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Author Topic: Estimating heat output - 5850  (Read 1866 times)
amazingrando
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August 19, 2011, 09:24:44 PM
 #1

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

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chungenhung
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August 19, 2011, 10:13:42 PM
 #2

assuming the card uses 200W.
one hr of 200w will produce 682 BTU per hour
http://www.mhi-inc.com/Converter/Energy%20Converter.htm

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SgtSpike
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August 19, 2011, 10:15:20 PM
 #3

A 5850 only uses about 150w.
amazingrando
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August 19, 2011, 10:20:08 PM
 #4

Kind of a stupid question, but if the GPU uses 150W, is all of it dissipated as heat?

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August 19, 2011, 10:25:10 PM
 #5

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.
amazingrando
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August 19, 2011, 10:30:38 PM
 #6

Thanks SgtSpike!

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Kermee
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August 19, 2011, 10:34:52 PM
 #7

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

mike678
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August 19, 2011, 10:47:56 PM
 #8

A 5850 only uses about 150w.
Overclocked 5850's generate more than this Wink 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.
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August 19, 2011, 11:07:16 PM
 #9

A dedicated mining rig can be used as a space heater - and one as efficient as machines sold as dedicated space heaters Smiley though more expensive Wink

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 Cool


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SgtSpike
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August 19, 2011, 11:15:27 PM
 #10

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 Wink 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.
mike678
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August 19, 2011, 11:18:26 PM
 #11

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.
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August 19, 2011, 11:43:33 PM
 #12

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.
Kermee
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August 20, 2011, 12:09:20 AM
 #13

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

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August 20, 2011, 12:25:52 AM
 #14

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 Cheesy

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 Smiley 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... Sad


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 Smiley

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mike678
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August 20, 2011, 12:31:16 AM
 #15

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.
amazingrando
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August 20, 2011, 01:43:32 AM
 #16

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?

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SgtSpike
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August 20, 2011, 02:38:39 AM
 #17

Catfish, that was a far lengthier reply than I was requesting - a simple "no" would have sufficed.  Tongue  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.  Cheesy

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.
scifimike12
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August 20, 2011, 05:55:32 AM
 #18

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 40°C 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.  97°F inside, while it's 71°F outside just ain't right. 
haploid23
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August 20, 2011, 11:28:27 AM
 #19

A 5850 only uses about 150w.
Overclocked 5850's generate more than this Wink 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

mike678
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August 20, 2011, 02:56:32 PM
 #20

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?
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