IRLML9301PBF as high side switching for 5 V, 2 A load from a 3.3 V uC?

Question

Circuit:

^{simulate this circuit – Schematic created using CircuitLab}

I would like to cut off the power to my APA102-LED-strip during uC sleep - as the standby current of these strips is relative high.

The strip is directly powered via the 5 V from the USB powerbank. Theoretically it can draw up to 2.4 A@5 V = 12 W. In normal usage it will be ~1 A (5 W) The uC (ESP32-S3) is a 3.3 V type. I think I need a high-side switch for the LED-power path.

So I need a P-channel MOSFET with 3.3 V logic level and ~3 A or similar. For this I found at my available/favorite reseller (reichelt.de)

• IRLML6402PBF datasheet
• IRLML9301TRPBF datasheet
• AO3413 datasheet

First two labeled as logic level - third just found by price sorting.

Am I correct with my assumptions that all of these would work? Am I missing something? Every time I have to select MOSFETs I have to relearn how to read these datasheets properly and find the correct VDSon information. So I am looking forward to some educational answers ;-)

Answer 1

I think I need a high-side switch for the LED-power path.

I agree with that. A low-side switch won't work because the clock and data lines would also need to be switched.

Am I correct with my assumptions that all of these would work? Am I missing something?

The design you have proposed is flawed. The flaw lies in how you hope to drive the p-channel MOSFET. The problem I see is that the p-channel MOSFET's source is connected to 5 volts but, the gate can be driven no higher than 3.3 volts hence, the MOSFET will not deactivate fully.

There are ways around this by using an extra transistor of course.

So I need a P-channel MOSFET with 3.3 V logic level and ~3 A or similar.

Not really; the source is tied to +5 volts and you can easily pull the gate down to 0 volts so, a 5 volt logic level would work. The problem you have is that your proposed design cannot properly deactivate the MOSFET.

Answer 2

As pointed out in another answer, the issue is about your OUT1 line that can't go high enough to interdict conduction in the MOSFET.

You can employ many solutions, like

• adding an NMOS as inverter at OUT1, with a pull-up at 5V (this will revert your signal polarity)
• adding a 5V powered buffer, like a 74LVC1G17, better if with pull-up to 5V (let's call it R2) and still a series resistor to OUT1 (say R1); keeping the ratio of R2/R1 around 8...10, just to promote the higher level
• using a high-side switch, which is designed right for that, and will have additional protection features; I just pick one randomly, BTS5090

Note that the second solution (the pull-up addition) could nearly work even with your PMOS alone, but it's... meh...

Answer 3

This is probably what you need:

^{simulate this circuit – Schematic created using CircuitLab}

Perhaps there might be a glitch during startup depending on how fast the MCU will output high compared to the R2 pull-up. Test and see.