Newb Asks: Which Side Is A Charger Rated by the AC side or the DC side?

[Gregski]

don't laff I honestly don't know, and maybe it don't matter as it's the same thing, I donno that's why I am asking

so if you take my first gen Chevy Volt charger and plug it into the wall here in the States it shows that it's pulling 12 amps AC at 112 volts (so 12x112=1,344 watts) so does that mean it's charging at 1.3 kW

~ OR ~

since it's pushing 3 amps DC at 378 volts (3x378=1,134 watts) so does that mean it's charging at 1.1 kW?

[shown below for demonstration purposes I do not charge it off the truck without liquid cooling it]

well of course I had to do it

[arber333]

Yeah... well it is a good marketing. Whenever you are selling a device you look at the AC side and brag about how much current it DRAWS from your outlet. That is your power draw. The opposite side would come in handy if you would have to declare power imprint for taxes .
In europe you would draw 230Vac x 12A = 2760W from AC for example (courtesy of single transformer output vs split transformer in US...).
While with that particular draw charger would push about 6A x 370Vdc = 2220W on the DC side. What is missing we chalk up to 540W losses (switching and resistance) in an inperfect world. Of course i made the numbers up for better clarity but when i had Volt charger here i remember i observed about 370W losses. The higher the load was the lesser losses were proportionaly....

About the best efficient charger i observed was Volvo V60 hybrid charger = Eltek 3000W charger. It had only about 220W losses...

[Gregski]

well sh!t when all else fails RTFM

"The Lear 3.3kW is a Controller Area Network (CAN) controlled charger suitable for charging battery packs of 200 to 420 vdc from single-phase 120vac or 240vac split phase power.

The charger will produce a maximum of about 11.5 amps charging current at voltages up to 420vdc. Maximum input power is 3300 watts with efficiencies of up to 92%"

well 11.5x420=4,830 or 4.8 kW so that even with 92% efficiency aint 3.3 kW (it's 4.4 kW) so as the next sentence implies they gotta be quoting INPUT POWER aka AC the juice that comes out of our walls.

but how can we create more power out of 3.3 kW going in, how can the output be 11.5 amps at 420 vdc? that has to be worded wrong, it's probably amps as high as 11.5 OR volts as high as 420

[Zieg]

Well the Thunderstruck charger is limited to 10 amps output, or 3kW, whichever is lower. Could something similar be going on here? I think it's reasonable that the output ampacity would only be so much, although I'm not quite sure why the total power output would also be capped. Maybe due to heat?

[royhen99]

Page 9 of the spec.

"For example, at the high end of the voltage range, 12.6 amperes at 420 volts would require 12.6x420=5292 watts of power. But we are power limited to 3300 watts. So at 420 volts, the maximum current that can be applied would be about 3300/420 or 7.85 amperes of current".

[Gregski]

Page 9 of the spec.

"For example, at the high end of the voltage range, 12.6 amperes at 420 volts would require 12.6x420=5292 watts of power. But we are power limited to 3300 watts. So at 420 volts, the maximum current that can be applied would be about 3300/420 or 7.85 amperes of current".

I think you are mixing output 420 DC volts with input 3300 watts AC in your final computation

I'ma gonna read that page before I sound even stupider

[arber333]

well sh!t when all else fails RTFM

That is allways a good thing...

well 11.5x420=4,830 or 4.8 kW so that even with 92% efficiency aint 3.3 kW (it's 4.4 kW) so as the next sentence implies they gotta be quoting INPUT POWER aka AC the juice that comes out of our walls.

NO NO NO!
As you are limited by the INPUT you cannot draw more than 3300W.
Pmax = 3300W x 0.92 = 3036W This is max power that can be achieved on the DC side
Then you observe your max voltage 420Vdc
Idc = 3036W / 420V = 7.2A and change Which is sort of the output that i saw when crossing 400Vdc.

but how can we create more power out of 3.3 kW going in, how can the output be 11.5 amps at 420 vdc? that has to be worded wrong, it's probably amps as high as 11.5 OR volts as high as 420

You cannot!
Max amps stated are for the case when DC voltage is lower which would mean that constant power should flow through.
Idc = 3036W / 320V = 9.5A and change...so the charger can adapt to different pack voltages tokeep the power constant, up to a point...