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But don't forget that this curve is showing only typical behavior and with 5V across the MOSFET. totally useless information, because 2A * 5V = 10W. the MOSFET would expire in a puff of smoke if we tried that.
So, I just want to be sure that I'm understanding this correctly since when I look at the datasheet, it says the transistor you linked to can handle a continuous current of 4.9A at 4.5V on the gate which is definitely a lot more than 10W... am I reading this and understanding you correctly?
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user52526
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I think DrAzzy is doing a good job of explaining how to read the data sheet. And there is also lots of good detailed information if you google "how to read MOSFET data sheet". if we just look at this particular MOSFET that I mentioned (ZXMN2F30FH), I can show how I go about finding out how much current it can handle:
first, find out how much power dissipation it can handle. to do this, I have to decide how hot I want to run the transistor. I usually set +125C as maximum. then I decide on the hottest ambient temperature. let us say +40C. 125-40=85C. then I find the figure for thermal resistance from junction to ambient, 131K/W. so the maximum power dissipation it can handle is 40/131=0.3W. then I estimate the resistance Rds(on) at 3.3V. for this transistor I have values for 2.5V and 4.5V, and I decide that the resistance at 3.3V will probably be close to the average of these two values, so I say that at room temp and 3.3V gate voltage the resistance is (0.065+0.045)/2=0.055 ohm. but at 125C the resistance will be higher. Fig 4 tells me that it is about 1.3 times higher at 125C than at 25C, so I will use 0.072 ohm for the resistance value. Now that I know R and P it is possible to calculate the maximum current I that the transistor can handle. it is sqrt(P/R) or 2.0A.this is the procedure that I normally use for quickly figuring out how much steady state load current I can safely run through a specific MOSFET.
2A is a lot less than 4.9A, is it not?
if the MOSFET is switching on and off there will be additional power loss especially is it is driven from a GPIO pin, so it may be necessary to use a gate driver between the GPIO pin and the MOSFET gate - especially if the MOSFET is larger than this SOT23 size device we have been discussing.
tage
if we look at the datasheet I linked to, Vgs(th) is the threshold voltage at which the MOSFET starts to conduct current (0.00025A). This threshold is between 0.6V and 1.5V, at room temperature. This is good information in the way that if we have 1.5V on the gate we should be able to run at least 0.00025A in the MOSFET and can expect that the drain voltage will be 1.5V or below. But most of the time our loads are larger than 0.00025A, are they not?
Now, the threshold voltage is not useful information if we want to control a real life load. If the load is for example 2A we will need a lot more gate voltage if we want the transistor to turn on. If we look at figure 3 we could be fooled to expect that we only need 2V gate voltage. But don't forget that this curve is showing only typical behavior and with 5V across the MOSFET. totally useless information, because 2A * 5V = 10W. the MOSFET would expire in a puff of smoke if we tried that.
The only section in the data sheet that gives us any solid promise is the Rds(on) value at 2.5V gate voltage. If we apply 2.5V at the gate and the load current is 2A we know that the resistance will be max 0.065 ohm, if the transistor is at room temperature, because the data sheet says so. the power dissipation then is 0.065 * 2^2 = 0.26W, and we can be sure that the transistor can handle this heat dissipation. the temperature rise will be about 34 degrees.
With bipolar transistors (npn or pnp) the common mistake is to look at the DC current gain value and forget that this is specified at a relatively high voltage drop. for switching loads we have to look at the saturation voltage, and read the values of base current and collector current and the saturation voltage, to get an understanding of what the transistor can do. in most cases an ordinary bipolar transistor is not suited for driving amperes of load current, because the required base current is so high that it is unpractical to drive from a GPIO pin. it can be very difficult to figure out how much current is going into the base with a certain base resistor, because the GPIO output voltage will sag as the pin current increases. and it is many times not well specified what current will be available from the pin. and in the same microcontroller, the drive characteristics for GPIO pins can be different.