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Other => CPU/GPU Bitcoin mining hardware => Topic started by: amazingrando on August 19, 2011, 09:24:44 PM



Title: Estimating heat output - 5850
Post by: amazingrando on August 19, 2011, 09:24:44 PM
Hi everyone!

Does anyone have a good idea of how many BTU's a 5850 puts out at full load?  Trying to estimate how much airflow I need for my rigs.

Thanks


Title: Re: Estimating heat output - 5850
Post by: chungenhung on August 19, 2011, 10:13:42 PM
assuming the card uses 200W.
one hr of 200w will produce 682 BTU per hour
http://www.mhi-inc.com/Converter/Energy%20Converter.htm


Title: Re: Estimating heat output - 5850
Post by: SgtSpike on August 19, 2011, 10:15:20 PM
A 5850 only uses about 150w.


Title: Re: Estimating heat output - 5850
Post by: amazingrando on August 19, 2011, 10:20:08 PM
Kind of a stupid question, but if the GPU uses 150W, is all of it dissipated as heat?


Title: Re: Estimating heat output - 5850
Post by: SgtSpike on August 19, 2011, 10:25:10 PM
Kind of a stupid question, but if the GPU uses 150W, is all of it dissipated as heat?
Yes.  All wattage eventually turns into heat.  Even fans blowing air... the air creates friction until it slows to a stop, and all of that friction turns into heat.  Same with light - as it is absorbed by the various surfaces it touches, it turns into heat.  The reflected light is what we see, but it all keeps bouncing around until it is all absorbed.  But the majority of wattage in a computer is used by transistors switching, which creates heat as well.


Title: Re: Estimating heat output - 5850
Post by: amazingrando on August 19, 2011, 10:30:38 PM
Thanks SgtSpike!


Title: Re: Estimating heat output - 5850
Post by: Kermee on August 19, 2011, 10:34:52 PM
CPUs, GPUs, uC's, and most other semiconductors are less than 1% efficient.

So yes, basically 100% of TDP comes off as heat.

Cheers,
Daniel


Title: Re: Estimating heat output - 5850
Post by: mike678 on August 19, 2011, 10:47:56 PM
A 5850 only uses about 150w.
Overclocked 5850's generate more than this ;) My rig is set up for 920/325 and its 5 cards. According to my kw reader its about 1kw so closer to 170w per card.


Title: Re: Estimating heat output - 5850
Post by: catfish on August 19, 2011, 11:07:16 PM
A dedicated mining rig can be used as a space heater - and one as efficient as machines sold as dedicated space heaters :) though more expensive ;)

The 5850, even overclocked to 400+ MH/sec, consumes well under 200W. The 'assume 200W' mantra comes more from conservative PSU sizing threads than it does from accurate measurements of actual card consumption at full load. It may not be 100% accurate, but as a good over-estimate to use when sizing PSUs, I'd say it's a decent over-estimate to use also when sizing cooling. After all, you are better off having more cooling than being right on the limit.

All it takes is a refined OpenCL kernel that pumps the GPU to consume more power, or new code that only works well if you put the memory clock back to its sky-high 'normal' values (virtually everyone who understands about overclocking also underclocks their memory to 300 or so (I use 275) to reduce power consumption - it's significant), and a 'cooling capacity at the limit' solution would be undersized.

Alternatively, you may upgrade a bunch of cards and have the same problem. To be 100% sure with current 6-pin PCIe power supply specs, assume 225W per card (75W from the PCIe slot, and 2x 75W from the two PCIe power cables) and you'll be fine for any card upgrade short of dual-GPU units requiring 8-pin power feeds 8)


SgtSpike - I like your description of the second law of thermodynamics :D Yup, so far, it appears that entropy always increases :( Heat death, anyone?


Title: Re: Estimating heat output - 5850
Post by: SgtSpike on August 19, 2011, 11:15:27 PM
CPUs, GPUs, uC's, and most other semiconductors are less than 1% efficient.

So yes, basically 100% of TDP comes off as heat.

Cheers,
Daniel
Do transistors actually move?  I'm not familiar enough with how they work to know the answer...

A 5850 only uses about 150w.
Overclocked 5850's generate more than this ;) My rig is set up for 920/325 and its 5 cards. According to my kw reader its about 1kw so closer to 170w per card.
This is true.  I am assuming a stock card when I say 150w, and even then, that's not really an exactly accurate number.


