Yes, got that after going back to read the spec. GaN system also offer some power transistor (FET?) for 5mOhm at 80A. We would need to drive at least 20A at peak and perhaps more to counter the inductance.
One stated use for this inverter is photovoltaic applications. To get
2 kW, you need twenty 100 W panels. For 450 Vdc bus about thirty five 57 W panels would be required. These panels need a lot of space.In practice for PV panels, you are going to need panel specific or panel group specific MPPTs to feed a common DC bus so that the inverter could be used.
On a sunny day (Wed, 23 Jul 2014 15:00:19 -0400) it happened rickman wanted to disagree for the sake of it:
If you have space of all that stuff in that picture it is supposed to power you have PLENTY of space.
Strawman.
No, there is a space space, you cant stick your fins (LOL) out of that, you need space for the electrotrics too.
Forced air (fan), not sure if it is allowed, but fans take up space too, not enough space left, air temperature?
Better specs at:
Time for some back of the envelope guesswork and my rotten math.
Google wants the box to be no larger than 40 cubic-inches. The largest surface area would be a sphere, but I don't think they'll go for a ball shaped inverter. A cube will suffice at about 3.4 inches per side. That's tiny.
Let's pretend that the inverter will be made mostly from copper for best heat conduction and painted black for best radiation cooling. The density of copper is 0.31 lbs/in^3. My guess(tm) is that the inverter will end up about 75% copper (by volume), so this thing will have: 40 in^3 * 0.31 lbs/in^3 * 0.75 = 9.3 lbs of copper. 9.3 lbs = 4.2 Kg including any surface wrinkles to simulate fins and increase surface area.
The 60C maximum surface temperature implies a: 60 - 25 = 35C maximum temperature rise above ambient. (The spec mentions using SiC devices. These can run at 500C, but Google only allows 60C.)
Specific heat of copper is 0.385 Joules/gram or 385 Joules/Kg. The means it will take 385 Joules to raise the temperature of 1 Kg of copper 1 degree C or: 385 Joules/Kg/C * 4.2 Kg = 1617 Joules/C needed to raise the temperature 1 degree or: 1617 Joules/C * 35C = 56,000 Joules needed to raise the temp 35C (per the specification).
At 95% efficiency and 2Kw of output, that allows about 100 watts of dissipation. 1 Joule = 1 watt * 1 sec 56,000 Joules / 100 watts = 560 seconds The spec says nothing about environmental conditions during the test except that the temperature will be somewhere between 15C and 30C. Assuming a somewhat adiabatic (insulated) system and still air, the inverter will run for: 560 sec / 60 sec/min = 9.3 hrs before going out of spec and overheating. Even if one makes the heat sink out of diamonds, with a specific heat of 0.519 Joules/gram, it will only run for 12.5 hrs before overheating. In other words, without secondary cooling or science fiction efficiency, this inverter is not going to run for 100 hrs without going overtemp.
Diversion: The web page claims that the purpose of this exercise is: "Making them smaller would enable more solar powered homes, more efficient distributed electrical grids, and could help bring electricity to the most remote parts of the planet." I don't understand this. How does a miniaturized solar powered inverter do any of these things when a picnic cooler sized device already does the job quite nicely. Am I missing something here or is this entire exercise a waste of time and money?
-- Jeff Liebermann jeffl@cruzio.com 150 Felker St #D http://www.LearnByDestroying.com Santa Cruz CA 95060 http://802.11junk.com Skype: JeffLiebermann AE6KS 831-336-2558
CF/LED lights and small electronic appliances do not have PFC, so lots of current is drawn during the peak voltage.
The required PF was just 0.7 to 1.0, so this is realistic for nearly fully loaded motors, but the range is not enough for motors on idle or small electronic loads.
On a sunny day (Wed, 23 Jul 2014 12:17:43 -0700 (PDT)) it happened snipped-for-privacy@gmail.com wrote in :
Why not use 230V 50 Hz, that saves a lot of work?
