18-09-2015, 09:57 PM
Not sure about your numbers and can't sleep 
, so here are my numbers: 
(I originally wrote this for another forum...)
You can use simple 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:Triode...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
, so here are my numbers: (I originally wrote this for another forum...)You can use simple 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:Triode...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
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