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I was doing some efficiency testing with the GS6 charging function using a Kill-A-Watt meter which takes the power factor into consideration. I'm not completely familiar with power factor but my understanding of it is basically the difference between the power supplied (volts times amps) and the actual power used (watts) in a percentage. So if the VA is 100 and the power factor is .85 then the watts is 85. The difference is unused power which you're actually not charged for. Someone please correct me if I'm wrong.
The numbers I got were: 1698va. 1460watts. 0.86 power factor. 23.7 dc amps. 57 volts dc. So by my math it was roughly 1351 watts out and the 1460 watts in for an efficiency of 92.5%. A standard test of volts times amps would have been 1351out and 1698in for a 79.5 efficiency. To me the number figuring in the power factor is a more accurate number being as that's what's actually being consumed/purchased.
Again my understanding of power factor may be completed wrong so please chime in and tell me why I'm wrong or if I'm correct on how it works.
Your PF understanding is correct
I never tested the efficiency, that is really good
One thing to be aware of is the voltage regulation in that mode is a boost converter and could actually be that good
I was doing some efficiency testing with the GS6 charging function using a Kill-A-Watt meter which takes the power factor into consideration. I'm not completely familiar with power factor but my understanding of it is basically the difference between the power supplied (volts times amps) and the actual power used (watts) in a percentage. So if the VA is 100 and the power factor is .85 then the watts is 85. The difference is unused power which you're actually not charged for. Someone please correct me if I'm wrong.
The numbers I got were: 1698va. 1460watts. 0.86 power factor. 23.7 dc amps. 57 volts dc. So by my math it was roughly 1351 watts out and the 1460 watts in for an efficiency of 92.5%. A standard test of volts times amps would have been 1351out and 1698in for a 79.5 efficiency. To me the number figuring in the power factor is a more accurate number being as that's what's actually being consumed/purchased.
Again my understanding of power factor may be completed wrong so please chime in and tell me why I'm wrong or if I'm correct on how it works.
Very interesting. 92% might be possible at lower loads (i.e. the ~1500W of the test), but it does sound quite high!
One possible way to quantify that would be to measure the generated heat loss. From a standard watts measurement of 1698-1351 = 347W of waste heat (= 1,184 BTU), vs 1460 - 1351 = 109W (= 371 BTU), the temperature rise in an enclosed environment by time could be used to ascertain where it actually falls. I would think that a >3x difference in generated heat would not be difficult to measure!
To my understanding (which could be incorrect!), the "power factor" is a ratio of how closely the amperage waveform follows the voltage waveform. Thusly a pure resistive load will get a power factor of 100%. Whereas a non-PFC (power factor corrected) SMPS will "cut" the top of the AC waves, with a very non-sinusoidal amperage waveform--and get a poor power factor reading. The total amount of power drawn remains the same--but it's pulled in a very uneven fashion.
I'd have to say your both correct on PF
The current peak on a lower PF means the amps are drawn at a lower voltage making the total watts drawn lower
The current peak on a lower PF means the amps are drawn at a lower voltage making the total watts drawn lower
No no......the amps are drawn at a HIGHER voltage.
I'm talking: AC mains -> Bridge rectifier -> 400v filter caps. Quick web search turns up a number of diagrams with the expected result:
A quote from an Ametek document sums it up as follows: "The Power Factor of an electronic load is a number between 0 and 1 and is an indication of how efficiently it makes use of the AC electrical supply but is not a measure of how efficient it is in terms of energy usage."
Yes with a passive bridge, but that's not what we have
Look up the SCR dimmer circuit
It works by shifting the part of the wave used to power the light
That counts as a PF shift
Also a reactive load will draw current only during voltage change causing the current to time shift the voltage
That's the most often used description of PF
In every case the current curve doesn't peak at the voltage peak
PF is watts over VA
92% might be possible at lower loads (i.e. the ~1500W of the test), but it does sound quite high!
That was the highest I tested it at but it seemed to get more efficient the higher I went. At the 2.5 amp setting it was 90.4%. What really changed was the power factor which was at 0.68 at the 2.5 amp setting.
My tests were very short so it didn't account for fan draw at higher loads from the extra heat.
So I think what's causing confusion is the definition of Power Factor.
Like the Ametek document noted, Power Factor does not represent (nor affect or change) the total power consumed. It merely represents how evenly the power is used in an AC scenario.