Title: Re: Estimating heat output - 5850
Post by: mike678 on August 19, 2011, 11:18:26 PM
This is true.  I am assuming a stock card when I say 150w, and even then, that's not really an exactly accurate number.
Yes I agree that's an estimate on my part but I have a semperon so 150w for mobo/ram/cpu/hd is a pretty safe bet.


Title: Re: Estimating heat output - 5850
Post by: SgtSpike on August 19, 2011, 11:43:33 PM
This is true.  I am assuming a stock card when I say 150w, and even then, that's not really an exactly accurate number.
Yes I agree that's an estimate on my part but I have a semperon so 150w for mobo/ram/cpu/hd is a pretty safe bet.
More like 50w if your sempron is just idling while mining...

Total, 200w would be my guess.

Then again, it also depends how efficient your PSU is... add 20% or more to those figures for the heat loss from the transformers in the PSU.


Title: Re: Estimating heat output - 5850
Post by: Kermee on August 20, 2011, 12:09:20 AM
Do transistors actually move?  I'm not familiar enough with how they work to know the answer...

In a sense, the physical 'gates' do for the switching. =)

Cheers,
Kermee


Title: Re: Estimating heat output - 5850
Post by: catfish on August 20, 2011, 12:25:52 AM
CPUs, GPUs, uC's, and most other semiconductors are less than 1% efficient.

So yes, basically 100% of TDP comes off as heat.

Cheers,
Daniel
Do transistors actually move?  I'm not familiar enough with how they work to know the answer...
Semiconductors... the definition of the word implies that the solid-state 'machine' has non-zero electrical resistance, and as you know, electrons bashing their way through non-superconducting materials will generate heat.

Besides, back on the thermodynamics level, the transistors are doing work ordering the universe (aka reducing entropy) and this can't happen without entropy *increasing* in the overall system (i.e. waste heat). Being simplistic again, I'll get bitch-slapped by a physics professor soon :D

As to how transistors work - a 'switching potential' controls whether the transistor has lots of resistance (i.e. blocks current), or 'low resistance' (i.e. lets current flow, though *actual* ohms of an open-gate transistor isn't on the tip of my tongue... if it's more than a plain wire then ohmic heating will result, as in all other cases). It's not a 'mechanical' switch, but allows a '1' or a '0' on the base terminal (low power) to either block or allow a much greater current to flow from the collector to the emitter (in typical computer use of transistors, as gates). When the switch is 'open' and the large current is permitted to flow between the collector and the emitter, the transistor can be modelled as a boggo resistor, with ohmic heat waste related to the resistance of the collector-emitter path when fully 'open'. I have absolutely NO idea of the average resistance of an 'open' transistor, but I'm assuming it's not zero...

However this is a transistor in its most simple 'digital switch' form. Modern ICs use fancy variants to reduce power loss, and that's beyond my elementary electronics knowledge. What I *do* know is that transistors in modern CPUs and GPUs have got so damn small that quantum tunnelling effects cause a leakage current, even when the 'switch' is 'off'. This leakage gets wasted as heat. Then you've got the very high switching frequencies of modern processors - hundreds to thousands of MHz - resulting in more instances where the widely used CMOS gates have both of their component MOSFETs conducting (almost a short) as the switch changes from one state to another. The main power consumption of CMOS gates is meant to be during this 'in-between' phase as the switch changes state.

I don't know the practical resistance of the transistors when fully 'open' though - and if you're pumping high currents through these supposedly-open gates, my intuition says that plain old prep-school 'electronics' must come into play - the simple heat loss due to running a current through a non-zero resistance. The interconnects and logic board traces feeding the CPU / GPU aren't heavy gauge wires either... these also must account for some resistance heating.

On top of that, one must consider the non-DC effects due to high switching frequencies - the concepts of impedance and capacitance will apply to these rapidly flip-flopping (sorry, not meant literally) currents within the processor. With incredibly thin, but longer than the transistor size (45nm?) by orders of magnitude interconnects, mismatched capacitance probably delays the switching speed of the gate. And the more time a gate is 'in-between' states, the more time it's in the 'power-consuming' phase.

