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I know all the theory behind the MOSFETs, how to calculate the Id current based on (W/L) and (Vgs-Vth) and etc. But, now that I need to project a real solution, I feel like all of this is kinda useless. There are so many types of mosfets - power, logic level, small signal - and so many parameters that I'm not sure of what they actually mean, I'm totally lost, I dont know where to start.

This is the circuit that I want to project. Its a very simple and common one:

schematic

simulate this circuit – Schematic created using CircuitLab

By my researches, I think the resistors are correct.

The MCU GPIO Voltage is 3.3V. The maximum current that goes to my load is 200mA.

What kind of MOSFET M2 and M1 need to be? I think M2 need to be a singal level or small signal idk, but how I know the right one? The data sheets that I looked describes an "Max and Min VGS Threshold", but what does that means?

M1 can be any P-Channel? There is anything else that I need to worry about besides the source current? Does the VGS Threshold matters in this case?

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  • \$\begingroup\$ What sort of power consumption can you tolerate in the off-state? If your requirement isn't too strict there are a lot of USB downstream port power switch chips which do this in a single SOT23-5 package. If you can switch the negative rail instead you can do it with a single FET and resistor. And if your load is inductive you need a diode... Typically in terms of what you will find vs. what you need, your main issues to watch for are the Vgs at which you get low loss for a substantial current. With a higher supply you'd also have to watch the PFET's maximum Vgs limit. \$\endgroup\$ – Chris Stratton Sep 28 at 16:53
  • \$\begingroup\$ The threshold Vgs is typically that to pass microamps, find the graph or datapoints giving the voltage at which you can get an amp through it. If you insist on through hole, you'll find your choices quite constrained; SOT23 would be a typical and easy to use SMD package for parts suited to your power levels. \$\endgroup\$ – Chris Stratton Sep 28 at 16:55
  • \$\begingroup\$ This will be almost all the time turned on, so power consumption in off-state is not a problem. Can you give me an example of one of this switches? What do you mean with switch the negative rail? \$\endgroup\$ – David Daminelli Sep 28 at 16:59
  • \$\begingroup\$ The RT9742 is an example USB power switch chip (make sure to get the version with an enable input, not the one that just does overload protection). A low side switch would be for a load where you can put the switch in the ground lead, in that case you can use a single NFET simplifying things. What exactly is your load? The common reason you could not use a low side switch would be if the load had communication lines to something else, but that often typically means you cannot switch the high side either, without taking all the communication lines low first. \$\endgroup\$ – Chris Stratton Sep 28 at 17:07
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    \$\begingroup\$ Does this answer your question? What to look for in the datasheet when choosing a MOSFET? \$\endgroup\$ – JYelton Sep 28 at 17:25
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Pick any of the SOT23-3 MOSFET rated 350mA or above, from a well known brand. Buy yourself a tape of 100 pieces. I don't think you have to worry about anything else in your application.

Most N-Mosfets, nowadays, have a gate threshold between 1.2 and 3V. P-Mosfet have typically a threshold of -1.2V. Which means the gate of a P-Mosfets must be at the same voltage as the source, practically, to be off. Mosfets with different, and therefore unusual parameters will be exceptions.

It's fine to use a P-Mosfet for M1, but you can also use a N-Mosfet to power your load by connecting M1 to the low side of the load. Instead of connecting it to 5V, you would connect it to GND (The "negative rail" Chris talked about above). Like this you would use only N-Mosfets. It's also generally more efficient to use N-Mosfets. Unless you have a good reason to use a P-Mosfet of course. I suppose M2 is a N-Mosfet.

R2 is a little bit weak. I would increase it to 100K.

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  • \$\begingroup\$ This is poor advice which handwaves away the key concerns. Far better advice which actually addresses them is provided at nominated duplicate, electronics.stackexchange.com/questions/71378/… \$\endgroup\$ – Chris Stratton Sep 28 at 19:54
  • \$\begingroup\$ @ChrisStratton Stratton I'm being practical, and from the given informations, the circuit is a very classical, low power, one. Nothing to worry about. Popular MOSFETS, available by the hundreds of thousands, if not by millions on online vendors, would fit. The only parameter he should care about is that it should be at least 300 mA since the maximum load is 200mA. You shouldn't be too concerned about a 200mA circuit using no frequency. The answer would have been very different with another circuit. Or 30 years ago. \$\endgroup\$ – Fredled Sep 28 at 20:12
  • \$\begingroup\$ Current is not the issue. Selecting something with a suitable gate voltage at the needed current is the issue. That is what the link explains, but you carelessly handwave away. And there's nothing here in your post to justify this staying open as distinct question instead of simply being closed as a duplicate of the already nominated existing one, which already has a far better answer. \$\endgroup\$ – Chris Stratton Sep 28 at 20:15
  • \$\begingroup\$ @ChrisStratton Stratton But... Chris, most Mosfets would meet the criteria. I would even say that if he finds one which doesn't, it will be something very special. \$\endgroup\$ – Fredled Sep 28 at 20:31
  • \$\begingroup\$ Actually no, most would not. Especially not through hole power MOSFET's as beginners will choose trying to avoid SMD. This site is chock full of questions from people who chose a MOSFET with the wrong specs for their need. An answer which fails to address key issues like the difference between the data sheet threshold voltage, and the voltage at which the FET is actually usefully conducting is a very poor response to a question about correctly selecting FETs. You then go on to dig into low side switching, while again ignoring the issues that make it likely not workable for the asker. \$\endgroup\$ – Chris Stratton Sep 28 at 20:33

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