BQ24650 - reverse polarity protection

Hi Mark,

If you want to reduce power loss, you can remove Q28 and D27, and flip Q22 so that the body diode basically replaces D27. Then you can turn on Q22, just like the BATFET feature in power path devices.

Also, it looks like your battery has a connection straight through to your adapter through the body diode of your high side FET. This could lead to boost back. You may want to move D26 between the adapter and the VCC pin to prevent that.

David, 

How are you? Below are my comments:

1) appreciate all your comments, actually I have reserved in my design a mosfet to do exactly what you described (see attachment). I just erased the extra components to show the configuration we were testing. In the schematic you can see two Mosfet, so I can place either Q22 or Q21. The small blue arc between Q22 and Q21 (and other places) is a footprint I made to allow me make a short circuit in the PCB so I can test more configurations with the same PCB. My only concern with the solution you are proposing is that in our system we have AC adapter, one internal battery and one external battery as power source, only one the internal battery has to be charged and it cannot be charge with the external battery (only with adapter). So I have to test all scenarios to see if this approach fits our requirements, my biggest concern is that when we are operating with both batteries, the Mosfet should be off so the internal battery does not get charged by the external. This means that the internal battery will be connected to the system through the Mosfet body diode, according to the spec this diode only support 2.3A as forward current. If the system drains more than this and the external battery cannot provide all this current the mosfet can get damaged...right?

charger circuit.PDF

2) Yes you are right, actually I also found this problem some days ago. I fix it adding by hand a diode between VAC/DC and the drain pin of Q24.

3) regarding my initial question, Do you have any suggestion for the reverse polarity protection? Any idea different than a PMOSFET and the PTC fuse and diode? The ideal solution will be the PMOSFET one since it will have the less power losses, but when we test it all seems to be ok when the charger was OFF. Then we test it with the charger ON and also everything seems ok, we notice the charger was trying to detect the battery but Q28 seems to be off according to the Vgs we measure (also there was no charge current). We leave the battery connected backward for some minutes to test it well, then we disconnect it and plug it back correctly we notice the system shut down. After checking we found out that the BQ24650 was damaged (it get hot as soon as the AC is plugged in).

Regards

Mark

Mark,

I am doing well, thanks for asking! Hopefully things are going well on your end.

Could you help me understand which fet is separating the internal and external battery? I am having some trouble picturing this set up. Is the external battery connected through the system voltage with the microcontroller? Or is the int bat pin the connection for the internal battery to be charged, and the battery connector goes to the external battery?

I am glad you came to a solution for #2. Which side of the diode is VCC connected to? If it is on the FET side, the battery can still power your IC. If it is on the adapter side, the chip will be asleep unless an adapter is plugged in.

In terms of reverse battery protection, you could just scrap Q28 and the associated resistor divider and replace it with a schottky diode. That is probably the simplest way to do it, and shouldn't have too much efficiency loss if your Vf is small enough (see here: www.ti.com/.../slva139.pdf ).

Hi David,

Hope you are doing well, here my comments: 

1) Attached a modified schematic, with this file I think you can see all three inputs (AC adapter, External battery and internal battery) in one single schematic and how they are connected to VSYS.

Schematic for TI.PDF

2) Looking at the layout, VCC is connected at the cathode side (also shown in the schematic). So this means the IC can be powered when the battery is plugged, Which are the consequences of doing so? Even though thanks for pointing this out, I will change the connection to the anode side on all PCB's.

3) Is good that the IC will be sleep if I modify the VCC connection, but since the battery can be plugged backward when the adapter is plugged in then the IC will be ON and it could get damaged as it already occur to us.

3) regarding your proposal, we already read the file you sent. the problem with this solution are two, from my perspective, correct me if I am wrong. First, even choosing a low Vf diode (0.4V for a Schottky) the power loss will be a greater than using a MOSFET. Second, the battery will be be able to been charged since the diode will block any current flowing in the opposite direction (I will only able to discharge the battery).

Regards

Mark

Hi Mark,

Thanks for the additional info, that does help clarify things a bit. In this case, it is probably ok to leave the VCC connections to the anode side. Usually we will want the connection on the cathode side so that the IC can still control any ACFETs or BATFETs, but they are absent here.

You are correct about point 3, I apologize for the poor suggestion. A PMOS is the correct way to go. Do you have another means of switching it on and off?

David,

Actually we have a MCU on board but we prefer to have analog control over this to avoid any FW malfunction. Do you have any idea why the charger IC got damaged when we plug the battery backwards? I would like to disable the charger if this happens, the problems is that when the battery is reversed the charger always detects a battery and try to charge it, this since the voltage drop to almost zero and never goes up so the battery absent algorithm detects a connected battery condition.

Any suggestion?

Regards 

Mark

Mark,

It is tough to diagnose something over a forum, but if I had to guess it was either a pin that exceeded it's maximum voltage rating (SRP-SRN maybe, maybe REGN or VCC), or a pin that sunk too much current.

You may be able to control your PMOS by replacing R60 (the 100k resistor at the gate) with an NMOS. It is a bit unorthodox, but maybe something like this:

Just make sure you choose an NMOS that is capable of handling the +/- Vgs = +/- Vbat. There are NMOS's that exist that can handle +/- 20V, so that shouldn't be an issue.

Mark,

It was also suggested to me that it was most likely too reverse current sinking into your SRP and SRN pins through the ESD protection diodes. You can place a 10 ohm resistor in series with the SRP pin and another in series with the SRN pin. This will limit the reverse current flowing into the pins, and is a cheap/simple solution.