I'm studying this circuit below which in fact uses a N-channel MOSFET in a power line, with source at the supply voltage and drain at the load.

Why the source is the input supply voltage and not the drain?

LTC4040 datasheet

enter image description here


1 Answer 1


Because there is a parasitic diode inside the MOSFET and if you put it the other way it would not allow the input voltage to be much less than the output voltage (even on standby).

The MOSFET is equally happy allowing current to flow either way, even though it seems to be operating 'backward' when the 4.5V supply is present.

enter image description here

Edit: See the waveform during transition to standby:

If the MOSFET was the other way around, the standby power would backfeed other loads that may be connected to the 4.5V input supply.

This is the feature mentioned on the front page of the datasheet:

Input Disconnect Switch Isolates Input During Backup

enter image description here

  • \$\begingroup\$ But why it needs much less input voltage than the output voltage? \$\endgroup\$
    – MrBit
    Aug 18, 2015 at 18:18
  • \$\begingroup\$ Please see edit. \$\endgroup\$ Aug 18, 2015 at 18:30
  • \$\begingroup\$ And as is logic Vin > 0 in backup mode isn't acceptable... Right? \$\endgroup\$
    – MrBit
    Aug 18, 2015 at 18:36
  • \$\begingroup\$ Right, in many cases it's not acceptable. \$\endgroup\$ Aug 18, 2015 at 19:11
  • 1
    \$\begingroup\$ @Ah, OK. I misunderstood what you meant. Got it now. Although I think the OP maybe was worried that you need a complete Vin black out for that chip to be usable. Instead it seems it handles brownouts as well, since it isolates completely what lies before the MOSFET if it detects a Vin under that PFI threshold, if I understood its operations well. \$\endgroup\$ Aug 18, 2015 at 21:00

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