Trying to wrap my head around values for class A operation of el34

[thehoj]

I read this thread http://music-electronics-forum.com/t10296/
Which discusses one way of getting the appropriate cathode voltages and other associated values..
I see that he arbitrarily selects 275V as the B+ in this scenario, and since we want 250V on the plate for class A operation, we need 25V on the cathode, and we want 100ma cathode current, so we select a (25/.1) 250ohm resistor.
And that makes sense to me.

I was just wondering though if 25V on the cathode is any better, worse or even different than using any other scenario to accomplish the same 250V on the plate.
What if I said I had 300V B+ from my PT, and so I calculated for having 50V on the cathode?.. which would mean I would need a (50/.1) 500ohm resistor.

Mainly what got me thinking about this initially, is looking at the typical operation information for the el34 http://www.drtube.com/datasheets/el34-sed2002.pdf It mentions:

Typical Operation
AF Power Amplifier, Class A1 (single tube)
Plate Voltage 250 V
Grid 2 Screen Voltage 250 V
Grid 1 Control Voltage* -14 V
Peak AF Grid 1 Control Voltage 14 V
Zero Signal Plate Current 100 mA
Maximum Signal Plate Current 105 mA
Zero Signal Grid 2 Screen Current (avg) 15 mA
Transconductance (nominal) 11,000 mS
Load Resistance 2000 Ohms
Output Power at 5% distortion 10 W

This mentions "grid1 control voltage: -14V" Doesn't that mean that ideally I would want 14V at the cathode? Which would mean I'd want a 264V B+ and then a (14/.1) 140ohm cathode resistor?

I've been reading as much information as I can on the subject, some of it is just going over my head though. Hopefully someone can help me out with this.

EDIT: I would love to see a schematic with voltages of a class A push-pull amp using el34's.. Fixed bias or cathode biased, doesn't matter. For some reason I can't find this anywhere..

[thehoj]

I guess my real question was, is 25V on the cathode is any better, worse or even different than using any other scenario to accomplish the same 250V on the plate.? ie) 14v on the cathode, or 50v on the cathode?

[BiBi]

Don't forget to factor in screen current.
If you don't understand then read the data sheet for an EL34Results of tube search
Or go to the Valve Wizard site How to design valve guitar amplifiers
or go to R. Aiken's site Aiken Amplification and find the section "last word on class A"
and here ya go EL34 class A schematic http://www.drtube.com/schematics/ai/ai800.gif that took me all of two google seconds to find!

[thehoj]

Don't forget to factor in screen current.
If you don't understand then read the data sheet for an EL34Results of tube search
Or go to the Valve Wizard site How to design valve guitar amplifiers
or go to R. Aiken's site Aiken Amplification and find the section "last word on class A"
and here ya go EL34 class A schematic http://www.drtube.com/schematics/ai/ai800.gif that took me all of two google seconds to find!

I've looked over the valve wizard and aiken site articles, and the part I'm having a hard time with is how to fully understand / graph plate curves and load lines, and determining "the region where the voltage swing is symmetrical and biased in the center of the range".
I'm beginning to think I should follow someone's proven design rather than try to design my own power amp and preamp sections.. The schematic that you so easily found and linked to is more of a hifi poweramp, in an ultralinear configuration. I'm looking for a guitar tube amp schematic in a push-pull class A configuration, I still haven't been able to find one.

[madkatb]

I'm with thehoj on this one. I also have a hard time wrapping my brain around Class A vs. Class AB. I thought I got it when reading Merlin's page on the biasing of a Vox AC30 where he shows the loadline as being two "parts", part in Class A and part in Class AB. And I understand the difference between the Classes as to which part of the signal is amplified. But I still don't know how to look at a schematic or voltages and recognise the difference. Can you look at a schematic and recognise its Class. Has thehoj looked at one already and not known that that is what he was looking at? Or can you only recognise Class by doing the math? I feel like I'm missing a "Eureka" moment.

[Jazz P Bass]

I'm beginning to think I should follow someone's proven design rather than try to design my own power amp and preamp sections.. The schematic that you so easily found and linked to is more of a hifi poweramp, in an ultralinear configuration. I'm looking for a guitar tube amp schematic in a push-pull class A configuration, I still haven't been able to find one.

How about Duncan Amps>http://www.duncanamps.com/pdf/blues112.pdf

[thehoj]

Interesting. I had looked on there earlier, but didn't notice that.

Now looking at it, I see that we've got 362V on the plates

And it's biased basically like this,
I=10.8/150= 72mA tot or 36mA per tube

P=.036*352 = 12.6w plate dissipation

So it's biased for 100% plate dissipation, but the plate voltage being at 352 seems high.. The only reason I say that is because of the comments on aiken's site:

Just because you are biased at max dissipation does not mean you are class A! You must be in the region where the voltage swing is symmetrical and biased in the center of the range, where plate current flows for all unclipped output.

And the datasheet for the el84 talks about class A operation being at 250V plate voltage.

