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I'm a new product designer trying to learn electrical engineering on my own.

I have a circuit that I need to control the power to a set of periphery circuit segments including a display, SD card, and a series of four i²c haptic driver segments, etc... all running from 3.3V on a single Li-ion cell.

Here are the specs: Vbat 3.0 - 4.2V Ipulse 5A (Power on inrush) Imax 2 (max continuous) Inorm 250mA (normal operation)

In my limited experience it seems very difficult to find switching with no voltage drop especially and even harder to find a logic level switching solution with acceptable current capacity. The best I could find was this MOSFET that appears to have a Vsd of 1v at Vgs(on). That would still leave me with ~2.3v, a volt below the min(and max). How do devices normally handle this issue? I'm beginning to think I need to bite the bullet and do a deep dive learning DC-DC boost converters, but I hate to add more power loss, more components, high frequency noise, and inefficiencies into the mix. Oh and my oscilloscope is questionably useful at best.

Short of a relay, are there any good options for switching the load with no realized voltage loss, or is the right thing going to be a DC-DC buck boost segment?

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    \$\begingroup\$ You can use a p-channel MOSFET for sure. But. it will need a few other components around it to make it work. Not sure what you mean by "segment". \$\endgroup\$
    – Andy aka
    Apr 21, 2023 at 17:32
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    \$\begingroup\$ Where in the datasheet did you see Vsd=1V? \$\endgroup\$ Apr 21, 2023 at 17:39
  • \$\begingroup\$ Andy aka: segment is a circuit block or a cluster of components that have a specific task. user253751: I apparently read the Vgs graph wrong. I thought the shown 1V was the minimum voltage drop like a diode or transistor. Turns out MOSFETS work very different from BJTs. \$\endgroup\$
    – Michael.G
    Apr 21, 2023 at 18:42
  • \$\begingroup\$ @Michael.G That's not a use of "segment" that I've ever heard before. Are you perhaps translating a word from another language? \$\endgroup\$
    – Hearth
    Apr 22, 2023 at 4:43
  • \$\begingroup\$ For "cluster of components that have a specific task," I might say "subcircuit." \$\endgroup\$ Apr 22, 2023 at 13:40

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I think you have mis-read something in the datasheet of the MOSFET you linked to, though I'm not sure what. Presumably this MOSFET's gate connects to some sort of digital output so it will be 3.3 volts \$V_{GS}\$ when on. On page 3, middle left graph, you can see that this is plenty to give the best \$R_{DS(on)}\$ of around 0.01 ohms, which is plenty low and gives a 0.1 volt drop at your 5 amp inrush number. Notice the graph goes up to 40 amps. 5 amps is no problem and 0.25 amps is definitely no problem.

Now, this is an N-channel MOSFET which means it is best used to switch the ground and not the power. The reason for this is that the MOSFET measures its gate voltage with respect to the source. To have a gate voltage of 3.3V that means the gate is 3.3 volts higher than the source. The source on an N-channel MOSFET is (usually) the more negative side. If you would try to use it to switch the power, the negative side of the MOSFET would connect to the power line you were switching and its voltage would go up as the MOSFET turned on. This would lower the gate voltage (compared to the source) and the MOSFET would reach an equilibrium where the lowered gate voltage was just enough to let some current through to get a lowered power line voltage. This can be useful for certain amplifiers (it's called a common-drain configuration, or source follower) but it's useless for switching power. If you built the circuit the way I think you did, this is probably what you are seeing.

The usual solution is to use a P-channel MOSFET instead. It's the reverse of N-channel. The source connects to the power source, the drain connects to the thing you are switching, and the gate voltage goes negative to turn it on. P-channel MOSFETs have somewhat worse performance, but still much better than you actually need.

Note: the "off" gate voltage has to match the source voltage. If the power source is 4.2V and the gate is 3.3V, that is -0.9V \$V_{GS}\$ which may allow some current to leak through the MOSFET - it might not be completely off. If you need that to work, then you need to add a little bit more circuitry.

