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I'm losing my math somewhere and can't figure out where. I'm planning the battery config for my DIY 48 V powerwall using harvested 18650 cells. Let's say my baseline power usage is 2500 watts/2.5 kW, so 2500 W / 48 V is roughly 53 amps of current I need to pull from the battery/inverter (let's ignore DC to AC conversion lost for now to keep things simple). The first phase of the battery installation will be 14s245p. To figure out how much current each 18650 cell needs to provide I divided 53 A by 245 cells to get roughly 0.217 A / 217 mA per cell. I thought this was right, but out of curiosity I tried another route and divided the total watts usage by 3430 (the total cells in the battery for the first phase) and I got roughly 0.73 watts per cell. This equates to roughly 0.195 A / 195 mA per cell (0.73 W / 3.7 V). I know 217 mA is close to 195 mA, but I imagine they would be the same if not very very close. I'm doing something wrong here; how do I properly calculate the amperage/current needed per 18650 cell in this case?

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03-22-2020, 09:49 AM
(This post was last modified: 03-22-2020, 09:53 AM by Maniac_Powerwall. Edited 2 times in total.)
Your Error is to calculate with 3.7V per cell, but 48V divided by 14 equals ~3.429V. When calculating 0.73W / 3.429V you get ~213mA.

I always calculate with 50V wich would be 3.65V per cell. Its easy to calculate and close enough to "real" in my opinion.

My Powerwall Build Thread
Maniac_Powerwall: 1x 14S100P 18650 LiIo, 6x 345W LG panels, MPP Solar PIP5048GK, Batrium Watchmon4 BMS

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Okay, I think I see my issue - I'm confusing A and Ah... to run the house for an hour I need 2500 Wh of energy and EACH of the 3.x strings has to have 53 Ah capacity. But, if I need 2500 W of power at any given moment then the WHOLE battery has to be able to put out 53 A and NOT EACH 3.x string...??