[Chuck H]

I am not big on all the math. I'm not an EE. But the class A concept seems pretty simple to me so I don't worry about being able to calculate finite details on paper (which never performs like an actual amp anyway)...

To offset excessive heat (sans fans and such) class A amps typically run lower than max Vp by as much as it takes to get into max diss at the tubes max rated wattage. The bias needs to be set so that cutoff and saturation are equadistant. Only in this way can you achieve maximum headroom in class A. Looking at tube charts should give the info needed to achieve this. Or, more accurately, approximate and then refine to this ideal. There are so many variables with power supply resistance and current, drive voltage and the final NFB loop (sometimes) that trying to find finite calculations seems like an excersize in futility. There will almost inveriably be adjustments needed anyway so why trouble with developing a "formula" when approximations are easy and adjustments will be needed for a final design anyway??? Not to mention that symetrical vs. assymetrical clipping will sound different and is subjuctive as far as tonal superlatives. IMHE circuit adjustments, listening tests and actual bench work are better for this sort of thing as long as you understand the limitations on the tube in question. At least as far as guitar amps go since overdrive is intentional and tonal coloration is a typical consideration outside of conventional amplifier "how to".

Just sayin'

Chuck

[Steve Conner]

The exact figures don't really matter. Just run'em with low plate voltage and lots of idle current, and you're in Class-A or thereabouts.

More specifically, with a single tube, you should set it up so that the maximum possible peak plate current is about twice the idle current. That's really just a different way of restating what Randall Aiken said above, that the tube's idle point should be in the centre of its load line. (How long is a piece of string? Twice the distance from the middle to the end.)

It's not super-critical, but if the peak capability is much less than twice the idle current, you're not getting as much power as you could. And if it's much more, it's going to sound really weird when you overdrive it, because it'll be running in Class-B with a single tube.

With a pentode or beam tetrode you can set the peak output current to whatever you want by adjusting the screen voltage. So, you can achieve Class-A operation at higher plate voltages than the datasheet example, by using an abnormally low screen voltage.

If you have it right, the current draw from B+ should not vary too much when the amp is taken from idle to full unclipped power output.

[thehoj]

Okay, this is making a bit more sense for me now..
I'm still a bit confused.

So my idle current is the current over the plate, and the peak output current is set over the screen..?

I understand that if I'm calculating cathode current, that's plate + screen current, but if say I have 300V on the plate, and 200V on the screen, and my cathode resistor biases the tube for 100mA at the cathode, how do I break that up into current over plate, and current over screen?

I must not be finding the right graphs.
Looking at the el34 datasheet I just have performance curves of plate current to grid voltage, and plate current to plate voltage.

The exact figures don't really matter. Just run'em with low plate voltage and lots of idle current, and you're in Class-A or thereabouts.

More specifically, with a single tube, you should set it up so that the maximum possible peak plate current is about twice the idle current. That's really just a different way of restating what Randall Aiken said above, that the tube's idle point should be in the centre of its load line. (How long is a piece of string? Twice the distance from the middle to the end.)

It's not super-critical, but if the peak capability is much less than twice the idle current, you're not getting as much power as you could. And if it's much more, it's going to sound really weird when you overdrive it, because it'll be running in Class-B with a single tube.

With a pentode or beam tetrode you can set the peak output current to whatever you want by adjusting the screen voltage. So, you can achieve Class-A operation at higher plate voltages than the datasheet example, by using an abnormally low screen voltage.

If you have it right, the current draw from B+ should not vary too much when the amp is taken from idle to full unclipped power output.

[Steve Conner]

The screen current doesn't really matter. I'm ignoring it as a first approximation here.

The screen functions as a kind of "Electron enticer". The more voltage you put on it, the more encouragement it gives to electrons on their way to the plate, so the more will come out. Some of them don't hit the plate, they smack into the screen, but they're a small amount that can be neglected. Therefore, the screen voltage determines the maximum current the tube can flow.

Tube datasheets don't usually have screen voltage as a variable on a curve. They have whole different plots for different screen voltages. 150, 200, 250 etc. If you look at the Vp vs. Ip plot, the flat portion of the Vg=0 line, right at the top of the graph, is the peak output current. There should be a few different Vp vs. Ip plots for different screen voltages, so choose the one that has the right max current on it, and that's your screen voltage.

An easier method might just be to adjust the screen voltage for best overdriven tone, or most symmetrical clipping viewed on a scope. That automatically compensates for saggy screen voltage due to a screen resistor, but the theoretical method gets really math-heavy once you start worrying about screen voltage and resistance.

[thehoj]

Okay, so looking at this datasheet http://www.drtube.com/datasheets/el34-sed2002.pdf
If I'm looking at the plate current vs plate voltage graph, each of the graph lines is for the screen voltage at 250V, so if I had my plate voltage at 300V, that lines up with the plate current being at 140mA.
So if I biased the amp for 140mA current over the cathode resistor, my idle current is 140mA and the max current would be just under 160mA? So that wouldn't really be ideal for class A operation?
I would want the idle current to be half of what the max current is?
If I look at that screen voltage line, and I wanted 80mA idle current, that lines up with like 40V on the plates? That doesn't seem right. I must be interpreting your information wrong.
Maybe I need to find a graph with different screen voltages?

