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For a hobby project I would like to switch between two power supplys. The two power supplys are two 6s LiPo batteries. If one battery is empty i want to immediatelly switch to the other battery. (the load is a motor) For switching a 90A, 2.2mOhm n-channel MOSFET with a highside driver will be used. I already did some research and made a first schematic drawing: enter image description here

The maximum current is about 50A and the battery voltage varies between 18V (empty) and 25V (full).

So if e.g. BATT. 2A is empty (18V) and BATT. 2B is full (25V) and Q1 is therefore switched off and Q3 on (to only use the full battery), there is about 25V potential at the source of Q3 (and hence 25V at source of Q1), but only 18V at drain of Q1. So there is a difference of -7V between source and drain of Q1 and the parasitic diode of Q1 between source and drain will probably conduct which leads to an short circuit (at over 100A battery current for a short time, so the MOFET will be destroyed).

How can I avoid that transverse current?

Thank you for any help in advance!

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  • \$\begingroup\$ Put a very high-current Schottky diode at the output of both battery packs. You'll lose about a volt, and dissipate about 50 watts, but that's the simple version. \$\endgroup\$ – WhatRoughBeast Oct 23 '18 at 22:44
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    \$\begingroup\$ Thank you for you comment, but loosing 50W would be unacceptable! \$\endgroup\$ – Florian Oct 23 '18 at 22:47
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    \$\begingroup\$ Two questions- what happens to the Vgs voltage if the motor shorts/stalls, does it exceed spec? You may want to use that sense pin to add an overcurrent to shut down gate drive before the FET SOA is exceeded, it's designed to use copper artwork tempco, so that's actually nice. The second item I'm checking is the heat dissipation requirements, you say a 50A load, is that steady state (high torque situation)? If so you may want to check the heatsink requirements. I^2*R ~=2500*.002 ~=4.5W, but junction to ambient is ~40K/W with their 6cm^2 heatsink example, you'll need something better (?) \$\endgroup\$ – isdi Oct 23 '18 at 23:28
  • \$\begingroup\$ @isdi thank you for your comment. Because I am going to measure the passing current later via shunt, I will add a Smith-Trigger which monitors the voltage drop and switches the MOSFET off if a defined level is exceeded. For that reason I disabled the built-in overcurrent protector. Yes temperature could be a problem... It is planned that the drain-pad will be enlarged and connected to the bottom PCB-layer with vias and I hope this i enough. A small fan is also an option. \$\endgroup\$ – Florian Oct 24 '18 at 6:32
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Battery protection circuits, like this one from TI use two FETs back to back to avoid this issue.

enter image description here

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  • \$\begingroup\$ Thank you for the answer. Do I understand right, that the second MOSFET is used to switch off the reverse current of the first MOSFET? So in normal operation mode the current passes through the parasitic SD diode of the second MOSFET? But the ~1V of this diode will also create a 50W loss? \$\endgroup\$ – Florian Oct 24 '18 at 6:47
  • \$\begingroup\$ Under normal battery and load/charge conditions both FETs are turned ON, so both diodes are bypassed. The reason for the dual gate control is for battery undervoltage, battery charging overvoltage/current, and load overcurrent/short conditions, and since those conditions have defined current directions, the appropriate FET is turned off and the other FET body diode is reverse biased. Note that the battery being protected in that diagram is the non-standard symbol on the left, to the world, the "battery terminals" are the pack+/- nodes. Inspect Table 1 on sheet 8 for further details. \$\endgroup\$ – isdi Oct 24 '18 at 12:15
  • \$\begingroup\$ What isdi said is the case. With the FET on, the drop across the FET is very small compared to the diode drop, so dissipation is acceptable. I've used this trick of connecting FETs 'backward' for reverse protection in battery powered applications, it works well. \$\endgroup\$ – Phil G Oct 24 '18 at 14:08
  • \$\begingroup\$ I do understand now, thank you! Thats a very good solution and the loss is about 1% when switched on and nearly 0% when switched off. The control of the MOSFETs will be not that easy but viable. \$\endgroup\$ – Florian Oct 25 '18 at 7:29

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