quentron.com

[Madebymonkeys]

OK, Rewind:
Do you see any real questions in there?

Ok, a simple one which I can't get my head around too well (although the answer could be quite simple)!:
How does I2R apply to quenco as a lossy system?
For example, with the 1cm^3, how does the quenco not go into thermal runaway if its powered by heat?
And if its powered by heat but it cools as it runs - where is the break even point where it fails to operate?!

It's looking more like a messed-up peltier to me with no practical method of maintaining a differential across it (cheaply and space efficiently).

As I implied, maybe I am missing something, if you can let me know the answers that would be awesome.

[Madebymonkeys]

Ok, a simple one which I can't get my head around too well (although the answer could be quite simple)!:
How does I2R apply to quenco as a lossy system?
For example, with the 1cm^3, how does the quenco not go into thermal runaway if its powered by heat?
And if its powered by heat but it cools as it runs - where is the break even point where it fails to operate?!

It's looking more like a messed-up peltier to me with no practical method of maintaining a differential across it (cheaply and space efficiently).

As I implied, maybe I am missing something, if you can let me know the answers that would be awesome.

FYI, I am not disputing tunneling electrons (everyone knows that's proven) - I am just unsure how this can be a practical source of power. I am considering a system rather than quenco in isolation (which tbh could do whatever it likes but if its not possible to consume usable power from it then it falls flat on its ars#).

[lumen]

Ok, a simple one which I can't get my head around too well (although the answer could be quite simple)!:
How does I2R apply to quenco as a lossy system?
For example, with the 1cm^3, how does the quenco not go into thermal runaway if its powered by heat?
And if its powered by heat but it cools as it runs - where is the break even point where it fails to operate?!

It's looking more like a messed-up peltier to me with no practical method of maintaining a differential across it (cheaply and space efficiently).

As I implied, maybe I am missing something, if you can let me know the answers that would be awesome.

I thought the same thing, with low voltage output and high current, there is going to be a lot of heat that in turn would produce more current.

Additional layers to increase the voltage would help solve much of the local heating but when viewed correctly, there is no real problem.

Suppose you have a low voltage quenco between two heat sinks and only a heavy copper wire shorting the circuit.
The ambient heat will provide electron flow which in turn produces heat in the wire and in the heat sink. This heat originally was consumed from the environment inside the box and serves to help cool the wire and heat sink which reduces the output. However nothing actually changed since the heat produced is indeed the same heat that was consumed.

The process in this case may trap some of the initial heat from the environment inside the box in a loop within the heat sinks and wire, but when disconnected the heat would again balance out to starting conditions.

If you had two boxes, one with the quenco and heatsink and the other with the shorting wire, then the heat form the quenco would accumulate in the other box and the quenco box would become colder until infinity, except the process would become slower and slower as it cools.

The reason is that eventually in the colder environment the random encounters between electrons will less often occur that will energize any single electron with enough energy to tunnel the barrier. This is where a thinner barrier will allow the device to operate to a lower temperature, but will also lower the working voltage of each layer.

Of course, Theory is the word at this time.

[Madebymonkeys]

I thought the same thing, with low voltage output and high current, there is going to be a lot of heat that in turn would produce more current.

Additional layers to increase the voltage would help solve much of the local heating but when viewed correctly, there is no real problem.

Suppose you have a low voltage quenco between two heat sinks and only a heavy copper wire shorting the circuit.
The ambient heat will provide electron flow which in turn produces heat in the wire and in the heat sink. This heat originally was consumed from the environment inside the box and serves to help cool the wire and heat sink which reduces the output. However nothing actually changed since the heat produced is indeed the same heat that was consumed.

The process in this case may trap some of the initial heat from the environment inside the box in a loop within the heat sinks and wire, but when disconnected the heat would again balance out to starting conditions.

If you had two boxes, one with the quenco and heatsink and the other with the shorting wire, then the heat form the quenco would accumulate in the other box and the quenco box would become colder until infinity, except the process would become slower and slower as it cools.

The reason is that eventually in the colder environment the random encounters between electrons will less often occur that will energize any single electron with enough energy to tunnel the barrier. This is where a thinner barrier will allow the device to operate to a lower temperature, but will also lower the working voltage of each layer.

Of course, Theory is the word at this time.

I was thinking more about the interface between the shorting wire and the quenco itself being a high (ish) resistance...and local to the quenco. What you said makes sense but I will need to digest the info a bit more!

I haven't done the math but for the cubic cm it looks like there will be many many thousands of watts dissipated from the junction - that's not ideal.

Cracking a few nm of material between two lumps of metal will become quite easy I think, once the differential between the two sides increases.

Let me digest a bit more.

[lumen]

I was thinking more about the interface between the shorting wire and the quenco itself being a high (ish) resistance...and local to the quenco. What you said makes sense but I will need to digest the info a bit more!

I haven't done the math but for the cubic cm it looks like there will be many many thousands of watts dissipated from the junction - that's not ideal.

Cracking a few nm of material between two lumps of metal will become quite easy I think, once the differential between the two sides increases.

Let me digest a bit more.

There is no temperature differential between the two sides. The quenco converts from an isothermal environment.

I do think because the electrons move from one side to the other that it will cause it's own differential and start to cool one side then loose efficiency.
The heat sinks from each side would need to be thermally connected and electrically isolated to maintain the same temperature on each side of the chip as close as possible.

I did some heat modeling on this and if you try to pull 5000W across 1 square centimeter area, there is a large temperature difference even in a solid copper heatsink.

Copper is just not conductive enough to supply ambient temperatures without the chip being 30F to 50F cooler even when trapped in solid copper blocks.
If the electrical junction did induce heat, at least the chip would be running in a more efficient temperature range.