MOSFET as a switch. Why does the voltage depend on the gate

cmosintegrated-circuitmosfetvlsi

Let's say I have an NMOS, the gate is connected to 5 volts, Vth is 0.7 volts and I want to pass a voltage of 7 using the NMOS as a switch. Can you tell me what will be the voltage at the source?
Will it be 4.3 or 6.3?

In my understanding, once a voltage greater than Vth is applied at the gate, the source and drain are effectively shorted, this means the voltage at the drain should appear at the source.

The reason I ask is because of the picture below. The picture suggests, irrespective of what the drain voltage is, the voltage at the source is always the gate voltage minus Vth

enter image description here

Best Answer

The illustration is correct. VGS must be above Vth for the NMOS to be ON, otherwise it will be OFF. Therefore in your circuit the maximum possible voltage at the source is VDD-Vth, otherwise the NMOS would be OFF. Note that in your circuit VGS=VDD-(VDD-Vth)=Vth, so it is ON, but the voltage at the source cannot increase further because it would turn OFF the NMOS (and the source voltage would decrease, turning it ON again). It therefore comes to an equilibrium where VGS=Vth.

The fact that there is three NMOS instead of one doesn't change this situation, because all 3 NMOS have the same VGS voltage.

If you want to use an NMOS as a switch and have the full VDD on the load you should place it in the low side as in the following picture.

schematic

simulate this circuit – Schematic created using CircuitLab