Single Ended Parallel Lead - review my planned overhaul

[martin manning]

The 0.707 converts peak voltage to RMS, which is used to calculate power. Refer to the Valve Wizard link for details. Notice that with the assumptions I made (center biased, and plate voltage swinging all the way down to 0V) the output power will be exactly half of the max plate dissipation. In reality it will be a bit less.

[GlideOn]

The 0.707 converts peak voltage to RMS, which is used to calculate power. Refer to the Valve Wizard link for details. Notice that with the assumptions I made (center biased, and plate voltage swinging all the way down to 0V) the output power will be exactly half of the max plate dissipation. In reality it will be a bit less.

I took the past day and read through the Merlin pages on the Single Ended section and while it was not at all an easy read (he like to write ahead of himself and then explain later in), I'm actually feeling pretty reassured actually I'm on the right path(s). I saw the formulas for loads and voltages.

While I didn't see anywhere the 0.707 figure was written, I'll have to take your word for it. I'm of the understanding that the 2x EL34 arrangement will make anywhere from 21-25w, depending on bias and load.

Another insight I picked up on was that the load is not only based on the Anode voltage and the rated power of the output tube.

I will actually be using a pair of Shuguang EL34Bs - a 30w tube, technically. And they are a "little lower impedance," which from a few quick calculations, rings true:

"Z = Va^2 / Pa
Where:
Va = Anode voltage.
Pa = Maximum anode dissipation."


And we also know to get the B+ (using full diode rectification)

We use the 300-0-300 config of the 600v/330a AnTek PT.

300 x 1.4 = 420v.




However let's call this 415v as 5H choke is being used, dropping 5v.

It might be even less considering there's also a 1.2k/3w dropper after the first filter cap in the B+ which will only help impedance matching really. But we'll stick with this for now.



Using the Merlin formula, input the following:

415^2 / 30w

= 5740.56

Halve that for parallel load:

= 2870, roughly


The 2500r/2.5k primary impedance of the AnTek OT is pretty damn close to that.

Another note from Merlin:

"A higher impedance or lower supply voltage would allow us to bias deeper into class A (i.e., hotter), for more second harmonic distortion. Using a slightly lower impedance would force us to bias colder, which tends towards a more 'raw' overdriven tone."

So this will mean a slight tonal change from the 5k OT and 4100r load before from using single 6550 tube.

So...should I rethink my Cathode Bias resistors?

I selected 330r simply due to the fact I noticed it on nearly every single or parallel EL34 designs. Lots of these were using 2.5k OTs and mid to high 300s voltage

Should I go higher value like 470r or 530r?


Edit: the original schematic lists the Cathode Resistor @ 430r and voltage across is 30.5v with the 346v pate-to-cathode.

[Helmholtz]

Without a schematic this is all guesswork, but..
Your PT might give a loaded B+ around 400V.
Biasing for 100% PD means an idle current around 65mA per tube.
This requires a cathode voltage of roughly 16V.
Means a common cathode resistor of 125R for 2 parallel tubes.

[GlideOn]

Not sure if this will help, but here's original schematic:

AX84_SEL_100616 (1)_240226_185735_2.jpg

[GlideOn]

Without a schematic this is all guesswork, but..
Your PT might give a loaded B+ around 400V.
Biasing for 100% PD means an idle current around 65mA per tube.
This requires a cathode voltage of roughly 16V.
Means a common cathode resistor of 125R for 2 parallel tubes.

Maybe not 125r shared, but certainly 250r as individual cathode resistors!

The advice I got in TDPRI was to have individual cathodes so as to prevent bias and thermal runaway issues.

However, the lower Cathodes you are calculating only work if there is a significant voltage drop to the screens.

As in the schematic I posted above, the B+ has a main 1k/2w dropper, followed by a 5k/5w Screen Dropper, followed by individual Screen Grid Resistors of 1k/5w (will be each).

I'll have a calculated 415v off the B+ (300 x 1.4 formula), but real-world may be around high 300s, low 400s after the 1.2k dropper, losses for inconsistencies, heat, etc.