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  • \$\begingroup\$ Thank you for the explanations! I will need some time to understand all of that but I will definitely look into the P channel MOSFETs. I am getting that I misread the datasheet. The fixed voltage test reading can be confusing. The Vgs graph seemed to show a 1V minimum drop but that is likely something else entirely. I will have to do more studying on MOSFETS, they aren't straight-forward in my mind at all yet. Your explanation of the G-S voltage relationship was very helpful. I may actually use this on the negative side of I can due to as I understand floating pins can cause current leakage. \$\endgroup\$
    – Michael.G
    Apr 21, 2023 at 18:28
  • \$\begingroup\$ That explanation is excellent. I've read it a few more times and it now makes a lot of sense (I think). I am just selecting components so far, nothing is layed out yet except for rough block diagrams. If I understand right I can set it up so VGPIO @ 3.3V > base of an NPN BJT + source is Vbat up to 4.2V > Gate of the PNP MOSFET with a pull up resistor between S and G right? That way the PNP gate is saturated untill the BJT is switched on and it drains the MOSFET gate charge to low. I wish I could have posted a picture here for you to make it easier to see what I mean. I'll test it in Falstad's. \$\endgroup\$
    – Michael.G
    Apr 21, 2023 at 19:04
  • \$\begingroup\$ @Michael.G Yes, that's exactly how you can make it work with a different control voltage from the power supply voltage. The MOSFET is called P-channel, not PNP, but who's counting? \$\endgroup\$ Apr 21, 2023 at 19:05
  • \$\begingroup\$ Awesome! Thank you very much. Gotcha I'll remember that; N- and P- channel. \$\endgroup\$
    – Michael.G
    Apr 21, 2023 at 19:13
  • \$\begingroup\$ It looks like they're misreading the Vsd rating (i.e. the forward voltage of the body diode) to be some equivalent of Vcesat in a BJT. \$\endgroup\$
    – Hearth
    Apr 22, 2023 at 4:44
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There are all-in-one ICs that are designed for this kind of power switching. Internally I believe they all use a P MOSFET. TPS22932 comes to mind.

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all running from 3.3V on a single Li-ion cell. ... Here are the specs: Vbat 3.0 - 4.2V

When I saw this, I thought "how is this going to work when the battery is at 3.0 Volts?"

What you probably want is a boost-buck or Sepic DC-DC converter. These can take in voltages above and below the target voltage, and still output a stable 3.3V. There are reference designs and even development boards for many to experiment with. That solves one problem.

Short of a relay, are there any good options for switching the load with no realized voltage loss

Terminology is important here. The only thing that could actually switch with zero loss would be a super-conductor. Everything else will be lossy. Question is, how lossy?

I wouldn't discard relays so quickly. A latching-type relay could be used, which only requires a short pulse to permanently "latch" into that state. (This type often has two coils - one to "set" it, and the other to "reset" it. Or you reverse the polarity of the coil.) Practical if it will not be switching frequently.

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  • \$\begingroup\$ Thank you for the comments. I should mention that this is a very small wearable device so space is a concern. I was also planning on the SoC limiting the battery by software to 3.3V as 'depleted’. As the SoC can run on power down a little below 3V, there is little concern about the control functioning even if the battery cannot run the haptics, SD, or display. Once the SoC reads the battery is charged enough to run it can re enable the power control. \$\endgroup\$
    – Michael.G
    Apr 21, 2023 at 18:34
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    \$\begingroup\$ @Michael.G It should be noted that you'll get a longer life out of the batteries if you use a DC-DC converter to boost the voltage to a constant regulated value. It's basically the principle of the "joule thief" circuit that beginners and hobbyists enjoy building. Of course this will be a design choice on weighing battery life vs literally the weight of the device itself. \$\endgroup\$
    – Ste Kulov
    Apr 21, 2023 at 21:44
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There are parts specifically designed for switching loads and a standard industry term is indeed "load switch". For example, here is the Texas Instruments product range page for their load switches.

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