Going from 65 TDP too 125 TDP - Impact on electricity bills?

[xfxrising]

Building a new pc and i was gonna get a phenom 960t and try to unlock it (Im on a strict budget and i can get this for £70 which is great Price/performance) and im 15 so ya know, living with parents and all and because we now have a 4th child in the family, money is a little tighter than before xD So i was wondering how much of an impact it will have on electricity costs compared to what im using at the moment.

I do a lot of video rendering so im spending like 4 hours+ a week at full load however 960t will probably finish these renders in like 1 hour and a half or less so im spending less time in load which should = less electricity being used over time right? And ill probably keep it at 4 cores during general use and unlock the other 2 when im editing/rendering so that im being more conservative on energy. May even see if i can underclock it if i need to

So im guessing my real question is how much it would add to the electricity cost per month using the 960t compared to what im using now? im guessing it won't effect it too much but i wanted to ask anyway just to make sure.

*Additional Question: Would i be able to unlock the cores on a Gigabyte GA-78LMT-S2P mobo? It supports EC AOD-ACC but in the description it only mentions overclocking and nothing about unlocking*

Sorry for the crappy writting and rambaling aha

 

[EarthDog]

Negligible... do the math... Find out how much your parents pay per KW/H and work it out. Chances are its certainly not anything your parents will notice. I cant imagine more than $1 /month if anything.

 

[xfxrising]

Negligible... do the math... Find out how much your parents pay per KW/H and work it out. Chances are its certainly not anything your parents will notice. I cant imagine more than $1 /month if anything.

pretty much what i was thinking, thanks

 

[ruff0r]

Negligible... do the math... Find out how much your parents pay per KW/H and work it out. Chances are its certainly not anything your parents will notice. I cant imagine more than $1 /month if anything.

Probably 30$ over 3 years more.

Price per KW/h 0.128$

Difference is 60W

60Wx3 hours = 0,180 kw/h x 0.128$ = 0,02304$ per day.


0,02304$ per day = 0,6912$ per month = 8,2944$ a year

 

[Xenturion]

Joking aside, as was said, the change in price each month should be negligible. If you were planning on having your rig crunch 24/7, that might be a little bit different. Even then, I wouldn't expect it to be more than a couple of dollars.

 

[Aquinus]

TDP is the measure of how much heat must be taken away from the CPU to keep it cool. TDP is cooling power. The actual usage of the CPU is going to be higher than the TDP because not all power is lost to heat. If you start overclocking that number changes and high TDP CPUs tend to eat more faster than lower TDP overclocked CPUs.

All in all, if you're not crunching I doubt it will change much. If the computer is under constant load all the time, I can see it easily adding up depending on the CPU. You won't know unless you put the current rig and the new on one a kill-a-watt but in general higher TDP CPUs use more power under load. Idle usages don't vary a whole lot (but do a litlte bit because SB-E has more components in the CPU,) between similar CPUs (like SB-E vs SB,) but loaded usages do.

Anything that uses a reeasonable amount of power constantly over time adds up long term.

 

[EarthDog]

The actual usage of the CPU is going to be higher than the TDP because not all power is lost to heat.

Actual power use is less usually.. it least it is on GPUs?

For example, look at the 650ti Boost. Its TDP is 140W, power use is 115W/127W(o/c). In the marketing materials they state it will hit 140W only using TDP apps (Furmark/Kombuster).

A bit confused at that statement I quoted.. but its likely me as I have been confused most of today at minimum.

 

[Tatty_Two]

Probably a chocolate bar a month

 

[Sasqui]

There was a great article a few years back (and of course I have no idea where i read it, lol) ... about how a faster processor that uses more wattage will save you money compared to a slower processor that uses less wattage.

The electirc power per compute cycle is typically less for a newer/faster CPU. Meaning a few things:

• You'll be running at full speed less of the time (tasks are completed faster while using less overal power for the same task)
• You'll probably have the computer on less, meaning the loss due to inefficiency will be decreased.

Viola!

 

[Xenturion]

Depending on what 65W processor you're referring to, the Phenom II might have dramatically improved power gating and throttling that will reduce power consumption when the system is idle. Older processors weren't quite as effective at reducing power.

As far as the motherboard is concerned, it does support Core Unlocking. Every Gigabyte motherboard from the era had their "Core Unlocker" logo on the box. I'd imagine since the transition to the FX series, they decided to remove the logo to reduce confusion as it only applies to Phenom IIs. If you download the manual from Gigabyte's site, there is a section in the 'Advanced BIOS Features' submenu labeled "CPU Unlock". In the picture it's disabled because they've got a FX-8000 cpu installed.

 

[xfxrising]

Depending on what 65W processor you're referring to, the Phenom II might have dramatically improved power gating and throttling that will reduce power consumption when the system is idle. Older processors weren't quite as effective at reducing power.

As far as the motherboard is concerned, it does support Core Unlocking. Every Gigabyte motherboard from the era had their "Core Unlocker" logo on the box. I'd imagine since the transition to the FX series, they decided to remove the logo to reduce confusion as it only applies to Phenom IIs. If you download the manual from Gigabyte's site, there is a section in the 'Advanced BIOS Features' submenu labeled "CPU Unlock". In the picture it's disabled because they've got a FX-8000 cpu installed.