I can certainly play around with those to get maximum power, (though they are pretty minimal to begin with), but I'd like to know exactly how we arrived at the 250r Cathode and voltage drops, because that is where I struggle to see the math adding up.

Keep in mind too that PT voltage is going up and OT load is getting halved from 5k single to 2.5k to accommodate the parallel load.

[Bergheim]

... I'd like to know exactly how we arrived at the 250r Cathode and voltage drops, because that is where I struggle to see the math adding up.

I didn't find a datasheet for the EL34B, so I'm using the 25W version as an example.
Also, I'm using this awesome site for easy load line work: https://www.vtadiy.com/loadline-calcula ... calculator
For this example I'm assuming 415V anode, 410V screen and 5000R output impedance (single tube).
To dissipate 100% (25W) at 415V, anode current has to be 25W / 415V = 60mA.
According to the calculated load line, the tube will pass 60mA with a grid voltage (bias) of about -33V in relation to the cathode. Since the input grid is grounded (0V) via the grid leak, the voltage at the cathode can instead be raised to +33V via a resistor in series with the cathode (maybe you know this stuff very well already, not trying to be rude). Lets say grid current is 10% of the anode current in a pentode, current through the cathode will be another 6mA, so the total current through the cathode is 66mA. The resistance that is needed to raise the cathode is therefore 33V / 66mA = 500R.

Note that plate voltage is measured from plate to cathode, so in this case actual plate voltage is 415 - 33 = 382 (same for the screen grid), so the tube will probably dissipate around 80-90% instead of the calculated 100%.

To make Helmholtz' numbers work out (-16V grid and 400V plate), the screen has to be as low as 250V to stay within 100% dissipation. Don't know if that was intentional, but I suspect he's using the same site I linked to and forgot to change the screen voltage, as the default screen voltage is 250V

[Helmholtz]

To make Helmholtz' numbers work out (-16V grid and 400V plate), the screen has to be as low as 250V to stay within 100% dissipation. Don't know if that was intentional, but I suspect he's using the same site I linked to and forgot to change the screen voltage, as the default screen voltage is 250V

You're right, I used the vtadiy calculator with the default 250V screen voltage.
Reason was, I hadn't seen a schematic and thought the OP would follow Martin Manning's advice to use a low screen voltage.
Now looking at the screen supply wiring of the schematic (assuming R6 is shared by both tubes) and using a screen current of about 9mA per tube (from the EL34 datasheet it's 15% of plate current), the screen voltage should be around 100V lower than Vak, i.e. about 280V.
From this the cathode resistor calculates as ~260R per EL34 for a cathode voltage of 19V.

[GlideOn]

... I'd like to know exactly how we arrived at the 250r Cathode and voltage drops, because that is where I struggle to see the math adding up.

I didn't find a datasheet for the EL34B, so I'm using the 25W version as an example.
Also, I'm using this awesome site for easy load line work: https://www.vtadiy.com/loadline-calcula ... calculator
For this example I'm assuming 415V anode, 410V screen and 5000R output impedance (single tube).
To dissipate 100% (25W) at 415V, anode current has to be 25W / 415V = 60mA.
According to the calculated load line, the tube will pass 60mA with a grid voltage (bias) of about -33V in relation to the cathode. Since the input grid is grounded (0V) via the grid leak, the voltage at the cathode can instead be raised to +33V via a resistor in series with the cathode (maybe you know this stuff very well already, not trying to be rude). Lets say grid current is 10% of the anode current in a pentode, current through the cathode will be another 6mA, so the total current through the cathode is 66mA. The resistance that is needed to raise the cathode is therefore 33V / 66mA = 500R.

Note that plate voltage is measured from plate to cathode, so in this case actual plate voltage is 415 - 33 = 382 (same for the screen grid), so the tube will probably dissipate around 80-90% instead of the calculated 100%.

To make Helmholtz' numbers work out (-16V grid and 400V plate), the screen has to be as low as 250V to stay within 100% dissipation. Don't know if that was intentional, but I suspect he's using the same site I linked to and forgot to change the screen voltage, as the default screen voltage is 250V

No not all all rude In fact quite the contrary as you are respectfully reinforcing my points by offering knowledge both redundant and new, but that is precisely the approach I need, for better or worse a little "hand-holding" to carefully analyze each step needing taken.