Or does it make sense to use both of the graphs.. If I look at the plate current vs grid voltage graph, and say I look at the 250V Screen line, and then follow that down to 100mA (since 100mA is more realistically what I want to be biased at) that graph line takes me to -14 grid voltage.. But I don't know how to extrapolate a -14 grid line on the other graph to find out the appropriate plate voltage..
I guess if I work with the lines that are there, and I take the 300V screen line, at 90mA that's -20 grid voltage. So if I look at the -20 grid line on the other graph, at 600V on the plates I'm looking at 50mA on the plate?

What am I doing wrong here..?

The screen current doesn't really matter. I'm ignoring it as a first approximation here.

The screen functions as a kind of "Electron enticer". The more voltage you put on it, the more encouragement it gives to electrons on their way to the plate, so the more will come out. Some of them don't hit the plate, they smack into the screen, but they're a small amount that can be neglected. Therefore, the screen voltage determines the maximum current the tube can flow.

Tube datasheets don't usually have screen voltage as a variable on a curve. They have whole different plots for different screen voltages. 150, 200, 250 etc. If you look at the Vp vs. Ip plot, the flat portion of the Vg=0 line, right at the top of the graph, is the peak output current. There should be a few different Vp vs. Ip plots for different screen voltages, so choose the one that has the right max current on it, and that's your screen voltage.

An easier method might just be to adjust the screen voltage for best overdriven tone, or most symmetrical clipping viewed on a scope. That automatically compensates for saggy screen voltage due to a screen resistor, but the theoretical method gets really math-heavy once you start worrying about screen voltage and resistance.

[thehoj]

Okay, so reading through the valve wizard stuff on setting up a single el34 for class A operation a few times, it's starting to sink in more and more. The Valve Wizard -Single Ended

I can basically assume that if I follow his calculations (and see where he gets them from the graphs he's using), I'll achieve the same results. He selects 300V plate voltage, which means my max plate current is 75mA, and I can see the load line he's marked on the first graph, and that the plate voltage at 300 with plate current at 75mA puts the bias point right in the middle, so the swing is from 0V to 600V and I'm right in the middle.

The section on screen voltage was confusing to me at first because with the second graph the Screen voltage line's placement indicates that when using a 470ohm HT dropping resistor we're dropping roughly 8V based on the amount of current draw through that resistor (Power tube and preamp) making our required screen voltage 250V (based on the second graph), but he ends up selecting 288V for the screen voltage because after the HT dropping resistor, and the screen resistor we're dropping roughly 12V. That just means we need to bias differently now though. Looking at the second graph again he extrapolates that at 288V the grid voltage should be about -17V. Calculating the cathode resistor needed based on the plate current + screen current (7:1 ratio) we'll want 86mA at the cathode, giving us 200ohm cathode resistor.

It makes sense when I type it out.

Now I'm just trying to translate this into a PP design. I should be able to apply the same configuration to both Power tubes in the PP configuration, each with their own screen grid resistor. But at the start of the article he talks about using the impedance calculation Zout=Va^2 / Pa. For his article this is 300^2/25 = 3600ohm. Now my question about this is, how does this translate to 2 el34's in PP configuration?
My Va is going to be the same, so 300^2, but do I double the plate dissipation for 2 tubes? Would it be 90,000 / 50 = 1800ohm?

EDIT: Looking at some datasheet's for el34 class A PP in triode mode it looks like the impedance goes up.. Is that the same for pentode mode? Would I be looking at more like 7200ohm?
If I take into consideration voltage for both EL34's added together (600V) and plate dissipation for both added together (50), 600^2 / 50, I do get 7200ohm..

Sorry that my posts are going all over the place.

[Steve Conner]

Again, maybe Merlin is reading this and can comment, but...

For PP, transformers are sold according to plate-to-plate load impedance, which should be 4 times the value you calculate in the single ended case.

But in Class A, both tubes are always driving both ends of the OT at once, so each one sees twice the impedance it would otherwise, because the other one is helping.

So overall, you want an OT with a plate-plate primary impedance of twice the value you would calculate in the single-ended case.

[Merlinb]

Steve is right. If we're talking genuine class A push pull amp, then it can be treated as two identical single-ended stages which happen to have their transformers wound on the same core. I imagine it like a SE schematic that has a mirror placed on the page, making a symmetrical image. Class A is the easiest kind to design!

So you can still use the formula to get the load for one valve:
Z=Va^2 / Pa
But you must double this value of Z to get your anode-to-anode transformer impedance.

Note that this is an idealised formula. By drawing load lines you might find that you want to tweak the impedance to truly stay in class A. Alternatively you might be stuck with an off-the-shelf impedance, in which case you will have to tweak the supply voltages instead. Or just admit that it doesn't really matter if you stray momentarily into class B!