C States and actual wattage - no more help needed

[Tweakin]

So,

When you manually set the core voltage for a cpu, this voltage will remain constant regardless of the C-State of the chip. My question is this:

Regardless of the input voltage, will the current (wattage) consumption be reflective of the chips usage, i.e. while in a lower C-State Condition the chip will require/consume less current and have a lower wattage requirement.

If this is true, what is that added benefit (if any) of using an auto voltage?

 

[Dman8777]

I'm not sure if this is correct but I assume most of the power usage for a processor is consumed during switching operations of the transistors (charging the parasitic capacitances). When the processor is idle and nothing is being switched, the power consumption will be significantly reduced regardless of the voltage level.

I'm not exactly sure what you mean by auto voltage but if it's like any modern proc, the voltage drops down during idle periods and the proc downclocks. At lower clockspeeds the logic levels are easier to detect and consequently, a logic high doesn't have to be as high as it would at higher clockspeeds. A lower voltage results in lower current spikes during switching and also decreases power loss through leaky transistors and such.

 

[Tweakin]

...I'm not exactly sure what you mean by auto voltage...

Auto meaning default regulation of the vcore voltage within the bios, manual meaning user input of a specific value.

 

[videogames101]

Easy formula:

Power_dynamic = Cap_total * Vdd^2 * Activity_Factor * Frequency

Reducing your activity (less switching because fewer active transistors), frequency (less switching across chip), or voltage (less energy per switch event) will reduce dynamic power consumption. Capacitance is a physical parameter which is determined by the chip design.

Generally when you reduce your frequency it allows for lower stable voltage as well.

So in a c-state, frequency is lowered to reduce power consumption, and because frequency is lower Vdd can be lowered without causing instability causing even further power savings.

Also, static power consumption is proportional to Vdd, so lowering Vdd reduces static power too.

 

[Tweakin]

Easy formula:

Power_dynamic = Cap_total * Vdd^2 * Activity_Factor * Frequency

Reducing your activity (less switching because fewer active transistors), frequency (less switching across chip), or voltage (less energy per switch event) will reduce dynamic power consumption. Capacitance is a physical parameter which is determined by the chip design.

Generally when you reduce your frequency it allows for lower stable voltage as well.

So in a c-state, frequency is lowered to reduce power consumption, and because frequency is lower Vdd can be lowered without causing instability causing even further power savings.

Also, static power consumption is proportional to Vdd, so lowering Vdd reduces static power too.

So why are so many users staying with "auto" voltage to gain better power savings?

 

[JoeRambo]

So why are so many users staying with "auto" voltage to gain better power savings?

It boils down to the following simplified explanation: C states or not, CPU has to execute certain amount of instructions even when idle. What it means in practice - CPU has certain amount of so called "unhalted" cycles. So if CPU was dropped to low multiplier and "nominal" clock is now lets say 800Mhz, it still has to wake up to execute those instructions.

And to (reliably) execute instructions during those cycles most efficiently you need just enough voltage to run that frequency your CPU is at now.

Auto voltage means that CPU simply knows that for reliable 800Mhz, it needs maybe 0.8V and requests that. Manual would run on maybe 1.2V ( cause it needs that voltage to reliably operate @4.5Ghz ). So if you have 200Mhz worth of unhalted cycles to execute, that 0.45V extra can turn into substantial savings.


EDIT: and before You worry too much, we are talking about several watts here on Haswell class CPU when idle, as long as C and package power states are manually enabled, you are going to be fine. You can burn substantially more for example with default settings on Asus MBs that run uncore at fixed 3X00Mhz on Haswell CPUs.

 

[TuxDave]

So,

When you manually set the core voltage for a cpu, this voltage will remain constant regardless of the C-State of the chip. My question is this:

Regardless of the input voltage, will the current (wattage) consumption be reflective of the chips usage, i.e. while in a lower C-State Condition the chip will require/consume less current and have a lower wattage requirement.

If this is true, what is that added benefit (if any) of using an auto voltage?

I'm not really sure what you're saying but:

Wattage is not the same thing as current. Wattage/Power is current * voltage. If you run a CPU at a lower voltage, the voltage term goes down AND the amount of current (to charge up the nodes) go down. That's why hand-wavy math shows wattage being related to voltage^2.

You only need to bump up voltage to hit higher frequency. So if you're running at lower frequency AND you didn't allow the voltage to automatically go down to take advantage of that lower frequency, you are needlessly wasting power. That's why on the first order of understanding, "auto-voltage" allows you to save power.

 

[Tweakin]

Ok, I give...I'm going to try and ask the same question in another manner...here it goes.

