# What will the MOSFET Drain to Source current be?

I would like to use a MOSFET to allow a microcontroller to connect and disconnect the power to a motor using the circuit shown below: The current drawn by the motor will change as the motor moves, varying from 100mA to 300mA. A PTC fuse will be put in line with the motor to prevent it drawing too much current.

Am I correct in saying the MOSFET will act as a switch in saturation region so long as the drain to source voltage is high enough (for a given gate to source voltage)? So for the below example, for a VGS of 6V, the MOSFET will be in saturation so long as the drain to source voltage is above 0.21V? If this is correct, am I also correct in saying the drain to source current will not decrease below the IDS stated on the above graph? For example using the below graph, if VGS=6V and VDS=8V(therefore the MOSFET is in saturation), then IDS will be 7.5A? Therefore it will try to force 7.5A through the motor? Or is it the case that the MOSFET can allow up to 7.5A to flow but if the motor only draws 300mA, 500mA, 1A etc, the MOSFET will allow this? • @DKNguyen do I not want it to be in saturation so the MOSFET can act as a switch?
– MRB
Nov 9, 2021 at 16:13
• You want to operate in the ohmic region
– Mike
Nov 9, 2021 at 16:22

NOTE: Saturation means different (nearly opposite) things for MOSFETs and BJTs.

As as switch, you actually want the MOSFET to be in the linear region on your graph. That's the region where it acts as a resistor (the I-V graph is linear there too just like a resistor's I-V graph). That's where Vds < Vgs. The saturation region in your graph is when the MOSFET is choking because it can't conduct any more current through it.

Picture it like this: When you drive Vgs hard, it is like you are opening the water valve so much that it passes the water through it so freely that all the pressure across the valve (Vds) drops nearly to zero.

If you don't drive Vgs very hard, the water valve only opens part way and it fills up and chokes the flow and pressure builds up across the valve (Vds becomes high).

• What is the issue with the MOSFET conducting the maximum amount of current though it? In the saturation region it has a low RDSon and therefore less heat generated from wasted power? Apologies for the follow up question, I want to ensure I am understanding fully.
– MRB
Nov 9, 2021 at 16:22
• @MRB That's exactly why. Low Vds means low heat loss. It also means that there is wiggle room in the current through the MOSFET which lets the load determine the current flow. If there was no wiggle room then the MOSFET is the one that determines the current flow through the load and now the current is controlled by the level of Vgs. That's what you do for an amplifier. Nov 9, 2021 at 16:23
• So in the linear region, the load dictates the current flow and in saturation the MOSFET dictates the current flow? In linear mode, the reason the load dictaes the current flow is because the voltage across the drain to source will change in order to deliver the current required by the load?
– MRB
Nov 9, 2021 at 16:25
• Yes. But it's not that Vds will change to deliver more current to the load. It's that the MOSFET can conduct more current through it as necessary. Vds just responds to that (i.e. if you increase Vds, more current can still flow through the MOSFET). YOu don't increase Vds directly though. It's more like you increase the voltage supply and it spills over to be across Vds and the load, and as more current flows through the load, the load voltage increases which reduces the Vds across the MOSFET to find an equilibrium current. Nov 9, 2021 at 16:27
• Okay so if the voltage across the drain to source changes in order to allow the current drawn by the load, let's say for example the motor in the first picture in the question requires 6V and draws 500mA. Let's say the supply is 7V. 1) where will the 'extra' voltage go? 2) let's say the motor draws more current, where will the MOSFET get the additional voltage it requires from?
– MRB
Nov 9, 2021 at 16:32