I admit there's many holes in my knowledge when it comes to this calculation stuff, lots of "it is known" - isms, but post like these are a revelation and you have no idea how helpful this is to read, so I thank you tons!

I kinda learned to run before I could walk with these seemingly elementary things and Merlin is...well, he is, and I am not.

One of those holes is having for example is having no idea the grid input was 0v referenced as that would indicate ground, but now I see it plain as day, so obviously linked to the Master Volume circuit in this amp whose' lug is grounded. Knowledge filled.

And the 10% rule for grid current is useful for getting the Cathode current, which is taken from the 60-65ma current needed to drive a 25w or 30w tube.

But for the sake of this current topic, we need to know the Screen Voltage, or better put choose the appropriate Screen Voltage for best performance.

You're right, I used the vtadiy calculator with the default 250V screen voltage.
Reason was, I hadn't seen a schematic and thought the OP would follow Martin Manning's advice to use a low screen voltage.
Now looking at the screen supply wiring of the schematic (assuming R6 is shared by both tubes) and using a screen current of about 9mA per tube (from the EL34 datasheet it's 15% of plate current), the screen voltage should be around 100V lower than Vak, i.e. about 280V.
From this the cathode resistor calculates as ~260R per EL34 for a cathode voltage of 19V.

I don't know how or why 250v is the default value for screens, but there's already a 5k/5w dropper coming off the B+1 right after the 1.2k dropper and choke. Using the 10% formula of the 60(ish)ma, were getting anywhere from 30-33v dropping, certainly not 150v + which would call for comically large wattage resistor and value if that were the case. I think the max Screen for EL34 is anywhere between 425v-500v so we should be fine, no?

Now with the screen voltage drop reciprocal of the cathode according to Merlin, we now have a Cathode near 500r as @Bergheim calculated.

Selecting which practical value resistors I should use now depends on the output tube and you guys are right, there's no input for an EL34B 30w which is surprising as they've been around for at least two decades now.

I've used the VTA calculator and while the graph is clear and legible, I find it frustrating as to why it doesn't allow you to manually input things like dropping resistors, cathode resistors all at once. It's bizarrely great and limiting at same time.

For giggles, input a 30w I do know of - the 6L6GC - and the results were a healthy raise in power with 66ma and a lower Cathode R to boot:

Screenshot_20240805-222831.png

The Shuguang tube does have indeed have a spec sheet as listed on this site and gives a clear indication listed as a "30w, tube," but hilariously the actual spec of max P is 25w...perhaps they typo'd but it doesn't inspire lots of confidence

http://www.analogmetric.com/goods.php?id=24

TungSol is another 30w B variant, there are also JJ/Tesla EL34L 30w, TAD EL34M 23w...plenty of EL34s not confirming to the 25w rating or impedance.

I have used TungSol EL34B and Shuguang/Ruby/Valve Art EL34B for over 10+ years now, not a single problem with any of them, slightly preferring the TungSol if want to record with them, but Shuguang being $20 and TungSol being $43, the beer money tube wins out for gigging.

So more or less I am using those Shuguangs with this Class A amp that will likely eat through them like candy.

Maybe I can get away with a lower Cathode R, considering they are "maybe" a 30w and lower load requirement.

Maybe I should error on side caution, go with the higher value?

[Helmholtz]

I don't know how or why 250v is the default value for screens, but there's already a 5k/5w dropper coming off the B+1 right after the 1.2k dropper and choke. Using the 10% formula of the 60(ish)ma, were getting anywhere from 30-33v dropping, certainly not 150v + which would call for comically large wattage resistor and value if that were the case.

I can't see a flaw in my calculation.
Actual screen current is best found from the EL34 datasheet.
This gives a screen current to plate current ratio of 14.9%.
With an assumed plate current of 65mA this means a screen current of 9.7mA per tube or 19.4mA total.
The 5k dropper will drop 97V and the individual 1k screen resistors will drop another 9.7V.
So screen voltage will be more than 100V less than B+.

Using the calculator, both plate and screen voltage should be referenced to the cathode potential of maybe 20V.

