making 221.5k in series.  Assuming your meter has an internal resistance of one megohm it forms a voltage divider of;

1000/(1000+221.5) = 0.81866558

49.6 * 0.819 = 40.6224V

If your mains is 6% high;

40.6224 * 1.06 = 43.059744

Well shucks, -43V.  What a coincidence.

There is a very important lesson here - your test instruments change what is being tested.  Mostly it's negligible, here it isn't.


What do you measure back towards the rectifier, i.e. point Z, and B+/SP1?  Are they high too?

First check your meter against a known voltage (or another meter), cheaper DMM's are not known for their accuracy.

What is the state of the battery in your DMM?  As the battery goes flat the meter loses accuracy.

But how far off is it?  440/413 = 1.0653753 or 6% high.  What is your mains voltage, 'coz a variation of 5% is quite common?

Did you read the Notes?
Particularly;

"LINE VOLTAGE 120VAC 60HZ"

Is it?  Exactly?  If not you have to correct for it.

Then;

"SUPPLY AND BIAS VOLTAGES MAY VARY +5%, -10%"

If you have a few percent error in your meter, a few percent mains voltage high, and take their error band, you could easily read 440V and still be within spec.

Did you have the controls set as specified?

We normally assume (and it is often stated in the notes) that voltages are measured at idle, however since they don't specify it is possible these voltages are under drive conditions which would explain why they are high at idle.

The control grid is grid one, where the signal and bias goes.  The grids that are connected to the HT via resistors are the screen grids.

Have you measured the output voltage at clip into a resistive dummy load and calculated the power?  Does it work out to be within 10% of rated power?

If it is you have demonstrated why professionals don't use "scattergun servicing" techniques.  Replacing a good set of output valves for $100 when it might only be a 5c resistor (or even a $20 12AX7) would either quickly make you go broke, or build a lousy reputation for replacing expensive components and still not fixing the fault complained of.

It's your amp and you're rich so you don't care?  Fine, but the bottom line is that it still ain't fixed.

Methodical, step by boring step, checking everything, and building up a mental picture from your voltage (or CRO) measurements.  When I get stuck I've found that printing out the circuit and writing the voltages found at each point on it can quickly expose what I've been missing.

Unless you are going to fit current sensing resistors in each output valve cathode so you can actually measure the voltage, thereby calculate the valve current, then with the supply voltage calculate the plate dissipation, I agree with DrGonz and Enzo - don't mess with the bias.
Title: Re: blues deville got the blues
Post by: ilyaa on March 08, 2014, 06:51:11 PM
when i take a signal from the preamp out, shouldnt it short out the power amp?

its not doing so - makes it difficult to hear if anything is changing because ive got two amps playing....

Title: Re: blues deville got the blues
Post by: DrGonz78 on March 09, 2014, 01:39:59 AM
Yeah just using the preamp out is not going to shunt off the output speaker. In fact, actually those amps use those connections like it is an effects loop as I recall. Do you have a dummy load that you can attach to the output of the amp to cut out the speaker?
Title: Re: blues deville got the blues
Post by: ilyaa on March 09, 2014, 04:48:28 AM
i dont have a dummy load yet - planning on making one soon.

what if i put a dummy lead into the power amp in? with nothing on the other end - will that be safe? wont that just fool the power amp into thinking its getting no signal in - therefore none out. and then i can take the preamp out and do with it what i wish. is that flawed thinking?
Title: Re: blues deville got the blues
Post by: J M Fahey on March 09, 2014, 05:23:17 AM
Quite close   :cheesy:.
Make a shorted plug and plug it in the Power Amp in, this will mute the Deville power namp.
It will not affect the Pre out.
Title: Re: blues deville got the blues
Post by: Roly on March 09, 2014, 09:49:36 AM
Quote from: ilyaa
when i take a signal from the preamp out, shouldnt it short out the power amp?

Not normally because the Pre Out often gets used as a signal pick off, say to the PA or a recorder, and you would want the whole amp to stay active.