Anyways I have a DC-DC converter on my solar panel that feeds my laptop with 18V DC eliminating the AC, yes that is with a 12V huge sealed lead acid now. I have a second DC converter for my eeePC (9.5V DC), and a third adjustable one that I use to charge all sort of batteries. The first 2 converters are < 7$ on ebay free shipping. I have some more for other stuff. I put some big 5 pin XLR connectors so things cannot be plugged into the wrong converter. They can, but then you get no power. The connectors is the most expensive part of the setup.
use the same outlet plug. We have another 100 years
The world is either 110V/60HZ or 220V/50Hz. So, we have to compromise with 100V/50Hz and both side can win and lose. 110V (or rounded to 100V) is mu ch safer than 200V. We couldn't agree on this for the past 100 years. Hop efully, we could during the next 100 years. LOL.
Yes, and that has...what do to with time domain impulse response of a dynamic system?
Tim
(Not taking the time to indent Google Groups, original follows below)
-- Seven Transistor Labs Electrical Engineering Consultation Website:
From Wiki:
"The total harmonic distortion, or THD, of a signal is a measurement of the harmonic distortion present and is defined as the ratio of the sum of the powers of all harmonic components to the power of the fundamental frequency. THD is used to characterize the linearity of audio systems and the power quality of electric power systems. Distortion factor is a closely related term, sometimes used as a synonym."
The distortion of the power output is a factor of the impulse response caused by the changing load.
Why not get rid of the 50/60 Hz AC distribution completely and go to complete DC distribution.
The need for AC has been big iron transformers and running induction motors. Today, a lot of motors, some of which are quite large, are driven by VFDs, thus requiring an AC/DC conversion at input with current harmonics issues and on the output side, there are voltage harmonics issues.
Having a standard, sufficiently high DC bus voltage and VFDs directly mounted to the motor, you can get rid of much filtering etc.
One problem with the quite low 50/60 Hz feed is that induction motor speed is limited to below 3000/3600 rpm, With a motor specific VFD, much higher speeds can easily be obtained, making it possible to use smaller motor dimensions.
In high voltage distribution, HVDC systems are used, mainly in submarine cables, in which the cable stray capacitances limits the length of AC transmissions in cables. With HVDC, there are no such limitations, making it possible to use buried HVDC cables also on land, thus avoiding huge HV overhead lines, with rights of way issues and fierce resistance by local population, when trying to build a new HV AC overhead line.
I'm thinking Musk took a stroll by Google and hinted to them about how cool it would be to run an inverter from his car's battery.
Tim
-- Seven Transistor Labs Electrical Engineering Consultation Website: http://seventransistorlabs.com
I would do a 8"x10"x0.5" wall mounted panel. Sound like they want it in-do or mounted, so the space constraint.
Most of the top efficiency inverters are 97%, so it might not be too diffic ult to achieve 97%. One listed above 98%, but remains to be verified. How ever, to get the last 1% or 2%, the inverter must be adaptively matched to the characteristic (impulse response) of the load. Namely, very fast A2D a nd DSP loop.
Yes. Our time. Their money.
Testing inverter voltage THD is not normally done with a changing load. First load to test with while designing an inverter is with a resistive load, then the load can be changed to something different which has to sink as well as source. The THD and P.F. is normally tested with a decent power meter and as far as I have seen only includes up to around the 40th or 50th harmonics or so.
It looks to me that they may be trying to emulate a PV array by requiring the 10 Ohm source resistance. 450 VDC is approximately the max power point voltage of an array applied to a 600VDC maximum V inverter system (80% of Voc)
They obviously want SiC and GaN semiconductors to be part of this which makes sense by seeing the company names listed.
I can only say good luck on this challenge.
boB K7IQ
To get the last 1% or 2%, the inverter must adaptively matched to the characteristic of the load, to minimize losses.
Valid or not, we have to play by their rules. A typical household appliance, the fridge, turns on and off. I am sure it will be included in the test, perhaps with a blender running as well.
