5 ms·

Conduct electricity with zero resistance. This has been possible at cryogenic temperatures (powers supermagnets, MRI etc) but running at room temperature _and p

Conduct electricity with zero resistance. This has been possible at cryogenic temperatures (powers supermagnets, MRI etc) but running at room temperature _and pressure_ means in theory that you don’t need expensive support machinery. Has been the holy grail for the field. If true, this specific case is only low current, but demonstrates such a thing is possible - almost certainly winning an instant Nobel Prize.

8n4vidtmkvmk 3y ago
Does this mean there would be no heat waste? And devices like phones and computers wouldn't get hot?
- alangibson 3y ago
  Yes, among many other things. There's many vast amounts of electricity wasted due to transmission losses. Magnets for fusion plants would become much cheaper.
- zparky 3y ago
  Correct
- _joel 3y ago
  and faster
- Tuna-Fish 3y ago
  Computers would still get hot, because the main cause of heat is not resistance in wires, it's that when a transistor switches with charge in the gate, you need to dump that charge, and when you do this it turns to heat. They would get less hot, because there are plenty of transmission losses too.
  - guerrilla 3y ago
    People were arguing about this earlier (some claimimg the opposite) but I didn't see a conclusion. Do you have an external citation handy?
    - hatsunearu 3y ago
      I was one of the people who said power is wasted in wires. When the charged stored inside a chip needs to change (like go from high to low), that energy associated with the charge needs to go somewhere. Currently most of the charge is dissipated in the wire and some of it within the transistor. If the wires have no resistance, the transistor will be the one dissipating the energy, not the energy simply disappears. So technically both options are right, but my position is the “technically right, but practically wrong” position lol That said if the interconnects are less resistive, the switching process becomes more efficient and less power will be wasted than the bare minimum required.
      - guerrilla 3y ago
        Ahh right. I didn't think of that. Thank you for explaining.
        hatsunearu 3y ago
        I wanted to add but my morning poop time was over so never had a chance to write this. If you have two capacitors, say C1 and C2, and say their capacitance is both C. Say C1 is charged up to 2V. The energy stored in that cap is 1/2 CV^2 = 2C. Say C2 is not charged up. The charge on C1 is Q = CV = 2C. The charge on C2 is 0. Say suddenly you connect C1 to C2 via a lossless wire. The charge on C1 and C2 must be equal before and after the connection since charge cannot be destroyed. After the connection, the voltage on both caps will be equal, and the charge on them will therefore be equal. Charge on each cap is 1C; voltage is therefore Q/C = V, V = 1V on both caps.. Now let's look at the energy on either cap. 1/2 CV^2 = 0.5C. The combined energy on both caps are 1C. Where'd the energy go? We just said we connected the capacitors with a lossless wire, so it can't be dissipated there, and capacitors by definition cannot dissipate power. The answer is that lossless wires cannot exist, and if you do the math more carefully, this time with real resistance and take the limit as R->0, you will see that the power-time integral of the wire (dissipated energy) will approach 1C. The same argument can be made for integrated circuits; as your resistance drops, more and more of the portion of loss will be dissipated in the "other" sources of loss. edit: superconductors are lossy at AC (but not as much as regular conductors get lossy at AC), and the capacitor connection is an AC phenomena so even with superconductors there will be a teeny loss even with superconducting wires. The rest of the loss will happen in other ways (EM radiation, dielectric loss, loss from capacitor resistance, etc)
- ulrikrasmussen 3y ago
  It is my understanding that computers would still need to somehow dissipate energy when they perform irreversible computations, and that will turn into heat. E.g. when you compute the AND of two bits, then the result is only one bit and you have to dispose of the remaining bit either as heat or as a garbage output signal.
  - scarmig 3y ago
    That's a thermodynamic limit, but we're not even close to hitting that yet. RTP superconductors still aren't going to magically make computers emit zero heat, though; there are other sources besides resistive losses. I was under the impression that other factors dominated, though a couple people responded yesterday to tell me that resistance is the primary source of heat. Not an expert in that area, would love for someone who does chipset design to clarify.
- briffle 3y ago
  One of the biggest use cases (for superconductors that meet their requirements) would be long distance power transmission, where as much as 30% of the power is lost over long distances. Essentially, (and very top level) you could produce 30% more power, without adding any more production capactity.
  - 0cf8612b2e1e 3y ago
    HVDC has losses of more like 3%.
- atq2119 3y ago
  There's the rather non-trivial problem of actually manufacturing chips with the superconducting material. But if that can be solved, then yes, it could make computing way more efficient.
zackees 3y ago
Also, the clock speed of CPUs are limited by resistance and heat. Both of those go to zero. We may see 100ghz CPUs and 3D stacked cpus with almost no need for cooling.
- lambdasquirrel 3y ago
  Wouldn’t you still get heat from current switching back and forth and transistors changing polarity?
- OkayPhysicist 3y ago
  At those frequencies, induction would be a huge headache.

