Google Offers a Million Bucks For a Better Inverter

In the full spec document, there are hints that, if the top contenders either meet/exceed all the specs, or miss all the same, they will judge based on related quantities.

Explicitly, the condition is, if none meet EMC -- but perhaps they will apply the same reasoning to other constraints, assuming no one achieves them (or everyone does).

In the full spec document, they specify loads and weighing by CEC.

In particular, you aren't going to do more than 30W by convection and radiation alone, and no more than 100W (= 95%) with forced air.

It's referred to as a power resistor. Presumably, they have a good bench supply of some sort (presumably, better than 1% voltage tolerance), and a bigass wirewound (which might not have as good a tolerance -- something to keep in mind).

No one makes them in high enough voltage (EPC's stop at 200V). The other companies that claim products in that voltage range, don't have any product available (sometimes not even the preliminary datasheets).

Likewise, no one makes SiC with low enough Rds(ON). This is a silicon project, through and through -- which is kind of silly given the interested parties.

If they had specified a lower voltage input, GaN would be actually available. But so would a smorgasboard of Si.

If they had specified a higher voltage input, SiC would be attractive -- but Si still competitive, and GaN completely out. And also such a high voltage (indeed, "medium voltage" by power line standards!) is... kind of unreasonable in the first place.

Tim

--
Seven Transistor Labs 
Electrical Engineering Consultation 
Website: http://seventransistorlabs.com
Reply to
Tim Williams
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GaN system promises 650V. So, either everybody get it or everybody fail.

Reply to
edward.ming.lee

How did you calculate the static losses for some specific efficiency ?

In a full bridge configuration, there are two switches in series. For

1 % losses , there can be total 1 % voltage drop across these two switches.

With 450 V DC bus voltage, that would allow 2.25 V across each switch. Even IGBT transistor should be able to do that, if you choose high current devices and run it well below maximum current.

Reply to
upsidedown

If that's strictly your only loss, sure. But then there's switching loss. And it's been a long time since I've seen a lossless inductor, capacitor, etc...

Tim

--
Seven Transistor Labs 
Electrical Engineering Consultation 
Website: http://seventransistorlabs.com
Reply to
Tim Williams

That's your 45W saturated loss. But getting in and out will have d450v/dt * 10A of additional loss.

I guess you can call that switching loss.

Even PCB traces and wirings get to be a problem at 10A.

Reply to
edward.ming.lee

****** ******

In fact the PV panel simulator has an unloaded output voltage of 450 V and internal resistance of 10 ohms. At 5 A load current, the _loaded_ output voltage is 400 V. Now that 5 A x 400 V = 2000 W which is the required output power, so with 100 % conversion the input power would also be 2000 W.

I guess that the original specification was for a 400 Vdc fixed source, but tho check for partial loading condition that PV panel model was used.

For full load calculations, 400 Vdc input should be assumed. Thus my _static_ loss calculations are slightly off, for 1 % losses, only 2 V would be allowed across each switch (not 2.25 V).

Since only one top transistor and the opposite bottom transistor is conducting at the time, the 5 A source current will flow through these two transistor generating 5 A x 2 V + 5 A x 2 V= 20 W loss in saturated state.

Shouldn't that be 5 A (the converter input current) not 10 A (output current) ?

Since you appeared to allocate several percents for passive component losses, I do not understand why you discard certain types of switches due some fractional percentage differences.

Reply to
upsidedown

5A straight pass through will not produce enough output power.

Output power is 170V (RMS of 220V) times 11.7A (RMS). They are measuring output power, not input power. Instantaneous current is over 10A.

Even if the converter is 99%, overall power efficiency could be less than 50%.

Reply to
edward.ming.lee

5 A x 400 V = 2000 W

Assuming for a moment PF=1 so 2 kW = 2 kVA

The output was supposed to be 240 Vrms single phase 2 kW, thus the RMS current is 8.33 Arms. With 100 % efficiency, I see no problem.

With 95-99 % efficiency, some slight adjustments needs to be done on the input side.

Yes for the output, but not for the input.

During the conversion process, some energy is stored in the LC low pass filter components.

Reply to
upsidedown

And the folding bridge FETs are on the output side.

Reply to
edward.ming.lee

Yes, you are right. That's what they say in the doc.

"In this configuration, the 240 V RMS AC output of the inverter is between terminals designated AC (Hot 1) and AC (Hot 2). These terminals connect to each other through an isolation transformer whose other coil is connected t o the AC load bank that will apply the load profile as discussed in the loa d profile section. The isolation transformer is center tapped on the invert er side and connected to ground. This has the effect of fixing both of the AC (Hot 1) and AC (Hot 2) lines to be 120 V RMS to ground, 180 degrees out of phase, similar to what would be found in a North American household."

But most North American households I know are 240V/120V PTP, not RMS.

Reply to
edward.ming.lee

I guess that I have missed something critical about the topology used.

At least I understood that we are talking about some full (H) bridge configurations, in which the switching is done at the input side and the low pass filter reactances will smooth/store the PWM signal to something similar to a sine wave.

Did I miss something ?

Reply to
upsidedown

Low pass filter does not store energy. The bridge FETs oscillate around the output sine wave, but fairly close to it.

Reply to
edward.ming.lee

I do not want to be unpolite, but WTF ?

Reply to
upsidedown

Den onsdag den 30. juli 2014 17.12.26 UTC+2 skrev snipped-for-privacy@gmail.com:

sure it does, a half bridge into an LC is basically a sync buck

-Lasse

Reply to
Lasse Langwadt Christensen

Never mind. It's 240V/120V RMS.

Reply to
edward.ming.lee

Sorry, what i mean is that the LP filter does not change the average current in and out. If we require 8A average current out, we need to drive 8A average in. Some of the time, it's more than 10A.

Reply to
edward.ming.lee

Den onsdag den 30. juli 2014 17.56.14 UTC+2 skrev snipped-for-privacy@gmail.com:

since it is really a buck converter the input and out currents depend on the voltages

-Lasse

Reply to
Lasse Langwadt Christensen

So, to get 2000W out at 240V, we need 8A average and perhaps 11A peak. Driving that with 2V across two iGBT (as someone suggested) would mean more than 40W in saturation.

Reply to
edward.ming.lee

This became a real problem in the mid '80s when offices started getting computers and other electronics in buildings that were wired for older electric typewriters and adding machines. This was written up in electrical contracting magazines after several large fires caused by an undersized neutral. I received the magazine, because I bough a lot of electrical supplies for my industrial electronics repair work.

We had the neutral overheat at the CATV headend I maintained in the early '80s. It was built in 1979, and the undersized neutral caused the breaker box to overheat. We had to bring an industrial electrician in at midnight, and we were down for about six hours to have new wires pulled to the meter, and the guts replaced in the breaker box.

--
Anyone wanting to run for any political office in the US should have to 
have a DD214, and a honorable discharge.
Reply to
Michael A. Terrell

That would require hiring an electrician to replace a failed power supply, in most places. It would also require the wiring be in conduit.

--
Anyone wanting to run for any political office in the US should have to 
have a DD214, and a honorable discharge.
Reply to
Michael A. Terrell

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