What Is The Optimum Load For A Triode?

[Diabolical Artificer]

I've read quite a few sources on the subject but I'm still struggling to find a relatively straight forward method for calculating this. Specifically I'm interested how it pertains to OP valve OPT primary Z in Class A & B, hifi rather than max power, but also in Class A common cathode gain stages.

Any light anyone can shed on the subject would be most welcome, Andy.

[Radio Wrangler]

It may be somewhat simpler, though less well-defined than you think, Andy.

First, you need a set of Ia/Va curves for the valve, several curves on the same plot covering a range of grid voltages.

Shade out the areas you don't want to run in because of ratings..... too high voltage, too high current.

Now shade out the areas you don't like. too curved, too non-linear, going into grid. current.

Into the area you've still got left, fit a straight line with a ruler. sloping downwards in current at higher voltage.

You have now fitted a load line to a single-valve class-A stage. The slope of your load line in volts/mA is the needed anode load impedance in Kilo Ohms. Your HT needs to be at the mid point of the swing.

A class A push pull amp is two valves in parallel, just running in antiphase. One increases in current as the other reduces. so if one bottle goes up 1mA the other drops 1mA and the total current change is equivalent to 2mA expressed in one side of the primary. Or you could say it's 1mA expressed across the entire primary. You get twice as many volts across the whole primary, encompassing both phases. Be careful that you don't use current from one view and the voltage from the other and you can work out your load line impedance. depending on your viewpoint you get either the Z per phase or the Z across phase to phase.

You have to sit down and get it clear in your head, visualise it working. forget all the common jargon statements, work it out from scratch. Others have been here before and sorted it out, but there's a lot of noise and dodgy science mistaken for short cuts. The most difficult bit is turning down things which look to be short cuts. Go the full route, the long way round. Enjoy the scenery, it can set you straight on a number of things.

What makes it click in your head might not be the same as in anyone else's

David

[thermion]

I’ve had the same thoughts on this too, data sheets for optimum load are fairly easy to find but distortion v anode load curves are much harder to find. For example I recently did a px4 amp, the optimum load is around 4000 ohms a-a, but lowest distortion is around 10k. I think all you can do is go off the published curves if you’re lucky enough to find them. Clipping points are pretty easy to plot. I’d like an AP distortion measuring set, but know very little about them.

[GrimJosef]

Originally Posted by thermion

... I’d like an AP distortion measuring set, but know very little about them.

For me the most important thing about them is the cost (they are pricey but extremely capable). For valve amplification they are likely to be over-specified and less expensive equipment will be more than adequate.

If you want to measure the distortion of a well-designed modern solid-state amplifier which, even in percentage terms, might have a few zeroes between the decimal point and the first non-zero digit, then something at the level of AP test gear will be needed.

A valve amp will be a hundred to a thousand times worse in terms of distortion. But a good one will still be of order ten times better than the speakers you're listening through and between ten and a hundred times below the threshold at which you can objectively hear the distortion (double-blind, controlled, statistically significant test).

Cheers,

GJ

[kalee20]

The optimum load for a triode depends on whether you are prepared to accept grid current or not.

If you are - and you can drive the grid as hard as is necessary - then it's just (quiescent anode - cathode volts) divided by quiescent anode current. Same as a pentode really.

If you don't want grid current, then it's rather more complex. Triodes don't have the attracting force of the screen grid to encourage current to flow, so on the upward swing of current, the anode voltage falls and lessens the 'pull' on electrons surrounding the cathode, just when you want it to be big. So triodes are relatively insensitive.

For a good triode output stage, you want the HT voltage to be as high as possible, without exceeding the valve's ratings. Then, for maximum power, there's an optimum quiescent current which depends on the valve's ra, and an associated optimum load. If this optimum current results in the anode dissipation figure being exceeded, just wind the current back back till it isn't, and use that current. You'll have less power, but % efficiency will be greater.

F Langford-Smith has a chapter about matching the load - it's very explicit. Many of the results are stated without proof, but I checked them and they're right.

[thermion]

I have a Marconi tf 3321 distortion factor meter here and it does for what I want, it’s just slow, all that nulling out of the fundamental & balancing etc .I’m only a hobbyist with no formal electronics training as such.