8n4vidtmkvmk 3y ago

Does this mean there would be no heat waste? And devices like phones and computers wouldn't get hot?

alangibson 3y ago
Yes, among many other things. There's many vast amounts of electricity wasted due to transmission losses. Magnets for fusion plants would become much cheaper.
zparky 3y ago
Correct
_joel 3y ago
and faster
Tuna-Fish 3y ago
Computers would still get hot, because the main cause of heat is not resistance in wires, it's that when a transistor switches with charge in the gate, you need to dump that charge, and when you do this it turns to heat. They would get less hot, because there are plenty of transmission losses too.
- guerrilla 3y ago
  People were arguing about this earlier (some claimimg the opposite) but I didn't see a conclusion. Do you have an external citation handy?
  - hatsunearu 3y ago
    I was one of the people who said power is wasted in wires. When the charged stored inside a chip needs to change (like go from high to low), that energy associated with the charge needs to go somewhere. Currently most of the charge is dissipated in the wire and some of it within the transistor. If the wires have no resistance, the transistor will be the one dissipating the energy, not the energy simply disappears. So technically both options are right, but my position is the “technically right, but practically wrong” position lol That said if the interconnects are less resistive, the switching process becomes more efficient and less power will be wasted than the bare minimum required.
    - guerrilla 3y ago
      Ahh right. I didn't think of that. Thank you for explaining.
      - hatsunearu 3y ago
        I wanted to add but my morning poop time was over so never had a chance to write this. If you have two capacitors, say C1 and C2, and say their capacitance is both C. Say C1 is charged up to 2V. The energy stored in that cap is 1/2 CV^2 = 2C. Say C2 is not charged up. The charge on C1 is Q = CV = 2C. The charge on C2 is 0. Say suddenly you connect C1 to C2 via a lossless wire. The charge on C1 and C2 must be equal before and after the connection since charge cannot be destroyed. After the connection, the voltage on both caps will be equal, and the charge on them will therefore be equal. Charge on each cap is 1C; voltage is therefore Q/C = V, V = 1V on both caps.. Now let's look at the energy on either cap. 1/2 CV^2 = 0.5C. The combined energy on both caps are 1C. Where'd the energy go? We just said we connected the capacitors with a lossless wire, so it can't be dissipated there, and capacitors by definition cannot dissipate power. The answer is that lossless wires cannot exist, and if you do the math more carefully, this time with real resistance and take the limit as R->0, you will see that the power-time integral of the wire (dissipated energy) will approach 1C. The same argument can be made for integrated circuits; as your resistance drops, more and more of the portion of loss will be dissipated in the "other" sources of loss. edit: superconductors are lossy at AC (but not as much as regular conductors get lossy at AC), and the capacitor connection is an AC phenomena so even with superconductors there will be a teeny loss even with superconducting wires. The rest of the loss will happen in other ways (EM radiation, dielectric loss, loss from capacitor resistance, etc)
ulrikrasmussen 3y ago
It is my understanding that computers would still need to somehow dissipate energy when they perform irreversible computations, and that will turn into heat. E.g. when you compute the AND of two bits, then the result is only one bit and you have to dispose of the remaining bit either as heat or as a garbage output signal.
- scarmig 3y ago
  That's a thermodynamic limit, but we're not even close to hitting that yet. RTP superconductors still aren't going to magically make computers emit zero heat, though; there are other sources besides resistive losses. I was under the impression that other factors dominated, though a couple people responded yesterday to tell me that resistance is the primary source of heat. Not an expert in that area, would love for someone who does chipset design to clarify.
briffle 3y ago
One of the biggest use cases (for superconductors that meet their requirements) would be long distance power transmission, where as much as 30% of the power is lost over long distances. Essentially, (and very top level) you could produce 30% more power, without adding any more production capactity.
- 0cf8612b2e1e 3y ago
  HVDC has losses of more like 3%.
atq2119 3y ago
There's the rather non-trivial problem of actually manufacturing chips with the superconducting material. But if that can be solved, then yes, it could make computing way more efficient.

zackees 3y ago

Also, the clock speed of CPUs are limited by resistance and heat. Both of those go to zero. We may see 100ghz CPUs and 3D stacked cpus with almost no need for cooling.

lambdasquirrel 3y ago
Wouldn’t you still get heat from current switching back and forth and transistors changing polarity?
OkayPhysicist 3y ago
At those frequencies, induction would be a huge headache.