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I have this battery: https://www.amazon.com/gp/product/B07SWBS55F/ref=ppx_yo_dt_b_asin_title_o09_s01?ie=UTF8&psc=1 which has 156W of output when all 3 output ports are added.

To avoid unnecessary conversions, I'd like to have 1 output of ~150W, so I opened it up and there's a battery pack containing the 18650 cells in 7s2d and a small PCB that all the original in/outputs are attached to, so the PCB is likely for regulating charging, stepping down outputs to the manufacturer's provided outputs, & keeping track of battery charge level. I wanted to keep the PCB for USB power out/charging/battery level purposes, but add a 3rd output that would simply be the 2 leads from the battery cell itself and bypassing that PCB entirely for a 24-25V output that would let me draw current limited only to the cells' actual rating. It'd also mean I eliminate another power conversion step & its associated lost efficiency.

Is this safe to use the battery cell output directly or is there any safety circuit I missed that would have to be added? (or is this going to severely reduce the life of the battery/how could I avoid that) Could I use the battery from my new port while it's being charged? Could I use both my new port as well as an output on the existing PCB (for example, plugging a phone in to the original USB output and using the new raw output simultaneously)? Would the charge indicator still be accurate?

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  • \$\begingroup\$ I'm assuming trying to modify the existing 24v5a output on the PCB to remove the 5a limitation isn't going to be practical in part because that same port is used to charge the battery as well, and any changes would have to take that into account... which I'm definitely not qualified for.... \$\endgroup\$
    – ieatpizza
    Feb 24, 2020 at 8:21
  • \$\begingroup\$ Supplier: Amazon, DataSheet: None, Verdict: no guarantees on the unit being used as intended (even if it were specified somewhere) and certainly no guarantees on modifying it in some way. Engineers like hard facts and data on equipment and components bought from reputable suppliers. \$\endgroup\$
    – Andy aka
    Feb 24, 2020 at 9:30
  • \$\begingroup\$ I agree it's a little vague, it just happened to be one of the only options for portable power banks over 150wh so I figured it's at least worth trying before giving up and figuring out how to build my own. :/ \$\endgroup\$
    – ieatpizza
    Feb 24, 2020 at 10:25
  • \$\begingroup\$ Reading the specs it seems the 24V output is actually directly from the battery (perhaps through an over-current protection device). \$\endgroup\$ Feb 24, 2020 at 10:48
  • \$\begingroup\$ This pack is NOT rated at 156W. It has a battery capacity of 156 Wh. Don't modify it, just use the "24V" output at no more than 5A (which at full charge 29.4V gives you 147W) \$\endgroup\$
    – user16324
    Feb 24, 2020 at 13:49

1 Answer 1

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I say "probably" 6 times below. This is because one can never be certain what an unknown manufacturer/designer has done BUT I'd expect my answers would be correct in most cases. YMMV.

Is this safe to use the battery cell output directly

"Safe" is in the mind, practice and ability of the beholder.
Yes, it's potentially safe, if you connect competently, do not short or discharge with too high a current or to too low a voltage.

or is there any safety circuit I missed that would have to be added?

You need to tell us the cell brand and model.
As the pack is rated at 156 Watts max then Ibattery max is in the order of I = W/V 156 W/(7 x 3.6) = 6.2 A (probably). If you limit current draw to under say 7A you are probably within battery specifications.

The cells hopefully have internal cell protection circuitry - but maybe not. Adding a gross overcurrent protector to the output "would do no harm".
If direct discharging you should never discharge a 7SxP pack to under 3V x 7 = 21 V and ideally a little higher.

... is this going to severely reduce the life of the battery/how could I avoid that)

As above. Treat the battery like an LiIon battery likes to be treated.

Could I use the battery from my new port while it's being charged?

Yes, but. It MAY have a "gas gauge" circuit to help judge capacity (but, probably not) and direct access would scramble that. But odds are it would be OK.
The charging termination may be somewhat confused by direct access discharge, but, probably not badly so.

Could I use both my new port as well as an output on the existing PCB (for example, plugging a phone in to the original USB output and using the new raw output simultaneously)? Would the charge indicator still be accurate?

Probably. Again, IF gas gauging is used it could get confused, but it most likely has not got that degree of sophistication. Remember that total watts is direct but usual loads.

The charge indicator is probably based on cell voltage and a little magic, so direct access not a problem.

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  • \$\begingroup\$ I'm assuming I'd have to rip the cell pack itself apart to find the manufacturer. There is a second PCB that's packed within the cell pack's wrapping though, I'm guessing that's the undervoltage/overcurrent protection built-in? \$\endgroup\$
    – ieatpizza
    Feb 24, 2020 at 9:30
  • \$\begingroup\$ @ieatpizza It's not compulsory to pull the pack apart. I probably wouldn't if it is well wrapped up. It is probably enough to assume the max current as above. That's a relatively modest rate but it does vary widely amongst models and brands. Yes - a PCB inside the cell wrappings is PROBABLY a BMS (battery management system). Solder carefully and sensibly. Don't heat the cells. ... \$\endgroup\$
    – Russell McMahon
    Feb 24, 2020 at 10:03
  • \$\begingroup\$ Perfect, will try it out. thanks! \$\endgroup\$
    – ieatpizza
    Feb 24, 2020 at 10:12

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