Question about how valves work.

[shane]

I had a spare QQV with a missing pin, so being curious I cracked open the envelope to see what was inside, and a thought occurred to me.

In my QQV amp, I have 280v on the anodes, and about 25v on the cathode. If the cathode resistor is 270R, that gives a current of 90mA shared between the two halves of the valve, which means that about 26w is being dissipated somewhere between the anodes and cathode. My question is, where? The anodes and cathode themselves are substantial chunks of metal which do not have significant resistance, and the space between them is empty except for a stream of electrons and the grids, so how does the power generated heat up the anode, which I presume is where the energy goes?

[ed]

http://en.wikipedia.org/wiki/Thermionic_emission

[Mike H]

This had me going round in circles for years, at one point I believed vacuum tubes can drop a Voltage without generating any heat, brilliant I thought, but of course they must do, V x I = Watts

The answer is here:

http://en.wikipedia.org/wiki/Vacuum_tub ... d_transfer

[Nick]

I think to understand the answer to the question you need to consider the electrons. The voltage between the cathode and anode has a voltage potential. The electrons emitted by the cathode are accelerated by the voltage gradient. the higher the voltage, the faster the electrons are moving. Like anything with mass, the faster they go the more energy they contain (consider being hit by a brick at 1meter/hour or 1meter/second). when the electrons hit he anode they are stopped dead, so the kinetic energy (energy of movement) is transferred to the anode and heats it up. Thats where the heat comes from. Take a lump of steel, hit it with a hammer for 10 minutes, and it will get warmer. The same thing happens to the anode. The electrons are very small, but they are moving fast, and there are a lot of them.

[shane]

Ah, light dawns. For some reason I thought electrons didn't have mass.

[pre65]

Ah, light dawns. For some reason I thought electrons didn't have mass.

Some are C of E.

[jack]

Ah, light dawns. For some reason I thought electrons didn't have mass.

Some are C of E.

[pendant-mode]Anglicans also do this, but generally call it Holy Communion - the term "Mass" is considered Anglo-Catholic[/pedant-mode]

http://en.wikipedia.org/wiki/Thermionic_emission

Can't sleep , so here are some numbers:

You can use simpler maths than that to show what's going on. What follows is an ideal-case solution, ignoring some of the losses that occur in a real valve...

The anode heats up due to it being bombarded with electrons from the cathode. If we know the number of electrons arriving and the energy they carry, we can work out the incident energy in the electron stream and thus how much power the anode has to dissipate.

The energy gained by an electron moving though a potential difference is E = QV, where Q is the charge on the electron and V is the potential difference. If the electron starts at rest (from the cathode), E is just the kinetic energy of the electron.
For any electron, Q = 1.6e-19 Coulombs (this is a constant)

NB: The definition of an Ampere is the passage of 1 Coulomb per second past a point (about 6.25e18 electrons)

So, as an example, lets look at the Ia vs. Va graph for an ECC83C http://en.wikipedia.org/wiki/File:Triod ... istic1.png. At Vg = 0 (fully conducting) we can read off that Ia = 5mA at Va = 225V. Assume Vk = 0.

Thus, in our example, the energy per electron, E, = QV = 1.6e-19 * 225 = 3.6e-17 Joules

...and the number of electrons per second, N, for a given Ia = total current/charge per electron = Ia / Q = 5e-3 / 1.6e-19 = 3.125e16 electrons/sec going from the cathode to the anode.

The total energy per second of the electrons impacting the anode is therefore the number of electrons per second x their energy which = N * E or Ia / Q * Q * Va. The "Q"s cancel out leaving us with the anode power dissipation being Ia * Va (as expected) or 5e-3 * 225 = 1.125 Watts.

The anode therefore heats up - there can be no heat loss though convection (we are in an vacuum) so conduction via the anode connection and black-body radiation (infra-red) are the only ways.

As an aside, given the kinetic energy of each electron from the E = QV equation above, we can determine how fast they are going.
Using the general-case kinetic energy equation E = 0.5 * m * v * v, we re-arrange to get v = sqrroot(2E/m).

The mass of an electron, m, is = 9.1e-31 Kg and we have E from the equations above (3.6e-17 J) so v = 8.9e6 metres per second (about 20 million mph ). The speed of light, c, is about 3e8 metres per second, so our electrons are doing about 0.03 c. If we say (guess) that the cathode->anode gap is about 5mm, then each electron takes approximately 560pS to make the trip... some things in valves happen quite quickly...

Ho, hum...

Edit: Just noticed that my example is outside the SOA in the datasheet - the maths is still valid, though

[Mike H]

Flippin' 'eck....

Some are C of E.