Electronic – MOSFET off state at certain voltage

For an N-Channel MOSFET, current is switched from drain to source. But a P-Channel MOSFET works in the opposite way - in a P-channel MOSFET, current is switched from source to drain. See this appnote from IRF. Also, the anode of the integral body diode in a power MOSFET is connected to the source of an N-Channel, but the drain of a P-Channel. See this excerpt.

Basically, when you have a positive voltage connected to a load, and you want to switch it on and off, use an N-Channel MOSFET between the negative terminal and ground. Allow current to flow by applying a positive voltage that will saturate the transistor (10-12 for power MOSFETs, 3-5V for logic level). Turn it off by pulling the gate down to the source.

When you have a load with the negative terminal grounded (which is usually preferable; don't muck with ground if at all possible!), and want to apply or remove a positive voltage, use a P-channel MOSFET. Pull its gate up to the source (which is connected to V+) to turn it off, or pull it to ground (through an open collector output if your logic signal is less than V+) to turn it on (So that Vg is 0, and Vs is, say, 12V, therefore Vgs is -12V).

Depletion mode mosfets are less common, and usually only available in N-Channel. For N-Channel depletion mode, the gate must be pulled below the source (which is often ground). Stick with enhancement mode for most switching applications unless you need something strange.

This schematic shows both (enhancement-mode) configurations:

To identify the source and drain, look at the side which the arrow is connected to. This is the source. If you've got a physical component, a diode test on a meter is useful both for finding the switched current direction (Apply positive voltage to the terminal which the diode test identifies as negative) and for a basic test (not a guarantee) that the transistor isn't burned up. To differentiate between N-Channel and P-Channel, look at the symbol: N-Channel pointing iN.

Your description is correct: given that \$V_{GS}>V_T\$, if we apply a Drain-to-Source voltage of magnitude \$V_{SAT}=V_{GS}-V_{T}\$ or higher, the channel will pinch-off.

I'll try to explain what happens there. I'm assuming n-type MOSFET in the examples, but the explanations also hold for p-type MOSFET (with some adjustments, of course).

The reason for pinch-off:

Think about the electric potential along the channel: it equals \$V_S\$ near the Source; it equals \$V_D\$ near the Drain. Recall also that potential function is continuous. The immediate conclusion from the above two statements is that potential changes continuously form \$V_S\$ to \$V_D\$ along the channel (let me be non-formal and use terms "potential" and "voltage" interchangeably).

Now, let's see how the above conclusion affects the charge in the inversion layer. Recall that this charge is accumulated under the Gate due to Gate-to-Substrate voltage (yes, Substrate, not Source. The reason we usually use \$V_{GS}\$ in our calculations is because we assume that the Substrate and the Source are connected to the same potential). Now, if the potential change along the channel when we apply \$V_{DS}\$, the Gate-to-Substrate voltage also change along the channel, which means that the induced charge density will vary along the channel.

When we apply \$V_{SAT}=V_{GS}-V_{T}\$ to the Drain, the effective Gate-to-Substrate voltage near the Drain will become: \$V_{eff}=V_{GS}-V_{SAT}=V_T\$. It means that near the Drain the Gate-to-Substrate voltage is just enough to form the inversion layer. Any higher potential applied to Darin will cause this voltage to reduce below the Threshold voltage and the channel will not be formed - pinch-off occurs.

What happens between the pinch-off point and the Drain:

The Gate-to-Substrate voltage in this region is not enough for a formation of the inversion layer, therefore this region is only depleted (as opposed to inverted). While depletion region lacks mobile carriers, there is no restriction on current flow through it: if a carrier enters the depletion region from one side, and there is an electric field across the region - this carrier will be dragged by the field. In addition, carriers which enter this depletion region have initial speed.

All the above is true as long as the carriers in question will not recombine in the depletion region. In n-type MOSFET the depletion region lacks p-type carriers, but the current consist of n-type carriers - this means that the probability for recombination of these carriers is very low (and may be neglected for any practical purpose).

Conclusion: charge carriers which enter this depletion region will be accelerated by the field across this region and will eventually reach the drain. It is usually the case that the resistivity of this region may be completely neglected (the physical reason for this is quite complex - this discussion is more appropriate for physics forum).

Hope this helps