Power dissipation in valves.

[martin_m0nxp]

Could you knowledge folk help me understand how power is dissipated within a valve?

I'm not talking about the power that goes into the heater, and I'm not talking about the parasitic losses (e.g. due to the resistance of the cathode or plate electrode).

To help me better phrase my question: Imagine a simple NPN BJT circuit - Emitter connected to 0V, collector via a load resistor R to 100V and the base biased such that the transistor is dropping about 50V (with 50V across the load resistor R). If the load resistor is 100 ohms then 0.5A would be flowing, and the power dissipated in the load would be 25W.

The power dissipated in the transistor would also be 25W as it would be also dropping 50V at 0.5A.

This is fine - I can imagine the process by which the silicon is acting as a resistance - the electrons are moving within a material putting energy into the material through collisions etc...

Now replace the BJT in the example above with a triode and bias accordingly to achieve the same current flow on the anode.

By logic the same power dissipation should be experienced in both the triode version and the BJT version (ignoring parasitic and heater loss).

My question is - physically where in the valve is this heat being produced? As there is a vacuum between the cathode and the anode we can't be heating that (you can't heat nothing right?). The control grid (as I understand it) controls the flow of electrons by producing a repulsive electric field, so it itself isn't involved in the bulk flow of electrons.

I just can't visualise what is going on....

[MeerKat]

Hi Martin,
I do not claim to be an expert in these matters but I think the simple answer here is the heat comes when the electrons from the cathode colliding with the anode, each electron causes heat generation when it bashes into the anode. I have seen anodes glowing red when excess current flows through the valve.

[broadgage]

Yes, apart from heater consumption and minor parasitic losses, most of the dissipation is in the anode as a result of the electrons colliding with it.

That is why the anode is large in comparison to the other components of a valve.
Most of the heat is radiated away except in very large power valves that employ water cooling.

[martin_m0nxp]

Okay - that makes some sense, but I can't help feeling there must be more too it (or maybe there isn't).

If it was all related to collisions with the anode then the power dissipation in the valve would be at it's maximum when the anode current was at it's highest, however in the BJT example the power dissipated in the transistor is it it's greatest when the drop is shared evenly between the load and the transistor.

By the simple maths of P=VI surely the same must be true of dissipation inside the valve?

(If you're wondering where all this curiosity has come from - having burnt my finger a second time on a UL41 I just can't help wondering how it's even possible to generate that much heat!).

[GrimJosef]

Originally Posted by martin_m0nxp

...

If it was all related to collisions with the anode then the power dissipation in the valve would be at it's maximum when the anode current was at it's highest, however in the BJT example the power dissipated in the transistor is it it's greatest when the drop is shared evenly between the load and the transistor.

By the simple maths of P=VI surely the same must be true of dissipation inside the valve?

Most of this is correct, but the first statement isn't. The power dissipation in the valve depends both on the anode current and on the energy with which the electrons arrive at the anode (you get hotter if you're hit by a fast-moving thing than if the thing is only crawling along). The energy of each electron on arrival depends on how much it's been accelerated by the voltage. So the power in the valve depends on current and voltage, exactly as in the BJT.

What burnt you in the case of the UL41, incidentally, was both the power available to heat your finger and also the temperature of the glass bulb which was what caused so much power to flow into your finger. A UL41 can dissipate 10W or so (anode + screen) and there's another 4.5W coming from the heater. A BJT dissipating 14.5W would definitely have to be on a heatsink to allow it to get rid of that power without getting too hot. In the absence of a heatsink it would become a 15W soldering iron. They burn too if you touch them.

Cheers,

GJ

[martin_m0nxp]

That still doesn't help though - surely in the 50% drop situation the electrons are moving slower than in the situation where there is little voltage drop across the tube (as they will have been partially deflected / slowed by the field on the grid?).

[Radio Wrangler]

The anode voltage sets up an elevtric field from cathode to anode. The heat of the cathode shakes loose electrons which being negatively charged experience a force from this field. The force accelerates the electrons towards the anode and the electrons acquire kinetic energy. Where does this energy come from? The field. As the negatively charged electron moves into more positive regions it cancels pare of the driving field, but if the anode is held at a fixed voltage, then more charge must leave it up the anode connection to maintain its potential. Think of it like a capacitor charging current, not to increase the potential on the capacitor, but to maintain the potential during an effective increase in capacitance.

So far, there has been no heat in the anode, but our electron has been getting fast. Electrical energy has been taken from the anode circuit and resides temporarily in the speed of the electron. When the electron hits the anode it is immediately decelerated and releases its kinetic energy as heat.

