Electronic – Why do N-channel MOSFETs have a voltage drop when load is connected on source side

mosfet

I read how MOSFETs work, although there is this thing I can't understand. To make things simpler let's take only n-channel MOSFETs into account. Normally, the load is connected on the drain side, like this:

Drain side load

When \$V_{GS}\$ is above \$V_{thres}\$ and high enough for the desired current to conduct, the MOSFET is in full saturation, and the effective resistance is low, making the \$V_{DS}\$ very small. This happens even if \$V_{DS}\$ is higher or lower than \$V_{GS}\$.

However, if you connect the load on the source side, like this:

Source side load

In order for the MOSFET to go into full saturation, it's required that \$V_{GS} > V_{DS} + V_{thres}\$. Why is that? The load is still in the same current path; does it somehow lower the effective voltage at the drain (in the first case) so that \$V_{DS} < V_{GS}\$ before current flows?

Best Answer

The difference between the gate voltage and the channel voltage needs to be above the threshold voltage for the MOSFET to conduct well. If the load is between the mosfet and ground, then the more current you push through the mosfet, the more voltage drops over the load and the less gate-source voltage there is.

In the first topology, it doesn't matter how much voltage is dropping across the load, the mosfet always sees Vg - Vs, where Vs is always 0. In the second case, Vs rises with the load current and in turn pinches the mosfet back towards "off". So in the second case, you have Vg - Vs where Vs starts at 0 and increases according to I_load*R_load.

You're effectively lowering Vgs by putting the load below the Nmos.

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