Velleman K4040 is Always Blowing Output Tubes

[KeithC]

I have a 15 year old Velleman 4000 that does the same. It's sat on a shelf for best part of 10 years now, gathering dust.

[Original Burnedfingers]

Is there any way you could post the schematic so we could have a look see?

[jane]

Here it is:

[gingertube]

GeorgeK,
G'day from Adelaide - I'm running a similar amp, home built using Plitron Toroidal Output Tranny and Power Tranny with +485V on the EL34 and biased at 45mA, that is, I'm running them even harder than you are in the K4040. I've had similar tube failure problems from time to time. Most of the EL34s I've lost have been cheap Chinese and JJ's.

I changed the screen resistors on the EL34 to 1K 5W. Do this to your amp, R93, R94, R95 and R96 to 1K 5W. For Ultralinear Mode 180R is just not enough and you may even be getting parasitic oscillation. I originally had 150R screen resistors and DEFINITELY had a oscillation, I saw it on the oscilloscope). If the 5W you get from the local hobby shop won't fit then the 3W Vitreous Enamel Wirewound Welwyn you can order from Farnell will do the trick (Cat No 950-3692).

I finally settled on "Winged C" Svetlana EL34s - not only were these more reliable than the JJs BUT they also sounded better - unfortunately they also cost a fair bit more. So give Arthur a ring (Evatco) and see what he wants for 2 quads.

The problem may also be associated with the large Rg1 values on the EL34s and EL34 grid current - unfortunately fixing that is a larger problem. A better quality tube with less grid (Leakage) current would alleviate that problem to some degree which is why the Winged C Svets might be better. If anyone decides to tackle a fix for that, then check out 12DW7 to see if the "12AU7ish" side is correct to suit the PCB layout for the concertina splitter (12DW7 is like 1/2 a 12AX7 and 1/2 a 12AU7 in one envelope). If so, double the current in the concertina dropping the 47Ks to 22K and halve the 220K and 100K resistors and pots (Oh and double the coupling caps).

Hope this helps,

Cheers,
Ian

[richwalters]

Im a bit suprised running a parallel ul o/p stage without snubbers with high gm tubes such as 34/6550, KT88 class with top flight ouput transformers. I use them especially when high capacitively loaded piezo uppers are used.
My immediate design instinct is to run a square wave through the amp at low sig level and see what the waveform quality is at 10Khz at the o/p tranny sec. I believe the amp uses pcb design, the layout is important for good output stage performance.
Ok my worst fears regarding the 83 but had this amp arrived on my bench I would butchered the phasesplitter and front end for something better....the aim is not to get excess slewing in preceeding stages before the output stage slew limits so it under performs with high distortion.
I'd always thought the 82 would be better than an 83 vis a vis in this application esp for driving a double power stage.

[gingertube]

Rich,
I copied this from a previous post (and edited it) COZ its worth repeating.
The 12AX7 concertina splitter is going to be much better than you might expect (Could certainly be improved thou')

FOR A CONCERTINA SPLITTER:
For equal loads on Anode and cathode

Zout = RL.ra/RL(u+2)+ra

The ra term on the bottom line is insignificant compared to RL(u+2) term so drop it. Then the RL terms top and bottom lines cancel leaving

Zout approx = ra/u+2

At typical values of u (>=20) U+2 approx = u, so simplify again

Zout approx = ra/u = 1/gm

For a 12AX7 that means Zout approx 650 Ohms

If driving output stage directly The equal loads on Anode and Cathode will NOT be guaranteed if:
1) Output stage strays out of Class A (When a tube cuts off it has no gain so Miller capacitance will change, particularly with Triode Mode Output, less so with Ultralinear and less so again in Pentode Mode)
2) Output Stage strays into grid current.


If the Anode load drops significantly then:

Zout cathode = (RL+ra)/(u+2) x ra/RL ; ra/RL term insignigicant so

Zout cathode approx = (RL+ra)/u+2 ;at usual values of u

Zout cathode approx = RL/u + ra/u = RL/u + 1/gm

That is it increases by fixed amount of RL/u


If the cathode load drops sifgnificantly then:

Zout anode = RLxRL(u+1)+RL.ra / RL(u+2)+ra

RL squared (u+1) is much larger than RL.ra and RL(U+2) is much larger than ra so simplify expression to

Zout anode approx = RLxRL(u+1)/RL(u+2)

and at reasonable values of u

Zout anode approx = RL

Summary:
As the loads on Anode and cathode become unbalanced then

Zout anode increases from 1/gm toward RL
Zout cathode increase from 1/gm toward 1/gm +RL/u

Thats why its suggested that low u is better when driving other than Class A Output Stage.

For a 12AX7 that means:
Zout anode increases from about 650R toward 47K
Zout Cathode increases from about 650R toward 1.2K

You can see why a high current (low RL) and low u solution would be better in the unbalanced load condition.

This analysis is based on the Morgan Jones treatment in "Valve Amplifiers" 3rd Ed.

For the K4040 Amp I would be increasing the output tube idle current to at least 50mA to give more Class A Power before the CLass AB transistion for just this reason.

By snubbers I take it you mean zobel networks between Anode and Screen connections on the Output Tranny???

Cheers,
Ian

[jane]

If you look at the concertina splitter on the K4040 schematic you will discover that it is not biased, the grid is connected to same level as the cathode. I have always been wondering if that is intentionally or a mistake by the designer.

[richwalters]

Ian, aes you are quite right. When I design around a concertina phase spl' I always use a buffer driver stage to drive o/p tubes. Williamson did it and so did the others. That takes care of the HF misbalance which the concertina has. My favourite is an ECL82 penny connected as triode in common cathode config and scrap the triode section. The load lines don't look ideal but all features, gain, noise perfom, thd, freq response and output Z are excellent.
As to the 4040, To drive parallel o/p stage with 415V B+ the driving voltage (g1) on each p-p stage grid would be around 35-40V that's too much for a ECC83 driver and without enough low Z. It strikes me that the output stage is actually controlling the phasesplitter !
C19,20,21,22 do seem very low in value.
I ought to post my design . It's in old microsoft designer so will have to manipulate it into another program.
:-more later
richj

[richwalters]

Ian# Gingertube,

Circuit dia of 30/ 150 watt power amp. Power supply uses Switchmode /PFC not shown.
Graph of thd 400Hz performance.
I can recommend this amp although the 7199 is getting on the rare list
The circuit is simple but amps of this class there is alot of safety watchdogs and warm up procedures thrown in.
Generalised;
s/n -70dB down full o/p
no measurable switchmode interference.


richj

[gingertube]

Jane,
Actually the K4040 concertina is biased - grid leak bias via R23 (the 820K).

richj,
Circuit Diagram?

Cheers,
Ian