Mosfet application doubt

[Veketti]

Dear All,

This is most likely stupid guestion, but I'll ask anyway to be sure. For example IRLML6344 chart "Typical output characteristics" have line for 4.5V Vgs but it is plotted to at around Vds = 0.5V. So the guestion is, if I have Vds = 12V and Vgs = 5V is it out of its operating range? My assumption is that it is ok as long as the Vds < 30V and Vgs < 12V it's ok.

Second guestion is, why didn't they plot it any further if it is ok to be operated with higher Vds Vgs combo?

Thank you in advance

[TimFox]

The curves for high V_GS are truncated because the drain current is too high for higher V_DS at higher gate voltage.

[Veketti]

Excellent, makes sense. So the conclusion is that Vds = 12V and Vgs = 5V is perfectly fine as long as the Id is within limits (<4A for continuous current)?

Just a side note. Maybe the Id 5A and 4A are wrong way around. Now higher current is lower temperature.

[fourfathom]

Excellent, makes sense. So the conclusion is that Vds = 12V and Vgs = 5V is perfectly fine as long as the Id is within limits (<4A for continuous current)?

No.  If you have a Vds of 12V and and an Ids of 4A that means the transistor is dissipating 48 Watts.  This is not going to end well for that tiny SOT-23 package with its specified maximum power dissipation of about 1W.

How do you plan to use this FET?  Typical applications are low-power linear or higher-current switching (where the Vds is near zero when current is flowing).

[magic]

Excellent, makes sense. So the conclusion is that Vds = 12V and Vgs = 5V is perfectly fine as long as the Id is within limits (<4A for continuous current)?

This means "never", because drain current will be tens of amps if Vds=12V and if it's limited to 4A then drain voltage will be very low (4A · RDS(on)).

This plot applies to linear operation, with simultaneous high voltage and not negligible current through the FET.

You are probably considering switching. When you switch the FET on, external load limits current and then Vds becomes very low. If the load fails to limit current (e.g. shorts out) then you end up in the uncharted corner (high Vds, high Vgs) and the FET doesn't survive for long.

[Microdoser]

Excellent, makes sense. So the conclusion is that Vds = 12V and Vgs = 5V is perfectly fine as long as the Id is within limits (<4A for continuous current)?

No.  If you have a Vds of 12V and and an Ids of 4A that means the transistor is dissipating 48 Watts.  This is not going to end well for that tiny SOT-23 package with its specified maximum power dissipation of about 1W.

How do you plan to use this FET?  Typical applications are low-power linear or higher-current switching (where the Vds is near zero when current is flowing).

Well if you use something like a GSFD8003, they have a rated power dissipation of 80W, of course at what voltage and current they can do that, and what cooling setup they would need is a different matter, and they are only rated at 3A, but they are a tiny TO-252 (DPAK) package. My point being that you can get far more than 1W rated tiny Mosfets.

[TimFox]

Also note that different maximum ratings, such as V_DS, I_D, and power dissipation are quoted by the manufacturer, but that does not mean that you can run the device at its maximum voltage and maximum current, simultaneously, if the resulting power dissipation exceeds the temperature-derated power rating with your heat sink (or none).
For switching MOSFETs, one typically runs near maximum current at very low voltage, and maximum voltage at very low current, and then looks at the SOA to see what happens during the switching transition from ON to OFF, respectively.

[thm_w]

Well if you use something like a GSFD8003, they have a rated power dissipation of 80W, of course at what voltage and current they can do that, and what cooling setup they would need is a different matter, and they are only rated at 3A, but they are a tiny TO-252 (DPAK) package. My point being that you can get far more than 1W rated tiny Mosfets.

Sure, but OP linked a SOT23 package.

[Veketti]

Ah, yes what a brainfart. I'm considering load switching and sure the Vds wouldn't be 12V. I made some calculations with tweaked values for better understanding. I checked the datasheet and according to that Rdson is roughly 22mohm for Vgs=5V. So in my calculations this circuit draining 4A Vds would be 0.088V and power dissipated by the FET 0.352W which is well between the limits. Is my calculations and understanding correct that this is ok?

[Microdoser]

Well if you use something like a GSFD8003, they have a rated power dissipation of 80W, of course at what voltage and current they can do that, and what cooling setup they would need is a different matter, and they are only rated at 3A, but they are a tiny TO-252 (DPAK) package. My point being that you can get far more than 1W rated tiny Mosfets.

Sure, but OP linked a SOT23 package.

SOT23 packages do go up to 5W that I can find, and a small design change, not being fixed on a particular package, gives much more freedom of choice. Their link was only an example, not a fixed choice.

Ah, yes what a brainfart. I'm considering load switching and sure the Vds wouldn't be 12V. I made some calculations with tweaked values for better understanding. I checked the datasheet and according to that Rdson is roughly 22mohm for Vgs=5V. So in my calculations this circuit draining 4A Vds would be 0.088V and power dissipated by the FET 0.352W which is well between the limits. Is my calculations and understanding correct that this is ok?

Ohms law is your friend when looking at things like dissipated power, yes. The dissipated power isn't the voltage x amps running through it (total power), but the voltage drop x current (power dissipated). A FET controlling 12V@4A isn't dissipating 48W, or there would be no power left over for components in the rest of the system. All the power would be turned into heat. If it has a resistance of 22mOhm when active, then you have calculated correctly and the dissipated power is closer to 1/3W.
 
As well as using Ohms law that way, you can also use P=I²×RDS(on)​