With a transformer voltage of 300V - 0V - 300V the loaded B+ should be a bit below 400Vdc (Merlin gives a ratio of 1.3 between Vdc and Vac). Actual value depends on PT regulation and filter capacitance.

From these results I used a B+ of 380V and a screen voltage of 280V with the calculator.
Also don't forget that the screen current increases the voltage drop across the cathode resistor.
This is something the calculator doesn't take care of.

I can't see a 1.2k dropper or a choke in your schematic. If you changed the supply wiring, things will be different.

[GlideOn]

I don't know how or why 250v is the default value for screens, but there's already a 5k/5w dropper coming off the B+1 right after the 1.2k dropper and choke. Using the 10% formula of the 60(ish)ma, were getting anywhere from 30-33v dropping, certainly not 150v + which would call for comically large wattage resistor and value if that were the case.

I can't see a flaw in my calculation.
Actual screen current is best found from the EL34 datasheet.
This gives a screen current to plate current ratio of 14.9%.
With an assumed plate current of 65mA this means a screen current of 9.7mA per tube or 19.4mA total.
The 5k dropper will drop 97V and the individual 1k screen resistors will drop another 9.7V.
So screen voltage will be more than 100V less than B+.

Using the calculator, both plate and screen voltage should be referenced to the cathode potential of maybe 20V.

With a transformer voltage of 300V - 0V - 300V the loaded B+ should be a bit below 400Vdc (Merlin gives a ratio of 1.3 between Vdc and Vac). Actual value depends on PT regulation and filter capacitance.

From these results I used a B+ of 380V and a screen voltage of 280V with the calculator.
Also don't forget that the screen current increases the voltage drop across the cathode resistor.
This is something the calculator doesn't take care of.

I can't see a 1.2k dropper or a choke in your schematic. If you changed the supply wiring, things will be different.

Id trust your math over mine, but be careful about solely referencing the original schematic as there will be some changes (I don't have an updated schematic, only layout). Allow me clarify the changes:


- 5H Choke added subbing the 100R after Mains Filter (5-6v drop expected per manufacturer).

- 1.2k/2w dropper unchanged. After the B+1

- 5k/5w dedicated Screen Dropper resistor, unchanged.

- 1k/5w Screen Grid dropper(s)pins 6 to 4 [will be each tube]


Point taken, but remind me again what the exact math was for determining the voltage drop of the 5k?

I know P = I x R...but, having the Anodes, Screens and Output Transformer connections shared/paralleled messes with my head a bit. Are we taking the combined current of the 2X EL34 now or just the individual?

[Helmholtz]

Allow me clarify the changes
- 5H Choke added subbing the 100R after Mains Filter (5-6v drop expected per manufacturer).
- 1.2k/2w dropper unchanged. After the B+1.
- 5k/5w dedicated Screen Dropper resistor, unchanged.
- 1k/5w Screen Grid dropper(s)pins 6 to 4 [will be each tube]

To be sure I'd like to see a drawing of the modified power supply.
Remember that any series DCR before the B+1 node will drop your voltages by 0.15V per Ohm.
So the 100R alone drops 15V.
This is assuming a total DC current draw of 150mA (plates, screens and preamp).

Point taken, but remind me again what the exact math was for determining the voltage drop of the 5k?
I know P = I x R...but, having the Anodes, Screens and Output Transformer connections shared/paralleled messes with my head a bit. Are we taking the combined current of the 2X EL34 now or just the individual?

I'm assuming that a single, common 5k screen dropper is used for both tubes, so the screen currents add.
The individual 1k screen resistors only see the screen current of a single tube.
(V= I x R, P= I² x R )

[GlideOn]

Allow me clarify the changes
- 5H Choke added subbing the 100R after Mains Filter (5-6v drop expected per manufacturer).
- 1.2k/2w dropper unchanged. After the B+1.
- 5k/5w dedicated Screen Dropper resistor, unchanged.
- 1k/5w Screen Grid dropper(s)pins 6 to 4 [will be each tube]

To be sure I'd like to see a drawing of the modified power supply.
Remember that any series DCR before the B+1 node will drop your voltages by 0.15V per Ohm.
So the 100R alone drops 15V.

