Ryan_m_b said:
Recently I've been in a discussion with someone online who claimed that the efficiency of a computer with regards to energy is not a linear relationship. I assumed that if you were to halve the power going into a computer chip (assuming it was designed to deal with that and wouldn't just turn off) it was logical it would only work at half the capacity. They claimed that there are some fundamental reasons why this isn't the case and that actually the relationship is computation rate in bits/sec = power^(3/4). In other words if you drop the power by to 50% you only drop the computation rate to 59%.
I can't get my head around this and unfortunately they heard it from someone else (that I'm reasonably certain is a credible source but is difficult for me to contact). I've done a lot of searching around online and can't find any details on this. Is it true? Is there any relationship like this?
Thanks!
Could be. But since your "someone" had heard it from a friend, he/she could be talking apples to your oranges. It even could be a description of the curve explaining the increased power consumption of CPU's that went tfrom 1Mz to 4 to 60 to 400 to GhZ from the 80's onward, with increased byte length thrown in also. A version of Moore's law. In which case it has little application to one particular machine.
For a gate,
The basic equation is Ohm's law, P=EI, applied dynamically to a reaction circuit. Since, at the the gate (CMOS), oe is just charging and uncharging a capacitor to change the bit from ON/OFF, this becomes P = CV
2f, where C is the capacitance, V the applied voltage, and f the frequency. If one relates f to bits/sec, than that is just the basic starting form to work with.
Of course there are other things to consider when manufacturing a chip. Such as gate density on a chip which goes by N
2, which would increase power consumption by the same rate per chip. One way to increase gate density is to make each gate smaller. Smaller gates means quicker turn ON/OFF by a factor of N, which means the chip could either use a lower supply voltage or an increased frequency with the same power usage.
In case you are wondering how P=EI translates to P=CV
2f for a capacitor, you can recall that for a capacitor, the energy stored suffers the following relationship,
E = 1/2 QV, where E is the energy, Q is the charge on the capacitor, and V the voltage.
with P=E/t, P=1/2 QV/t = 1/2 QVf ( Or, = IV with I=Q/t )
Of course another half of energy from the power supply is dissipated in the resitive elements (in the chip) when the capacitor is charging or discharging, so we do get the full P = CV
2f when we know that Q = CV.
I just thought I would write that down for you as an addition to the thread.