If the default "auto" voltage of any current intel chip is 1.075 while in windows, it will throttle down to, lets say .075 as the freq of the chip drops down due to C-State.

Same scenario but now I "manually" set the vcore to 1.075 and while in windows, it remains at this voltage even though the freq of the chip has dropped down due to C-State.

My question is this...which one costs less to run over a years time. I'm not asking about OC'ing, under-volting or Ohms or Kirchhoff's law or any other physics...just a simple EE question, which one will cost me less to run.

 

[Tweakin]

...and before You worry too much...

I'm ok...it was just a question. I believe that over a years time the "auto" would save your a few pennies, but I keep finding posts about "efficient" overclocking by using the auto setting and establishing a proper "offset" to ensure you don't overshoot the desired range, and I don't think it is "efficient" as the "Auto" range will normally cause an excessive amount of voltage to ensure stability, which is normal operation.

So, after a year I believe both settings would be almost the same when you paid your electrical bill...that's all.

 

[JoeRambo]

The reality is that most CPUs are spending time being either idle, or running way below available power. So 2-3 watts saved when idle, multiplied by thousands of hours spent idling, would need hundreds of hours and @load savings in the 10+ watt range to offset it. There are 8760 hours in a year, so even if you run 24/7 saved or burned watt or two will not exactly matter (<-polar bears might not agree).

Still, I find it completely irrelevant, cause currently most are running either locked Intel CPU on auto, or are probably running Haswell K CPU on manual (cause auto causes ridiculous voltages in AVX/FMA loads).

 

[TuxDave]

Ok, I give...I'm going to try and ask the same question in another manner...here it goes.

If the default "auto" voltage of any current intel chip is 1.075 while in windows, it will throttle down to, lets say .075 as the freq of the chip drops down due to C-State.

Same scenario but now I "manually" set the vcore to 1.075 and while in windows, it remains at this voltage even though the freq of the chip has dropped down due to C-State.

My question is this...which one costs less to run over a years time. I'm not asking about OC'ing, under-volting or Ohms or Kirchhoff's law or any other physics...just a simple EE question, which one will cost me less to run.

If we're making up numbers, let's run with this:
Power = C * V^2 * F. Let's make C = 1

Auto-voltage on:
-------------------------------
Freq = 1, Voltage = 1, Power = 1 (V^2 * f)
Freq = 0.1, Voltage = 0.5, Power = 0.025

Auto-voltage off:
-------------------------------
Freq = 1, Voltage = 1, Power = 1
Freq = 0.1, Voltage = 1, Power = 0.1

That's your EE answer without going into the finer details of the CPU architecture.

 

[Tweakin]

...most are running either locked Intel CPU on auto, or are probably running Haswell K CPU on manual (cause auto causes ridiculous voltages in AVX/FMA loads).

That was my point! Throw the math out the window as there are too many variables to include in the algorithm. If the auto voltage swings way above and way below "nominal" voltage, essentially this is the same as establishing a manual voltage somewhere in the middle.

Hence, at the end of a years time the difference in electricity used amounts to the change I have in my pocket...so the term "efficient" overclocking is just a misguided term.

That was my whole point. Whew!

 

[TuxDave]

That was my point! Throw the math out the window as there are too many variables to include in the algorithm. If the auto voltage swings way above and way below "nominal" voltage, essentially this is the same as establishing a manual voltage somewhere in the middle.

Hence, at the end of a years time the difference in electricity used amounts to the change I have in my pocket...so the term "efficient" overclocking is just a misguided term.

That was my whole point. Whew!

So..... not really a "simple EE" question.

 

[coffeejunkee]

Well, speaking just from experience, I see the same idle power when c-states are enabled, regardless of lower vcore or not. I think it is because c-states simply turn off large parts of the cpu, so it doesn't matter what vcore is used. My voltcraft might not be sensitive enough for short spikes though, obviously I see some slight 2-3% activity every once in a while but the number doesn't really change much.

What you do give up though, is EIST. Every time your cpu is active it will use the fixed vcore you set, while EIST might choose to use a lower one. Also, when overclocking you might need to turn off c-states for stability reasons. In that case offset vcore does make a difference in idle power use.

Another factor is possibly more stress on the vrm's when using fixed vcore. I think I heard one of the Asus guys say this, but I lack the knowledge to confirm if true or false.

 

[videogames101]

It will cost less to run by allowing the cpu to downclock and downvolt automatically, unless you spend much greater than normal time under load AND you set a manual voltage below the nominal for the given operating frequency.

That's my take anyway.