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Aug 01 2013

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The ‘Free Lunch’ Power for your tube Amplifier

free lunch

free lunch

The ‘Free Lunch’ Power for your tube Amplifier

(This is a continuation of an earlier post: Tetrode/Pentode amp owners – a “must” read ).
In the beginning, I wish to thank Mr. Dennis of:


for his kind support in shedding some light on various aspects of the OPTIMISED ELECTRON STREAM© Technology, for without his help this text would not come to be. I will be extensively quoting him during the course of the text.

But before we specifically dig into the details of just one very specific application of OPTIMISED ELECTRON STREAM© Technology, and mainly that bit regarding our „magic” diodes, I would like to just stress the fact that what I will be pondering about now is just one small little fragment of a much wider portfolio of solutions offered within the OPTIMISED ELECTRON STREAM© suite. Whats more, within this one exemplary application, we may have three different “case” scenarios:

There are three separate and independent operating conditions

1. Triode

2. Pentode / Tetrode / Beam Power Tetrode

3. Ultra-linear/Distributed Load

The text as below applies only to the mode no. 2. – the Pentode / Tetrode / BPT operation.
The amplifier circuit topologies 1).  and 3). are outside the scope of this current article, but I will probably dig into these somewhat similar (but not identical) cases at a later date.

Finally, just to remind you, the technology is free for DIY use only. Commercial use requires prior written agreement / consent from OPTIMISED ELECTRON STREAM©.

OK. now come the diodes … A good starting point is to start with the obvious:

The diode is a “semi-conductor”. That is what it was called from the outset of its invention. That is to say it is a unidirectional conductor. .. It conducts only in one direction.

For the purposes of OES Optimized Electron Stream (C) applications, the DC current flow is deemed to flow in the direction of the arrow on the diode.

OK, so far so good. Current flows in the direction of the arrow. Seems obvious. Or … is it? Just for the sake of completeness, let me remind you that there is a so called „arbitrarily agreed” direction of current flow ( from „+” to „-” ) and most of present day schematics and electronics are drawn and constructed according to the assumption that the flow takes place from „+” to „-”. This is a result of the fact, that a long long time ago people that were facing lack of understanding of basic electricity concepts had to agree upon some “standard”. So somebody tossed a coin … and there you have a nice „arbitrarily agreed” direction of flow. From “+” to “-”. And from then on, we live with it happily ever after.

But, it just so happens, as it later turned out, that there is such a thing called an electron, and guess what. Shit happens. The electron is the „thing” that actually moves around and it is responsible for „carrying” current. It just so happens that the electron has a NEGATIVE charge … So basically, … they flow from „-” to „+”.

As to keep in clear from the very outset, let’s just call this the „electron flow” so as not to confuse it with „current” (which goes in the ‘opposite’ direction).

Who cares, you say ?

Well – your tubes care, for that matter. Because there are physically electrons being emitted from you hot cathodes and gathered by the anodes ( the destinations!), anodes which are subjected to a high positive voltage, one that ‘lures’ the free electrons within the tube toward the anode. So, to make a long story short … In „Tube-Land” … we think in ‘opposite terms’. Upside down, so to speak.

Down Under  :)

The actual „flow” of electrons is what we care about and analyze at length. Look at those tubes glowing. It is not an “arbitrarily agreed” current direction, that is glowing, believe me.

The physical electron flow takes place in the opposite direction to „the agreed current”.

If you look at the „arrow” within the pictograph of the silicon diode, assuming it is forward biased and conducting, the „current” goes according to the direction of the arrow ( i.e. the „flow” of electrons takes place in the OPPOSITE direction).

When a diode conducts, meaning it is polarized with DC voltage according to the direction of the arrow, the current flows freely through the diode, but there will be a small voltage drop upon the diode, of about 0,5 V. For simplicities sake, lets assume that this drop is fairly constant and little dependent of the value of the current ( a fairly good approximation ). But what’s more important, a 0,5V voltage difference in the wake of hundreds of volts flying around here and there within tube-land, … for simplicity’s sake let’s just forget this 0,5V difference and consider it as not meaningful.

