Enphase 5p Battery specs - What is AC round-trip and DC round-trip efficiency? - Page 1
Forum section: electronics / general

author: DougSpindler
author_rank: Super Contributor
timestamp: March 28, 2024, 03:52:21 pm
content: Just purchased and installed an Enphase 5p battery. Could someone please explain what is meant byAC round-trip efficiency 90% (AC to the battery to AC at 50% power rating)DC round-trip efficiency 96%The 5p only has AC terminals for charging and load so not sure why they list DC efficiency. (There is no way to DC charge.)I'm guessing the RT AC efficiency means if 1kWhr is used to charge the battery I will only get 900 Whrs back when discharging. (Provided the discharge load on the batteries is at 50% of the continuous capacity load 3.84 kWhr.)Thankshttps://enphase.com/download/iq-battery-5p-data-sheet

author: Siwastaja
author_rank: Super Contributor
timestamp: March 28, 2024, 04:18:58 pm
content: Yes, you got it right.QuoteJust to offer more detailed information for those interested. Indeed it's not very relevant for end user if the only use case is through the AC inverter-charger. Battery chemistry and voltage ranges is similar extra information; just for reference.

author: DougSpindler
author_rank: Super Contributor
timestamp: March 28, 2024, 05:07:48 pm
content: Quote from: Siwastaja on March 28, 2024, 04:18:58 pmThank you so much. I've learned a lot from the people in this forum.So if my calculations are correct, Enphase configured the batteries to provide 8 amps during "normal" discharge cycle.3,840Whrs / 240v = 16Ahrs.Applying the 50% setting 8 amps per hr at 240 volt. Or 16 amps at 120 volts.It's a 5kWhr battery with a 15% reserve leaving 4,250Whrs useable. (Assume new battery, no degradation.)So if I had a 1,500 watt 120 volt resistive heater, (assuming 100% duty cycle) current draw would be 12.5a.Runtime (assuming 100% draw from the batteries and no additional heat loss from the higher current) would be 4,250/1500 = 2.83 hours.Are my calculations correct?

author: 5U4GB
author_rank: Super Contributor
timestamp: March 30, 2024, 11:21:34 am
content: Another thing you're going to run into is tare loss, which is the cost of keeping the electronics running. There's a goodEnphase-specific thread that covers this.

author: DougSpindler
author_rank: Super Contributor
timestamp: March 30, 2024, 02:42:50 pm
content: Thanks for sharing.Still collecting data and trying to understand how to read Enphase’s software but I have noticed the batteries are periodically during the day drawing 100W.I have the newer 5p battery and and trying to figure out how to get the parasitic loss from the batteries.

author: 5U4GB
author_rank: Super Contributor
timestamp: March 31, 2024, 08:29:11 am
content: The 100W figure sounds like the tare loss. I've got 5Ps as well and that's the figure for the inverters and other electronics surrounding them.

author: DougSpindler
author_rank: Super Contributor
timestamp: May 14, 2024, 05:12:28 am
content: Would you please explain what tare loss is.

author: 5U4GB
author_rank: Super Contributor
timestamp: May 14, 2024, 12:57:48 pm
content: Quote from: DougSpindler on May 14, 2024, 05:12:28 amSee the link inthis post, that explains it pretty well.

author: DougSpindler
author_rank: Super Contributor
timestamp: May 14, 2024, 04:01:54 pm
content: Thank you

author: DougSpindler
author_rank: Super Contributor
timestamp: May 14, 2024, 04:04:59 pm
content: tare loss appears to be the same as a parasitic loss.