[Tuna-Fish]

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]

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]

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]

Ahh right. I didn't think of that. Thank you for explaining.

[hatsunearu]

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)

[dougmwne]

Citation: thermodynamics.

https://en.m.wikipedia.org/wiki/Landauer%27s_principle

[EvgeniyZh]

We are very far from Landauer limit tho

[guerrilla]

I get it now. Thanks.

[calf]

Where was the argument, may I see it?

I vaguely remember switching current and leakage current as sources of heat, but mircoelectronic circuits has been a while for me...

[scarmig]

I believe they're referring to this comment thread I started:

https://news.ycombinator.com/item?id=36881808&p=2#36884542