I'm trying to design a high-side switch from a Battery Charger to a Battery. I'm using a P-Channel MOSFET, but I'm not confident I fully understand what will happen if the source is floating. Here's the circuit:


And here's what I think will happen. For simplicity I'm assuming T1 and T2 are a perfect switch when on, the diode in T2 drops 0.6V, and CHARGER max voltage is 12.8V when connected.

  1. If ACT is low and CHARGER is floating, T1 is off. For T2, Vg and Vs=11.4V or so, thanks to the reverse diode in the MOSFET, so Vgs=0V and T2 is off.

  2. If ACT is high and CHARGER is floating, T1 is on. For T2, Vg=0V and Vs=11.4V - the current flows from the 12V battery, through the reverse diode, R7 and T1 to ground. So Vgs=-11.4V and T2 turns on, which makes Vgs=12V - this makes Vgs=-12V and T2 stays on.

  3. If ACT is low and CHARGER is 12.8V, T1 is off. For T2, Vg and Vs=12.8V, so Vgs=0V and T2 is off

  4. If ACT is high and CHARGER is 12.8V, T1 is on. For T2, Vg is 0V, Vs=12.8V, Vgs=-12.8V and T2 is on.

Really I just need to confirm that if CHARGER is floating, there is no damage potential to T2.

  • \$\begingroup\$ I don't believe that 114uA will damage T2. \$\endgroup\$ Commented Dec 23, 2014 at 17:00
  • 1
    \$\begingroup\$ Your questions are too confusing. Let me make it simple. The circuit is OK as long as the charger is connected and also if CHARGER+ is floating. But you should verify Vgs max for T1 is > 12.8V. Please note that if you connect a load to net "CHARGER+" the load can draw power from net "+12V" if the PFET body diode is forward biased regardless of whether ACT is high or low. There is a limit to how much current a body diode can pass without damage. If this issue is a problem for you, you can add a second PFET in series, but with the body diode facing the opposite direction for total isolation. \$\endgroup\$
    – user57037
    Commented Dec 23, 2014 at 17:19
  • \$\begingroup\$ Thank you mkeith and PkP.. That's me happy then, as T1 is a BSS138 with Vgs(max)=60V, and CHARGER+ will always be > than the +12V net. \$\endgroup\$
    – Mike B
    Commented Dec 24, 2014 at 12:19
  • \$\begingroup\$ @Mike B, Oops. I think I meant Vgs for T2 should be greater than 12.8V. Double check it. A lot of power FET's are only rated for 8V. By the way, for the BSS138, Vgs max is 20V. Maybe you were looking at the Vds max specification? \$\endgroup\$
    – user57037
    Commented Dec 24, 2014 at 23:43

1 Answer 1


The source is never floating. It's either connected to the charger or to the battery via the channel (when the FET is conducting) or via the substrate diode (when the FET is in cut-off). And since your gate is connected to the source with a resistor, the FET can never start to conduct on its own.

Even if there were no substrate diode, there would be no ambigousness: whenever the gate would have sufficient charge to open the channel, the source would then be connected to the drain via the channel (this is how FETs work in CMOS ICs).


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