I have an AC 180 and an AC 200 P. I would like to connect them together to increase my total battery capacity, even though I understand that I may have to plug power draws into both at times. The ac180 Has Bluetooth and so therefore I can monitor the level of 1. So is there any reason why I cannot use the 10 amp DC output on my 180 to charge my AC 200? This will allow me to always know how much is in the combined system because when the 180 gets down near zero the AC 200 should still be full because it’s being charged by the 180. If this is viable, isn’t this more efficient than Daisy chaining through the AC port because this is just DC to DC with no conversion loss?
The plug and socket will probably get excessively warm transferring 10 amps during long durations
Thanks. What about using ing the ac200 25 amp dc port to charge the ac180? Is that more efficient and viable way rather than using the ac port and converting back to dc? It’s unfortunate if the DC ports can’t actually handle the amperage that they’re rated for. I’m wondering how to determine other than just trial and error if using the port actually gets hot.
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The 30 amp rv outputWould definitely be a better choice. It will not get hot since it will only draw enough 12 V power up to the amperage limit of the device being charged. Plus you’re only going to see approximately 100 W of charging input due to the low voltage. But it will work at that charge rate
Thanks. 30 amp not available. This is a ac200p. My experience with using the 15amp AC ports there is a pretty massive conversion efficiency loss. I may just try the DC port and report on its viability because slower but less loss is still more net energy available and that’s what you need when you’re out in the wild for days at a time.
I am referring to the 30 amp DC output. There is no free lunch however because the high amp regulated DC output consumes energy in the regulation process.
Sorry for prolonging this, but below is what blueetti says the ports are on the AC 200 P, do you mean the 25 amp port dc port? It just seems to me that using the DC port- whether 10 amp or 25 amp over an AC port is always more efficient, but I may be wrong.
There are:
- Six AC 110V power outlet ports.
- One DC 12V/25A.
- One DC 12V/10A.
- Two DC 12V/3A.
- Four 5V/3A USB type-A.
- One PD 60W faster charge port accepting 5V/3A, 9V/3A, 12V/3A, 15V/3A or 20V/3A charging.
- Two wireless charging areas on the top of the power station that will deliver a single output charge up to 15W.
Hello
To find out which is the most interesting in terms of energy efficiency between:
-discharge 12V 10A max via the 12V DC output of your AC180 into your AC200P;
-or discharge using your AC200P’s AC-DC adapter from an AC socket of your AC180
the answer is not obvious because as Mr. Scott-Benson says 12V is a very low voltage, and with high intensity (10A) you will have thermal losses and then you have to increase the voltage (with new losses)…
But perhaps you can conclude simply by doing an experiment: go with your AC180 full and your AC200P empty, and discharge, until your AC180 is empty, in one case from the DC12V socket of the AC180 and in the second case from the AC socket of the AC180 (with the AC200p’s AC-DC adapter) and you will see the result on the AC200P and you can compare. (You can also look at the instantaneous power coming out of your AC180 and that entering the AC200P but it will be less precise.
PS : to monitor AC200P, you can put a wifi smart socket (or bluetooth such as TUYA, SMARTLIFE compatible model) on the used AC socket of the AC200P and another smart socket on the AC socket of the AC180 to which the T400 or T500 charger of the AC200P and you can do a calculation (Smart sockets often have the function of electric meter).
Yes, the DC 12V/25A port is what I am referring to. It takes a certain amount of power to transform DC to AC power and it also takes a certain amount of power to generate a regulated DC voltage. It takes a little more power to generate a high output (25 amps) DC regulated voltage. What I am saying is that I would not assume that this high amp DC output port will be al that efficient for your intended purpose due to the electrical losses produced by regulating the voltage. If the DC output was coming direct from the internal battery at its native voltage, then yes I would agree it would be more efficient.