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I have this NVB5860NL Power MOSFET.

In the datasheet, it is given as the Gate Threshold voltage is 3V maximum. But that is for a condition of drain current in microamps.

But, my condition is, I require a continuous drain current of 5A from a 10V source with Gate Source voltage as 10V.

Whether can I use this part? (Mainly asking for the threshold voltage and the maximum drain source voltage)

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    The easy way is to check the datasheet for the Vgs values used for the Rds(on) measurements. If that Vgs <= your drive voltage < Vgs (abs max), you're good ( assuming you have a 2 Ohm load between the FET and that 10V supply. Otherwise see Bimpelrekkie's answer) –  May 25 '21 at 10:46
  • "continuous drain current"... Does that mean you want to make a constant current source, or you want to switch a 5A load? For the latter, look at guaranteed max RdsON @ Vgs spec in datasheet. – bobflux May 25 '21 at 11:25
  • I'd like have a constant current source. –  May 25 '21 at 11:29
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    *I'd like have a constant current source* So I guessed that correctly, see my answer :-) – Bimpelrekkie May 25 '21 at 12:32

1 Answers1

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In the datasheet, it is given as the Gate Threshold voltage is 3V maximum. But that is for a condition of drain current in microamps.

The Threshold voltage \$V_t\$ is the voltage (applied between gate and source: \$V_{GS}\$) above which the MOSFET starts to conduct.

Issue is, what does "start to conduct"mean, 1 nA flowing or 100 mA? So here the manufacturer has choosen a current \$I_D\$ = 250 uA which they force through the MOSFET using a current source and then they measure \$V_{GS}\$ and define that as the \$V_t\$.

The 250 uA is just a choice they made. Using 100 uA or 1 mA might have worked just as well.

This only gives us an indication of what \$V_{GS}\$ we need to apply to make the MOSFET turn on fully. Conclusion: \$V_{GS}\$ needs to be significantly larger than 3 V.

I require a continuous drain current of 5A from a 10V source with Gate Source voltage as 10V.

You can't do this by using the MOSFET on its own. There is no predictable relation between \$I_D\$ and \$V_{GS}\$ as this relation is extremely temperature dependent and different between transistors.

You cannot simply apply \$V_{GS}\$ = 10 V and expect 5 A to flow. In reality, much more current will flow, so much in fact that maintaining the 10 V supply across the MOSFET will destroy it (\$R_{DS,on}\$ at \$V_{GS}\$ = 10 V = 2.4 mohm so in theory 10 V / 2.4 mohm = 4167 A would flow).

If you want to draw a constant 5 A from a 10 V supply using this MOSFET you will need to construct a constant current source circuit which regulates the \$V_{GS}\$ of the MOSFET such that 5 A will flow. Here's an example of such a circuit:

schematic

simulate this circuit – Schematic created using CircuitLab

Here the 5 A current flows through the MOSFET and the resistor. If the current is 5 A there will be 1 V across the resistor. The feedback loop with the opamp and MOSFET regulate the voltage across the 0.2 ohm resistor to 1 V (since Vref = 1 V).

Note that a lot of power (almost 50 W) will be converted into heat inside the MOSFET so it will need a sufficent heatsink.

Bimpelrekkie
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  • Thank you for the detailed answer. I have 2 follow up questions. 1. What happens to the excess voltage that I apply above the Vt. Suppose the threshold voltage is 3V and I apply 5V, what happens to remaining 2V? What would be the voltage between the GS assuming S is GNDed? 2. When you say, "So here the manufacturer has choosen a current ID = 250 uA which they force through the MOSFET using a current source and then they measure VGS and define that as the Vt." - How can you allow current to follow, when you don't have the required threshold voltage? First Vt has be fixed and then only current ? –  May 25 '21 at 12:10
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    *what happens to remaining 2V?* I call that voltage the "overdrive voltage", at \$V_{GS}\$ = \$V_t\$ a conductive channels (between drain and source) starts to form in the MOSFET. When you apply more than just \$V_t\$ then that channel becomes "thicker" and more conductive. You can see it as is that 2 V is used to make the conduction better, so it lowers the on resistance of the MOSFET. For the "best" = lowest on resistance (\$R_{DS,on}\$) you need to apply as much as possible, for example \$V_{GS}\$ = 10 V like is used in the datasheet when specifying the value of \$R_{DS,on}\$. – Bimpelrekkie May 25 '21 at 12:17
  • *How can you allow current to follow, when you don't have the required threshold voltage?* You can do this with the circuit I included, just make R1 not 0.2 ohm but 1 V / 250 uA = 4 kohm. Then with \$V_{ref}\$ = 1 V, \$I_D\$ will be 250 uA and with a multimeter you can measure between drain and source of the MOSFET. The **opamp** takes care of applying the correct \$V_{GS}\$ to make 250 uA flow. As the current is that small, we can then say \$V_{GS}\$ = \$V_t\$ so we know \$V_t\$. – Bimpelrekkie May 25 '21 at 12:20
  • Thank you for the clarifications. So, Even if I apply more than the threshold voltage, say if Vt is 3V, and I apply 5V between GS, the actual voltage between VGS would be 5V, right? –  May 25 '21 at 12:23
  • *What would be the voltage between the GS assuming S is GNDed?* And that (grounding the source) you cannot do! If you do that we will not know what the current through the MOSFET will be. That current **MUST** be measured and that's why R1 (0.2 ohm) is there. OK, you could connect the source to ground and add a current measurement resistor in series with the drain **but that results in a much, much more complex circuit**. As a beginner, you really do not want to go that way. This solution (current measurement resistor in series with source) is very robust and "just works". – Bimpelrekkie May 25 '21 at 12:25
  • *First Vt has be fixed and then only current ?* Actually **both at the same time**! For that you need a circuit and that's the circuit from my answer. It uses a **feedback loop** which have the nice property that it will settle itself at a point where the \$V_{GS}\$ will be what it needs to be to get the current that you want (5A) flowing. And if the MOSFET heats up, no problem, the feedback loop will compensate for that and change \$V_{GS}\$ such that still 5 A is flowing. – Bimpelrekkie May 25 '21 at 12:30
  • Why is the I=5A included in that image? –  May 25 '21 at 12:30
  • *Why is the I=5A included in that image?* Because that's where the 5 A is flowing. It is not a current source! The current there is 5 A. – Bimpelrekkie May 25 '21 at 12:30
  • Let us [continue this discussion in chat](https://chat.stackexchange.com/rooms/124661/discussion-between-newbie-and-bimpelrekkie). –  May 25 '21 at 12:31