Thanks! And thanks to everyone else of course to, best forum ever lol

 

[Aquinus]

Actual power use is less usually.. it least it is on GPUs?

For example, look at the 650ti Boost. Its TDP is 140W, power use is 115W/127W(o/c). In the marketing materials they state it will hit 140W only using TDP apps (Furmark/Kombuster).

A bit confused at that statement I quoted.. but its likely me as I have been confused most of today at minimum.

Every company calculates TDP differently. nVidia is probably giving you peak consumption, not TDP. A circuit can't generate more heat than the power it takes in, that's physics. So unless its a space heater, I doubt it will be using >95% of its power as heat.

TDP is thermal design power. It represents how much cooling power in loaded operation would be required to take heat away from the GPU/CPU to maintain a safe usable temperature. Heat in this case is measured in watts, a unit of power over time as opposed to joules which is instantaneous power.

 

[EarthDog]

Regurgitating the definition isn't answering my question...

Its more than common for a tdp, no matter if its peak or average use, to be more than actual use. 3770k is 77w tdp and at stock igpu or not pulls less than that outside of ibt.

 

[Aquinus]

Regurgitating the definition isn't answering what I quoted... No worries though.

Learn to read.

Every company calculates TDP differently. nVidia is probably giving you peak consumption, not TDP. A circuit can't generate more heat than the power it takes in, that's physics.

 

[EarthDog]

Edited...

Still dorsnt answer shit. I haven't seen a tdp close to actual useage. Again I know the definition.

 

[Aquinus]

Still dorsnt answer shit. I haven't seen a tdp close to actual useage. Again I know the definition.

That's because TDP doesn't represent actual usage. If you understood any of that it would make sense. TDP is also a spec number, not a real number.

http://www.intel.com/content/dam/doc/white-paper/resources-xeon-measuring-processor-power-paper.pdf

You say you know the definition, but I still don't think you get it. Here, read a little more.

...and once again TDP:

TDP (Thermal Design Power)
Intel defines TDP as follows: The upper point of the thermal profile consists of the Thermal Design Power (TDP) and the associated Tcase value. Thermal Design Power (TDP) should be used for processor thermal solution design targets. TDP is not the maximum power that the processor can
dissipate. TDP is measured at maximum TCASE.1. The thermal profile must be adhered to to ensure Intel’s reliability requirements are met. Note: Different processors SKU’s have different TDP’s. At the time of this writing, Intel® Xeon® processors for 2 socket servers (5600 series) are available with a TDP specification from 40W up to 130W depending on the particular SKU1.

 

[EarthDog]

You said power use of said part will be more than the tdp... That's the only point I am confused about. Not the definition... Not anything else. Just confused about what I quoted you on above.

We know that isn't true.. look at you reviews that measure CPU power use. Its less than the tdp right?

 

[Aquinus]

We know that isn't true.. look at you reviews that measure CPU power use. Its less than the tdp right?

Not all tasks that put that CPU at 100% causes the CPU to consume the maximum amount of power possible.

Most reviews I see, the total system power is usually well over the TDP, but that is because most reviews give you power consumption of the entire rig.
http://www.guru3d.com/articles_pages/core_i7_3770k_and_3750_review_with_z77,12.html

I think Cadaveca might have done some review measuring the current off the 12v EPS on motherboard reviews, but I could be wrong.

 

[EarthDog]

Yes. Testing that is done right, like Dave's...clearly system power is not the power consumption of the CPU only and can be over a CPU's tdp.

 

[silkstone]

TDP is the measure of how much heat must be taken away from the CPU to keep it cool. TDP is cooling power. The actual usage of the CPU is going to be higher than the TDP because not all power is lost to heat.

Where else is that energy going to go, if not heat? Light? Sound? Movement? I am pretty sure that all the energy used by a CPU is converted to heat.

 

[phanbuey]

Where else is that energy going to go, if not heat? Light? Sound? Movement? I am pretty sure that all the energy used by a CPU is converted to heat.

I wouldn't think so as it takes impedance / bad conductivity to generate heat from electricity... Im pretty sure that the transistors consume most of the electricity in their operation, and eventually it flows out of the cpu into the maiboard to communicate with other system components / dissipated some other way.

I don't know but I would bet that CPU makers rely on designs that ensure as little energy as possible is converted to heat.

Either way older cpu's tend to be much less power friendly, as was said, so OP may actually save money lol.

 

[silkstone]

I wouldn't think so as it takes impedance / bad conductivity to generate heat from electricity... Im pretty sure that the transistors consume most of the electricity in their operation, and eventually it flows out of the cpu into the maiboard to communicate with other system components / dissipated some other way.

I don't know but I would bet that CPU makers rely on designs that ensure as little energy as possible is converted to heat.

It is all converted to heat one way or another. You are pushing electrons down a 22 nm "wire," I would say that the impedance would be quite high even though it is a semi-conductor.