Point taken, but remind me again what the exact math was for determining the voltage drop of the 5k?
I know P = I x R...but, having the Anodes, Screens and Output Transformer connections shared/paralleled messes with my head a bit. Are we taking the combined current of the 2X EL34 now or just the individual?

I'm assuming that a single, common 5k screen dropper is used for both tubes, so the screen currents add.
The individual 1k screen resistors only see the screen current of a single tube.
(V= I x R, P= I² x R )

Yes, shared 5k Dropper. Separate 1k Screen Grid Resistors.

No more 100R resistor.

Again, this amp does not exist nor has ever existed in this iteration as far as I'm concerned, so I don't have a drawing of a schematic as I'm not sure I could pull it off accurately. I am not an electrical engineer, just a hobbyist with passing knowledge of this.

What I do have however is a detailed, scaled layout seen in post #1, labeled SEL V2.pdf

Here it is once more:

AX84 SEL V2.pdf

[GlideOn]

Right, so with combined screen currents 65ma + 65ma = 130ma or .130A

5k or 5000R


P = I x R

.130A x 5000R

P = 650v


That exceeds the total voltage of the amp. That can't be right.

Or is that the dissipation of the resistor, which if correct is also not right?

What am I missing here??

[Bergheim]

Right, so with combined screen currents 65ma + 65ma = 130ma or .130A

5k or 5000R


P = I x R

.130A x 5000R

P = 650v


That exceeds the total voltage of the amp. That can't be right.

Or is that the dissipation of the resistor, which if correct is also not right?

What am I missing here??

The 5k screen supply dropper only passes current to the screens, not the plates. As Helmholtz is pointing out, screen current is about 15% of plate current at idle (no signal/sound). So with 60mA plate current, screen current is 9mA. The screens of both tubes will draw current from this node, so 18mA. Voltage drop at idle will be 5k * 18mA = 90V. In the layout, it looks like the screen supply node is fed via the 4.7k/47uF preamp node. So this 4.7k resistor also has to pass the screen node current in addition to the preamp current. If preamp current is 6mA (1mA per triode section), current through that resistor will be 18 + 6 = 24mA, and the voltage drop across it will be 4.7k * 24mA = ~113V. Power dissipation in that 4.7k will be 113V * 0.024mA = 2.7W, so it has to be at least a 5 watter as well.
Total voltage drop from B+1 plate supply to the 5k screen supply node will be 113 + 90 = 203V. If plate supply voltage is 380V per my earlier calculations, 380 - 203 = 177V. A pretty low figure, but it will work.
Pentodes draw a lot of screen current when overdriven, easily up to 5 times (90mA in this case) of the idle current. Almost 11k (4.7k + 5k + 1k) in series with the screens is quite a lot, and theoretically, the voltage at the screen supply will be dragged right down to zero during overdrive. This won't actually happen, since zero volts at the screens equals zero current through the plates, but it will drop very low and cause a considerable compression effect when the amp goes into overdrive. I'd consider connecting the 5k screen supply dropper to the B+1 supply instead of the 4.7k preamp supply. That way you'll get an actual screen voltage of about 290V (380 - 90), and you can keep the 4.7k preamp resistor at 1 watt.

[GlideOn]

Okay, 90v drop with the 5k and your math checks out! I had right intentions but wrong application!

I see now the error I made - I had put the 4.7k value as a desire to lower preamp voltages, but minor visual error it should go further down the B+ line, not replace the 1k I'm not a huge fan of the funky stock layout, but I have to use what I have ).

I'd like to keep maximum voltage at the power tubes, hence the whole reason of replacing transformer and doubling up on tubes!

Better yet as you suggest, I can omit the B+2 dropper altogether and connect the 5k/5w to the B+ line for max voltage potential, close if not dead on to that 415v figure.

I see now a way I can take that same 4.7k/2w and put it in between B+3 and B+4 filters for preamp voltage dropping. I'll just jumper the 5k/5w over to the immediate right instead of up/down orientation, then simply use new jumper wire to reconnect the B+ line to the preamp.

I'll make an update layout tonight to illustrate that change.