5 ms·

You're right that the resistance is where the heat is dissipated, but lowering the resistance does not actually change the amount of heat. Transistor switching

by kayson 3y ago

You're right that the resistance is where the heat is dissipated, but lowering the resistance does not actually change the amount of heat. Transistor switching can be modeled as a step input to an RC circuit [1]. If you integrate the power through the resistor to infinity, you'll see that the value of the resistor drops out. Intuitively, you might think of it like this: to charge a capacitor (or transistor) up to a certain voltage, you need a fixed number of electrons. That number of electrons will always pass through the resistor and generate heat based on their energy. Even if you change the resistor value, its still the same number of electrons, and the same amount of energy. What does change with resistance, though, is the time over which the power is dissipated. In practice, you have to make sure the resistors are small enough such that you can achieve your desired clock speed. There are actual resistive losses too, but they're mainly related to power delivery. [1] https://en.wikipedia.org/wiki/RC_circuit#Time-domain_considerations https://en.wikipedia.org/wiki/RC_circuit#Time-domain_conside...

moffkalast 3y ago
Ok but again, why do we need the resistors at all? That's what limits the speed at which the capacitor can discharge, so with theoretically 0 resistance you'd get immediate discharge and could go to infinite frequencies, or more realistically as far as the speed of electrons allows for consistent gate switching. To add a bit of troll physics here (but I'm told computers using this sort of principle actually exist), why not then channel those electrons to a boost converter that pipes them back into VCC, recycling most of the current? Theoretically a 99% power usage improvement, minus what the converter loses to heat, and that can be as low as 10%.
- thfuran 3y ago
  There's no such thing as superconducting semiconductors.
- kayson 3y ago
  It's not like the resistors are a component that is explicitly added. Every conductor has some resistance - the tiny wires that connect transistors together, the transistor itself, etc. Ideally, yes, you want those resistances to be as low as possible. But in practice there are design trade offs that happen if you do so; it's a balancing act. As the other commenter mentioned, there are no superconducting semiconductors, so the transistor is out. There has been some research into super conductors for the wires, but for the time being there is nothing that's easily integrated into existing manufacturing processes. Re: channeling electrons - what you've described doesn't quite make sense. Fundamentally, if you're taking an electron at ground or 0V potential, and changing it's potential to VCC, it requires energy that comes from somewhere. The battery (or power supply) is doing exactly that. As the electrons flow back to the ground, the battery "recharges" them up to VCC potential. What you can do, though, is put circuits in series between supply and ground. That way the electrons flow through the "top" circuit, do their thing, then flow through the "bottom" circuit. There's no free lunch though, as the voltage across each circuit will be reduced. Nonetheless, this is a common technique for low power analog circuits, and one I've used in the past. It's just not practical or worth it in digital circuits like a CPU.

by kayson 3y ago

moffkalast 3y ago

Ok but again, why do we need the resistors at all? That's what limits the speed at which the capacitor can discharge, so with theoretically 0 resistance you'd get immediate discharge and could go to infinite frequencies, or more realistically as far as the speed of electrons allows for consistent gate switching. To add a bit of troll physics here (but I'm told computers using this sort of principle actually exist), why not then channel those electrons to a boost converter that pipes them back into VCC, recycling most of the current? Theoretically a 99% power usage improvement, minus what the converter loses to heat, and that can be as low as 10%.

thfuran 3y ago
There's no such thing as superconducting semiconductors.
kayson 3y ago
It's not like the resistors are a component that is explicitly added. Every conductor has some resistance - the tiny wires that connect transistors together, the transistor itself, etc. Ideally, yes, you want those resistances to be as low as possible. But in practice there are design trade offs that happen if you do so; it's a balancing act. As the other commenter mentioned, there are no superconducting semiconductors, so the transistor is out. There has been some research into super conductors for the wires, but for the time being there is nothing that's easily integrated into existing manufacturing processes. Re: channeling electrons - what you've described doesn't quite make sense. Fundamentally, if you're taking an electron at ground or 0V potential, and changing it's potential to VCC, it requires energy that comes from somewhere. The battery (or power supply) is doing exactly that. As the electrons flow back to the ground, the battery "recharges" them up to VCC potential. What you can do, though, is put circuits in series between supply and ground. That way the electrons flow through the "top" circuit, do their thing, then flow through the "bottom" circuit. There's no free lunch though, as the voltage across each circuit will be reduced. Nonetheless, this is a common technique for low power analog circuits, and one I've used in the past. It's just not practical or worth it in digital circuits like a CPU.