On Wed, 23 Jul 2014 22:15:27 -0700 in sci.electronics.design, Jeff Liebermann wrote,
A sphere would be the *smallest* surface area. They specify a rectangular box. And the largest surface area is something thin and flat.
Great question ! And some nice heat-time calculations.
If this ~IS~ supposed to be an inverter operated strictly by PV input at 450V, the sun-hours available won't even be at 9 hours before the insolation (irradiance) reduces and power source availablility drops so that helps there, although even grid tie inverters are measured as if they are continuous duty or until the temperature stops rising.
boB K7IQ
More than that. They want something like a tablet or closed laptop. Perhaps something that can be shipped with priority mail.
A good inverter will have good surge capacity, fast voltage regulation with changing loads and decent short circuit/overload/overtemperature protection. I'm not sure if they mentioned that in the specs or not ?
boB
If Google wanted a specific form factor, they should have specified it. The problem with an 8x10x0.5 package is that half the dissipative surface area is wasted on where the bottom of the package sits on the floor or is mounted to the wall. I would do better with a cube, which only wastes 1/6th the surface area.
I know that Google is testing it essentially suspended in space. From:
So, the box loses 0.25 in^2 times 4 posts for a total of one square inch out of a total surface area of: Area = 6 * 3.4^2in = 69 in^2 A loss of 1/69th in irradiative area is not too bad. However, in my never humble opinion, it's quite unrealistic. Whether a cube or your slab, it's going to be mounted flat against something that's going to prevent radiation, block convection, and possibly offer come heat removal by conduction. If it's mounted on an insulated wall board, it's an insulator and there's no heat conduction.
That's the hard way. Each cycle of 60 Hz is quite repetitive. The load may change with the applied instantaneous voltage, but most of the cycles and load conditions look exactly like their predecessors. It would only be necessary to measure the load for the current cycle at as many points as practical, and adjust the subsequent voltage waveforms so that they fit the predicted load for maximum efficiency. If the load happens to change, one or two cycles might be screwed up, but after readjustment, the waveform is back to where it started. Trying to instantaneously track what is probably a fairly stable load is overkill.
That was NOT a rhetorical question. I'm quite serious. How is miniaturizing such an inverter going to provide the three goals that Google specified? "Making them smaller would enable more solar powered homes, more efficient distributed electrical grids, and could help bring electricity to the most remote parts of the planet." I don't see the connection, but maybe I'm missing something.
Also, what's with the 450 VDC input in series with 10 ohms? I don't know of any alternative energy power source that produces a constant output voltage. Well, maybe a steam turbine under ideal conditions powered by a nuclear reactor. Still, even these have voltage ranges.
And the source series resistance is way too high. Presumably it's the generator, solar array, wind generator, steam generator, battery, etc source resistance plus the input wiring resistance. At 100% efficiency 2000 watt load, at 450VDC, would require: 2000 watts / 450V = 4.44 amps Through a 10 ohm series resistor, that will burn: P = I^2 * R = 4.44^2 * 10 = 198 watts That 198 watts is going to be burned somewhere. This thing is suppose to be 95% efficient and it already has wasted an extra 198 watts in the source? I think someone missed a decimal point or two on the 10 ohms Rs value.
13:30 AM. Time to fall over.-- Jeff Liebermann jeffl@cruzio.com 150 Felker St #D http://www.LearnByDestroying.com Santa Cruz CA 95060 http://802.11junk.com Skype: JeffLiebermann AE6KS 831-336-2558
On a sunny day (Thu, 24 Jul 2014 09:02:30 +0300) it happened snipped-for-privacy@downunder.com wrote in :
I agree, but IIRC the problem with DC is in the switches. Once the connection is 'opened', the arc is not extinguished. Semiconductor switches would work, but not sure if I would rely on that for my life.
Mechanical switches would have to be much bigger for the same voltage as DC swicthes, with perhaps reduced lifetime too?
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