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Author Topic: JMP 2100 not making power  (Read 7353 times)

Roly

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Re: JMP 2100 not making power
« Reply #15 on: April 02, 2014, 05:52:48 AM »
Quote from: ilyaa
should i trust the selector? if not, can i measure resistance on output transformer leads to verify?

a) Yes.

b) mmmm doubtful.  The OPT is like a gearbox, and the speaker is the load on that gearbox, the "torque" of the load being "reflected" by the ratio back to the OP valve anodes.  The resistance you measure on the OPT secondary is just the resistance of the winding.  This can be useful for sorting out unmarked windings because a higher impedance winding will have more turns, more wire, and thus more resistance, but the resistance doesn't tell you anything about the ratio of turns between the primary and secondary, only respective steady-state mid-band AC voltage measurements between primary and secondary can do that.

The turns ratio of a transformer is the same as the voltage ratio.

The impedance ratio is the square of the turns or voltage ratio.

Thus a transformer with a turns ratio of 4:1 will also have a voltage ratio of 4:1 but an impedance ratio of 4 x 4 = 16:1.

To a first order approximation; a transformer doesn't have an inherent impedance, what you see on one side is the impedance connected to the other side "transformed" by the transformer impedance ratio.

Typical conditions for 6CA7/EL34 are;
54W out, 3.5k plate-plate, typical load ZL = 8 ohms, meaning an impedance ratio Zp/Zs of 437.5:1, from a turns/voltage ratio Vp/Vs of 20.9:1.


Here's some stuff to paste in your hat about Voltage Followers

In voltage follower stages, valve/FET/transistor/op-amp, the input is to the grid/gate/base/+input and the output is from the cathode/source/emitter/output, the anode/drain/collector generally being connected directly to the supply.

The follower has unity gain (strictly slightly less, ~ x0.999).

The input impedance is high, typically megohms.

The output impedance is low, typically ohms.

The function of a follower is to act as a buffer to isolate a high impedance source from a low impedance load.  You will find voltage followers driving tonestacks, as here, and in Line Out stages to drive long high capacitance cables without suffering treble loss.
If you say theory and practice don't agree you haven't applied enough theory.

ilyaa

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Re: JMP 2100 not making power
« Reply #16 on: April 02, 2014, 11:46:03 AM »
Quote
Typical conditions for 6CA7/EL34 are;
54W out, 3.5k plate-plate, typical load ZL = 8 ohms, meaning an impedance ratio Zp/Zs of 437.5:1, from a turns/voltage ratio Vp/Vs of 20.9:1.

where do you get the 3.5k?

Quote
respective steady-state mid-band AC voltage measurements between primary and secondary can do that.

whats that?

as for voltage followers, i do understand the role/function of a buffer. what i dont understand, is why my measurements seem to give me gain across the cathode follower.  1 V p-p in (from anode of the first triode) and 30 V p-p out (off cathode of the cathode follower)
« Last Edit: April 02, 2014, 11:47:22 AM by ilyaa »

Roly

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Re: JMP 2100 not making power
« Reply #17 on: April 02, 2014, 01:46:46 PM »
Quote from: ilyaa
where do you get the 3.5k?

It is a typical figure for 6CA7/EL34's derived by drawing a load line on the anode characteristic curve of the valve in question, between the maximum current point on the Y-axis to the maximum voltage point on the X-axis, the slope of which represents a particular load value.  It's one of the many sets of conditions that the valve manufacturer helpfully calculates for us, and which the vast majority of users follow.

This is a typical load line (which is for a pair of EL84's);



The red line is the load line running from a supply of 288V x2 = 576V (actually drawn 285V and 550V) to a maximum cathode current of 90mA, the slope being for 3.2k (per side) which is a quarter of the final 12.6kplate-to-plate, the required bias being -10.5V.

The green lines show the idle voltage and current conditions and just intersect with the maximum anode dissipation curve of 12 watts (which you normally have to plot yourself).

