CoinTerra announces its first ASIC - Hash-Rate greater than 500 GH/s

[stex2009]

Energy usage for this machine? anyone!
http://cointerra.com/product/terraminer-iv-2ths-networked-miner-january-batch/

[Supercomputing]

Energy usage for this machine? anyone!
http://cointerra.com/product/terraminer-iv-2ths-networked-miner-january-batch/

I estimate somewhere between 1.1 kWh and 1.3 kWh per unit.

CPUs (2,000) x (0.55) = 1,100 W
Power supply loss = 100 W
Cooling and other components = 100 W

[Supercomputing]

I still don't see why they should be the same price.  It wouldn't be fair to the batch #2 buyers who deliberately took a spot in line after batch #1 customers knowing they will get a better price for waiting.  If you were to make the prices they same, you would have January people cancelling out so that they can jump into the December queue.. that wouldn't make sense.

I see your point if we were dealing with Butterfly Labs. But I am betting that Cointerra with their expertise, can assembly these units very fast when they get to that point and queue order will not mean much at that point.
How do you justify close to an $8,000 difference per unit for those who will receive their mining units literally days apart?

Because the early adopters get the privilege of knowing they helped crowdfund someone else's business.   The later adopters get closer to a true market price.   First batch people received the right to pay for all the fixed costs of the company.
That should make you feel good.

Very good answer and I hope that someone does not end up with a raw deal.

[iCEBREAKER]

Energy usage for this machine? anyone!
http://cointerra.com/product/terraminer-iv-2ths-networked-miner-january-batch/

There is no way to tell what energy usage will be for a machine using chips which have not yet been completely designed.

Energy usage is a physical characteristic, which requires physical simulation to estimate.

Until tape-out is complete, the physical simulation of the finalized design cannot be assumed to have actually happened.

Cointerra's specs are at best ballpark estimates and at worst wishful thinking.

[DeathAndTaxes]

Energy usage for this machine? anyone!
http://cointerra.com/product/terraminer-iv-2ths-networked-miner-january-batch/

I estimate somewhere between 1.1 kWh and 1.3 kWh per unit.

CPUs (2,000) x (0.55) = 1,100 W
Power supply loss = 100 W
Cooling and other components = 100 W

Well if 0.55 J/GH is the chip power consumption throw in another 10% loss for DC regulators (so another 110W or so).   ATX power supply is 12V.  No ASIC runs at 12V so all ASICs have converters to drop the 12V down to ~1V used by the chip.   Nothing is 100% efficient, 90% efficiency is a good place to start.

[aerobatic]

Energy usage for this machine? anyone!
http://cointerra.com/product/terraminer-iv-2ths-networked-miner-january-batch/

I estimate somewhere between 1.1 kWh and 1.3 kWh per unit.

CPUs (2,000) x (0.55) = 1,100 W
Power supply loss = 100 W
Cooling and other components = 100 W

Well if 0.55 J/GH is the chip power consumption throw in another 10% loss for DC regulators (so another 110W or so).   ATX power supply is 12V.  No ASIC runs at 12V so all ASICs have converters to drop the 12V down to ~1V used by the chip.   Nothing is 100% efficient, 90% efficiency is a good place to start.

i did some similar calcs in one of my earlier posts... i suggest around 1500 watts for regular use and 2000 watts if allowing some room for over clocking & overvolting !

[VolanicEruptor]

Jesus, I'll have to dedicate a 15 AMP breaker just for mining 

[aerobatic]

Jesus, I'll have to dedicate a 15 AMP breaker just for mining 

yes, a 2 TH mining rig - even the most power efficient one - is probably going to max out a household circuit, i think thats why they've got two power supplies, which would allow you to spread the load across two circuits in a home environment (and unnecessary in a data centre environment)

[DeathAndTaxes]

Jesus, I'll have to dedicate a 15 AMP breaker just for mining  

Use 240V.  No seriously for any serious miner you likely should consider a dedicate branch and if you are going to do that look into 240V.  NEMA L6-30R outlet.  Good for 5.76 KW.  As an added bonus ATX power supplies tend to be 1% to 2% more efficient at 240V vs 120V. 

[aerobatic]

Jesus, I'll have to dedicate a 15 AMP breaker just for mining  

Use 240V.  No seriously for any serious miner you likely should consider a dedicate branch and if you are going to do that look into 240V.  NEMA L6-30R outlet.  Good for 5.76 KW.  As an added bonus ATX power supplies tend to be 1% to 2% more efficient at 240V vs 120V. 