Now, most of you have built, or at least seen the concept of, a simple DC power supply, consisting of transformer, a rectifier, and some filtering capacitors to get rid of the AC ripple component, the one that ‘rides’ upon the rectified DC originating from the diode rectifier. Lets assume for a while that the „speaker output transformer” is a source of AC voltage that we wish to … rectify:

So what happens is that the diode allows DC current to flow from the power supply to the Screen Grid, (as if it is a solid conductor) thereby energizing the Screen Grid and allowing the valve to operate normally, but at the same time BLOCKS the current flowing through the valve from passing in the opposite direction – ie out of the diode back into the power supply – as is the case in a conventional pentode or beam power tube amplifier

To better understand the above, we need to be clear of the existence of two totally separate, parallel worlds that live in your tube amplifier.

There is the ‘DC world’ and the ‘AC world’.

The former, the ‘DC world’,  is fairly easy to grasp, according to the intuitive WYSIWIG principle (What You See Is What You Get – on your DC high voltage range of you digital volt meter, that is). Just make sure there is no input signal (silence on the input jack). Now take your voltmeter, set it to a high voltage on the DC range and do some voltage measurements around your amp. Here, there, … everywhere. What you see is in essence a set of DC voltages, that are designed and implemented by appropriate resistor networks in such a manner so as to „force” the tubes to work at very specific, predetermined points of their characteristics. We could say that these are the „idle” operating points, or the points at which each tube is supposed to „sit upon” when there is no signal present and hence nothing for the tube to do.

The parallel world to this, is the ‘AC world’ – a world of AC signals, such as music, or ripple, or otherwise. These AC signals „ride” upon the DC conditions. The AC signals constitute „deviations”, if you will, voltage variations, that take place AROUND the points of DC equilibrium, around the centrally positioned „idle” operating DC points. The AC is „riding” upon the DC, causing the tubes to deviate from their „idle” positions, as defined by the DC conditions.

Now, imagine that you are … a Ripple.
Or better: a piece of beautiful music.


Now, … do you remember the very basic function of the DC voltage FILTER within the power supply ? Well yes, you say, its primary function is to provide a „smooth” DC. Why smooth ? Easy. So that my speakers don’t hum and don’t drive me crazy. The more capacitors and filtering within the filter, the better it is for my music, you say. OK. That is correct.

So, now imagine that you are an AC signal, or a ripple, whatever, …. and you are looking straight at the HIGH voltage connection, i.e. The „B+” of your amplifier’s power supply. Tell me, WHAT do you SEE ?

OES example 2

To be frank with you, as an AC voltage, you are totally NOT interested in the fact that this B+ pin is at say, 400V DC. That does NOT concern you. You are “AC”, not “DC”. But alas, what you ARE indeed concerned about, though, are those HUGE capacitors in that filter within the DC supply. Capacitors, which are supposed to FILTER you to GROUND.
And the bigger they are, the harder you fall. To GROUND.

Yes. You said that. Your own words. At the very outset. The B+ of you power supply, in AC terms, is as good a ground connection as any other. If you are an AC ripple/signal, and you are touching the B+ ….. you are basically GROUNDED. You do not exist.

The B+ pin   is an AC GROUND !!!

This is a very important phenomenon in solving the “magic diode” riddle, and a not so immediately obvious one.  Why is it important ? Have a look once again at what Dennis said:

…. but at the same time BLOCKS the current flowing through the valve from passing in the opposite direction – i.e. out of the diode back into the power supply – as is the case in a conventional pentode or beam power tube amplifier.

Hey, something does not click here … or does it ?

Why on earth would a signal from the second grid, which is at a lower voltage than the high voltage pin of the power supply, at all WANT to flow „against” the direction of the arbitrarily established general direction of current flow ?  Nonsense, in would seem.  The S2 grid is looking into a higher voltage, and we just said earlier that current flows from “+” to “-”, … so inserting a diode does NOT change a thing – it is simply conducting all the time – a S2 grid current flowing from a high voltage supply (B+) to S2 …

Well … not exactly. Is the other end of the diode REALLY at a “higher voltage” level ?
Not in AC terms….

Why would current or voltage want to flow from the Screen grid pin back into the power supply ? Electricity ALWAYS seeks the easiest path to return from whence it came – ie the source. It also always seeks to return to ground – i.e. to the earth itself.

