# Trying to understand behavior of p-FET in lipo charger circuit

I am building a lipo charging circuit based on Microchip's AN1149: http://ww1.microchip.com/downloads/en/AppNotes/01149c.pdf. It seems to have all the functionality that I need to build a circuit that can charge a lipo while also powering the main circuit.

The part that I am trying to understand is the p-FET that switches between battery source and USB source for the system load. This corresponds to figures 6 and 7:

The following are my questions:

1. Is this rationale for figure 6 correct: Vd = 4.2V (lipo), Vg = 5V, Vs = 5V -> Vgs = 0V so FET is off, body diode is reverse biased so no current flows through the FET.

2. Is this rationale for figure 7 correct: Vd = 4.2V (lipo), Vg = 0V, since anode of D1 = 0V, body diode of FET conducts, so Vs = 3.2V (assume 1V forward voltage). Once body diode starts conducting, Vgs = 0-3.2V = -3.2V -> Q1 turns on (logic level FET), and eventually settles at Vgs = -4.2V since the body diode will be bypassed by the conducting FET.

3. I would like to use this circuit to drive some tiny DC motors, probably pulling a maximum of 2A (only off the lipo, external power unplugged). If my rationale for #2 is correct, will the system load see a voltage of 4.2V (minus losses in the drain-source resistance), or does the forward voltage drop of the body diode come into play? Minus the drain-source resistance, is there any difference between the state in figure 7 (external power unplugged, running off battery) and directly powering the load off the lipo battery (no charging circuit)?

4. I found part Si4497DY (sorry, not enough reputation to post another datasheet link). It is logic level, has very low Rds, and can easily handle the current I'm using. It seems like it'd be reasonable for Q1, but I don't really have any experience hooking up a p-FET in this manner. Is there anything that I'm overlooking here?

Thanks!

## 1 Answer

Some answers:

1) Sounds good to me :-)

2) This is less complex than you think, since Vg is 0 the PMOS will just conduct. The Drain-Source diode does not come into play as the PMOS will simply short it (as you noted). Note that when a MOS transistors is on it will conduct a current from Source to Drain but also from Drain to Source. It is basically a symmetric device (disregarding the diode for a moment).

3) No the diode in the PMOS is shorted, the PMOS will conduct the current. The only difference between figure 7 and directly powering from the battery is indeed only the Rds_on of the PMOS. Which is low, so you probably would not notice the difference.

4) This looks like a fine choice to me, an alternative could be the SI2301 which is cheap but also very small (SMD type housing).

• Can you explain #2? I could easily be confused here, but I thought that conducting and not conducting in the FET is dependent on Vgs. Vgs = 0 --> not conducting in FET, Vgs < -Vth --> conducting through FET. If this is the case, wouldn't Vs = 0V (no current through load)? Oct 20, 2015 at 23:49
• Correct but this is a PMOS so the gate needs to be on a lower voltage than drain/source for it to conduct. And that is what happens here. Also drain and source are interchangable ! It depends on the voltage what we call drain or source. Since in this case the battery is on the left side, normally called drain, in this case it acts as the source ! I understand your confusion, just remember that a MOSFET is a symmetrical device, you must have seen pictures like: en.wikipedia.org/wiki/MOSFET and noticed that drain and source are identical. In a discrete MOSFET bulk and source are shorted. Oct 21, 2015 at 7:30
• The diode would be useful in preventing current from being wasted on the pulldown resistor when on battery power. May 14, 2016 at 6:10