BQ76952EVM: OK

Hi Sandeep,

The question is more about thermal design than the FET.  Many FETs will handle a high current.

Consider CSD19536KCS, It has 150A maximum current.  RDSON with 10 V VGS is 2.7 mOhms.  But at 100 A you will need to dissipate 27 W.  It is within the capability of the part, but you must be able to get the heat out of the battery.  If you have a large metal chassis and cooling air you can more easily dissipate this than if you have a small plastic battery enclosure. Often you will parallel several FETs to reduce the effective resistance reducing the heat in the battery and spreading it across multiple devices.

When you parallel FETs you may need a small individual gate resistor or ferrite bead to isolate the gates from each other to avoid oscillation during switching.  Check with industry FET references. You will want a common resistance to set the switching speed.  If you have a protection event at 100A or short circuit you want to turn off quickly to not heat the FET excessively but not so fast that you excite the cell inductance and produce a damaging inductive transient.  You could model the effect with a simulator, or test experimentally.  As you increase the gate capacitance with larger or more FETs you will want a smaller resistance.

Since you indicate that -ve switching is required, and the built in gate drivers for the BQ76952 are high side, you will need an external gate driver.  You will also need a power supply of approximately 12V to run the gate driver.  The DDSG and DCHG outputs can be used to control a gate driver, See http://www.ti.com/gatedrivers for information on gate drivers.  Individual gate drivers are usually rated in amps for fast switching for applications like motor control or switching power supplies, for a load switch you will want a gate resistor to slow the FET switching.

The charge gate driver output of a standard gate driver will typically not like to be pulled above its supply such as when the discharge FET is off.  An isolated driver is a high performance way to solve this but requires an isolated power supply.  With a simple driver you will need to block current back into the driver with a diode.  With current blocked the RGS would turn off the FET, you may want to enhance the turn off speed with a transistor circuit.  Although it is controlled by a different BMS device, see the application note  https://www.ti.com/lit/pdf/slua773  for an example of paralleling low side FETs. In Figure 6 R13 would be omitted and D1 would block current into the charge driver.  Q9 would assist in turning off the charge FETs, it is shown as a low VGS P-ch FET, but you can use a PNP, or JFET. 

With low side FETs consider how you want to interface other features of the device.  If you want to use load detect configure a circuit to pull down LD when the PACK- is sufficiently high.  For wake up consider pulling down the TS2 pin or  pulling up LD.