Everyone talks about the anode being heated by electron bombardment, but the supplying of the kinetic energy gets glossed-over. All of the energy released as anode heat has to be conveyed to the electron before it ever contacts the anode.

David

[Craig Sawyers]

A coulomb is the amount of charge flowing, and an amp is a charge of 1 coulomb flowing for 1 second.

Our one coulomb is the number of electron charges, or 1/(1.6 x 10^-19) = 6.25 x 10^18 electrons per second. And that is 1 amp.

So a typical valve with an anode current of 10mA will have an electron flow of 6.25 x 10^16 electrons per second. A lot of electrons flowing from cathode to anode!

And the (kinetic) energy gained is equal to the total charge times the (anode) voltage - say 200V. Note it is independent of the cathode-anode spacing.

So 6.25 x 10^16 x 1.6 x 10^-19 x 200 = 2W

Or, equivalently and more simply IV = 0.01 x 200 = 2W

The reason that the glass envelope gets hot with relatively small powers is the anode is heated by that 2W in vacuum. So the heat generated can only be dissipated by radiation. So it gets into a thermal equilibrium when 2W flows in from the electrons, and 2W flows out, usually in the infrared, via Stefan Boltzmann. Since most of that that is absorbed by the glass envelope, it heats up. Which cools mainly by convection - hence ventilation slots on valved gear casework.

Craig

[Craig Sawyers]

A beermat calculation of anode temperature if 2W is dissipated and the anode are is 2cm^2 with an thermal emissivity of 0.5, gives an anode temperature of ~400C.

That is an overestimate, since that is a dull red heat. I'll try and refine that calculation!

Craig

[Argus25]

I have always ignored the kinetic energy of the electrons or the energy produced by collision with the anode in terms of understanding the dissipation. The energy dissipated is all about the electric field and the charge that moves across that field.

The electric field potential or voltage is an energy density, which can be expressed in joules per Coulomb. So if a Coulomb of charge crosses a field of 1 volt then one joule of energy has to be applied if the charge is forced against the field, and one joule of energy is given to the charge if it goes with the field and in that case the charge acquires energy from the field. The released energy in the latter case as the charge arrives at the anode terminal can only be as heat, light or other invisible EM radiation. So a valve could get hot, have a red anode and a blue glow.In the case of a transistor it's mainly heat, but then of course there is the LED. What happens to that energy and how it raises the temperature depends on the usual thermodynamic principles of conduction, convection (outside the bulb) and radiation etc.

One very interesting thing that I did not know in my early work with valves was the factor that limited the maximum dissipation of a particular valve. It is the dissipation where the temperature of the metal parts (typically the anode) is below a value that was used in manufacture and out-gassing of the valve. If the temperature goes higher than that, gasses are released and the valve is poisoned.

[Craig Sawyers]

The kinetic energy of the electrons is the only mechanism for energy transfer in a valve. And it is electric-field independent, and depends only on the anode voltage.

How else can electrons, in this case behaving as classical charged bullets, interact with the anode other than via kinetic energy? There is nothing esoteric going on here.

Think about it this way. If the cathode/anode distance increases but the anode voltage stays the same, the electron has further to accelerate - but acquires precisely the same energy.

And a Joule per Coulomb is a Volt - nothing to do with energy density (which is Joules per cubic metre).

Craig

[GrimJosef]

Originally Posted by martin_m0nxp

That still doesn't help though - surely in the 50% drop situation the electrons are moving slower than in the situation where there is little voltage drop across the tube (as they will have been partially deflected / slowed by the field on the grid?).

Sorry, but that's wrong too. If there is 'little voltage drop across the tube' then there is nothing to accelerate the electrons. They will be moving faster, not slower, when there is 50% of the supply voltage across the tube than when there is little voltage across the tube.

Perhaps an easier way to see your way through this would to start out from the proposition that any device with 50V across it and 0.5A flowing through it (DC of course) will be dissipating 25W, and then see if you can come up with a reason why this is not what's happening in the valve ? (Answer: there is no such reason.)

Cheers,

GJ

[martin_m0nxp]

Sorry, but that's wrong too. If there is 'little voltage drop across the tube' then there is nothing to accelerate the electrons.

Don't be sorry - I posted the question to be educated!

The above was the crucial bit I was forgetting about - at the point when the maximum current is flowing the voltage drop across the tube is at it's least, so therefore less acceleration potential.