I really don't know enough about the detail of electronic engineering, it's not my field. How basic transistors work is fairly simple (I've only described the functionality in digital circuits - transistors are analogue devices too and the relationship between the base potential and the collector-to-emitter resistance doesn't have to be binary - in fact the 'basic' transistor is not, and the simplest use of a transistor is as an amplifier), but in the absence of superconductivity, an integrated circuit is pushing a load of current through a complicated semiconducting machine that *must* have resistance / impedance... and ohmic heating is a basic process that anyone can understand, regardless of what trickery takes place between the 12V PSU rail fed into the processor and ground :) And, of course, on the subject of thermodynamics and information theory, these processors are *performing calculations* / concentrating information / reducing entropy - aka doing work. If total system temperature remained the same, then you'd have a system that reduced entropy, which ain't allowed. So the total entropy increases by converting the electrical energy into heat energy - all energy ends up as heat at the end... :(


This is just my Friday-night-too-much-cider 'understanding' - any physicists or EEs around? Please put me straight if I'm talking shite and misleading people, I'm always up for learning stuff :)


Title: Re: Estimating heat output - 5850
Post by: mike678 on August 20, 2011, 12:31:16 AM
More like 50w if your sempron is just idling while mining...

Total, 200w would be my guess.

Then again, it also depends how efficient your PSU is... add 20% or more to those figures for the heat loss from the transformers in the PSU.
I'll just post exact specs.

MSI 890FXA-GD70
corsair 1200watt gold series
5xsapphire hd 5850
2x1gb ram
this hd (http://www.newegg.com/Product/Product.aspx?Item=N82E16822136770&Tpk=22-136-770)
Sempron 130

At 920/325 I get around 1kw. I'm too lazy to find exact numbers feel free to figure it out if you want.


Title: Re: Estimating heat output - 5850
Post by: amazingrando on August 20, 2011, 01:43:32 AM
I'll just post exact specs.

MSI 890FXA-GD70
corsair 1200watt gold series
sapphire hd 5850
2x1gb ram
this hd (http://www.newegg.com/Product/Product.aspx?Item=N82E16822136770&Tpk=22-136-770)
Sempron 130

At 920/325 I get around 1kw. I'm too lazy to find exact numbers feel free to figure it out if you want.

With how many GPUs?


Title: Re: Estimating heat output - 5850
Post by: SgtSpike on August 20, 2011, 02:38:39 AM
Catfish, that was a far lengthier reply than I was requesting - a simple "no" would have sufficed.  :P  That said, thanks for writing all of that!  Makes sense that transistors simply increase resistance to force the electricity to another path though.  No physical movement involved.  So, if that's the case, then Kermee is wrong - it's not "basically" 100% of electricity is direct heat loss - it IS 100% electricity is direct heat loss.  Unless you happen to have glowing-hot semiconductors, which, probably won't last long anyway.  :D

More like 50w if your sempron is just idling while mining...

Total, 200w would be my guess.

Then again, it also depends how efficient your PSU is... add 20% or more to those figures for the heat loss from the transformers in the PSU.
I'll just post exact specs.

MSI 890FXA-GD70
corsair 1200watt gold series
sapphire hd 5850
2x1gb ram
this hd (http://www.newegg.com/Product/Product.aspx?Item=N82E16822136770&Tpk=22-136-770)
Sempron 130

At 920/325 I get around 1kw. I'm too lazy to find exact numbers feel free to figure it out if you want.
That system cannot possibly be drawing 1kw of power, unless you have multiple GPU's.  I calculate it would use about 230w, PSU efficiency included.


Title: Re: Estimating heat output - 5850
Post by: scifimike12 on August 20, 2011, 05:55:32 AM
Don't forget that the cooler your card is, the same will be for your room.

I have my 5870's mining continuously in my basement, both barely hit 40C on a hot day.  But if I were to add a few MSI 5770 Hawks' @ 85% fan speed, my basement begins to warm up a bit, and so do my cards.

I tried setting up a simple build in my room and couldn't bare the heat after a few nights.  97F inside, while it's 71F outside just ain't right. 


Title: Re: Estimating heat output - 5850
Post by: haploid23 on August 20, 2011, 11:28:27 AM
A 5850 only uses about 150w.
Overclocked 5850's generate more than this ;) My rig is set up for 920/325 and its 5 cards. According to my kw reader its about 1kw so closer to 170w per card.
it actually generates less. 5850's TDP is 150w, that means it's designed to dissipate 150w of heat. the actual usage, even at 100% load, will draw less than 150w. overclocking the core (however underclocking the memory) might put you close to the TDP, but it shouldn't go over unless there's heavy overvolting.

and also note that AC and DC power are a little different. the rated 150w TDP is based on DC pulled from the PSU, when you read 1000w your meter, that's AC power, which is pulled from the wall. in this case, you need to take into account the PSU efficiency before claiming it's 170w of real DC power draw