Putting a plug into the Main In however you would typically want the preamp de-selected so the main amp could be used as a slave to another preamp, or the return from your Fx loop, so as JM says simply plug a shorted plug into the Main In and the main amp input will then be disconnected from the preamp due to the switch contact on the Main In socket, and shorted and silenced by your shorted plug while the preamp signal will still be available at Pre Out.
Title: Re: blues deville got the blues
Post by: ilyaa on March 14, 2014, 04:59:26 AM
alright so i did some mixing and matching and have come up with the frustrating conclusion that the problem is maybe in both the pre and the power amp (is there a possibility of it being kind of between the two? the phase inverter??)

IF i take the preamp out and connect it to another amp, i get the same kind of boxy and congested, even grainy sounding break-up.

IF i play my guitar directly into the power amp in, the amp sounds better but still has an element of the problem - a kind of unpleasant breaking up/distortion that verges on crackliness.

basically it seems like more of the preamp acting up, but its hard to be sure that the power amp isnt also a part of the problem, somehow....

additonal detail i forgot to mention: some months ago the amp was doing this weird thing where it would make and sustain a loud hum whenever id hit an F#. it would sustain the note in a loud, hum similar it 60Hz power stuff, but not exactly (it would just hold the F#!). the third preamp tube (phase inverter?) would amplify the noise if i tapped it (microphonic?). i figured it was the problem so i switched it out for a new one. the problem went away BUT strangely enough when i switched the old tube back in (out of curiosity) it did NOT come back, nor was it microphonic anymore.

ill try some tube swapping just to see if that changes anything, but otherwise:

does it make sense to troubleshoot the phase inverter given the symptoms?
Title: Re: blues deville got the blues
Post by: DrGonz78 on March 14, 2014, 06:47:42 AM
Perhaps it's the phase inverter solder joints? Perhaps the whole amp needs some soldering re-flowing. The fact that you have problems on both sides of the amp makes it hard to conquer and divide. On the preamp side could be solder joints too, but I would be checking the coupling caps as well. Make sure you don't measure DC voltage on the pots that follow the coupling caps. Like after C2 on Volume R7 or after C9 Master Vol R21. I am sure other's here will have better advice soon.
Title: Re: blues deville got the blues
Post by: Roly on March 14, 2014, 07:44:09 AM
Quote from: ilyaa
boxy and congested

I asked about a dummy load full power test because output valves, like car tyres, wear out.  "Congested" could be the OP bottles getting tired, but I always test before replacement, so you need to get on to building that dummy load.

(http://i1341.photobucket.com/albums/o743/Roly49/Australian%20Guitar%20Gear%20Heads/29112011003.jpg) (http://s1341.photobucket.com/user/Roly49/media/Australian%20Guitar%20Gear%20Heads/29112011003.jpg.html)
{constructional shots also on Photobucket}


Valve microphony is not the only sort; it is possible for e.g. a valve socket/pin to be mechanically sensitive, but a microphonic valve does not self-cure, it has a clear high tinkly quality when the glass is gently tapped.  Pops and splats are dirty/loose socket connections, not microphony.
Title: Re: blues deville got the blues
Post by: g1 on March 14, 2014, 12:34:40 PM
  Along with resoldering the tube sockets like Dr.Gonz suggested, I would also suggest resoldering the main filter caps and the 2 power resistors in the low voltage power supply.
The power supply is the only thing really common to both the preamp and the power amp, so if indeed the problem is in both sections, the power supply could be the culprit. 
The only other thing common to both is the speaker.
But you said the clean channel is fine.  If the clean channel is fine, then going straight in to the power amp should also be fine.
Do more testing to see if the problem occurs when straight in to the power amp or when using the clean channel.
Title: Re: blues deville got the blues
Post by: ilyaa on March 19, 2014, 04:14:08 AM
alright heres where im at:

did some resoldering - power supply, phase inverter - even put in new preamp tubes and a new phase inverter tube - problem persists.

after some extensive mix/match (including other amps and speakers) testing today, im pretty convinced the problem is in the deville's power amp.

if i play the amp's clean channel, it sounds good, but it still breaks up in a boxy,jagged,unpleasant kind of way. the drive channel, on the other hand, is not only quieter, but also kind of hollowed out (no low end) and much much much more jagged/boxy. not pleasant tube saturation but kind of gnarly, clipping, farty and grainy kind of distortion.

