Quick Question on mosFETs - Page 1
Forum section: electronics / beginners

author: Falcon69
author_rank: Super Contributor
timestamp: March 19, 2016, 05:40:16 pm
content: Okay, looking at the data sheet, it does not say it is logic level for the mosfet.The Gate Threshold Voltage says 1.1v - 2.5v with a typical amount of 1.7v.Does that mean the gate will require a total of a minimum of 1.1v to turn on, and to get it to fully turn on I would need to supply 2.5v or more for it to turn on fully......no matter what the source voltage is? Or rather, in an N-Ch it would be the drain voltage, or the power supplied to the circuit the mosfet is turning on.Also, if I don't supply the absolute rating of 2.5v......does that just mean that the circuit being switched will only be able to draw less amps? For example, at 2.5v the it can draw 295mA continous, but at 1.1v it would only be able to draw, say, 100mA?Here's the data sheet.http://www.mouser.com/ds/2/308/NTJD5121N-D-119599.pdfAlso, anyone know of another way to switch on 16 circuits (LED's) simultaneously with individual in/out (source/drain), with just one jumper or switch?

author: rx8pilot
author_rank: Super Contributor
timestamp: March 19, 2016, 05:48:44 pm
content: The threshold is where the device starts to turn on. The resistance of the device will go down as the gate voltage goes up until it is saturated which is in the neighborhood of 10v. This device will operate reasonably well at 5v logic, but rather resistive at 3.3v.Sent from my horrible mobile....

author: mariush
author_rank: Super Contributor
timestamp: March 19, 2016, 05:49:20 pm
content: Look at Figure 1 and Figure 2 of your datasheet for the answers you want.As for 16 leds, see specialized led drivers like TLC5928 or maybe have a look at chips like ULN2003A (not as efficient as mosfets but cheap .. or ULN2803A for the 8 darlington version)

author: Kilrah
author_rank: Unknown
timestamp: March 19, 2016, 05:50:04 pm
content: It means it may start to conduct anywhere between 1.1 and 2.5V, but typically around 1.7V. That would change from one part to another.Seeing the "On?Resistance versus Gate?to?Source Voltage" graph it's a part that you'd want to drive with about 5V if you want to use it as a switch. If your voltage is too low then the resistance of the MOSFET increases, reducing current in your load but much more importantly increasing the heat dissipation in the MOSFET, which could blow up as a result.

author: madires
author_rank: Super Contributor
timestamp: March 19, 2016, 05:53:16 pm
content: All the answers are in Figure 2 Transfer Characteristics of the datasheet. Based on that I'd say that the MOSFET is fully turned on at 3.3V. At a lower V_GS it operates in resistive mode.

author: Falcon69
author_rank: Super Contributor
timestamp: March 19, 2016, 05:59:05 pm
content: okay, thank you, my sorce voltage is from a logic gate, so 5V.Hmm, ULN2803...i have some of those on hand. That may work.

author: Falcon69
author_rank: Super Contributor
timestamp: March 19, 2016, 07:03:06 pm
content: well, I tried. On the breadboard.The ULN2803 will not work, as it grounds all the led's through a single ground. I would need them to be isolated.

author: rx8pilot
author_rank: Super Contributor
timestamp: March 19, 2016, 07:38:54 pm
content: You may want to consider a device with lower Rds(on). That is the resistance between the DRAIN and SOURCE pins when the device in 'on'. For a DC switch, the lower the better and they get very low. This one is around 2 ohms with a 5v measured from gate to source voltage. You can get a device that is far better than that and it makes a difference when driving LED's - even low current ones. You don't need exotic at all, just something <250mOhms (.025 Ohms) or so. When looking at the specs - look at the resistance (Rds(on)) at the the gate voltage you will provide - 5v. Most of them will call out 4.5v and 10v like the one you linked but they usually have a graph to you can get an idea of where you would be at 5v.Mosfets can be fragile - the die will blow up before the outside even gets hot in some cases. For a switch, the goal is to get from off state to on state as quickly as you can. So if the signal is coming from a logic chip with a limited current capability - use a resistor from the logic to the gate that will limit the current going into the gate pin of the FET to whatever the logic device can handle. The gate pin itself does not pass current like a bipolar transistor but instead it behaves like a capacitor. The gate pin will pull a lot of current up until it is 'charged' where it will pull nothing at all. During the time that the gate is charging - the FET is very resistive and will dissipate a lot of heat. If this happens quickly - who cares. If it happens slowly, the FET may over heat and die.Hope this is helpful....