Title: Re: Estimating heat output - 5850
Post by: mike678 on August 20, 2011, 02:56:32 PM
That system cannot possibly be drawing 1kw of power, unless you have multiple GPU's.  I calculate it would use about 230w, PSU efficiency included.
I updated the post there's 5 5850's I mentioned it a little earlier in the thread and I assumed it was known already. Plus I wrote that when I was drunk lol

it actually generates less. 5850's TDP is 150w, that means it's designed to dissipate 150w of heat. the actual usage, even at 100% load, will draw less than 150w. overclocking the core (however underclocking the memory) might put you close to the TDP, but it shouldn't go over unless there's heavy overvolting.

and also note that AC and DC power are a little different. the rated 150w TDP is based on DC pulled from the PSU, when you read 1000w your meter, that's AC power, which is pulled from the wall. in this case, you need to take into account the PSU efficiency before claiming it's 170w of real DC power draw
I have a CORSAIR Professional Series Gold AX1200 (http://www.newegg.com/Product/Product.aspx?Item=N82E16817139014)

How would I go about calculating watts used by each card then?


Title: Re: Estimating heat output - 5850
Post by: bcforum on August 20, 2011, 03:09:11 PM
CPUs, GPUs, uC's, and most other semiconductors are less than 1% efficient.

So yes, basically 100% of TDP comes off as heat.

Cheers,
Daniel

I'm puzzled by your comment. What actual work is being performed by the 1% here? Even the air that is moved isn't being stored at a higher pressure, so after you run your 1KW rig for 1 hour you have increased the heat in your room exactly the same as a 1KW heater.



Title: Re: Estimating heat output - 5850
Post by: Kermee on August 20, 2011, 07:17:46 PM
I'm puzzled by your comment. What actual work is being performed by the 1% here?

Transistor gating/switching for the 2.15 billion transistors in a 5850.  There is 'work' being done still.

Like I said... it's less than 1%.  You cannot apply 'high school' level of physics in entropy with semiconductors.

It 'work' efficiency wasn't an issue, then we wouldn't need to continue to shrink dies sizes.

Cheers,
Kermee


Title: Re: Estimating heat output - 5850
Post by: bcforum on August 21, 2011, 04:57:01 PM

Mustn't feed the trolls.


Title: Re: Estimating heat output - 5850
Post by: mike678 on August 21, 2011, 04:58:54 PM
it actually generates less. 5850's TDP is 150w, that means it's designed to dissipate 150w of heat. the actual usage, even at 100% load, will draw less than 150w. overclocking the core (however underclocking the memory) might put you close to the TDP, but it shouldn't go over unless there's heavy overvolting.

and also note that AC and DC power are a little different. the rated 150w TDP is based on DC pulled from the PSU, when you read 1000w your meter, that's AC power, which is pulled from the wall. in this case, you need to take into account the PSU efficiency before claiming it's 170w of real DC power draw
I have a CORSAIR Professional Series Gold AX1200 (http://www.newegg.com/Product/Product.aspx?Item=N82E16817139014)

How would I go about calculating watts used by each card then?
Any one have an answer to this?


Title: Re: Estimating heat output - 5850
Post by: catfish on August 21, 2011, 05:38:28 PM
OK then - honest question from a point made earlier.

Calculating power used by a graphics board should be easy, with my limited understanding. It's a DC circuit, running with 12V feeds, so assuming you can measure the current flowing on each of the feeds into the GPU board (two direct feeds from PSU and the feed from the PCIe slot itself), and assuming your 12V feed is a clean 12V (if not, I suppose you could measure the exact '12V' potential at all three points), prep school physics says that the power used (converted / dissipated as heat / available to do work / etc.) is simply the voltage multiplied by the total current. If the current is measured in amps then multiplying this by the voltage gets you power in watts.

OK, so getting a representative overclocked 5850 mining at full load, then measuring the input current and voltages (the different inputs could have different potentials... after all, the two PCIe power feeds are usually fat cables straight to the PSU, with minimal resistance and hence minimal voltage drop... but the PCIe *slot* feed of up to 75W has to come from the ATX / logic board connector, and then across a bunch of PCB traces... my guess is that this route (and the thinner-gauge PCIe pins) will have a higher resistance and hence a larger voltage drop), multiplying them together will give you a total power draw from *one* typical 5850 card.

Let's say the measurement and calculation gives 175W. I've got 5 slots, so if I filled them all with 'typical' 5850s, I'll need 875W from the PSU purely on the 12V supply, ignoring the CPU and other loads.