Totally agree... all power supplies seem to work better at 240 volts... and some are even designed to go higher (for datacenter use)

-- Jez

[greenbtc]

Jesus, I'll have to dedicate a 15 AMP breaker just for mining  

Use 240V.  No seriously for any serious miner you likely should consider a dedicate branch and if you are going to do that look into 240V.  NEMA L6-30R outlet.  Good for 5.76 KW.  As an added bonus ATX power supplies tend to be 1% to 2% more efficient at 240V vs 120V. 

Totally agree... all power supplies seem to work better at 240 volts... and some are even designed to go higher (for datacenter use)

-- Jez

Almost everything is more efficient at 240V. In fact, I don't know of anything that isn't...but I'm sure there is something.

[VolanicEruptor]

So I can double up on watts on a 15A if I switch to 240V too?  Would 14 gauge wire still be enough?  (I know the code may differ slightly between countries, so just theoretically).  

edit: I think I know the answer is yes, but it still throws my brain for a loop trying to imagine why increasing the voltage won't melt the wire, and yet you can still deliver more power with the same gauge of wire

[DeathAndTaxes]

So I can double up on watts on a 15A if I switch to 240V too?  Would 14 gauge wire still be enough?  (I know the code may differ slightly between countries, so just theoretically).  

edit: I think I know the answer is yes, but it still throws my brain for a loop trying to imagine why increasing the voltage won't melt the wire, and yet you can still deliver more power with the same gauge of wire

Simple answer is yes.   Wire guage is based on current.  By switching the circuit to 240V you effectively double the wattage of the circuit.   Remember in many cases there are multiple outlets on the same branch circuit so you want to make sure you change all the outlets (or at least mark them it wouldn't be code but it would keep you from doing something stupid).

You will need to rewire the branch at the breaker panel.  In the US a 120V circuit is wired neutral, hot, and ground.  240V is two hots and ground (no neutral).  

So
120V 15A circuit = 1800W * 80% (derate for continual load) = 1440W usable.
240V 15A circuit = 3600W * 80% (derate for continual load) = 2880W usable.
Same wire double the power.

[Minor Miner]

So I can double up on watts on a 15A if I switch to 240V too?  Would 14 gauge wire still be enough?  (I know the code may differ slightly between countries, so just theoretically).  

edit: I think I know the answer is yes, but it still throws my brain for a loop trying to imagine why increasing the voltage won't melt the wire, and yet you can still deliver more power with the same gauge of wire

Simple answer is yes.   Wire guage is based on current.  By switching the circuit to 240V you effectively double the wattage of the circuit.   Remember in many cases there are multiple outlets on the same branch circuit so you want to make sure you change all the outlets (or at least mark them it wouldn't be code but it would keep you from doing something stupid).

You will need to rewire the branch at the breaker panel.  In the US a 120V circuit is wired neutral, hot, and ground.  240V is two hots and ground (no neutral).  

So
120V 15A circuit = 1800W * 80% (derate for continual load) = 1440W usable.
240V 15A circuit = 3600W * 80% (derate for continual load) = 2880W usable.
Same wire double the power.

can you explain why when you get into 460 three phase that 1.73 number comes in all of a sudden?   I cannot for the life of me understand that.

[VolanicEruptor]

So I can double up on watts on a 15A if I switch to 240V too?  Would 14 gauge wire still be enough?  (I know the code may differ slightly between countries, so just theoretically).  

edit: I think I know the answer is yes, but it still throws my brain for a loop trying to imagine why increasing the voltage won't melt the wire, and yet you can still deliver more power with the same gauge of wire

Simple answer is yes.   Wire guage is based on current.  By switching the circuit to 240V you effectively double the wattage of the circuit.   Remember in many cases there are multiple outlets on the same branch circuit so you want to make sure you change all the outlets (or at least mark them it wouldn't be code but it would keep you from doing something stupid).

You will need to rewire the branch at the breaker panel.  In the US a 120V circuit is wired neutral, hot, and ground.  240V is two hots and ground (no neutral).  

So
120V 15A circuit = 1800W * 80% (derate for continual load) = 1440W usable.
240V 15A circuit = 3600W * 80% (derate for continual load) = 2880W usable.
Same wire double the power.