I do derive great pleasure from vintage electronics and that’s the main thing.
Terry Bateman’s BVWS amp articles are always a good read, that’s where I saw the AP plots. I’ll have a read of David’s post above about load lines and see if I can get my head around it

[kalee20]

I'm in a better position to respond, here is a link to the RDH, look at pages 555 onwards (=pdf pages 595 onwards).

The results are a bit 'design by numbers' though if you look at the idealised load lines, and work out what the hard limits are, at which point any increase in signal results in distortion, you can get the sort of insight that David indicates in post #2. If you don't, then just use the results.

[Radio Wrangler]

The key thing is to get playing with these things.

If you're just starting to design things, forget the more complex varieties, forget push-pull, forger classes after A in the alphabet. Go for a single-ended class-A stage. You may not want one as your goal, but it's where you should start. Everything else is built on this, everything else is described in terms defined in one of these stages. Walk, do not run. The scenery is better, more illuminating.

(Tommy Cooper voice ON) Pick a triode, any triode! Well, one you have found a set of curves for. (Tommy Cooper voice OFF) Print several copies of those curves, you're going to write and draw all over them and you'll keep changing your mind and starting again. Sometimes you'll want to go back to an earlier idea, so don't screw them up in disgust, write a time and date on each one so it has a name and a place in an order. Write some comments to let your future self be reminded what you were thinking when you did each one.

Start playing. It is supposed to be fun, so therefore it's playing. Pin the tail on the donkey? no it's pin the load line on the triode.

So let's draw a load line. If your curves have anode curent on the Y axis from 0 going up to however many mA upwards, and anode voltage on the X axis, going from 0 rightwards to lots of volts, then your load line is a straight line sloping downwards as you look rightwards.

Just draw one, you don't need to put it anywhere specific just yet You're not trying to guess anything, yet. Plant one anywhere and we'll see what it does. Saves a lot of money and time over keeping building new amplifiers to find out.

So you've got a randomly chosen load line drawn over a set of characteristic curves. Now let's evaluate it.

This tells what happens with a real resistor loading an inductively biased anode. Assume L is very big, so the resistance dominates the AC loading while the inductor provides lossless DC power. The energy stored in the inductor when the anode voltage swings below HT provides the energy needed when the anode swings above HT. This doesn't need a specific value of inductance, it works just so long as you have enough. Later we can look at its effects at the low frequency end and pick L to get enough bass extension, but that comes later. The load line resistance FORCES a relationship between anode current and voltage. You get to pick your HT voltage which puts where the quiescent voltage is. You can see the load line going into the danger area of excessive anode voltage, but on your graph this is only a mathematical construct. It knows nothing of valve destruction, flashing over transformer windings and other drama. You have to choose where to bottle out.

Now look at the left end of the load line. The valve curves are dropping in this area. The load line wants a high current, but the valve curve say it can't deliver. The valve curves win. Again, you have to choose how far you can push it. You now have some first-pass limits on what you could try with an amplifier design.

But the anode curves are a family, done with diferent grid voltages, and you can see where each member of the family of curves crosses you load line. Dot the crossings so they stand out.

In a nice, happy world, you should get them crossing at nice equal intervals on your load line.... evenly spaced. No? you don't? They ger a bit closer together at one end, and spaced out at the other? Great! you are now seeing the distortion.

You feel it's a bit too much so you decide not to swing your amp quite so far into low anode voltage and avoid some too-close crossings. There still will be distortion, but not as much.

You now have a load line and you have chosen limits on the anode voltage swing. The load line transforms this into limits on anode current swing. Pick the middle of the length of the load line you want to use and this point tells you the HT voltage and quiescent current to use.

Now have a think about the anode current swing across the used segment of your load line. Is it running into trouble by asking too much at one end? or is it not making full use of what that valve could do?

Grab a fresh print of the curves and draw a new load line with a different slope, maybe shifted up/down a bit, maybe shifted left/right a bit. Evaluate it the same way you did the first one.

Rinse and repeat.

You are feeling your way around, discovering how things interact and slowly fitting a load line to the valve curves and you fears of voltage-induced breakdown versus distortion.

You can learn a lot just using printed copies of the curves, a ruler and a pen. You are making the raw design decisions that you've only ever seen distilled into a simple recommended anode load resistance. You now see the scenery, feel the pulls of various limitations and compromises.

Designing valve amplifiers with output transformers is a very graphical thing to do.

