How is this P-Channel MOSFET Being Used (and what's this other component)? - Page 1
Forum section: electronics / beginners

author: ruthsarian
author_rank: Contributor
timestamp: November 15, 2017, 09:10:26 pm
content: Attached are pictures of the PCB from a flashlight I have which has a feature that detects when it's standing on end at which point it turns off the flashlight led and turns on another set of LEDs to engage a 'lantern' mode. It's got a microcontroller and what I assume is an accellerometer/gyroscope with its markings scrubbed off and some transistors to control turning on the 3 different banks of LEDs depending on what mode and orientation it is in.What I'm not sure about is how the power is being delivered to the circuit. It runs of 2 C-size batteries connected in series. The positive terminal of the batteries connects to the spring seen on the right side of the PCB in the images. The negative terminal connects through the metal case of the flashlight and up to the edge of the LED PCB then down into the main circuit.With that in mind, I see the only thing connecting the positive terminal to the rest of the circuit is a SOT-23 marked "A19T" which the internet tells me is a P-channel MOSFET. The positive terminal from the batteries is going into the DRAIN of the fet. The SOURCE goes to the back of the PCB and into a pair of inductors and the GATE is connected to ground with a capacitor across gate and source to ... smooth out the power?My first question(s) is ... how is power going from DRAIN to SOURCE in the fet? Is it actually going through the body diode? If that's the case, what's the purpose in doing that? Is this some kind of reverse polarity protection? Why not use just a diode if that's the case?My second question is what are those two components marked "40_1c"? My searches have turned up nothing. Are they mosfets? (and if you do know what it is, how'd you find out?)Which leads me to my last question. There's two inductors on the back that connect through to the two diodes on the front (marked 'SS22') that then connect to the two '40_1c' components. Is this some a boost converter? Or is this some kind of voltage regulation and filtering? And why the pairs of components connected in parallel? To spread out the load/heat dissipation, I'm guessing.Thanks for any insight.

author: Peabody
author_rank: Super Contributor
timestamp: November 15, 2017, 09:24:00 pm
content: I'll take on the first question. Yes, it is indeed used as reverse battery protection. The MOSFET will conduct in either direction when turned on. With the gate grounded, current will flow through the MOSFET body diode to the source, and when the voltage gets high enough the transistor will turn on, bypassing the body diode. But with reverse battery installation, the transistor remains off, and the body diode also blocks any current. This is a pretty standard method of reverse polairy protection, and it's used because in normal operation there is virtually no voltage drop across the MOSFET, whereas a diode would of course drop some material fraction of a volt.

author: jcw0752
author_rank: Contributor
timestamp: November 16, 2017, 04:46:00 am
content: I think you are seeing a boost converter with with current regulation. This would allow the unit to meet the probable junction voltage of 3V on the white LEDs and maintain a regulated current as the battery voltage drops. At some point the boost converter will not get enough voltage as the battery drains and the circuit will stop working but you will have a much better utilization of the battery energy. It would be interesting to measure the voltage across the LEDs as the battery voltage is slowly lowered. This way you can find the point at which the circuit stops working. Pretty cool circuit for a flashlight.John