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The set-up

For a project, I have a set-up where the power to the main load comes from either a AC-DC adapter input (24V) or a battery pack (12V to 16.8V - 4S4P 18650 Li-ion cells). The battery pack is connected through a BMS module.

Both power supplies are connected to a switching circuit that 'selects' the right source to use (DC if available, battery otherwise), using the LTC4416-1 chip.

The DC input is also connected to a charging circuit using a DC-DC buck converter with CC/CV limiting to the BMS/battery pack.

The problem

For safety, I want to put a reverse current blocking protection between the buck module and the BMS/battery. (To prevent current from flowing back if the DC plug is pulled and thus the buck has no power.)

I've read/seen a lot of posts/articles/datasheets/instruction videos/etc about possible options, but can't seem to find the right solution for my situation (supporting charging current that starts @ 4.08A - 16.8V).

Options:

  • Specially tailored (Schottky) diodes --> simple/affordable, but a lot of wasted power and heat
  • Dedicated ICs (like a load switch) --> can't find one for my specs; either too little maximum current or too low maximum voltage
  • High side switch using P-MOSFET(s) or N-MOSFET(s) --> can't find/figure out the correct diagram/circuit logic for this, since they all handle reversed polarity protection, which is not the case here..

My question is: what IC or (MOSFET) circuit can I use to block the reverse current?

NB: Since the whole assembly will be mounted on a panel, there's no reversed polarity issue, only reverse/leak current from the battery to the buck module.

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How to block reverse current from battery to charger output?

I assume here that you are already using the recommended layout of the LTC4416 AND that V1 is the AC-DC supply AND that V2 is the battery supply for your Power Path Switch.

Your Power Path Switch circuit will look something like this:

enter image description here

Since V1 is also used to drive the Charger DC-DC convertor all you need is to use the G1 gate drive from the LTC4416 to drive another load switch pair.

schematic

simulate this circuit – Schematic created using CircuitLab

If V1 (the AC-DC power supply) is ON, then the output of the Charger is connected to the Battery.
If V1 is OFF and you are on battery power then G1 is high and G2 is low. With G1 high the output of the Charger is disconnected from the Battery.

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  • \$\begingroup\$ Well spotted ! Can I just fork the G1 signal or will that throw off the IC? (Or reduce the signal current too much?) \$\endgroup\$ – Philip Jun 10 at 5:15
  • \$\begingroup\$ @Philip Just use the signal from G1. The FET gate voltage is a maximum of V1-8V so the drive to M1/M2 will be less, but should be sufficient for the task. If you are concerned then select a FET with a lower VGS(th) of about 3V maximum. If the charging current is moderate then you might even be able to use very small FETs like the AO3401. \$\endgroup\$ – Jack Creasey Jun 10 at 13:28
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An ideal diode would be the best solution with a suitable low RdsOn FET.

enter image description here D4185 RDS(ON)< 20mΩ(VGS= -4.5V)

$3.19enter image description here

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ Yes, I did look into those, but could not find a suitable (and affordable) one, and kind of lost my way between ideal diodes, e-fuses, load switches and hot swap switches, etc. But I will have another look at them, thanks! \$\endgroup\$ – Philip Jun 10 at 5:21
  • \$\begingroup\$ They have low cost boards for BMS that do this too. \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Jun 10 at 6:13
  • \$\begingroup\$ @Philip Go to ebay, search for "ideal diode". Should turn up a lot of small boards, from about €4-5, usually rated for up to 28V and 8A, some a bit more expensive with higher voltage/current. \$\endgroup\$ – JimmyB Jun 10 at 17:27
  • \$\begingroup\$ Do you know which IC they use on those, to drive the MOSFETs? \$\endgroup\$ – Philip Jun 10 at 18:57
  • \$\begingroup\$ my 1st link showed you one example. The PNP thermal and electrical balance error determines how ideal the crossover is. \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Jun 10 at 19:07
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You have presented a number of solutions, but skipped over the most obvious one.

You're using a DC adapter. This implies you have 120 VAC (or maybe 240 VAC) mains for the charging current. Just find a normally open relay with a coil the is rated for the same as the mains current and has contacts that can handle the DC current. Connect the coil to the same mains power as the DC adapter. Wire the relay contacts so they complete the circuit when the relay is closed.

If mains power is lost for some reason, the DC adapter will be disconnected from the buck module in your diagram.

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  • \$\begingroup\$ Indeed an obvious solution I didn't consider. Only problem is I'm trying to avoid relays, since the panel will be used on video productions and any click sounds can disturb audio recording. (But maybe I'm overly cautious.) \$\endgroup\$ – Philip Jun 10 at 2:15

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