if i plug in a diffeent preamp direct into the deville's power amp, i get the same problem - gnarly, boxy, congested distortion

these are new powers tubes, as well.

testing wise, the only voltage that looks off is the phase inverter cathode voltage - its about 10 volts too low. could that be causing the symptoms? seems like it could, since the phase inverter directly feeds the power amp....

where is the cathode bias on that tube coming from???
Title: Re: blues deville got the blues
Post by: Enzo on March 19, 2014, 10:05:25 AM
Did some resoldering?   Did you specifically resolder all the pins of all the power tube sockets?   Not touch up the ones that looked suspicious, I mean solder every darn one of them?

My schematic says 33v on the PI cathode, and you get 23?  Pins 1 and 6 are the plates, what is on them?   They both ought to have about the same voltage.  What voltage?  I don;t know, somthing like 250-270 comes to mind  But what I don't want to see is zero volts nor do I want to see 400v on those pins.

And when taking readings, did you do the power tubes?  And was there good B+ on pin 4 of each as well as pin 3?
Title: Re: blues deville got the blues
Post by: ilyaa on March 20, 2014, 02:25:27 AM
Quote
Did you specifically resolder all the pins of all the power tube sockets?

i resoldered the phase inverter pins and i THINK i did the power tubes but i dont remember now. i will go over the whole tube board again, though, just in case!

and ill do those readings -

Roly:

Quote
could be the OP bottles getting tired, but I always test before replacement

what do you mean test? is there a way to tell the power tubes are bad other than with your ears? any actual, measurable indicators?
Title: Re: blues deville got the blues
Post by: Roly on March 20, 2014, 12:05:01 PM
Quote from: Roly
could be the OP bottles getting tired, but I always test before replacement
what do you mean test? is there a way to tell the power tubes are bad other than with your ears? any actual, measurable indicators?

Well I have a dummy load with a built-in wattmeter which I find very handy, but a cobbled up bunch of power resistors, voltmeter and calculator to apply the Power Law works too.

P = ERMS2/R

Adjust for the onset of sine clipping and measure the voltage across the load.

Power bottles generally come in two classes, adequate, and clearly stuffed.  If a "60 watt" amp makes 50+ watts I'm happy.

I'll give you a protip; the human ear is a truly lousy judge of output power because it has inbuilt Automatic Gain Control.  I have been asked to look at big valve amps that were still loud but "sounding a bit weak" to the owner that have turned out to be literally making only a watt or three.  This deception can be compounded by preamp gain and speaker efficiency.  The only reliable way to know if a valve amp is making power is to measure it.


Title: Re: blues deville got the blues
Post by: J M Fahey on March 20, 2014, 01:33:59 PM
Agree.
FWIW I have personally measured 100W Twin Reverb, still with their Factory Installed 80's 6L6, and Marshall JCM900, still with their 80's EL34 .... and both were still putting out around 40W RMS.
Complain was: "it does not have "authority" any more, I no longer drown the drummer, something must be wrong, although amp still sounds and plays perfect" .
So there you have a clear example of tubes which needed replacing.
In that case, human ear was acceptable, because it was not "measuring" power (it can't tell 30W one day from 50W other day) but because it was *comparing* , as in "the amp is louder than the drums, played at the same time ... or not"
Similar example to: you can't tell temperature of a bucket of water with your hand with any precision beyond warm/cold/hot/unbearable , but you can quite accuretely *compare* 2 buckets, by immersing your hand back and forth.
Title: Re: blues deville got the blues
Post by: ilyaa on March 20, 2014, 01:46:27 PM
so the procedure is:

plug in dummy load - turn on amp - feed signal (knobs all the way up? halfway up?) - measure voltage across dummy load?
Title: Re: blues deville got the blues
Post by: J M Fahey on March 20, 2014, 02:00:10 PM
*Basically* yes, tone controls on 5, master on 10, feed 100mV 1KHz at input (or 400 or 440 Hz), start raising volume until it starts clipping, then back up alittle.
Scope is best.
If not available, an old trick is to put a piezo tweeter in parallel with load, start of clipping can be heard through it as an annoying buzzy sound, raise/lower volume to zero its start.
Then measure voltage across load and do the math.
Where to get those tones?
Download the MP3 and play through some portable MP3 player or your phone.
Make a headphone>guitar plug cable.
You can join both (stereo) headphone wires into a single (mono) one because signal is mono and headphone outs 99% of the time have a series resistor.
400Hz: http://users.skynet.be/fa046054/home/P22/track55.mp3
1KHz: http://users.skynet.be/fa046054/home/P22/track57.mp3
For the full Monty: http://www.testsounds.com/
Warning: not a virus but they offer a get rich quick or earn money without working scheme, simply ignore it.
Title: Re: blues deville got the blues
Post by: ilyaa on March 21, 2014, 01:21:42 AM
yeah i have a scope to put on there...

am i measuring DC volts with the meter? or p-p volts that the scope is showing? i assume the latter....
Title: Re: blues deville got the blues
Post by: Roly on March 21, 2014, 04:57:17 AM
I normally feed in a fixed signal and use the amp volume to find the clip point, but it amounts to the same thing.

Quote from: ilyaa
am i measuring DC volts with the meter?

ilyaa!!!  What is the amp outputting to the speaker?  {musta bin a late night post I guess  :loco }


Peak to peak, half for peak, divide by root(2) for RMS, P = ERMS/R
Title: Re: blues deville got the blues
Post by: ilyaa on March 21, 2014, 11:51:10 AM
Quote
ilyaa!!!  What is the amp outputting to the speaker?  {musta bin a late night post I guess  :loco }

hehehe....yayaya i was just making sure!

i know: DC on the speak = death
Title: Re: blues deville got the blues
Post by: ilyaa on March 22, 2014, 02:00:32 AM
Quote
a cobbled up bunch of power resistors, voltmeter and calculator to apply the Power Law works too.

as far as a dummy load goes:

can i just wire together a bunch of power resistors in parallel?? if i am just interested in power output.

say i wanted 100W @ 4 ohms, would 10 40 ohm/10W resistors do the trick? do i need to simulate inductance somehow? or is that not crucial to this application?

(and the reason i asked about DC volts across the load is cause you said voltmeter up there ^^.....)
Title: Re: blues deville got the blues
Post by: Roly on March 22, 2014, 09:02:40 AM
Quote from: ilyaa
say i wanted 100W @ 4 ohms, would 10 40 ohm/10W resistors do the trick? do i need to simulate inductance somehow? or is that not crucial to this application?

(and the reason i asked about DC volts across the load is cause you said voltmeter up there ^.....)

Yes.  No.  No.  They will get hot so an old metal computer power supply case (blah blah).

A secure connection is vital for valve amp testing.

"Voltmeter" = Multimeter on AC range.

(http://i1341.photobucket.com/albums/o743/Roly49/Australian%20Guitar%20Gear%20Heads/200Wfan-blowndummyload_zps88a60c59.jpg) (http://s1341.photobucket.com/user/Roly49/media/Australian%20Guitar%20Gear%20Heads/200Wfan-blowndummyload_zps88a60c59.jpg.html)

(http://i1341.photobucket.com/albums/o743/Roly49/Australian%20Guitar%20Gear%20Heads/28112011.jpg) (http://s1341.photobucket.com/user/Roly49/media/Australian%20Guitar%20Gear%20Heads/28112011.jpg.html)

(http://i1341.photobucket.com/albums/o743/Roly49/Australian%20Guitar%20Gear%20Heads/28112011002.jpg) (http://s1341.photobucket.com/user/Roly49/media/Australian%20Guitar%20Gear%20Heads/28112011002.jpg.html)

(http://i1341.photobucket.com/albums/o743/Roly49/Australian%20Guitar%20Gear%20Heads/28112011003.jpg) (http://s1341.photobucket.com/user/Roly49/media/Australian%20Guitar%20Gear%20Heads/28112011003.jpg.html)

(http://i1341.photobucket.com/albums/o743/Roly49/Australian%20Guitar%20Gear%20Heads/29112011.jpg) (http://s1341.photobucket.com/user/Roly49/media/Australian%20Guitar%20Gear%20Heads/29112011.jpg.html)

(http://i1341.photobucket.com/albums/o743/Roly49/Australian%20Guitar%20Gear%20Heads/29112011001.jpg) (http://s1341.photobucket.com/user/Roly49/media/Australian%20Guitar%20Gear%20Heads/29112011001.jpg.html)

(http://i1341.photobucket.com/albums/o743/Roly49/Australian%20Guitar%20Gear%20Heads/29112011003.jpg) (http://s1341.photobucket.com/user/Roly49/media/Australian%20Guitar%20Gear%20Heads/29112011003.jpg.html)

Note: the circuit and this build are only similar, not identical, but I'm sure you will get the idea.