This is a load line for a classic 12AX7 triode stage;



Here the load line is in blue and runs between the supply voltage of 300V at (A, valve cut off) and the maximum valve current of 3mA at (B, valve saturated), the slope representing an anode load of 100k.  The Q or idle point is a little above half the HT supply voltage.  Illustrated is a 1V change in grid voltage from -1V to -2V producing a change of 0.6mA in the valve current, resulting in a 60V change in the voltage across the 100k load.

Note the different general shape of pentode vs. triode curves.


Quote from: Roly
steady-state mid-band AC voltage measurements between primary and secondary

Steady-state; a continuous sine wave of unchanging amplitude.

mid-band; away from the natural high and low frequency roll off of the transformer, typically 1kHz or 440Hz.

AC voltage measurement; with suitable AC voltmeter or CRO.

between primary and secondary; the ratio of the voltages measured on each side of the transformer, Vprimary/Vsecondary = voltage ratio = turns ratio = the square root of the impedance ratio, i.e. square the voltage ratio to get the impedance ratio.

E.g.
If a transformer is known to have a voltage ratio of 21:1 then the impedance ratio will be;

21 * 21 = 441:1

If we connect an 8 ohm load to the secondary then the impedance across the whole of the primary (plate-to-plate) will be;

8 * 441 = 3528 or 3.528kp-p


Quote from: ilyaa
what i dont understand, is why my measurements seem to give me gain across the cathode follower.  1 V p-p in (from anode of the first triode) and 30 V p-p out (off cathode of the cathode follower)

Well since that is actually impossible there must be another reason.

The stage before should provide a gain of about x30 between its grid and its anode (which is tied to the grid of the follower), so it's Sydney to a brick that what you think is the preceding stage anode is actually its grid that you are looking at.  {Or perhaps you are the first person to break Ohm's Law ...

 um ...   ???

 no, you have confused which pin you were probing.}


ed typo
« Last Edit: April 02, 2014, 11:10:06 PM by Roly »
If you say theory and practice don't agree you haven't applied enough theory.

J M Fahey

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Re: JMP 2100 not making power
« Reply #18 on: April 02, 2014, 08:23:02 PM »
To make sure, read first AC signal at the cathode follower's own cathode, then measure signal at its own grid (not the other triode's plate).
Use a high impedance, AC coupled meter, such as a scope or a good multimeter .

g1

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Re: JMP 2100 not making power
« Reply #19 on: April 03, 2014, 12:16:34 PM »
Agree with both above.  Looking back at how you called out the stages, you have to look at V2 as two stages, not one.

It appears you have sufficient drive at the power tubes and reasonable output power.  New power tubes properly biased will probably get you a bit more power out which should be in the ballpark of normal for this model.

One thing to note is that power tube grids can't be driven beyond the bias voltage, they will just clip the signal.  So for example if your bias is -30V, anything beyond 60Vp-p at the power tube grids will get clipped, even if the PI is capable of more output level (like you saw with power tubes removed).
« Last Edit: April 03, 2014, 12:25:33 PM by g1 »

ilyaa

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Re: JMP 2100 not making power
« Reply #20 on: April 03, 2014, 01:20:38 PM »
oh so in the first half of V2, that anode goes to the grid in the second half of V2? is that right? so there is voltage gain there to feed the cathode follower?

didnt realize that tube was two stages.