I understand that, thank you.. I know you are correct, but it is still counterintuitive to me as to how you can send the same current through the wire yet somehow get double the power.  
As a thought experiment I tried exaggerating the scenerio. what happens if you use 24,000V?  Does that mean you can use 288000W through a 14 gage wire, as long as you stay at 15A?  Obviously not, but I cannot see why the situation would be different.

... sorry for sidetracking I just want to get a better understanding of this.  I've always known the formula W = V x A but I've never been able to fully grasp it in real life operation

[DeathAndTaxes]

I probably won't do it justice but the lines are out of phase.  AC power is a waveform.   The phases don't peak at the same time they are offset by 30 deg.   The 1.73 comes from 2 cos(30 deg).

This diagram might help.



If you look at Phase-B it peaks at right below the letter "e" in Phase-B.  Now draw a vertical line down to the Phase-A.  You notice at the same point where Phase B has a positive peak, Phase A isn't at the negative peak, it is slightly less.   How much less well about 0.73x the peak.  If the distance from peak to centerline is 1 the distance from one phase to another when one phase is peaking is 1.73.  Using voltage, image the x-axis as ground (0 V) and each phase peaks at 277V.  So you can mark the top and bottom of the graph as 277V. If the peak voltage is 277V then the voltage between any two phases is 1.73*277V=480V.  A to B = 480V, B to C = 480V A to C = 480V.  A to GND = 277V. B to GND = 277V.  C to GND = 277V.

As for "why"?  It is more efficient in terms of power vs size and number of conductors.

BTW I believe it is 480V not 460V.  Each of the phases has a line to ground voltage of 277V.  They are tapped line to line producing 1.73*277V=480V between conductors.

[Epoch]

As a thought experiment I tried exaggerating the scenerio. what happens if you use 24,000V?  Does that mean you can use 288000W through a 14 gage wire, as long as you stay at 15A?  Obviously not, but I cannot see why the situation would be different.

... sorry for sidetracking I just want to get a better understanding of this.  I've always known the formula W = V x A but I've never been able to fully grasp it in real life operation

Yes, as strange as it may seem, you can easily transmit 288000W (15A, 24kV x 80% derating) through 14 gauge wire (though at that voltage you'd need to separate the conductors further to prevent arcing. That is a separate issue but let's ignore that).

The reason for this is that the power is not dissipated in the wire; it is dissipated in the load. If you can grasp that, you are well on your way to understanding what is happening here.

Your formula W = V x A is correct. The total power dissipated in the entire circuit is 288000W, that is true. But the key thing here is that 287999W of that is dissipated by the load (random device, or space heater, or ASIC mining farm, etc.) while the remaining 1W is dissipated in the wire.

The wire itself has very low resistance so the voltage drop across any part of it is very low. So applying the formula: V (very low) x 15A = a very low wattage. The wire hardly heats up at all. The bulk of the 24kV voltage drop happens across the load, and the power there is V (large) x 15A = a very large wattage. It is the load that has to withstand that huge amount of power, not the wire.

This is how a small wire can transmit very high powers. In fact, in the power transmission industry people try to crank up the voltage as high as possible in order to transmit more power along a fixed-gauge wire. That is why high power AC transmission lines are measured in kilovolts, or even hundreds of kilovolts. Transmission lines don't use 120V or even 240V because they would never be able to transmit enough power to be useful.

[VolanicEruptor]

Thanks Epoch, I just sent you a PM with one last question because I didn't mean to hijack this thread. 

[Dabs]

Fat wires are better. 220 to 240 volts is better. I live in a country where everything is almost 220 to 240 volts, and every wire I've seen is noticeably bigger, even if they actually need a thinner wire.

Only appliances imported from 110 volt countries are 110 volts, and they tend to fry when plugged in a 220 volt outlet. hehehehe.

[aerobatic]

So
120V 15A circuit = 1800W * 80% (derate for continual load) = 1440W usable.
240V 15A circuit = 3600W * 80% (derate for continual load) = 2880W usable.
Same wire double the power.

In Europe we have electric heaters that run at 13 amps (3,000 watts).  some of our kettles, and sandwich makers run at 3,000 watts... so i think thats the usual max for a household circuit (13 amps at 240 volts)

and of course, our ovens have a 30AMP circuit.

-- Jez