HTH
Title: Re: blues deville got the blues
Post by: J M Fahey on March 22, 2014, 09:03:00 AM
Yes, 4 ohms total impedance, 100W dissipation is fine.
Most (99%) amps are tested with resistive loads.
Inductive/reactive ones, simulating complex impedance shown by real world speakers, are needed to test complex short protection circuits, which may seem to work with resistive loads, and self trigger, with *horrible*  buzzy/farty sound when driving speakers.
Title: Re: blues deville got the blues
Post by: ilyaa on March 22, 2014, 11:11:00 AM
coool - ill make one today.

is the diode section just for powering the fan? ooooo i see is it just stealing a bit of the load's AC, rectifying it, filtering it, and powering the fan?? so clever!
Title: Re: blues deville got the blues
Post by: phatt on March 23, 2014, 06:47:35 AM
Why yes,,,any dummy can build one,,
 but it takes very clever people to build a better dummy load. :P
Phil.
Title: Re: blues deville got the blues
Post by: Roly on March 23, 2014, 10:11:30 AM
...and for my next trick, a linear scale wattmeter using germanium diodes.

(http://www.ozvalveamps.org/optrans/fenderbm40w.jpg)

{after I pull mine apart to see how I did it}
Title: Re: blues deville got the blues
Post by: ilyaa on March 25, 2014, 07:56:03 PM
resoldered power tubes.

the clean channel sounds good, actually, except it gets kinda congested when it starts to break up.

but the drive channel still sounds really bad. hollowed out and totally unclear.

i get ~200VDC on pin 1 of the PI and ~400VDC on pin six. so thats bad. should i check the resistors feeding them, coming from point Y on the schematic?
Title: Re: blues deville got the blues
Post by: g1 on March 25, 2014, 09:01:47 PM
  Sounds like V3A is not conducting.  Are both sides of V3 heaters lit?  Have you resoldered that socket?  Or tried another tube there?
Title: Re: blues deville got the blues
Post by: ilyaa on March 25, 2014, 10:08:41 PM
ive tried another tube(s) and resoldered that socket.

both heaters are lit (im pretty sure - just two little glowing points on either side of the tube near its socket - the two preamp tubes and the PI are all lit up like that - that's the heaters, right?).



Title: Re: blues deville got the blues
Post by: Enzo on March 26, 2014, 05:49:59 AM
If the [late resistors were open, you'd get no voltage there at all.   The fact one side is real high tells us that side is not conducting.  So make sure that cathode is connected, and look in the tube to see that BOTH heaters are running.
Title: Re: blues deville got the blues
Post by: g1 on March 26, 2014, 11:23:36 AM
 Recheck DC volts at PI tube pins 1,3,6, and 8.
As Enzo said, if it was one of the resistors, there should be no voltage at the plate.  But might as well check their resistance, there are 3 of them: R50, R51, & R52.
Title: Re: blues deville got the blues
Post by: ilyaa on March 26, 2014, 02:31:18 PM
resoldered that whole part of the board.

checked voltages:

pin 1 - ~200VDC
pin 3 - ~22VDC
pin 6 - ~400VDC
pin 8 - ~22VDC

the resistances of those plate resistors are looking alright - except that r51 and r52 are both reading around 108K - when r51 should be 91K. is that difference crucial? by the looks of it, someone has changed r51 before. should i remove the board and test them out of circuit? r51 is dropping ~200VDC and r52 drops almost nothing.

Title: Re: blues deville got the blues
Post by: Enzo on March 26, 2014, 10:10:45 PM
The pin 6 side is not conducting.  Did you look at the heaters like I asked to see if BOTH heaters in that tube are running?