The energy doesn't just flow out of the CPU. I'm not sure how they calculate the power requirements of a CPU, but I'm fairly certain that the motherboard has it's own power supply and so the energy usage of the CPU would be pretty independent of the mainboard.

http://en.wikipedia.org/wiki/Forms_of_energy

 

[Aquinus]

It is all converted to heat one way or another. You are pushing electrons down a 22 nm "wire," I would say that the impedance would be quite high even though it is a semi-conductor.

The energy doesn't just flow out of the CPU. I'm not sure how they calculate the power requirements of a CPU, but I'm fairly certain that the motherboard has it's own power supply and so the energy usage of the CPU would be pretty independent of the mainboard.

http://en.wikipedia.org/wiki/Forms_of_energy

Yeah, but all circuits don't lose all their energy to heat when it starts moving electrons. For all of the power entering to be released as heat would make it a space heater.

Let's assume for a moment that 77-watts is the TDP for the 3770k. The formula for heat is that Q is directly proportional to the resistance and the square of the current. You make it sounds like the power that enters the motherboard stays there and does something until the system can "eat it." That is not the case.

As electricity powers a circuit some power is lost due to impedance as the electron travels, however by the time that electron is done doing work on any given circuit the electron returns to ground and gets put back on the power line in the opposite direction.

So no, a computer is not a perfect space heater and if it were, you would want a new computer. Unless you really like leakage, in that case you might have a good LN2 chip, but in general less leakage is better and not all CPUs have the maximum amount of leakage possible in any circuit.

I'm pretty sure when my computer is running full power that if it was converting it all to heat, 450-watts under load would warm up the room within minutes. It does not.

Every CPU has a different amount of leakage, so claiming that all of it is used as heat is inaccurate and wrong since different CPUs generate more or less heat during operation, which by itself means that it is not consistent.

 

[silkstone]

Yeah, but all circuits don't lose all their energy to heat when it starts moving electrons. For all of the power entering to be released as heat would make it a space heater.

Let's assume for a moment that 77-watts is the TDP for the 3770k. The formula for heat is that Q is directly proportional to the resistance and the square of the current. You make it sounds like the power that enters the motherboard stays there and does something until the system can "eat it." That is not the case.

As electricity powers a circuit some power is lost due to impedance as the electron travels, however by the time that electron is done doing work on any given circuit the electron returns to ground and gets put back on the power line in the opposite direction.

So no, a computer is not a perfect space heater and if it were, you would want a new computer. Unless you really like leakage, in that case you might have a good LN2 chip, but in general less leakage is better and not all CPUs have the maximum amount of leakage possible in any circuit.

I'm pretty sure when my computer is running full power that if it was converting it all to heat, 450-watts under load would warm up the room within minutes. It does not.

Every CPU has a different amount of leakage, so claiming that all of it is used as heat is inaccurate and wrong since different CPUs generate more or less heat during operation, which by itself means that it is not consistent.

Power is measured in Joules per second. It is the rate at which work is done. Work is done when energy is transformed. If a system is using 100 W of electrical power, it is literally changing 100 J of energy per second into a different form. That from must be electrical to heat + some magnetic & radiated energy.

There is no 'leakage' energy doesn;t just dissapear. It is either transformed or it isn't.

I'm also not exactly sure what you mean by space heater. They transform electrical energy, primarily, into radiated heat in the infrared spectrum (80%) whereas a cpu will be converting it into conductive (and to some extent convective) heat. You are mistaken that it would heat a room within minutes. Radiated heat and conducted heat act differently.

Also, Semi-conductors act differently from standard materials. some further reading: http://nopr.niscair.res.in/bitstream/123456789/8335/1/IJPAP 44(7) 543-547.pdf
Semi-conductors are non-ohmic conductors and so you cannot equate Q= I^2.R. But even if it did, remember that Power is also equal to I^2.R (P=I^2.R)
P=Q=I^2.R - Electrical energy is transformed into heat energy at an efficiency of 100% for an Ohmic conductor.
http://en.wikipedia.org/wiki/Joule_heating
http://en.wikipedia.org/wiki/Power_(physics)

I imagine that you are right in that not 100% of the energy is transformed into heat. But, your reasoning doesn't make sense. There will be some radiant energy produced by the transformation, but I am guessing that this would be minimal in comparison to the amount of heat energy.

 

[RCoon]

You said power use of said part will be more than the tdp... That's the only point I am confused about. Not the definition... Not anything else. Just confused about what I quoted you on above.

We know that isn't true.. look at you reviews that measure CPU power use. Its less than the tdp right?

The laws of physics dictate that in order for a certain amount of energy to come out (in this case TDP(HEAT)) an equal or more amount of energy must come in(ELECTRCITY). That means the TDP will change depending on how much power is being drawn by the CPU in order to perform cycles. So if the CPU is generating 95watts of TDP, the CPU MUST be drawing AT LEAST 95watts of energy, but of course 100% of energy isnt converted to heat in a processors case, so in may actually be drawing 110watts of power and the process of the CPU using the energy has a biproduct of heat. More often than not, at full load and at full TDP (ie Maximum heat coming from the processor), the processor is likely drawing way more than the TDP wattage.