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I purchased an N-channel MOFSET: FQP6N40C

I'm trying to use a 3.3V microcontroller output to activate this MOSFET to control a 12V power source. However, when applying the 3.3V signal, the MOSFET fails to activate. Applying a 5V signal however does activate the MOSFET.

Looking at the "On Charactericstics" below, shouldn't a 3.3V signal suffice to activate the MOSFET? Or am I misunderstanding this metric?

enter image description here

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1 Answer 1

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No, a 3.3V signal won't turn this FET fully on (in fact, you'd be lucky if it conducted any current at all). Look at the conditions defined for the threshold voltage: "Id=250µA". This means that when you apply the threshold voltage to the FET (as defined by the manufacturer), a current of 0.25mA flows from drain to source (as long as the drain-to-source voltage is big enough). It might also vary a lot depending on the FET's actual threshold voltage, which is only loosely specified as being somewhere between 2V and 4V.

You have to go well past a FET's threshold voltage to turn it fully on. In your case, the manufacturer only defines the FET as "on" when its gate is at or above 10V as you can see from the specification of Rds(on). There is no guarantee about the FET's on-resistance for a gate voltage of less than 10V, therefore this FET is unsuitable even for 5V operation.

The typical characteristic graphs in the datasheet tell a more complete picture. Take a look at figure 1 (on-region characteristics) and figure 2 (transfer characteristics). You can see that a FQP6N40C isn't even fully on at 5V gate voltage - it actually needs at least 7V, typically, and possibly more depending on batch-to-batch variation.

Note that a switching FET is only "fully on" when it's in its ohmic region for all permissible drain-source currents: This means it behaves like a simple resistor from drain to source as long as you stay within its maximum ratings. This is the case at 10V gate voltage for the FQP6N40C, as can be seen in figure 1. The relationship between drain-source voltage and drain-source current is (roughly) linear all the way up to 10A. At lower gate voltage (i.e. 5V), the FET can still conduct but it's not fully on. As you go above a certain drain-source voltage, it enters saturation and the current will not increase much even if you increase the drain-source voltage dramatically. This is very much undesirable for switching FETs as it causes huge conduction losses and possibly destroys the FET - but it is actually the preferred (even required) operating region for analog designs. In simpler terms: At 5V, the FET is still "somewhat on" but not "fully on". It might even be "on enough" depending on your application but batch-to-batch threshold voltage variation might come to bite you. Its maximum drain-source current is dramatically reduced if you don't operate it at the manufacturer's recommended gate voltage.

You'll have to find a FET that specifies a maximum on-resistance at 3.3V gate voltage (or lower). These will typically have a threshold voltage of around 1V (maximum).

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    \$\begingroup\$ Looks like I misinterpreted the data sheet. Good lesson for me to learn! \$\endgroup\$
    – Izzo
    Jan 22 at 22:58
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    \$\begingroup\$ Suitable FETs are sometimes labelled as "logic-level" parts. \$\endgroup\$
    – Hearth
    Jan 22 at 23:09
  • \$\begingroup\$ @Jonathan S. Im having trouble understanding your answer, if I look at figure 1, when Vgs is 5V and Vds is 40V, the mosfet's drain current is between 10^0A = 1A and 10^1A = 10A, so how is 5V not enough to turn the mosfet on? \$\endgroup\$
    – Steve
    Jan 23 at 0:02
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    \$\begingroup\$ @Steve A switching FET is fully on when it's in its ohmic region at maximum allowed current: That means it behaves like a simple resistor between drain and source. At 5V gate voltage, however, that's not the case for this FET: The drain current saturates at 3A and stays roughly constant even if you increase the drain-source voltage. As you can see in figure 1, saturation happens in the 5V, 5.5V and 6V curves, but not at 6.5V and above. Therefore, the specific FET that was characterized in this figure is fully on at 6.5V (and I rounded it up to 7V). It's "somewhat on" at 5V but not "fully on". \$\endgroup\$ Jan 23 at 0:33
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    \$\begingroup\$ @Steve Exactly. The reason for this is that the FET will be destroyed if it ever enters the saturation region during steady-state conduction in a switching application. Let's take the 5V case for this FET as an example: You're switching 300V with it and the (inductive) circuit wants to push 5A through it. But at 5V, the FET will only pass 3A before it saturates. That means it'll suddenly have 300V (or more) across it with 3A passing through it. That's 900W of dissipation in a TO-220 FET, which is more commonly known as "surprise fireworks". \$\endgroup\$ Jan 23 at 12:49

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