This is a guitar amp, none of the values are very critical, those resistor values can be pretty much anything and it will make sound.
Title: Re: blues deville got the blues
Post by: ilyaa on March 26, 2014, 10:41:34 PM
both heaters are running, as far as i can tell.

just two glowing points, one on either side of the tube, right?
Title: Re: blues deville got the blues
Post by: Enzo on March 27, 2014, 08:49:51 AM
yes.   Then I worry that your tube socket is not making contact with pins 6 or 8,
Title: Re: blues deville got the blues
Post by: ilyaa on March 29, 2014, 03:36:26 PM
Enzo!!

my god what a mess - pin 6 of the PI on the tube socket itself is *somehow missing* the actual metal part that the tube pin is supposed to make contact with!

i tried a few rigs but the only one thats worked, i think, is a blob of solder on the tube pin just so it reaches up to the connector.

and, voila, i get 33 v on the cathodes and about 240V on the plates....

i dont want to run the amp like this, but now i know that this is the problem....

now i assume ill just have to change that socket but i really dont want to do that! what a pain. any success stories of rigging something up in a situation like this? or has anyone ever encountered a situation like this?! i tried some metal molex like connectors but its just not cutting it.....
Title: Re: blues deville got the blues
Post by: phatt on March 29, 2014, 10:31:35 PM
I can't remember the type of valve socket in those amps but I'd guess it's pcb mounted but you maybe able to desolder that one pin and replace it.

That of course means finding some other sleeve which will fit. Tricky one :(

Some pcb mount valve sockets are rather weak, due to the simple fact they make the metal thinner every year trying to extract every last cent of profit.
Phil.
Title: Re: blues deville got the blues
Post by: Roly on March 30, 2014, 06:37:50 AM
Can't say for PCB mount, but with conventional sockets the contacts are normally quite removeable, and one scrounged from another (old) socket should get you going.
Title: Re: blues deville got the blues
Post by: Enzo on March 31, 2014, 01:30:07 AM
These legs are kinked, if I recall, and so bending them straight to get them out of one socket and them rebending them once in the new socket is likely to break one.

Quote
Some pcb mount valve sockets are rather weak, due to the simple fact they make the metal thinner every year trying to extract every last cent of profit.


Oh, let's not extrapolate to motives and stuff.   The fact is that the vast majority of these sockets works just fine its entire life.   All was needed to happen was catching the socket pin with a tube pin when installing a tube at some point and snapping the socket pin.


Pull the tubes, pull the screws, flex the ribbons to get the socket board outside the chassis far enough, 9 solder joints cleared and the socket is out.  The reverse that to install.  No big job.
Title: Re: blues deville got the blues
Post by: ilyaa on April 07, 2014, 03:12:50 PM
k all good.

put in a ceramic socket.

all voltages look good.

EDIT/FOLLOW UP: on the blues deville schematic, it says 13.6VAC on the output. wouldnt that only be giving me 20 something watts output power into 8 ohms?? (as per the schematic.....)? yet below that it says 58 watts rms.....can anyone clarify that test point for me?

amp sounding good and loud and punchy again. i think i maybe like the clean channel more now than i did - there is still something about the drive channel that i think i didnt really notice before, but now that im listening to things a bit closer than i used to, i realize the drive is a little bit character-less and even has a blown out sound. cant quite tell if its an actual issue or just the overdrive's natural character....

i might be back asking about that in a bit, but for the time being at least the amp is back to full power and doing what it should!!

thanks!
Title: Re: blues deville got the blues
Post by: Roly on April 08, 2014, 10:38:46 AM
13.6V in 8 ohms

P = E2/R

P = (13.6^2)/8 = 23.12 watts


58 watts in 8 ohms

58 = E2/8

E = root(58 * 8 )

E = (58 * 8 )^0.5 = 21.54VAC

I can't see how the test voltages given can be for full power.  Adjust your 1kHz signal level so that you just get output stage clipping into your 8 ohm dummy load, and you should have (about) 21.5VAC across it (and a very hot load).
Title: Re: blues deville got the blues
Post by: ilyaa on April 08, 2014, 11:34:20 AM
i have a question about this dummy load and DIY dummy load construction in general.

