I want to charge a battery bank from a car alternator. Either side may have a higher voltage than the other, e.g.:

  • Battery bank can be 14V while alternator-side is 12V (engine off, solar charging)
  • Battery bank can be 12V while alternator-side is 14V. (engine on, solar not charging)

I thought of using a relay, but the battery may be able to draw a higher current than the alternator can support. Therefore, switching to control the current is essential.

The current is very high, e.g. 100A, and therefore fast switching is needed to reduce heat build-up. By using MOSFETs in parallel, the current and consequent heat is reduced significantly.

To reduce jolting action on the alternator when switching on, and therefore reduce stress and increase lifetime, I imagine the switching ought to be high-frequency. Maybe 1000Hz is sufficient?

I'm interested in using NPN, because they're much cheaper. I'll use a cheap DC-DC boost circuit to provide 25V for driving the gate.



simulate this circuit – Schematic created using CircuitLab


1) Should the body diodes face towards eachother or away from eachother?

2) In the diagram, the connection is: Battery -> Vd -> Vs + Vs <- Vd <- Alternator

Will Vs be floating when both MOSFETs are off? E.g. is it possible that Vs becomes 0V and therefore violates the Vgs max rating?

I was thinking that by facing the body diodes towards eachother, Vs will never be less than the highest voltage minus the drop over the body diode.

Also, I'm wondering if current has to be able to flow from Vs? Again, in the current schematic, with body diodes against eachother, I don't see how Vs would allow any current to flow.

3) I'm in particular worried about violating the Vgs< +-20V rating.

  • When Signal=5V then Vg=25V, max Vs=14V, max Vgs +14V.
  • When Signal=0V then Vg=0, Vs=(0-14V?), max Vgs -14V.

Is this correct?

  • \$\begingroup\$ How will this charge your battery when alternator voltage is less than battery voltage. Maybe you need to explain functionality a bit more. What it looks like you are designing is a MOSFET solid-state-relay - go look it up because it might help you realize that a floating source connection may prove problematic. \$\endgroup\$
    – Andy aka
    Commented Jan 6, 2017 at 15:38
  • \$\begingroup\$ Sometimes the engine/alternator is not running. The car battery will then be at ~12.3V and may drain the battery bank. That's why it needs to be bi-directional. When the engine/alternator is running, the voltage will be at ~14.0V. I wrote this in the beginning of the question. \$\endgroup\$ Commented Jan 6, 2017 at 15:45
  • \$\begingroup\$ As mentioned, a solid-state-relay would not allow me to control current. \$\endgroup\$ Commented Jan 6, 2017 at 15:47
  • \$\begingroup\$ Floating source problem could be alleviated simply by reversing the MOSFETs? Battery-Vs-Vd-Vd-Vs-Alternator ? \$\endgroup\$ Commented Jan 6, 2017 at 15:47
  • \$\begingroup\$ "The car battery will then be at ~12.3V and may drain the battery bank. That's why it needs to be bi-directional. " So you never want to charge the car battery from the battery bank? That seems like a good reason the switch can be uni-directional so it will block the reverse current from flowing. \$\endgroup\$
    – John D
    Commented Jan 6, 2017 at 15:59

1 Answer 1


1) You can face the body diodes either way and it will work, but I usually have the source terminal of both MOSFETs connected together. That way your gate-source driver is only between two points rather than 3. For NMOS this would mean the diodes point away from each-other.

2) To turn on the MOSFET you only need to develop a voltage between the gate and source. You don't need any sustained current to flow through the source. Some small amount of current will flow between gate and source when the gate is initially driven as the gate capacitance charges.

2/3) If you use a proper gate driver with the sources tied together you won't have to worry about the gates floating because the driver will always determine the gate source voltage. One way to drive NMOS FETs is with an isolated supply that was made for gate driving, combined with a gate driver chip.

For example you could use this floating power supply...


Vin = 9V to 18V. Dual isolated output voltages +15V and -5V with shared floating ground. Available on Digikey for $16.75.


With this gate driver chip


Available for $2.87 on Digikey.


When using MOSFETs in parallel you have to worry about the fact that they don't all switch on at exactly the same time. So one of them always takes a lot of current for a short period just following a switch, which stresses it and leads to early failure. Also they will share current, but it won't be 100% equal so some will get hotter than others. You might have an easier time just using one MOSFET that is rated for 100A. For example you could use this one...


It is available for $19.60 on Digikey.


It is rated at 660A continuous current provided that you can cool it. It has a 0.85 milli-ohm on resistance. So at 100A the drop across it would be 85mV and 8.5W of heat would be generated.

The device has a large isolated pad on the back side that has a junction to pad thermal resistance rated at 0.144 C/W. The pad has screw holes for mounting. So you could theoretically put that isolated pad right on the frame of the vehicle to sink out as much heat as you need.

IXYS corporation sells other similar MOSFETs if you need a different package style.

I see you have a 1 ohm resistor as over voltage protection for your alternator. If your alternator can really put out 100A then you probably need a much bigger resistor.

  • \$\begingroup\$ The main advantage of paralleling, is that the current gets divided, e.g. P = (I/number of MOSFETs)^2 x Ron. For 2 MOSFET, power is 50w. For 6 MOSFETs, power is 5.5w. The MOSFET you recommended has a very high Ron, so there is no benefit of a single large current. \$\endgroup\$ Commented Jan 6, 2017 at 19:14
  • \$\begingroup\$ I agree about peak current capability. The AP99T03GP can handle 200A continuous. So the problem is mainly if current is unequal and results in a lot of heat. But I'm hoping that the high heat will increase resistance, leading to a natural equilibrium of equally distributed current. \$\endgroup\$ Commented Jan 6, 2017 at 19:17
  • \$\begingroup\$ I read just now about 1EDI60N12AFXUMA1, but it seems the main advantage of using a driver chip is that it provides galvanic isolation. Are there any other advantages over the schematic I posted in the question? \$\endgroup\$ Commented Jan 6, 2017 at 19:19
  • \$\begingroup\$ How much should the OVP resistor be? I'm worried that too high resistance will cause some of the over-voltage to leak into the main circuit. \$\endgroup\$ Commented Jan 6, 2017 at 19:22
  • 1
    \$\begingroup\$ @user95482301 You are correct that increases in resistance with temperature result in natural current balancing. But it will still be somewhat unequal. \$\endgroup\$
    – user4574
    Commented Jan 6, 2017 at 21:27

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