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Since there is a threshold voltage needed to switch on a E-MOSFET, as compared to no threshold voltage needed to switch on a D-MOSFET, I would deduce that a E-MOSFET has lower noise than a D-MOSFET.

Is my reasoning correct? Is there less noise in an E-MOSFET than in a D-MOSFET, and if so, why ?

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Since there is a threshold voltage needed to switch on a E-MOSFET, as compared to no threshold voltage needed to switch on a D-MOSFET.....

That isn't correct...

For any MOSFET, there is always a gate source threshold voltage. That threshold voltage is the voltage required between gate and source that typically causes a small amount of drain current to be conducted. Some MOSFETs might use a drain current of 100 μA and others might use a drain current of 250 μA as their reference point. Others may use some other lowish value.

Either way, this small (but not zero) drain current is the value at which the gate-source voltage is defined as \$V_{GS(THRESHOLD)}\$.

Take the BSP29 from Infineon as an example. It is a depletion-mode, N-channel MOSFET and, in its data sheet, you'll see this: -

enter image description here

And, if the gate-source voltage falls more negative it turns off even more: -

enter image description here

And you can also see this in the table above. Look at drain-source cut-off current.

With \$V_{GS}\$ at -3 volts, drain current is typically 0.1 μA with 240 volts DC between drain and source i.e. it properly turns off with the correct control voltage on the gate. And, you can take the gate-source voltage to even more negative values. This device has a limit of -20 volts.

I would deduce that a E-MOSFET has lower noise than a D-MOSFET. Is my reasoning correct? Is there less noise in an E-MOSFET than in a D-MOSFET, and if so, why ?

If there is "less noise" in an E-MOSFET than a D-MOSFET, it has nothing really to do with threshold voltages. Both have threshold voltages.

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  • \$\begingroup\$ do you agree that in a depletion MOSFET, in absence of polarisation of the gate, there is a channel ? While in a enhancement MOSFET, in absence of polarisation of the gate, there is no channel. \$\endgroup\$ Dec 4 '21 at 12:43
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    \$\begingroup\$ @MathieuKrisztian not sure what you mean but, if you mean if Vgs is 0 volts (for depletion mode), then the device will be conducting from drain to source as per the graphs in the data sheet. Is this what you mean? For an enhancement mode device, the channel will be pretty much non-existent. \$\endgroup\$
    – Andy aka
    Dec 4 '21 at 13:19
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    \$\begingroup\$ The channel doesn't exist in any significant way (D-MOSFET) when the gate source voltage is more negative that 3 volts. Think along these lines: a depletion and and enhancement mode MOSFET are exactly the same except the depletion device has got an invisible "bias" voltage that keeps it still conducting when Vgs is 0 volts @MathieuKrisztian \$\endgroup\$
    – Andy aka
    Dec 4 '21 at 13:54
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    \$\begingroup\$ No, when the D-MOSFET gate source voltage falls to a wholesome negative value (-3 volts for instance), the channel is unformed in any significant way. Just look at the lower picture in my answer and review what the drain current is when Vgs is -2 volts for instance (it's going to be less than 10 uA and a lot less than 10 uA for a J banded device) @MathieuKrisztian \$\endgroup\$
    – Andy aka
    Dec 4 '21 at 14:10
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    \$\begingroup\$ thank you for your explanation \$\endgroup\$ Dec 4 '21 at 14:37
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I don't think so.

If you look at modeling of the various noise sources in MOSFETs, for example, in this paper, there is no term in any of the equations that involves just Vt, only Vgs-Vt.

So, whether Vt is positive or negative has no effect on the noise, according to those models.

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