(http://i1341.photobucket.com/albums/o743/Roly49/Australian%20Guitar%20Gear%20Heads/200Wfan-blowndummyload_zps88a60c59.jpg) (http://s1341.photobucket.com/user/Roly49/media/Australian%20Guitar%20Gear%20Heads/200Wfan-blowndummyload_zps88a60c59.jpg.html)


we have ten resistors in parallel in series with ten more resistors in parallel. how can their power dissipation be additive if all of the current flowing through the load has to pass through the first ten resistors before it passes through the second ten? meaning, wont the first ten resistors have to bear the full brunt of the power, anyway? what am i misunderstanding here?

let's say 72 watts going in to 8 ohms (and lets say these are all 40 ohm resistors to make it even and easy)

P = I^2 * R
72 = I^2 * 8
9 = I^2
I = 3A

that 3 amps splits ten ways as it enters the load (im ignoring the rectifier/fan cause my dummy load doesnt have one), so 300mA into each resistor of the first parallel combo:

P = I^2 * R
P = 0.3^2 * 40
P = 3.6 W each resistor
x10 = 36 W dissipated by the first group and the same by the second group
x2 = 72 W

now it makes sense to do it this way rather than just having one row of ten 80 ohms resistors, because in that case each resistor would be burning up 0.3^2 * 80 = 7.2 Watts instead of only 3.6 W.

the math makes sense to me. what doesnt make sense is that the series current (ALL of it) passes through one group and THEN the next. seems like it should ALL be dissipated in the first group seeing as it ALL passes through there. how does the current know there are ten more resistors to go? i guess the relationship here between voltage current resistance and power cannot be quite intuitive....

and say i built a 4 ohm dummy load with one row of ten 25 ohm 10 W resistors and one row of ten 15 ohms resistors and put those two rows in series (strictly hypothetical ;) ) and i wanted to make this into an 8 ohm dummy load and i just added a 10 W 4 ohm resistor in series with both rows, right at the front or the end, that would be a bad idea:

let's say the same 72 watts, so 3A coming in -

that 3A passing through a row of ten 25 ohm resistors would only dissipate 0.3^2 * 25 = 2.25W per resistor, but when it passed through that added 4 ohm resistor it would dissipate 3^2 * 4 = 36W and bye bye resistor and dummy load and maybe OPT....

i guess after doing all this math i understand what's going on pretty well. the current itself only depends on total resistance, but the POWER dissipation depends on total current AND on how that current is divided/passes through the load.....


Title: Re: blues deville got the blues
Post by: g1 on April 08, 2014, 12:26:49 PM
  The 13.6VAC shown on the schematic is what you should get when you use the specified input signal shown at the input jack, with the control settings described in the notes section.  All AC voltages shown are under these same conditions.
  The 58W is the rated output at full power, with what ever signal level required.
Title: Re: blues deville got the blues
Post by: Roly on April 09, 2014, 05:33:51 AM
Quote from: ilyaa
how can their power dissipation be additive if all of the current flowing through the load has to pass through the first ten resistors before it passes through the second ten?

Intuitively; because they are all equal the voltage will divide equally between the top half and the bottom half.

You can simplify this to two resistors in series, the ten in parallel above are 39r/10 = 3.9r, and similarly the bottom ten, so you effectively have two 3.9r resistors in series, 3.9+3.9=7.8 or roughly 8 ohms.  Since they are all equal the current will split equally through all of them, so the main current will split into ten times I/10 resistors, one-tenth through each resistor.

Because the current splits equally they will all equally share the power, and since we have 20 * 10W resistors that gives the whole load a capacity of 200 watts.


I've noticed that you seem to be having a bit of trouble generally with the idea of a voltage divider, so...

A single resistor R across a voltage source E (such as a battery) gives a current I according to Ohms Law;

I = E/R

Now if we divide that resistor into two equal parts the voltage on the mid point will be half the total applied voltage.

Similarly if we make the resistors unequal the voltage will distribute inversely as their resistance.

Say we connect a 1k and 2k resistor in series.  If the 1k is the upper or more positive end the voltage at their join will be 2/3rds of the supply.  If we invert them it will be 1/3rd.