So the answer to my original question (phrased in a single paragraph for my future benefit) - power is dissipated in a tube in a two-stage process. First the energy is used to accelerate the electrons. This kinetic energy is then converted to heat when they strike the anode - the heat leaving the anode mostly through IR radiation which is absorbed by the glass. The power lost through this is therefore governed by the potential difference from cathode to anode and the current, so just as in a BJT there will be a peak dissipation at 50% conduction.

Thank you all for your wise words!

[ms660]

Some available valve data will show a max anode power diss. power curve in their Ia-Va curves, here's some valve data for the UL41:

http://www.r-type.org/pdfs/ul41.pdf

If you scroll down to the 3rd graph you will see the power curve (9 watts) for the max. anode dissipation, any point on that curve is the product of Ia*Va as would be expected.

Small power bottles in radios etc get very hot, basically too much going on in a small envelope.

Lawrence.

[Craig Sawyers]

Turns out my beermat calculation of anode temperature is not far out. There is chapter and verse on this in the RCA 1962 Electron Tube Design book (downloadable). Schade has an entire chapter on electrode temperatures in valves. Lots of theory dealing with everything from heat flow in grid wires to coupling to the glass envelope, and laminar/turbulent air flow over the glass envelope.

Anyhow, he calculates the temperature at the anode of a 6L6G and gets 425C, as compared with the measured temperature of 415C. The screen grid supports get up to a stonking 525C calculated (measured 515C).

So the internal temperatures in an operating valve are very high. And explains why the glass envelope temperature is so high. Which begs the question regarding how modern valved power amps have the valves exposed, which simply has to be against regulations regarding accessible hot surfaces!

Craig

[Argus25]

It does pay to remember though that the work done by the field is the same regardless of what goes on in the "black Box" which is the device under consideration, be it a valve. Transistor or resistor.

It is always the product of the charge that passes the two points (for a valve the cathode & and the anode) and the potential across the two points at the time and of course for dynamic processes and complex waveforms that would have to be calcualated, but for sine waves and DC it's easy. This remains true regardless of the processes; there could be electron oscillations emitting EM waves, heat from collisions etc. The power dissipated is always calculated from the external circuit currents & voltages around the "black box" not from the variable processes inside it.

(PS: one can consider voltage as an energy density if you consider coulombs as a volume or amount of electrons which they are)

[GrimJosef]

Originally Posted by Craig Sawyers

... the glass envelope temperature is so high. Which begs the question regarding how modern valved power amps have the valves exposed, which simply has to be against regulations regarding accessible hot surfaces!

I guess the same question arises with 100W filament bulbs in larger lamps (admittedly these are becoming a thing of the past) and also with electric irons. And soldering irons, now I come to think about it.

Quite a few modern valve amps do come with valve cages, although I've never seen an interlocked one of these.

I don't know if the approved test houses test for this. In the case of manufacturers who self-certify perhaps we won't get a ruling until someone either reports them to Trading Standards or takes them directly to court.

Cheers,

GJ

[broadgage]

I have seen high power valves in which the anodes glowed a fairly bright red.
The envelopes were made of Pyrex, ordinary glass would have softened.

Some high power anodes were made of graphite as this material stands higher temperatures than metal.
The drawback is the amount of outgassing that graphite produces, prolonged and therefore expensive high vacuum pumping is required whilst over running the valve, before it can be sealed.

[turretslug]

I'm sure there was a class of medium-power (hundreds of watts anode dissipation) valves whose anodes were made of something like molybdenum that actually needed to be run at dull red or above to ensure effective gettering, these types running too hot for the barium-type getters of lower-power valves.

I'd wondered about the de-gassing of graphite anodes, they sound as if they'd be quite problematic for both absorption and adsorption.

[kalee20]

Yes I have heard the same as turretslug, my memory says tantalum but molybdenum rings a vague bell too. These metals tend to absorb gas when hot, so 'self-getter.'

As for the OP's post, heat is produced at the heater (obviously) and also at areas which electrons hit, which is usually the anode and screen-grid.

Each electron carries a bit of energy, which is released when it hits the anode. Think of it as giving the surface a minute thump, which sets the atoms at point of impact jangling as they are constrained by their elastic bonds in the solid. Continuous impacts eventually set the whole structure in random motion - hey presto, it's getting hot.

The total energy received is proportional to the number of impacts per second (the current) and to the energy of each impact (how fast the electrons were travelling, which depends on the voltage that has accelerated the electron).

Occasionally, electrons go places they're not supposed to - in a CRT a wrongly-adjusted magnet can make them hit the glass of the neck of the tube, which then heats up, possibly to the point that it cracks.