For example, a lot of Chinese HF inverters like to rate their products in "KVA", i.e. 5kva inverter. But this neat trick allows them to use power factor in the opposite way to what's discussed here--where low-power factor loads REDUCE the maximum load before it shuts down. In other words, a load with a PF of 0.5 will overload the inverter at 2,500W. And a load with a PF of 1.0 will overload the inverter at 5,000W.
In theory, this is quite simple to understand, especially in line with a bridge rectifier circuit. Say the inverter's good for 5kva @ 220v = 20.8A RMS. But if the load is pulling 5kw of power--but only using 1/8th of the waveform, this means that the current drawn during that "peak" is going to be at least double or quadruple (in order to get the total required wattage)--and most "kva" inverters are HF inverters which can't do much of anything for surge power anyway.
I've never seen PF numbers quantified to "reduce the drawn power", but rather always in context of "reducing the maximum power capabilities" of an inverter.
Think about it this way: if the GS inverter with a PF of 0.8 means that the "volts * amps" reading of 1698 should be calculated to 80% = 1,354W "actual power"...THAT means that I should be trying to make the GS inverter have the absolute worst power factor possible. Because let's say if the GS inverter's PF is 0.1, then "volts * amps" of 1698 * 10% = 169.8W input but still 1,384W output -> poof, we've achieved over-unity free energy out of an inverter! (Let's not mention the 300W of heat generated in the process.)
That was the highest I tested it at but it seemed to get more efficient the higher I went. At the 2.5 amp setting it was 90.4%. What really changed was the power factor which was at 0.68 at the 2.5 amp setting.
I'd be curious if you provided screenshots of the XFMR AMPS 'scope channel at the various currents...whether I should adjust the charge "sine" table for a better PFC result.
Worth noting that the "MaxAmp" settings will eventually encompass more than just charge; they're intended to be the "limit" for the AC input that the inverter tries not to exceed (I know, I know...the manual...) There's a "Charge Percent" that allows you to specifically drop charge current lower.
OP is using external metering for the calculation
All the explanations of PF are actually correct and explaining different scenarios
I'd be curious if you provided screenshots of the XFMR AMPS 'scope channel at the various currents...whether I should adjust the charge "sine" table for a better PFC result.
- 15 amps
-
10 Amps
5 Amps
2.5 amps
Worth noting that the above listed amps are the set amps however the actual amps differ slightly. I maxed out the calibration to get it close.
That aligns perfectly with the SCR dimmer concept
Wow I need to adjust the charge sine table so badly...! That's not very close to a balanced result...
however the actual amps differ slightly. I maxed out the calibration to get it close.
As long as you weren't adjusting the calibration to change the actual consumed current in charge; the charge regulation should leave a small headroom just for safety's sake. If properly calibrated, a 15A heat gun on the output of the inverter in pass-thru (no charge) should register properly on both the input and output current sensors.
As long as you weren't adjusting the calibration to change the actual consumed current in charge
Oops, that's exactly what I did.
the charge regulation should leave a small headroom just for safety's sake.
Does that mean that during charging it should use less currant than what it says? It was actually using a bit more current than what it was showing. I don't remember the numbers off the top of my head but it was getting 14 gauge wires very hot when set to 12.5 amps.
Either way I'll calibrate it as you mentioned and go from there.
The numbers I got were: 1698va. 1460watts. 0.86 power factor. 23.7 dc amps. 57 volts dc. So by my math it was roughly 1351 watts out and the 1460 watts in for an efficiency of 92.5%. A standard test of volts times amps would have been 1351out and 1698in for a 79.5 efficiency. To me the number figuring in the power factor is a more accurate number being as that's what's actually being consumed/purchased.
I'm very confused. I don't know of any power company that bills residential customers in a way where power factor matters. They bill you by the KWH - which is often termed "real power". If you are a large business customer, they will further penalize you for poor power factor - but that's usually some kind of multiple added to the per-KWH rate - it doesn't change how much power you consumed - that remains in KWH.
Anyway, real efficiency would be KW DC / KW AC. (Real power out / real power in) The power factor might partially explain why the efficiency is good or bad, but it doesn't tell you what the efficiency is.
The 80% efficiency is much more likely to be correct. My 24v GS6K sees charge efficiencies down into the 60s at 3000W or so. Keeping it around 1000-1500 usually lands comfortably in the mid-to-high 70's.
Be careful charging at higher amps -- I've found that to be one of the only real weaknesses of the GS inverters. You push the charge setting too high, and you may burn up the traces on the mainboard.