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Author Topic: blues deville got the blues  (Read 14083 times)

g1

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Re: blues deville got the blues
« Reply #30 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?

ilyaa

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Re: blues deville got the blues
« Reply #31 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?).




Enzo

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Re: blues deville got the blues
« Reply #32 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.

g1

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Re: blues deville got the blues
« Reply #33 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.

ilyaa

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Re: blues deville got the blues
« Reply #34 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.


Enzo

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Re: blues deville got the blues
« Reply #35 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.

ilyaa

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Re: blues deville got the blues
« Reply #36 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?

Enzo

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Re: blues deville got the blues
« Reply #37 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,

ilyaa

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Re: blues deville got the blues
« Reply #38 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.....
« Last Edit: March 29, 2014, 03:42:38 PM by ilyaa »

phatt

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Re: blues deville got the blues
« Reply #39 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.

Roly

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Re: blues deville got the blues
« Reply #40 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.
If you say theory and practice don't agree you haven't applied enough theory.

Enzo

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Re: blues deville got the blues
« Reply #41 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.

ilyaa

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Re: blues deville got the blues
« Reply #42 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!
« Last Edit: April 07, 2014, 04:42:05 PM by ilyaa »

Roly

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Re: blues deville got the blues
« Reply #43 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).
If you say theory and practice don't agree you haven't applied enough theory.

ilyaa

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Re: blues deville got the blues
« Reply #44 on: April 08, 2014, 11:34:20 AM »
i have a question about this dummy load and DIY dummy load construction in general.




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.....