Once you're comfy with this you can try tetrodes and you'll see the dreaded kink in the characteristics, robbing you of the desire to swing towards the left. You feel disappointed with the power you can get and the distortion. So you see why the beam tetrode/kinkless tetrode was a bit of a saviour. and you dare use more of the terrain on the curves.

Only when you are really comfy with these single ended class A structures should you even think about other classes and push-pull. But when you come to them, you'll have a lot more knowledge of the foundations and they won't be such a jump into the unknown.

David

[Diabolical Artificer]

Into the area you've still got left, fit a straight line with a ruler. sloping downwards in current at higher voltage

With you up to this sentence, but I can see several flaws in that method, see attached, pic 1. The valve is a 2C34, just happened to have it on the table. Shading out all the areas you mentioned David gives a total power OP of 1.2w {(Va max-Va min) x (Ia max-Ia min)} /2 = 80*0.03/2 = 1.2w. Z pri= 130v/20mA = 4k3*2= 8k6?

The above is a bit of an extreme example of David's methodology and possibly I've misunderstood parts of it. I've drawn a few load lines & studied the subject a bit & always understood the biasing of a Class A stage not to be center biased and Va can exceed Va max because of the inductance of a OPT. Also whilst not ideal a valves AC conditions as part of it's loadline will go into non linear regions (or we lose OP power) and in Class B, the loadline can, and does, go above max power dissipation.

The optimum load for a triode depends on whether you are prepared to accept grid current or not.

If you are - and you can drive the grid as hard as is necessary - then it's just (quiescent anode - cathode volts) divided by quiescent anode current. Same as a pentode really.


The optimum load for a triode depends on whether you are prepared to accept grid current or not.

If you are - and you can drive the grid as hard as is necessary - then it's just (quiescent anode - cathode volts) divided by quiescent anode current. Same as a pentode really. If you don't want grid current, then it's rather more complex

Indeed, when the grid starts pulling current the triode becomes a diode on part of the cycle, so you get more power OP but more distortion & compression.

As mentioned I've looked at numerous sources to get the answer to my question. Answers I got were Z= 2*rp or 4*rp, also Z=(Va/Ia)*4 and lastly Z=(Ea/Ia)-(2*ra). I'm currently wading through RDH4 - I have the actual book, extracting the knowledge I want, I'll keep plodding on though.

Usually, in the past I've used a valve datasheet which gives all the specs you need EG Va, Ia quiescent, Vg1, RL a-a & power OP, built it, tested it, tweaked it. This approach works ok for common beam tetrode/pentode OP valves & smaller common signal triodes but for valves that you want to run at other parameters or have very limited info on their datasheets, it doesn't work & I was after digging deeper but not down to atomic level : ) A middle way as it were.

Re testing for distortion I have an HP3904B which is very useful & sufficient for testing valve circuits but after extensive testing I sometimes end up with loads of data, but have trouble deciphering it. There often isn't a straight forward if I do this to X, Y does this that's where you need to draw graphs. As David says enjoy the journey look at the trees, a map would be nice though.

Thanks for all your replys, still haven't assimilated post # 8 but I suspect there's no easy formulae or answer so it's a matter of donning the heavy lead hat & jumping into the treacle.

Andy.

[RichardGM]

One of the limits on Andy's graph is marked max DCI. Should that not be what it says, the maximum DC, with the current with a signal swinging above and below that value?

[julie_m]

Originally Posted by RichardGM

One of the limits on Andy's graph is marked max DCI. Should that not be what it says, the maximum DC, with the current with a signal swinging above and below that value?

That 40mA figure looks a lot like the maximum amount of current a speaker transformer can handle .....

When I designed and built a small valve amplifier from scratch, for a fun learning experience, I used a laptop computer with a Velleman K8055 experimenter board. I also had to build some simple level-shifting circuitry on breadboard, with a simple potential divider to measure the anode voltage (which I was adjusting manually) via one of the ADC channels, an op-amp to boost the voltage across a current-sensing resistor in the cathode into the other ADC channel and another op-amp to invert and amplify one of the DAC outputs to apply a suitable negative voltage to the grid.

[Radio Wrangler]

There seem to be many definitions of what 'Maximum' means. Philips data sheets used to be quite good about explaining what they meant in them. Here, we're trying to design a good amplifier. We don't want our peaks to exceed peak ratings, we don't want our averages to exceed average ratings. With audio signals it usually works out that if we get the peak ratings right, the averages won't be too bad, but we still have to check them.. Ratings don't tell you what quiescent voltages and currents to use, but you often in valve data sheets get those stated as 'operating conditions'.

