For a hobby project I would like to switch between two power supplys. The load are two motors which can also be switched on and off separatly. The two motors are BLDC controllers in reality. I am only in developing phase so no test arrangement was done.

I developed the following circuit, which I am not sure about if it works flawlessly: Schematic

  • U4-7 are highside MOSFET drivers with an integrated charge pump to achieve voltages above VCC.
  • R7, R8, R15, R16 are pull-down resistors to avoid floating potentials on the driver input and R1, R2, R9, R10 have the same function but at the output of the drivers.
  • Q1-4 are my n-channel power MOFETS (V_GS_max = +/-16V)
  • R3, R6, R11, R14 are restricting the maximum gate current
  • and R4, R5, R12, R13 I think are necessary to ensure that V_GS is never above 16V despite the source potential is floating (e.g. if Q1 is on and Q3 is off)

Some facts:

  • Maximum power-supply voltage: ~25V
  • Maximum current: ~50A through Q1/Q2 and ~25A through Q3/Q4
  • Q1 and Q2 can not be on at the same time (software-wise)
  • Q3/Q4 are PWM driven the first second after startup (to reduce the inrush current of the BLDC capacitors)

So my questions are:

  • are there any considerations for stable operation?
  • are there any maximum ratings which could be exceeded (especially V_GS or crossflows)?

I would appreciate your advices!

EDIT: I updated my schematic drawing with new MOSFET drivers some unnecessary pull-ups removed and some additional capacitors for voltage stabilization: enter image description here

The problem with transverse currents if one battery has a lower voltage than the other is in a new thread: MOSFET-switching between two power supplys: how to avoid transverse current?

  • \$\begingroup\$ Welcome to EE.SE. Unless you need a dual 'AND' function to drive high voltage at high speed, what is this circuit for? \$\endgroup\$ – Sparky256 Oct 21 '18 at 22:41
  • \$\begingroup\$ Thank you! This curcuit is for an uninterrupted power supply: if BATT2A is lost or too low BATT2B is used. The two motors can be switched on and off for overcurrent protection. \$\endgroup\$ – Florian Oct 21 '18 at 22:43
  • \$\begingroup\$ You are going to need to do something with the back EMF from the motors during switching. Schottky catch diodes at a minimum. \$\endgroup\$ – Dean Franks Oct 21 '18 at 23:06
  • \$\begingroup\$ Your gate driver is actually a high-side load switch. It has a maximum output voltage of VCC. All your N-FETs are going to operate in saturation. \$\endgroup\$ – sstobbe Oct 22 '18 at 0:01
  • \$\begingroup\$ Dean Franks, no the motors are already driven by a BLDC-controller and i only supply this controller. @sstobbe, ok thats very bad, I was not sure about that driver... do you know a suitable Highside-Driver? \$\endgroup\$ – Florian Oct 22 '18 at 6:49

If you wish to use n-channel MOSFETs as static switching elements, you will need to create an auxiliary power supply rail. To drive a high side n-MOS into triode you need bring the gate 5 V to 10V above the supply rail you're switching.

Some gate-drivers include a bootstrap capacitor to operate as a charge pump. However this requires that the load is being switched periodically and not a static output.

So, you can either generate an auxiliary power supply rail above VCC with a charge pump or boost converter.

Or, Implement your design using p-channel MOSFETS. Just make sure not to over-drive the gate beyond its rated Vgs (a zener clamp works well here).

You will also want to include bulk and decoupling capacitors as well.

  • \$\begingroup\$ Thank you for your answer! I know that V_GS has to be over V_CC for a highside n channel MOSFET. Today I found a perfect driver for my FETs: the LT1910 from Linear Technology. (and this driver has an included charge pump with timer for contioius operation) P channel MOSFETS are not designated due to their drawbacks. Where do I need bulk and decoupling capacitors? \$\endgroup\$ – Florian Oct 22 '18 at 13:51
  • \$\begingroup\$ Your present design would have also over-driven Vgs to - 24V. The body diode of Q1 or Q2 would conduct into the alternate battery pack as well. All the ICs should have decoupling caps. \$\endgroup\$ – sstobbe Oct 22 '18 at 14:10
  • \$\begingroup\$ In which state is V_GS at -24V? Do you mean if one battery is lower than the other one, the s-d diode will conduct? But how can i avoid that? I guess you mean all outputs... but why should i need decoupling capacitors technically? \$\endgroup\$ – Florian Oct 22 '18 at 14:38
  • \$\begingroup\$ Say Q1 had been turned on. Both Q1 and Q2 have their sources equal to VCC. Now you are pulling the gate of Q2 to ground, so Vgs = -Vcc (upwards of 25V you listed in your question). Those are all good precise technical questions that do well on this site. Ask a question :) \$\endgroup\$ – sstobbe Oct 22 '18 at 14:47
  • \$\begingroup\$ Ok I understand, but V_GS will never be pulled to ground because the driver IC pulls V_GS to +Vcc if the input is low. Did not want to spam the board, but if this question is still unsolved for me after overthinking I will probably place a new qustion for that topics. So do you see other possible problems except of the one mentioned? \$\endgroup\$ – Florian Oct 22 '18 at 15:09

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