J M Fahey

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Re: JMP 2100 not making power
« Reply #21 on: April 03, 2014, 02:16:43 PM »
Yes, first triode supplies voltage gain, second buffers that output.
ThereĀ“s your 30X gain which sounds very reasonable.

ilyaa

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Re: JMP 2100 not making power
« Reply #22 on: April 07, 2014, 02:49:03 PM »
alright, we're good to go:

there had been a mysterious knob on the back panel of the chassis that i was just ignoring - that was the presence knob, rewired just like the schematic.

i removed the weird master volume mod and rewired the normal presence knob back to stock (left that extra knob on the back disconnected).

got some 6550s and put them in and biased them (they biased right no problem). one of the screen grid resistors blew up while i was burning them in, but i replaced it and things seem to be fine. thinking it was just old and ready to go.....

everything else seemed to be fine. amp sounds great! loud and really detailed sounding, with all kinds of surprising harmonics that come through with dynamic playing. especially once you pass a sweet spot on the volume and things start to break up. it was putting out about 13-14 Vrms across 4 ohms (so almost 50W) - i think we're all good!

thanks, guys - another amp back to life. and i learned some stuff!


Roly

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Re: JMP 2100 not making power
« Reply #23 on: April 08, 2014, 10:20:28 AM »
Not thrilled about the screen resistor burning up - that is not a good sign.

This time you encounter a different aspect of guitar amp servicing, The Mad Modder.  Many times we stand there looking at some modded amp chassis, scratching our heads and wondering "What WERE they thinking?".  Sometimes you can get a clue as to what they had in mind, wrong though it may have been, but other times it just plumb evades you.  Generally the safest thing is to go back to original.

An oldie, but an extreme goodie;
https://www.youtube.com/watch?feature=player_embedded&v=yDJibf8uZnQ
If you say theory and practice don't agree you haven't applied enough theory.

ilyaa

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Re: JMP 2100 not making power
« Reply #24 on: April 08, 2014, 10:52:59 AM »
i think it may have burned up because i flipped the amp on without waiting for it to warm up when i was burning the tubes in - could the surging power/current have done the resistor in?

Roly

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Re: JMP 2100 not making power
« Reply #25 on: April 09, 2014, 04:00:40 AM »
What "surging current" @ilyaa?  If you plan on making a hobby/career out of repairing amplifiers you will have to get coldly calculating and seriously rational about your circuit analysis.


Let's go back to some basics.

How does conduction occur in a valve?  Electrons, right?

And where do these electrons come from?  The heated cathode, right?

How many electrons come off a cathode that is cold?  None, right?

So how could any current, surge or otherwise, occur before the cathode has heated up?

The simple fact is that it can't, and that you are another victim of the widespread "standby switch mythology".

Since I'm inherently lazy, you will find one of my demolition jobs on standby switching mythology here;

http://www.guitargear.net.au/discussion/index.php?topic=40125.msg442774#msg442774

If you give it a bit of thought you might conclude that standby switching actually causes "surges"!

In a typical guitar amp the standby switch opens the HT rail somewhere, and the result is that all the bypass electros downstream from there are discharged through bleeders or hot preamp valves.  Now when the standby switch is closed again all this capacitance has to charge up, and the result is a surge current through the power supply, which in general isn't good for anything (least of all the standby switch itself).

Cold valve with HT applied from a solid-state rectifier.  No heat, no electrons, no conduction.  But as the cathode slowly heats up so the electron emission slowly starts and the valve slowly comes into conduction.

Pre-heated valve, up to temperature with the standby switch open (as widely advised).  Lots of electrons, lots of potential conduction.  So then you close the standby switch and suddenly apply peak HT volts from the supply to the output valves, and WALLOP! a surge of current goes through the output valves until the power supply is discharged back down to its loaded working voltage.

Now which situation do you think might be more damaging for the valves?

Well it's a trick question because valves are well able to cope with either situation.

If you examine the history of guitar amps you discover that many of the early circuit arrangements were lifted from Ham radio AM modulators, and that these often had standby switches to idle them when the Ham station was in receive mode.

This obviously doesn't apply to guitar amps, but the standby switch seems to have made the transition in spite of that, and its only apparent function being to disable the amp between sets during breaks without disturbing the amp controls.

For a lot of people (who would like to impress with their ...ahem... grasp of the technical) this alone wasn't enough, so a whole bunch of Old Wives Tales were born to make the standby switch more "techie" than it actually is.

