# Saturation Points for Mosfets; driving gate with 3.3V output

I'm still new to this so please go easy on me :)

I'm looking at using N channel Mosfets to act as a switch between my microcontroller (ESP8266 breakout board) and LED Strip. Here's the schematic:

When I was reading about how mosfets and transistors work, I saw that for it to be saturated, the potential voltage between gate and source pins (or base and emitter in transistors) needs to be higher than the value listed under V_GS(th). Could I do this with the 3.3V output of a ESP8266 GPIO pin? The datasheet for the two mosfets I was considering using are slightly confusing me.

I'm asking because under the max value they only use tiny currents. IRLB8721, they use a drain to source current (Id) of 25μA and the max Vgs (needed for mosfet to turn on right?) is only 2.35V. For the IRLZ44N it's 2V with 250μ. So are these logic level Mosfets? Would they work?

Another value I'm not to sure about is Vds. They set it equal to Vgs in the values above, but for figure 3 where you can see other (higher) Id currents, they set it to 15 and 25V. Is this relevant?

IRLB8721:

IRLZ44N:

I would be really grateful if someone could help me clarify this.

On a side note, is it necessary to use resistors between the microcontroller and gate? How would I do this, and for what function?

Thanks so much!

• fwiw, i use irlz44n with 8266s all the time, works well, but i've never pushed 15A either... Dec 30, 2017 at 19:47
• How much do you push? They can't turn on fully, so I'm a little worried about them acting strange and getting too hot if I push a few amps. Dec 31, 2017 at 11:31
• the most i suppose is 10m of 5050 60/m LEDs, whatever those draw. the to220 is room temp on that at 50% pwm... Dec 31, 2017 at 15:42
• How much current does your LED strip need/how much current did you plan to drive at each MOSFET? Feb 20, 2022 at 23:33

There's no need to use a resistor between MCU pin and FET gate.

The 8721 quotes a Vgsthmax of 2.35v at 25uA drain current.

The 44 quotes a Vgsthmax of 2.0v at 250uA drain current.

Both appear somewhat marginal for a good conduction current at 3.3v, though the 44 is clearly able to conduct more current at a lower voltage, so you're more likely to get away with it using this one. Depending on how much current you want to sink, don't be surprised if they don't turn on fully, and get hotter than you expect.

You might be better with bipolar transistors, they are easy to turn on fully with 3.3v. Here you would need a series resistor to limit the base current.

• Thanks for your help - perhaps it would be easier to use bipolar transistors; since I'm not too sure if the mosfets would work well with 3.3v at the gate, especially since I'll be using way more than 250uA. Would the commonly tip120 darlington pairs do the job? Or should i stick to a simple single npn bjt transistor such as the tip31? Dec 30, 2017 at 18:47
• tip120 would be fine, 3.3v will turn on a darlington easily. IIRC with gain >=1000, you won't need much base drive for reasonable currents. But remember the VCEsat of a darliington will be higher than for a simple BJT, around one volt. Dec 30, 2017 at 19:00
• Ok cool. I'll stick to a standard transistor to reduce the power loss. I was set on using MOSFETs due to their sometimes superior qualities, and was even considering using drivers, that I didn't look at the classic bjt transistors. I'll just use these to make it easier... Dec 31, 2017 at 11:32
• Or just use another MOSFET designed for lower logic levels ... Feb 20, 2022 at 23:34

The gate threshold voltage is pretty much irrelevant to most applications. Note that this is usually defined where the FET conducts only a little current. That's not useful when you want to use it as a switch to conduct lots of current.

The main specs you need to look at are Rdson, and at what gate voltage that spec applies. Things you do with this:

1. Multiply Rdson by the maximum current when the switch is on. This tells you how much voltage the FET will drop. Make sure that amount of voltage loss is acceptable to your load.

2. Multiply the square of the current by Rdson. That is the power the FET will dissipate when on. Make sure that is within what the FET can do, and that your mechanical setup can deal with the heat.

3. Make sure the circuitry driving the gate guarantees at least the gate voltage that Rdson is specified with.

If any of the above don't check out, you need a different circuit or a different FET.

• The first two aren't a problem, the third is what my initial question concerned - whether or not I could get the Mosfet to turn fully on with what the microcontroller driving the gate outputs (3.3V). I'm still confused at the numbers in the datasheet, as I mentioned. They give the Rdson as resistances with Vgs set to 4V, 5V and 10V. My microcontroller doesn't give this. However right below that they quote the max Vgs as 2V? Dec 30, 2017 at 17:43
• Is it because the 2V Vgs is only for 250uA, while for the lowest voltage they specify for Rds(on) at 4V is for 21A? For driving an LEDstrip with up to 1A, I wouldn't be able to use this mosfet? Dec 30, 2017 at 17:48
• @Cas: If Rdson is only specified for 4, 5, and 10 V on the gate, then the answer is you don't know what will happen with only 3.3 V. This FET is inappropriate for this application. Get a different FET. Dec 30, 2017 at 18:49
• Yep that's what I was thinking - thanks for helping Dec 31, 2017 at 11:04

You need a logic-level MOSFET that is compatible with 3.3 V levels.

The ones you have posted aren't well suited because their datasheets don't include the required characteristics.

Take for example the datasheet of the PSMN4R3-30PL, for comparison. In figure 8 it specifies the R_DSon for V_GS(V)=3.3 as 7 milli-Ohm or so, which is quite low. (basically, lower R_DSon -> lower temperature/more efficient -> better) If the datasheet doesn't specify this value (in a plot or in table ) for V_GS <= VCC_MCU then you should look for a part that is designed for lower logic levels.

Figure 3 shows the safe operating area. For your use case the DC curve is relevant since you supply current continuously. At V_DS(V)=12 you see I_D=7A. Thus, with that part you could drive up to 7 A into your LED strip.

I don't know much about such LED strips so I don't know if that would be sufficient.

You need a gate resistor to limit the current when charging the gate. You can look at the datasheet of your MCU to see the maximum current a GPIO pin can drive. Higher current means faster charging, at a small time scale. A common design choice is to also pull-down the output pin.