This is why high voltage power lines have constant leakage back to earth through the air (capacitive coupling), resulting in transmission losses. The higher the voltage, the greater the losses. However I digress.

The voltage at the B+ terminal is at a particular DC potential.

When both Plate and Screen are fed from a common B+, the voltage at the Plate is always lower than that of the Screen Grid – because of DC resistance in the output transformer – especially in Class A.

Hence there is a reverse polarity between Plate and Screen compared to Cathode and Screen. However notwithstanding that electrical configuration, the electrons are traveling fast through the mesh screen and cannot stop their travel, continuing onto the Plate. As previously noted, the Screen Grid offers only a partial anode to the electrons and can therefore only collect those electrons directly striking it.

So why do those electrons collected want to conduct directly back to the B+?

Because it is easier. The basic principle is that where there is a potential difference then current will flow.

If in the case of professional amplifiers where the Screen Grid DC voltage is substantially lower than that of the Plate one might assume there will be no DC current flow back to the B+, however the test is this: “Is it easier for the electrons to conduct directly back to the DC Screen B+ Supply, or travel on to the Plate then through the output transformer to arrive at the same point – i.e. AC ground ?”

Now all the above refers to the DC circuit.

If we look at the AC circuit – then you can see that under positive signal conditions the AC voltage at the Screen Grid will be significantly higher than the B+ DC  (= AC Ground).  This means that ALL and ANY AC signal, appearing at the Screen Grid, will immediately divert to ground.

Please consider that there is a modulated stream of electrons flowing up through the tube … they are flowing to any positive potential they see above them. So they flow both to the anode, as well as to the Screen Grid (G2), which are both positive with respect to the cathode.

Now, … here comes the CATCH.  I just said “modulated”.  When I said „modulated” – that implies that there is an AC signal. But what did we just say about AC signals and how they treat the B+ high DC voltage output of your high voltage supply ? This output is a direct AC connection to ground !

So now it turns out, that yes, the diode HAS perfect sense and it DOES actually BLOCK something. It blocks the AC signal current flowing from Screen Grid …. to the power supply, meaning to GROUND ( a very high voltage ground, but none the less ). Now, sit awhile and think about it.

You have a hot cathode. It is emitting a stream of electrons, as per „original” design of your amplifier. Let’s say that the cathode is emitting 100% of all electrons that are passing through the tube. What’s more, lets assume that your amplifier is designed in such a way, that within the specific voltage conditions, the output tube is at the maximum reasonable and allowable power that allows for it’s reasonably long period of usability and healthy life. So, the 100% of the flow of the electrons is directly related to the power output you get.

Now, …. let’s think. Where do they all go ? They go up, towards two possible destinations: The Anode, and the Screen Grid. Both of these destinations bear a fairly high positive potential that attracts the electron stream towards them.

For simplicity’s sake, lets assume that the gross amount of the electrons, normally about 70 ~80 % goes directly to the massive anode. Lets also assume that the rest of the electrons are „attracted” on the way to the Screen Grid, the wires of which are „along the way” to the anode. Within a „traditional” amplifier, about 20~30% of the electrons go directly to the Screen Grid.

Ok. so far, so good. Or maybe not so good at all ?

The 70% of the electron stream flows through the anode to the load (the transformer), and then they flow directly to GROUND ( a ground that is 400VDC higher, than what you would normally expect, but YES, for AC signaling purposes, it IS a GROUND ).

But what happens to the 30% of the electron stream that flows out of the screen grid ? If you have a traditional pentode connection, that would imply that they flow via a resistor to B+. If we are looking at the AC component contained within this elecron flow, we could say, that 30% of the signal flow goes through a resistor directly to a GROUND ( a ground that is 400VDC higher, than what you would normally expect, but YES, for AC signalling purposes, it IS a GROUND ).


You have just tossed 30% of you usable music output power into the waste bin !!!!

My question to you is as follows: Since you keep on complaining about the very low output power of your amplifier, or why does it start to clip or blip or otherwise, why does it not „manage” to „drive” those demanding speakers you purchased recently, why is it so „weak” … then why do you purposefully throw about 30% of your free lunch directly into the waste bin ?

