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I have the datasheet of a MOSFET but I can't tell if it's a logic-level MOSFET or not.

I read that if an Rds(on) for Vgs = 5 V is specified, then it should be a logic-level MOSFET.

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2 Answers 2

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It isn't.

It doesn’t claim to be in the datasheet. It DOES claim to be a Power FET however.

"Logic Level" is a nice vague advertising term which doesn’t have a strict definition. However, most engineers would look for a FET which is fully “ON” at whatever votlage your logic is running at, often 1.8V. This FET doesn’t even start switching until 2V.

The threshold voltage is not the switching voltage. If it start switching at 2V, you don’t want to be using 2V to switch it, you want something more like 5V. The graph on page 6 (figures 5 and 6) are useful. Fig 5 shows current through at different gate voltages, fig 6 shows the impact of gate voltage on current. Both of these graphs suggest that 8V on the gate is required to get full performance out of these FETs.

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No, there is no Rds(on) defined at Vgs=5V, therefore you cannot anticipate how it will react at that voltage.

Find a mosfet that has Vgs = 5V included in the testing conditions.

You can read more here

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  • \$\begingroup\$ Allright thanks! How about this one: pdf1.alldatasheet.com/datasheet-pdf/view/171118/…? On page 4, there is Vgs=5V included \$\endgroup\$
    – Qstudent
    Commented Oct 11, 2022 at 7:31
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    \$\begingroup\$ Fig 5 and fig 6 in the datasheet will let you calculate an Rds ON for 5V. \$\endgroup\$
    – Puffafish
    Commented Oct 11, 2022 at 7:32
  • \$\begingroup\$ So thnat means if I have 5V switching voltage from an arduino for example, I would need Rds, which I can figure out from figure5 and 6? \$\endgroup\$
    – Qstudent
    Commented Oct 11, 2022 at 7:38
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    \$\begingroup\$ @Qstudent Yes. This one has a maximum resistance Rds(on) of 0.05 ohm when you use a 5V signal at the gate. Without a heatsink, if ambient temperature is assumed to be 60C, you can use this mosfet to pass a maximum continous current of 3A. If you need more current than that, you need to find a lower Rds(on) at 5V or use a heatsink (needs a lot of calculation). \$\endgroup\$
    – Anas Malas
    Commented Oct 11, 2022 at 7:45
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    \$\begingroup\$ Careful here, Anas Malas and @Qstudent. A 5 V logic gate driving high does not output 5 V, it outputs a voltage within a range e.g. 2.0..4.7. And supply voltage VDD will not be 5.0000 V, that has a tolerance. Typically the lower output voltages are while under load but the output at min. load current will not be 5 V. So (a) the output range must be taken from the driving IC's datasheet and then (b) the lowest supply rail considering its tolerance must be used as VDD. Can't just use 5 V if you want a reliable design. \$\endgroup\$
    – TonyM
    Commented Oct 11, 2022 at 7:51

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