There is a saturation voltage but it is not the same thing.Is there a saturation voltages in above FET?
Be careful. "Saturation" means something different for BJTs than it does for FETs. But so many people confuse the two and use saturation to mean the same mode of operation for both devices so you never know what someone is talking about when they say saturation and FET in the same sentence.
For BJTs, saturation is when it is fully on as a switch and forward-active means the BJT is operating as a linear device.
For FETs, saturation is supposed to mean that the device is operating as a linear device but a lot of people (as you are doing) carry over the term "saturation" from a BJT and use to refer to a FET being "fully enhanced" or fully turning on as a switch. So the reason you don't see a saturating voltage in the FET datasheet is because you are using the wrong term for the thing you want.
Operating as a linear device: Forward-Active or Active region
Operating as an ON-switch: Saturation region
Enhancement mode FET (the most common kind of FET):
Operating as a linear device: Saturation region
Operating as an ON-switch: Ohmic or triode region
Answer to your question:
Notice that the ON-switch state for a FET is called ohmic region. That's because it acts like a low value resistor when it is fully-on/fully-enhanced (in the equivalent mode of saturation in a BJT). Therefore, a FET has no equivalent to the "saturation voltage" in a BJT. Even if there was, it would not be called the saturation voltage because saturation doesn't mean the same thing for FETs as it does for BJTs.
A BJT acts more like a diode when fully-on (in saturation). A FET acts more like a resistor when fully-on (ohmic region or fully-enhanced for enhancement mode devices, fully-depleted for depeltion mode devices)
FETs can also exist as depletion mode devices (such as JFETs) in addition to enhancement devices but they are much much rarer. Enhancement devices are off by default and conduct more current when you apply more gate voltage. Depletion devices are the opposite. Depletion devices are on by default and conduct less as you apply more gate voltage.
1. Yes, if it is a 5V PIC because both of your listed MOSFETs have a rated on-resistance at Vgs = 5V. But you won't be able to switch at high frequency since the limited output current of the PIC pin limits the speed that the MOSFET on and off at (kHz or faster).Excellent information Thank you very much.
In my LED matrix I need to drive the 8 cathode rows via N channel MOSFETs. Earlier I planned to use 2N7000, but due to current demand I switched to STN3NF06L FET.
I have attached the datasheet for reference.I have below questions.
1. Can I drive the above FET from direct PIC micro pin?
2. Do I need a Gate to Source pull down resistor? (I need to omit that , PCB has no more space )
My Load is 32 LEDs per row that is 20mA X 32 = 640mA
The datasheet for the STN3NF06L Mosfet says that its maximum on-resistance is 0.12 ohms if it stays cool when its gate-source voltage is 5V. Then its maximum voltage drop is 640mA x 0.12 ohms= 0.08V.
1. Yes, if it is a 5V PIC because both of your listed MOSFETs have a rated on-resistance at Vgs = 5V. But you won't be able to switch at high frequency since the limited output current of the PIC pin limits the speed that the MOSFET on and off at (kHz or faster).
2. It's safer to have one but you don't necessarily need one.