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I was just reading a blog entry regarding the power conflict of the Arduino and trying to understand everything, when I came to a point where I could not understand the working principle of the circuit vs the explanation of the MOSFET I found online anymore: enter image description here

In the drawing above it shows that the gate voltage of the MOSFET is 5V and that it does not conduct in that state which also makes total sense, since a barrel connector is plugged into the device and it sources the power from the voltage regulator and not the USB. I checked out the datasheet for the FDN340P and it turns out being a P-channel Enhancement type, which is also visible in the schematic. Now I wanted to read more about MOSFETs and the first overview I saw looked like this:enter image description here Here it says under P-channel MOSFET enhancement type that it does not conduct if the voltage is 0 and that it conducts the higher the voltage is. This does not make sense to me considering the first schematic ... I also saw a second overview stating the same. I would be very happy if anyone could explain this to me and maybe the four types of MOSFETs in general since it is not clear to me after this. English is not my first language so I am sorry for any mistakes

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Depletion mode MOSFETs are "normally on" devices. When a voltage is applied to the DRAIN and SOURCE terminals, current immediately flows through the MOSFET; no additional "control" signal is required to get the current flowing. With depletion mode devices the goal is to stop the flow of current. This is done by applying a voltage to the MOSFET's GATE and SOURCE terminals, which forms a depletion region in the DRAIN-to-SOURCE current channel, which increases the resistance in the DRAIN to SOURCE channel, which reduces the flow of current from DRAIN to SOURCE. When the GATE-to-SOURCE voltage becomes sufficiently large, the depletion region becomes so large that it completely closes off (or "pinches off") the DRAIN-to-SOURCE current channel, the current in the channel stops, and the transistor is OFF (or "cutoff"). The GATE-to-SOURCE voltage that effectively stops the flow of current in the DRAIN-to-SOURCE path is called the pinch-off voltage.

A goofy analogy for this would be a garden hose that is connected to a water spigot, and the other end of the hose is open to the air. When you turn on the water spigot, water immediately flows through the hose; nothing else is needed to make the water flow through the hose. If you pinch the hose's walls (not the end of the hose) with your hand, the flow of water slows, and if you squeeze the hose's walls very tightly you "pinch off" the flow of water and effectively turn OFF the watering system.

Enhancement mode MOSFETs are "normally off" devices. When a voltage is applied to the DRAIN and SOURCE terminals, current DOES NOT flow through the MOSFET; an additional "control" signal is required to start the current flowing. With enhancement mode devices the goal is to start the flow of current. Here is a YouTube video that explains, in very simple terms (and better than I can here), the theory of operation of an enhancement mode MOSFET: Working of Transistors | MOSFET

A goofy analogy to this would be a garden hose that is connected to a water spigot, and the other end of the hose has a sprayer attachment that blocks (stops) the flow of water until you squeeze the sprayer's handle. When you squeeze the sprayer's handle you "enhance" the flow water through the hose and sprayer and turn ON the watering system.

For a beginner, the N-type and P-type designations mainly describe the voltage polarities from DRAIN-to-SOURCE and from GATE-to-SOURCE that turns these devices on and off.

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The explanation is badly written (or there is context elsewhere, not posted).

It is correct that the MOSFET doesn't conduct with 0V "on the gate" but doesn't define what it means by that.

It SHOULD say the MOSFET doesn't conduct with 0V between gate and source (or, Vgs = 0V)

Now look at your schematic : yes, the gate voltage is 5V ... and the source voltage is ... well it's connected directly to the 5V rail. So Vgs = 5V - 5V = 0V.

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