4 ms·

Exactly. If you increase the clock frequency you will get more ops/s out of the CPU, everyone can see that easily. The only problem is that the anmount of elect

by nom 10y ago

Exactly. If you increase the clock frequency you will get more ops/s out of the CPU, everyone can see that easily. The only problem is that the anmount of electrons you have to push through the transistors has a quadratic* dependence on frequency. However, It's also clear what gravypod is trying to say. The architecture itself has a greater influence on the "processing power" nowadays because the clock has reached physical limits and can't be increased much further without getting expensive (see IBM Power Sytems). Bottomline is that you can't compare CPUs by frequency alone, but it was never any different, the architectures 15 years ago just sucked so hard that all of them were too similar regarding instructions per cycle. I think we can blame AMD for that line of thinking, they had to push hard against Intel during a time where everyone used clock frequency to compare CPUs. *edit: Sorry I remembered it wrong, the commonly used power equation is P = C x V^2 x f (capacity, voltage, frequency). Power is linear dependent on the frequency!

static_noise 10y ago
> The only problem is that the anmount of electrons you have to push through the transistors has a quadratic dependence on frequency. I don't see how this is ultimately true. The amount of electrons needed to flip a capacitor is constant given that the voltage is constant. The more often you flip, the more current you need per time. The amount of electrons needed to flip a capacitor is proportional to the voltage. So, lower voltage means a) less power per electron and b) less electrons. Therefore the power is quadratic with the operating voltage. Third, if you want to increase the frequency, at some point, you have to increase the voltage or you get errors. Then you increase frequency and voltage at the same time and your power consumption shoots up.
- nom 10y ago
  Hmm maybe I'm wrong. I vividly remember an equation for the power consumption of a CPU, and it definitely had a f^2 in it. However, I currently can't derive it anymore (and it was definitely a simplification). I'll try to pull up the slides later. It's probably outdated, the influence of the individual factors may have changed. In the meantime I can only offer this [1] stackexchange post. Power of a CPU is definitely quadratic in f, or even worse. [1] http://physics.stackexchange.com/questions/34766/how-does-power-consumption-vary-with-the-processor-frequency-in-a-typical-comput http://physics.stackexchange.com/questions/34766/how-does-po... edit: I probably remembered it wrong and I now think that that the equation was P=C x V^2 x f. However, transition power effectively has an exponent much greater than 2, but that's apparently caused by the increased die temperature (according to the post on Anandtech).
  - static_noise 10y ago
    That linked thread on Anandtech digs in quite a bit. The real physics involved is, of course, more complex than my simple model. Instead of a square dependence, they find the 5th power of the supply voltage. I believe this has to do with how the transistors are designed which creates a lot of wasted current at higher voltages. The square dependence would show with perfect transistors.

by nom 10y ago

static_noise 10y ago

> The only problem is that the anmount of electrons you have to push through the transistors has a quadratic dependence on frequency. I don't see how this is ultimately true. The amount of electrons needed to flip a capacitor is constant given that the voltage is constant. The more often you flip, the more current you need per time. The amount of electrons needed to flip a capacitor is proportional to the voltage. So, lower voltage means a) less power per electron and b) less electrons. Therefore the power is quadratic with the operating voltage. Third, if you want to increase the frequency, at some point, you have to increase the voltage or you get errors. Then you increase frequency and voltage at the same time and your power consumption shoots up.

nom 10y ago
Hmm maybe I'm wrong. I vividly remember an equation for the power consumption of a CPU, and it definitely had a f^2 in it. However, I currently can't derive it anymore (and it was definitely a simplification). I'll try to pull up the slides later. It's probably outdated, the influence of the individual factors may have changed. In the meantime I can only offer this [1] stackexchange post. Power of a CPU is definitely quadratic in f, or even worse. [1] http://physics.stackexchange.com/questions/34766/how-does-power-consumption-vary-with-the-processor-frequency-in-a-typical-comput http://physics.stackexchange.com/questions/34766/how-does-po... edit: I probably remembered it wrong and I now think that that the equation was P=C x V^2 x f. However, transition power effectively has an exponent much greater than 2, but that's apparently caused by the increased die temperature (according to the post on Anandtech).
- static_noise 10y ago
  That linked thread on Anandtech digs in quite a bit. The real physics involved is, of course, more complex than my simple model. Instead of a square dependence, they find the 5th power of the supply voltage. I believe this has to do with how the transistors are designed which creates a lot of wasted current at higher voltages. The square dependence would show with perfect transistors.