You do like free lunch, don’t you ? Well obviously, if you do, you need to find a way so as to AVOID tossing 30% of what you „truly have” into the waste, from the very outset, without even trying to „use” it.

If we could only find a way, so as to „keep” the screen grid polarized with the correct DC voltage, so as to maintain the tubes basic „equilibrium” or operating point, but at the same time STOP the useful AC component from „running away” via Screen Grid to AC ground. We need to force the „signal” component to flow through our load, i.e. Through the transformer, which will pass on (most of) it’s energy to the speakers. The best way of doing that is to „BLOCK” the signal from „escaping” via the Screen Grid, via resistor (or other non value adding route) to AC ground (via the B+ of the power supply). If we BLOCK this escape route, we in essence shall achieve the result that the whole of the 100% of the signal component carried by the flow of electrons through the output tube goes directly to the anode, with no losses upon the way.

It is my humble understanding, that the OPTIMISED ELECTRON STREAM© Technology, in the form of that „simple” magic silicon diode, does exactly that.

… Consequently the normal Screen Grid current is now added to the Plate Current and supplied to the load – which is the output transformer. Thus the total Cathode Current remains the same as in normal operation but the signal current available to the load is increased, thereby increasing power output .

Now, since this solution is so good, why wasn’t it implemented in the old days ? Well that is fairly logical. In the old days, they didn’t have silicon diodes and silicon transistors !!

Any form of regulated power supply based on valves would be ridiculous in terms of space, cost and reliability, whereas a 50 cent silicon diode, which has more or less zero voltage drop, does the job magnificently.

So, there you have it. A nice easy modification for you amplifier, which is based on a tetrode or a pentode.

But … What are the implications ? Risks ?

Lets say, that you had 10 Watts worth of pure SE music power from an output tube of type „X”. These 10 Watts were previously „implemented” by using 70% of the electron stream flowing through the output transformer and speakers. You do remember, that within your „standard” textbook application, you normally tossed 30% of your „signal” into the waste bin, don’t you ?

Now, If we can achieve a result being that the anode current is now constituted of 100% of the electron stream, that would imply an increase of 100/70 = 1.428. This implies an increase of power by a (theoretical!) factor of 1.428 x 1.428 = 2.03 , or speaking otherwise, 203 percent !!!  If you previously had 10 Watts of music power, be prepared for a theoretical 20,3 Watts now. (Power is proportional to the square of the current, hence the 1.428 x 1.428 … and just to cool you off from the very outset: No, the practical power gain will NOT be so optimistic. But it will be there).

Ok, ok, hold it, you say. Those tubes cannot „give out” so much power. „Are we not risking that we are now over-stressing the output tubes ?”

No. We are not. We are only „using” what is „Rightfully Ours”.

We „Claim our Possessions”.

We are only taking back that part of our free lunch that we previously threw into the waste bin. The tube is exactly at the same DC conditions, as it was before, prior to the „diode modification”. It is exposed to exactly the same operating point, so the amount of power lost in the form of heat on the tube remains exactly the same. In terms of sustained power losses, taking into consideration the sheer thermal mass of the anode electrode, nothing has changed. Calculate a long term average of a sinus signal. It will turn out to be zero, irrespective of it’s amplitude. The AC component that is now riding at a higher amplitude on the anode of your anode, but thermally averages out to zero, so it is a non-issue. There is a little more heat from the higher DC component that is now lost on the anode (higher current), but previously it was lost in the form of heat on the screen grid, which is inside the ‘perimeter’ of the anode. Thermally speaking, nothing changes whatsoever. Instead of 70% + 30%, you now have 100% + 0%. Which one of the two do you prefer ?

Nothing has changed. The only thing that did change is that we have an OPTIMISED ELECTRON STREAM© flowing through our load (transformer + speakers).

Call it a „A reasonable 100%” if you wish.

This text is to be continued, … because we have only just touched the tip of an iceberg.

For now, try to ponder about what has been written, get your soldering iron hot, do the modifications, and in the mean time I will be thinking about a continuation of these ponderings.

{  I reserve the right to re-edit this text at my discretion, because I am in the process of verifying some of my finesse  happy-go-lucky interpretations directly with the “source”.  }

Best Regards,




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