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I'd like to add a second battery (36V 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 36V 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 N channel mosfets I plan to use are rated for 80V, 150 A and have a resistance from drain to source (RDS) of only about 3 milliohms (see https://www.mouser.ch/datasheet/2/427/sup60020e-1595623.pdf). 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.

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

enter image description here

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 36V 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 (VGS) of 20 V.

Now I have a few questions about this circuit:

1) when the switch is e.g. 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 with 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 occur on the Motor controller and battery BMS)

4) Any change recommended on that circuit?

Thanks a lot for your help!

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    \$\begingroup\$ "My goal is to use one battery at a time, not using both in parallel" why? \$\endgroup\$ – Bruce Abbott Sep 29 '19 at 3:22
  • \$\begingroup\$ Battery 1 is 6 years old with lots of internal resistance while battery 2 would be brand new. I agree that using both batteries in parallel would be great but I want to reduce the complexity as much as possible and have an error proof system (I'd still need a switch anyway with ideal diodes). I use this bike everyday to commute and don't want to always check that both batteries are at the same charge etc. For me a simple switch system would be plenty enough: use battery 1 when I do not need a lot of range, and use battery 2 when I need a longer range. Anything wrong with my circuit? \$\endgroup\$ – kilou Sep 29 '19 at 4:09
  • \$\begingroup\$ Is there a specific reason why you want to use a MOSFET? What you have in mind seems like a perfect application for a relay (automotive-grade). There are some latching-type relays which are pulsed into position and stay that way even when you turn the driving circuit power off. And the power dissipation on the closed contacts is much lower. \$\endgroup\$ – Lorenzo Donati -- Codidact.org Sep 29 '19 at 11:13
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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.

RthJA 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 RDSON 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 RDSON 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):-

schematic

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.

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  • \$\begingroup\$ Thanks @Bruce Abbott for the very detailed answer and the updated circuit!! Very appreciated! However he complexity of the circuit is increased with 4 FETs etc. It's likely needed to work correctly but this goes beyond my very limited skills. Someone else on another forum recommended to avoid FETs and just go with a simple battery switch such as this one from Blue Sea Systems: bluesea.com/products/6008200/…. It is only rated for 32VDC but it should be okay for my 36V battery (charges up to 42V). Looks easier for me to install. \$\endgroup\$ – kilou Oct 2 '19 at 18:07
  • \$\begingroup\$ That switch should work fine provided you don't try to switch while the motor is running. Switch voltage rating is mostly based on the ability to extinguish the arc that occurs when breaking current flow. \$\endgroup\$ – Bruce Abbott Oct 4 '19 at 2:12

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