Bitcoin Forum
December 08, 2016, 02:46:43 PM *
News: Latest stable version of Bitcoin Core: 0.13.1  [Torrent].
 
   Home   Help Search Donate Login Register  
Pages: « 1 [2]  All
  Print  
Author Topic: Estimating heat output - 5850  (Read 1864 times)
bcforum
Full Member
***
Offline Offline

Activity: 140


View Profile
August 20, 2011, 03:09:11 PM
 #21

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.


If you found this post useful, feel free to share the wealth: 1E35gTBmJzPNJ3v72DX4wu4YtvHTWqNRbM
Advertised sites are not endorsed by the Bitcoin Forum. They may be unsafe, untrustworthy, or illegal in your jurisdiction. Advertise here.
Kermee
Full Member
***
Offline Offline

Activity: 140



View Profile
August 20, 2011, 07:17:46 PM
 #22

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

bcforum
Full Member
***
Offline Offline

Activity: 140


View Profile
August 21, 2011, 04:57:01 PM
 #23


Mustn't feed the trolls.

If you found this post useful, feel free to share the wealth: 1E35gTBmJzPNJ3v72DX4wu4YtvHTWqNRbM
mike678
Full Member
***
Offline Offline

Activity: 168


View Profile
August 21, 2011, 04:58:54 PM
 #24

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?
catfish
Sr. Member
****
Offline Offline

Activity: 270


teh giant catfesh


View Profile
August 21, 2011, 05:38:28 PM
 #25

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.  Huh

...so I give in to the rhythm, the click click clack
I'm too wasted to fight back...


BTC: 1A7HvdGGDie3P5nDpiskG8JxXT33Yu6Gct
bcforum
Full Member
***
Offline Offline

Activity: 140


View Profile
August 21, 2011, 06:57:52 PM
 #26


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


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.  Huh

[/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 will help clear things up. If not, try 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.


If you found this post useful, feel free to share the wealth: 1E35gTBmJzPNJ3v72DX4wu4YtvHTWqNRbM
Pages: « 1 [2]  All
  Print  
 
Jump to:  

Sponsored by , a Bitcoin-accepting VPN.
Powered by MySQL Powered by PHP Powered by SMF 1.1.19 | SMF © 2006-2009, Simple Machines Valid XHTML 1.0! Valid CSS!