My question is - at what point do I have to take the nature of AC into account? Is it only an issue when comparing 'power-from-wall' readings (in the USA, from 'kill-a-watt' type devices - in the UK I've not seen specific brand name power analysers, I just use Maplin power meters and extension multiplugs with power analysers built in)??

Residential AC is typically single phase, isn't it? Are the mains-electricity power meters measuring power the same way (volts x amps)?? If so, then the RMS issue comes into play, and you can't assume (in the UK) 240V times the current to be the power. Or can you?

Sine-wave alternating current and RMS voltage makes things annoyingly confusing when switching to DC output. Is this a complete non-issue, and when a PSU says their 12V rail will handle, say, 850W, and my mains power meter says I'm pulling 650W from the wall, I've got a nice fat buffer and am safe and efficient? Or is the rated 850W based off 'peak' power supplied in AC form, whereas in reality only RMS is continuously available to convert to DC, and my 'buffer' may not be anywhere near as large as I think?


I'd like someone who really knows the detail of AC conversion to DC, and how mains power meters measure, to educate me here... Haploid23 - you said that reading 1000W from the wall is a measurement of AC power... is that peak (non-continuous), needing an RMS calc, or are 'watts' always 'watts' regardless of AC, DC, single or three phase, etc??

I buy UPS backups for my servers, and have an emergency generator, and notice that *these* appliances often mess about with 'VA' ratings instead of good old *watts* - this suggests that 'watts are watts', and the 'VA' is just a scam to make the power supply seem more powerful (since it's going to be peak, not RMS, voltage - surely)??

But WTFDIK.  ???


Title: Re: Estimating heat output - 5850
Post by: bcforum on August 21, 2011, 06:57:52 PM

Let's say the measurement and calculation gives 175W. I've got 5 slots, so if I filled them all with 'typical' 5850s, I'll need 875W from the PSU purely on the 12V supply, ignoring the CPU and other loads.

PSUs are most efficient when operated somewhat below their rating, this also gives some margin for unforeseen circumstances. If you ignore the CPU,
 you'll need  a ~1200 watt PSU (875W / 0.75). 0.75 is a derating factor which keeps the PSU from running close to the limit.

http://www.anandtech.com/show/2624/3 (http://www.anandtech.com/show/2624/3)


My question is - at what point do I have to take the nature of AC into account? Is it only an issue when comparing 'power-from-wall' readings (in the USA, from 'kill-a-watt' type devices - in the UK I've not seen specific brand name power analysers, I just use Maplin power meters and extension multiplugs with power analysers built in)??


In theory, all AC power meters will report the same number that you are being billed for. Check with Maplin to see how well their meters compare to the utilities.


Residential AC is typically single phase, isn't it? Are the mains-electricity power meters measuring power the same way (volts x amps)?? If so, then the RMS issue comes into play, and you can't assume (in the UK) 240V times the current to be the power. Or can you?

Sine-wave alternating current and RMS voltage makes things annoyingly confusing when switching to DC output. Is this a complete non-issue, and when a PSU says their 12V rail will handle, say, 850W, and my mains power meter says I'm pulling 650W from the wall, I've got a nice fat buffer and am safe and efficient? Or is the rated 850W based off 'peak' power supplied in AC form, whereas in reality only RMS is continuously available to convert to DC, and my 'buffer' may not be anywhere near as large as I think?


I'd like someone who really knows the detail of AC conversion to DC, and how mains power meters measure, to educate me here... Haploid23 - you said that reading 1000W from the wall is a measurement of AC power... is that peak (non-continuous), needing an RMS calc, or are 'watts' always 'watts' regardless of AC, DC, single or three phase, etc??

I buy UPS backups for my servers, and have an emergency generator, and notice that *these* appliances often mess about with 'VA' ratings instead of good old *watts* - this suggests that 'watts are watts', and the 'VA' is just a scam to make the power supply seem more powerful (since it's going to be peak, not RMS, voltage - surely)??

But WTFDIK.  ???

[/quote]

This is where things fall off the rails. Since current and voltage are sinusoidal you can't simply multiply them unless it is a pure resistive load (it isn't.)

Maybe http://en.wikipedia.org/wiki/Power_factor (http://en.wikipedia.org/wiki/Power_factor) will help clear things up. If not, try http://en.wikipedia.org/wiki/Volt-ampere (http://en.wikipedia.org/wiki/Volt-ampere)

In general, your PSU manufacture will quote an efficiency number, typically the absolute highest measurement they can coerce out of a golden unit at the exact right temperature and age. For the sake of argument, we'll pick 80% efficiency. Your 875W to power the GPUs will draw ~1100W from the outlet.