Need help using N-MOSFETs to switch on either one battery or the other

Question

I'd like to add a second battery (36 V Li-ion) to an ebike to increase its range. My goal is to use one battery at a time, not using both in parallel with e.g. ideal diodes or something.

I'd like to keep it as simple and as safe as possible and just use a 2-way switch to choose which battery the motor will pull current from. The 36 V ebike motor draws about 20-25 A max. so I figured out it would be best to use 2 N-channel MOSFETs to control which battery is being used.

The SUP60020E N-channel MOSFETs I plan to use are rated for 80 V, 150 A and have a resistance from drain to source (R_DS) of only about 3 mΩ. This means that at 25 A the mosfet will dissipate 1.875 W. This should be okay without heat sink because the MOSFET can be used without heatsink up to 3.5 W at an ambient temperature of 35°C.

I'm a noob in electricity/electronics so I tried to educate myself about MOSFETs and came up with the following circuit:

The idea is that when the switch in the center of the diagram is in the I position, battery 1 will power the motor. On the other hand, when the switch is in the II position, battery 2 will power the motor. However, battery 1 will always be on the bike so it would be used to control the gate on the MOSFETs. The two resistors R1 and R2 are used as a voltage divider in order to drop the input voltage of battery 1 from 36 V down to about 10 V in order to power the gate of the 2 MOSFETs since these MOSFETs have a max. voltage from gate to source (V_GS) of 20 V.

I have a few questions about this circuit:

1. When the switch is in position I, the MOSFET on the left will allow the motor to run from battery 1. However, this also closes the path from the positive terminal of battery 2 to the negative terminal of battery 1, meaning that the motor would also be running from battery 2 as well. Is that correct? The same occurs when the switch is in position II. This allows battery 2 to run the motor but the positive terminal of battery 1 becomes connected to the negative terminal of battery 2 so that the motor would also be running from battery 1 at the same time. Does this mean there is a huge issue with my circuit?

2. What happens if the two batteries don't hold the same charge, e.g. battery 1 is at 30 V and battery 2 is fully charged at 42 V? I want to use a switch to properly isolate each battery from the other (I don't want to run the 2 batteries in parallel with e.g. ideal diodes). Would this circuit allow any cross-current to flow from the battery with the highest charge to the battery with the lowest charge? In other words, is it safe?

3. Should I be concerned about voltage spikes and inrush current when switching either battery on? What should I do to prevent such inrush currents from damaging parts in the circuit (above all, I want to avoid any damage to the motor controller and the battery BMS)

4. Any change recommended on that circuit?

Answer 1

This means at 25A the mosfet will dissipate 1.875 W. This should be okay without heat sink because the mosfet can be used without heatsink up to 3.5 W at an ambient temperature of 35°C.

R_thJA of the SUP60020E is 40°C/W, so at 1.875W the junction temperature would be 75°C above ambient, ie. 110°C at 35°C. However at this temperature R_DSON is 1.5 times higher than at 25°C, so the power dissipation is 1.5 times higher causing temperature to rise by another 38°C. But now R_DSON is 1.8 times higher, so...

Result - one burned FET. Put a heat sink on it!

What happens if the two batteries don't hold the same charge e.g. battery 1 is at 30 V and battery 2 is fully charged at 42 V?

The negative terminal of battery 2 will be 12V lower than battery 1. When FET 2 turns on it will pull FET 1's Drain voltage down below the source, turning on the body diode and connecting the battery negatives together. This will cause a large current to flow until battery 1 charges up to about 0.7V less than battery 2.

To fully isolate the batteries you need back-to-back FETs, so when one FET's body diode is forward biased the other one is reverse biased.

Next issue is FET 2's Gate will be at 19.1V when turned on, dangerously close to the maximum rating. However FET 1 only gets 7.1V, which is not enough to turn it on fully. If you raise this voltage then FET 2's Gate will go over 20V. This problem can be solved by putting a 10~15V Zener diode between each Gate and Source, and removing R2 so the Zener diode limits the voltage.

You also need pull-down resistors on the FET Gates to prevent them from 'floating' and turning on unintentionally when the switch is off.

The final circuit would look like this (imagine SW1 and SW2 are a single ON-OFF-ON switch):-

^{simulate this circuit – Schematic created using CircuitLab}

This circuit is fully symmetrical, so it will work the same no matter which battery has the higher voltage.

Should I be concerned with voltage spikes and inrush current when switching either battery on?

LTspice says when charging 470uF (typical ebike controller bulk capacitance) the turn-on power surge peaks at 24V and 153A, and lasts for less than 100us. This is within the safe operating area of your FETs. The relatively high resistances combined with large Gate capacitance should slow FET turn on/off speed to reduce voltage spikes.