In the circuit below Q1701 is part of a charging circuit in an iPhone 6.

Since there are no datasheets available for the main components in the circuit I've been trying to figure out the operation.

With a charger disconnected:

  • S = 0V
  • G = 0V
  • D = 0V

Charger connected:

  • S = 5V
  • G = 4.29V
  • D = 5V

So this is where it leaves me confused, if a P-channel MOSFET turns on when it sees 0V at the gate (which I thought was the reason R1710 has it connected to ground) how is it turning on with 4.29V at the gate?

Part of the schematic


Charger connected:

S = 5v G = 4.29v D = 5v

How do you get G = 4.3 V? The gate is connected directly to ground, and the gate-source diode is reverse biased so it plays no role. The gate will be at 0 V here. (Unless there's something else connected to the REVERSE_GATE node that you haven't shown in the schematics you shared)

if a P-channel mosfet turns on when it sees 0v at the gate

A P-channel (enhancement mode) MOSFET turns on when it sees a gate-source voltage more negative than its threshold voltage.

The voltage relative to the circuit reference (ground) is irrelevant.

From comments,

why even have that mosfet if it's always going to conduct having the gate at ground? What's the benefit

This is a reverse protection circuit.

Note that current will normally flow from drain to source, rather than the usual direction for PMOS of source to drain. This gets the body diode oriented the right way for a reverse protection circuit.

Now if you accidentally reverse the power supply, you'd be connecting the drain to -5 V. Then Vgs will be positive and the FET will not conduct; also the body diode will be reverse biased and will not conduct. This prevents any power being delivered to the rest of the circuit, so it won't be damaged.

  • \$\begingroup\$ There are no other netlists for "reverse_gate" on the entire schematic. My only guess is maybe that net goes to the USB cable itself? So you're saying as long as the voltage at the gate is lower than the source it will turn on? Even if only a few milivolts lower? \$\endgroup\$ – user140123 Feb 23 '17 at 20:03
  • \$\begingroup\$ If V_gs is more negative than the threshold voltage, it will (start to) turn on. If V_gs is just a few millivolts negative, it likely won't pass more than a few nanoamps of leakage current. \$\endgroup\$ – The Photon Feb 23 '17 at 20:05
  • \$\begingroup\$ That's strange then, how can the design be of any substantial use? \$\endgroup\$ – user140123 Feb 23 '17 at 20:06
  • \$\begingroup\$ But in this case, Vgs should be about -5 V, so I don't see why you're worried about the "just a few millivolts" question. \$\endgroup\$ – The Photon Feb 23 '17 at 20:06
  • \$\begingroup\$ The design is useful because Vg = 0 and Vs = 5 V, so Vgs = -5 V. \$\endgroup\$ – The Photon Feb 23 '17 at 20:07

Some p channel mosfets can turn on a reasonable amount with just a few hundred millivolts between source and gate. 0.7 volts doesn't sound unreasonable especially if the drain current is low. Try looking at Alpha Omega's offerings to see what is possible. The AOC2413 looks good in this respect.


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