So when you hear these myths, just remember that the valve guitar amp is just about the only valve appliance that ever had a standby switch, and that you will look in vain at millions of mantle radios, radiograms, tape recorders, b/w TV's, &c&c, and never see a standby switch.
If you say theory and practice don't agree you haven't applied enough theory.

ilyaa

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Re: JMP 2100 not making power
« Reply #26 on: April 10, 2014, 03:03:34 AM »
first of all, that mod video is insane. i can hardly believe its real.....

what might have burned the screen resistor? normally is that a sign of a bad tube, pulling too much current?

Roly, what you say about standby switching makes a lot of sense - I never thought about it that way before. I read some of the other links and see that the standby myth has plenty of debunkers.

a related question: what about the standby switch and cap DIScharging? with that sound city ive been working on, i see that if i turn the standby switch off (open) before i switch the amp off the caps drain quite slowly. BUT if i switch the amp off and leave standby on (closed) they drain much more quickly. i guess thats because with the standby switch closed, the caps can drain through all kinds of stuff throughout the amp, but with it open, they can only drain through those rather large bleeder resistors. ive also noticed though, that if i leave it closed when i turn the amp off and let the caps drain, they start to charge back up if i open it again (even if they have drained to 0V....)....whats charging them, in this case??

Roly

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Re: JMP 2100 not making power
« Reply #27 on: April 10, 2014, 01:27:48 PM »
There are a limited number of ways of being right or sane, but there are an unlimited number of ways of being wrong or insane.  If you go on fixing amps the chances are one day you will encounter something that tops this.

I once had somebody bring me an amp kit they had built.  Generally techs try and avoid new builds like this because any and everything could be wrong with something that has never worked.  At least with an amp that has been working it's generally only one thing wrong.

Well this lady didn't know how to solder, so she had used rubber cement for all the joints.  After I got over my initial shock and took a close look she had followed the instructions carefully and actually made quite a neat job of it.  In the end I took the chassis outside, fired up a high powered soldering iron, and simply soldered through all the glue.

"Stinks like hell, the neighbours complain,
I don't give a hoot what you say,
I don't give a hoot what you say"

http://www.youtube.com/watch?v=eIVOt4fbs0U

With all the joints properly soldered it worked just fine first time.


Fried screen resistors aren't all that common except around 6CA7/EL34's were the screens are prone to collapse and short to the cathode.  6550's aren't terribly common here and I've never had any particular problem with them, so I can't say.  The only guess I can make for a self-curing fault like this is perhaps the anode connection in the socket was a bit dodgy and went open, but that's very much a shot in the dark.  A resistor may go open by itself, but it won't cook and burn up by itself.  Maybe one of the other guys with more experience of 6550's will comment.


Yes, your observation about cap draining time is correct; when the hot valves are left connected they will also drag the supply down.

And now you make the astute observation that caps can regain charge after fully discharging.  There is some debate about the exact mechanism but it is likely to be related to dielectric relaxation.

Basically when you discharge a capacitor to zero volts you don't actually remove all the stored charge, some remains trapped within the insulating layer.  If the cap is then un-shorted this charge will appear as a small voltage gain as it leaks out of the dielectric to the conductive plates on its surface.  The degree to which this happens depends of the dielectric material and where accuracy of stored voltage is important, such as in instruments, then high quality caps with specially formulated dielectrics that exhibit minimum "dielectric relaxation".  Electrolytic caps are built to store the most charge in the least space so normally have the greatest dielectric relaxation effect.

This is why I leave my discharge lamp clipped on when I'm working on an amp that has just been powered up.  The low resistance of the lamp will just let the residual charge leak away, and if I should happen to forget to remove it it will light up to remind me with no harm done.

Discussion here;
http://blog.shuningbian.net/2005/04/self-charging-behaviour-of-electrolyte.php
If you say theory and practice don't agree you haven't applied enough theory.