author: T3sl4co1l
author_rank: Super Contributor
timestamp: March 19, 2016, 09:28:03 pm
content: Fundamentally, a MOSFET isn't a switch. A switch is strictly on and off (within various meanings of "on and off", depending on what kind of switch you really mean..). A MOSFET is a linear device with varying levels of "on-ness".A more useful definition of a MOSFET is, a variable constant current sink, with a backwards diode, and which looks like a variable resistor when the voltage is low. The current is controlled by Vgs.A reminder that there is no absolute voltage. Vgs is always gate to source voltage, and Vds is always drain to source voltage. It doesn't matter if you stick a resistor under the source, or any other kind of network or source, because gate and drain are still relative to source, period. If you measure gate-to-ground voltage instead, and source-to-ground voltage is nonzero, you've failed reading comprehension, because you measured something no one told you to measure...What the datasheet tells you is, for this-much drain current, you need Vgs of so-and-so. Vgs(th) is measured at the given Id. The current is small, usually 1mA or less. Think of this as the point where it becomes sensitive; much less, and current just stays more and more zero. At Vgs = -20V, you can't get any more zero than zero* (and you'll be measurably there already by 0 or -1V, as it happens). But going up in voltage, it becomes more and more sensitive, and draws more and more current (or exhibits a lower and lower resistance, to a point).*Not actually zero, but leakage current. Usually in the nA to pA. They rate it at 1uA or something like that, but only because it takes time to precisely measure very small currents. Almost no one is going to care if a power-switching transistor leaks 1uA, anyway.The intent of the Vgs(th) figure is, this is the offset where it becomes sensitive. Your circuit needs to accommodate this variation, in order to use currents around this level; or for switching purposes, you need to deliver enough excess beyond this point to ensure it looks like a "switch".Consider a faucet. It goes from off to on over a variable range. The amount of pulling or turning needed to go from the fully-seated or fully-off position to a dribble is the threshold. It's different for every faucet, as I'm sure you've noticed (and some are more irritating than others). Likewise, it's different for every transistor, and varies with temperature by the way. (Transistors don't happen to exhibit the even more annoying characteristic that many faucets do, where it goes, say, off -- dribble -- full cold -- still cold -- a little OHGODSUDDENLY HOT -- still hot -- more hot -- end stop.)Low voltage MOSFETs (<= 20V) are more sensitive, so can be reasonably rated for "switch" duty with fairly low voltages, like 2.5 or 3.3V logic levels. For these, Rds(on) at Vgs(on) is just the same as any other device, but with smaller numbers.Anything 30-600V that says "logic level" will reach its rating at 4.5V, nothing less. (And anything for power or speed will do much better with >8V anyway, so "logic level" should never be used as an excuse for poor drive in a power application.) Due to Vgs(th) variation, the amount of Rds(on) (or Id if pulled into the linear range) at, say, 4.0 or 3.5V, is not easily defined, because it's not apparent how close the transistor is to Vgs(th) just by looking at it.Tim

author: Falcon69
author_rank: Super Contributor
timestamp: March 19, 2016, 10:18:49 pm
content: thanks guys, clears it up. Since I would only be powering one LED with each mosFET, sounds like I would be fine connecting the gate to 5volts via a jumper and 100ohm resistor (with a 1M ohm resistor from gate to ground), and it should turn the LED on/off just fine. The power for the LED will actually come from a XOR gate (it can output + or - 25mA). The LED will actually only be running at about 60uA. Yes, uA. Currently i have them connected through the Logic Gate then to ground with a 4k02 resistor, and they are lighting just fine. But I wanted to put a switch in-lne to turn them on or off or make it select a different color LED (which is my purpose for the mosFET low side switch).Thank you Tim and rx8pilot, very well explained.and yes, some faucets are a pain....like the ones with the motion sensor. You get them to turn on, but then you move your hand to start washing with soap, and the damn things turn off!