This is similar to connecting a (linear) pot across our battery.  If the pot is set to the mid point the voltage on the wiper will be half of the battery voltage.  If set to 10% then the wiper voltage will be one-tenth of the supply; to 90% it will be 9/10ths.

Now if we have a fixed resistor and a pot (or a valve, or a transistor) in series the voltage on the join will depend on the relative value of the pot to the fixed resistor in series, and if we consider a valve or a transistor to be basically an electronically variable resistor in series with a fixed resistor across a supply, we should then see that the voltage on the join of the resistor and the device will depend on the incoming signal to the control element, grid, base, or gate for a FET.

What is going on in a simple voltage divider, or active amplifier stage, is essentially Ohms Law applied to two resistances in series, the Voltage Divider.

(http://web.mit.edu/rec/www/workshop/voltage-divider.gif)

The voltage across the bottom bit is equal to the bottom bit divided by the whole, what proportion it is.

The output is the R2th bit of the total, R1 + R2;

R2 / (R1 + R2)

...and since this R2th bit of the whole resistance is also the R2th bit of the whole applied voltage the output voltage is the R2th bit of the input voltage;

Vout = Vin * R2/(R1+R2)


So now we tackle a slightly more complicated case;

(http://www.paleotechnologist.net/wp-content/uploads/2012/05/loaded_voltage_divider.jpg)

The first step is to reduce this four resistor network to a two resistor network;

R2 and R3 are in series, so they simply add together to make a single resistor;

6k + 500r = 6.5k

Similarly we combine the parallel pair R1 and Rload to make a single effective resistor.  The long form general case for this is;

1/Rtot = 1/R1 + 1/R2 + 1/R3 + ... 1/Rn

The short form for only two resistors in parallel is;

Rtot = (R1 * R2)/(R1 + R2)

so we can do it either way

1/3500 = 0.00028571

1/10 = 0.1

0.1 + 0.00028571 = 0.10028571

reciprocal
1/0.10028571 = 9.9715104 ohms

{a reality check is that we expect 3.5k in parallel with 10 ohms to be a bit less than 10 ohms, which this is}

Alternatively

(3500*10)/(3500+10) = 9.97150997

{notice that my wordprocessor calculator has given slightly different results in the 5th decimal place, but this is of no practical importance in general electronics.}

So now we have a simple two resistor divider consisting of 6.5k above and 9.9715 ohms below, or 9.9715ths of the whole 6500+9.9715;

9.9715/(6500+9.9715) = 0.00153173

The applied voltage is 10 volts, so the voltage at the output is;

10 * 0.00153173 = 0.0153173, or a bit over 15mV.


There are two laws named after Russian dudes that state (what should be) the bleedin' obvious, and we've just used one of them above, Thévenin's theorem.

Essentially Thévenin's theorem (http://en.wikipedia.org/wiki/Th%C3%A9venin%27s_theorem) says that a circuit of arbitrary complexity consisting of voltages, current sources, and resistances, can be reduced to a single voltage and resistance (which can be very helpful trying to decompose complex circuits).

Norton's theorem (http://en.wikipedia.org/wiki/Norton%27s_theorem) is similar except we end up with a current source and a resistor.


Kirchhoff's current law (KCL) - Kirchhoff's voltage law (KVL).  Kirchhoff's circuit laws (http://en.wikipedia.org/wiki/Kirchhoff%27s_circuit_laws) should be similarly intuitive; that the sum of all currents at a node is zero (the currents flowing out equal the currents flowing in - or you would have an accumulation or deficit of electricity).

The voltage law is equally simple, the sum of all voltages around a loop (circuit) is zero (all the drops equal the supply voltage - or again you would either have a surplus or shortage of electricity).

In both of these cases of course the currents and voltages have signs, so e.g. when the current inflow (+ve) is added to the current outflows (-ve), or the supply voltages (+ve) are added to the voltage drops (-ve) the sums are zero.


Now hopefully you will see that KCL applies to the central node in your dummy load, and that using Thévenin your multiple resistors can be reduced to just two in series, or just one effective resistor of ~8 ohms at 200 watts if that's what you need.

HTH