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I'm designing a circuit that integrates both USB Power and a Li-Ion battery when USB is unplugged. For the battery charging/protection/boost part, I'm following the schematic given in this Instructable.

The below has been completely edited to provide a more full schematic.

Currently, my project has a representative schematic like below which mostly parallels the GreatScott! schematic linked above. However, the key differences are replacing the 5V/GND terminals with an actual resistive 5V Load that draws 300mA [please assume this]. Additionally, a power line runs from the 5V of the USB [with power filtering using a ferrite bead and decoupling caps omitted] directly to the load via a schottky diode [omitted by accident], to power it when the USB is connected instead of the battery.

Note that BAT+ and BAT- on the image below (right in the middle near the protection IC block) are two solder terminals for a single-cell Li-Ion polymer battery, outputting 3.7V nominal, 4.2V max.

enter image description here

My main concern here is that is actually appears the BAT- terminal of the battery is never explicitly connected to ground. Thus, when USB power is removed, does BAT- ever form a complete circuit in this case? The only instance where I even see BAT- here is in teh Protection IC section [lower middle of the picture].

I'm aware that BAT+ should never be shorted to ground, since that would just be a short-circuit. However, where does the BAT- terminal go? The path of the BAT+ makes absolute sense, going into the Boost IC after a toggle switch. The BAT- connection is confusing me on the other hand.

Clearly, when USB is plugged in, the USB's ground becomes the functional GND for the circuit, and it is complete. However, the BAT- terminal never appears except right in the protection IC. Therefore, I'm confused how BAT+ and BAT- form a complete circuit w/ the load in the event the USB is disconnected.

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  • \$\begingroup\$ why would the negative terminal have to be connected directly to ground? ... if the original circuit works correctly, then why would you make changes to it? \$\endgroup\$
    – jsotola
    Jul 21 '20 at 2:00
  • \$\begingroup\$ @jsotola I'm actually not entirely sure how original circuit even manages to function, since the ground pad (on the right under the 5V pad) connects directly to a functional GND. However, I can't figure out where the Battery's negative terminal connects to this functional ground in a way that ends up completing the circuit. \$\endgroup\$
    – Felix Jen
    Jul 21 '20 at 2:50
  • \$\begingroup\$ the battery positive is not connected to the Vout ... why would the battery negative have to be connected to ground? \$\endgroup\$
    – jsotola
    Jul 21 '20 at 3:35
  • \$\begingroup\$ Connecting BAT- to GND would defeat the purpose of Q1. \$\endgroup\$ Jul 21 '20 at 11:46
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This is a common way to integrate a battery protection IC. The protection IC controls the two mosfets in Q1. If the battery is overdischarged, it turns off the right mosfet, preventing current from flowing from ground to the negative battery terminal (i.e. prevents the battery from discharging). If the battery is overcharged, it turns off the left mosfet, preventing current from flowing from the negative battery terminal to ground (i.e. prevents the battery from charging). In each of these cases, it allows current flow the other way (OD mode allows charging, OC mode allows discharging).

This IC also has other functions, like detecting overcurrent and short circuit, so it can also disconnect the battery if it sees those. It's also common to integrate a temperature sensor (although this IC doesn't allow it) to monitor the battery too.

Edit: to answer the actual question - No, do not tie the battery directly to ground, since that would make the protection IC useless and give you an unprotected li-ion battery, which is always bad.

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  • \$\begingroup\$ Thank you for this very on-point answer. This definitely makes sense. A follow up question on the actual MOSFET itself, it seems like a the Dual-FET shares a common drain, which isn't connected to anything external, only internally. Is that an accurate assessment, leaving the common-drain pins floating? \$\endgroup\$
    – Felix Jen
    Jul 21 '20 at 3:15
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    \$\begingroup\$ That is correct \$\endgroup\$
    – BeB00
    Jul 21 '20 at 4:08
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It would definitely help if you have the schematic for your entire project (and highlight the area that you want the exchange to lookout).

Anyways, it is important to notice that in the schematic above, BAT is not tied to ground. Quick glance, it does look like Batt is shorted to ground but clearly, that doesn't make sense. If BATT was shorted to ground, then BOOM. You have a fire on your hands. NEVER EVER SHORT YOUR BATTERY, ESPECIALLY IF IT IS A LITHIUM BASED ONE!!!!

Anyways, if you look a little closer (and read up the datasheet for the FS8205) you can figure out that the battery is connected to the discharge pin of the IC, not ground.

Now look at the picture of GREAT SCOTT's schematic, notice the diodes? They are not facing the same direction. One diode is facing to the right (the one connected to the MOSFET with source BATT). The other is facing to the left.

Esiantily, BATT is unable to go from the right MOSFET to the left MOSFET (because of the diodes and also how the MOSFETS are oriented). This can be easily seen in this poorly drawn diagram I made:

enter image description here

Now something else to point out. If you trace the rest of the path for BATT, it leads to pin 1 of the IC. This pin is called the "MOSFET gate connection pin" which is meant to discharge the battery if it is overcharged (for specifics, read the datasheet).

Overall, JUST DON"T SHORT BATT TO GROUND. THAT'S A TERRIBLE IDEA. USE COMMON SENSE!!

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  • \$\begingroup\$ Thanks for the answer! Following your suggestion, I've actually significantly edited my oirignal question to include a more representative full schematic, using a placeholder LED+resistor for the load. To clarify a little bit, I'm specifically worried about BAT- which is the negaitve terminal of the battery. I get that I should never short the positive terminal to ground [short circuit] or the +/- terminals together [also short circuit]. However, I'm at a loss on where to even connect the negative terminal to. \$\endgroup\$
    – Felix Jen
    Jul 21 '20 at 2:48
  • \$\begingroup\$ @kevin BAT- is clearly the negative battery terminal, which you always want connected to ground (or through a protection IC like this). You have misread the datasheet. The overcharge pin does not "discharge the battery if it is overcharged", it prevents the battery from being charged any more (but still allows it to discharge). The overdischarge pin does the opposite, stopping the battery from being discharged if it is overdischarged (but still allows it to charge). \$\endgroup\$
    – BeB00
    Jul 21 '20 at 2:57
  • \$\begingroup\$ @kevin Also, There is no way for current to flow from BAT- through the mosfet to pin 1, since that would mean current flowing from the mosfet source to the mosfet gate, which can't happen (unless your mosfet is broken). \$\endgroup\$
    – BeB00
    Jul 21 '20 at 3:06

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