David

[kesrmk69]

I have always used rule 3-5 * Ri (Ri = internal impedance of the triode).
Wondering why 3-5x but not 1.5x or not 10x.
With pentodes 0.1x * Ri.

[kalee20]

Definitely not 0.1 x ra for pentodes!

An ideal pentode has infinite ra, and 0.1 x infinity is still infinity!

You need to load a practical pentode with (Va - 25) / Ia, where Va is the anode voltage (less anything 'lost' across the cathode resistor) and Ia is the standing anode current. The -25V comes about because the anode doesn't pull right down to zero relative to cathode, but has a 'knee' at about 25V. If you have the curves, you can refine this.

[Radio Wrangler]

Originally Posted by kesrmk69

I have always used rule 3-5 * Ri (Ri = internal impedance of the triode).
Wondering why 3-5x but not 1.5x or not 10x.
With pentodes 0.1x * Ri.

Old, wet-finger, rules of thumb. A starting approximation if you have nothing better to go on, but you have to do the work with the curves and load lines to see whether they are any good in your instance. Wasting an output transformer is nowadays a very expensive mistake.

Andy was asking about how amplifiers were designed, how design decisions were made and anode load impedances etc decided upon. For this, you have to build some experience with curves and load lines until you've got your eye in and can see limitations. Line fitting to an area of curves is the foundation everything else is built on. There are rules of thumb, there are recommended bias and load conditions in device data sheets, but to appreciate them and understand them, you have to get comfy with curves and load lines. This is where a lot of the founding assumptions are made.

David

[kalee20]

Originally Posted by kesrmk69

I have always used rule 3-5 * Ri (Ri = internal impedance of the triode).
Wondering why 3-5x but not 1.5x or not 10x.

Or even, why not exactly Ri (which I prefer to call ra), because after all that is what the Maximum Power Transfer theorem says...

It's because the MPT doesn't usually apply - it is valid for a fixed generation voltage. But in practice the input to the output valve's grid is not fixed - you can make it whatever you want, by having a bit more gain upstream of the output valve or a bit less.

The optimum load is almost always the load that results in the maximum power for the circuit's operating conditions, when you have a completely free hand in the amplitude of drive voltage. And when you have that load, you then work out how much drive you really do need.

Distortion kicks in when the anode current just reaches zero at the signal peaks in one direction, or when the grid voltage just hits zero at the other signal peak. The optimum load makes these happen at the same signal level, so that no capability is 'wasted.' In practice the limits themselves are not 'hard' limits and the valve characteristics are not quite straight lines, but curve softly at the extremes, though as a first guess they can be considered so.

The combination of a limit in one direction that depends on the anode (the hitting zero current limit) and in the other direction being a grid limit (where the anode current and anode voltage are linked by the triode's ra) does result in an optimum load which depends on supply voltage, quiescent current, and anode resistance ra, as given in F Langford-Smith's book. The result is greater than 2 x ra, so your rule of thumb 3 - 5 times ra may well, by coincidence, not be far from the ideal.

[Richardgr]

I had thought there was a magic number linked to rp that was used, and see it described above.
When looking I came across this reference to a Paul Joppa article ....
Load impedance and operating points for single ended amplifier stages

I think that is part 1 of 3 articles. Part 2 and Part 3 are here.

[kalee20]

That article makes the point that (for sine waves) the efficiency of a Class A amplifier is limited to 50%. It doesn't quite go on to say that triodes are inefficient, because it's hard to get the anode pulled down to the ideal 0V, but that is so. Pentodes and BT's can do better (at the anode efficiency) but of course they lose out overall because you also have 'wasted' screen-grid current.

It also emphasizes that the optimum load depends on operating conditions as well as ra.

And - the article also mentions that if, in an effort to improve efficiency and power output you remove the constraint of not letting the grid go positive and you give yourself a gutsy drive circuit, you can do better and approach the 50% efficiency limit. But if you make the grid go more positive than the anode, the grid current rises sharply. That's something I'd not thought of, it's another constraint, so thank you for putting it up, Richard!

[Diabolical Artificer]

I'm reading & trying to assimilate all your replies as well as doing some testing & further reading. I'll post a more comprehensive reply soon, please bare with me, there's a lot to digest.

Andy.