According to this data sheet at 6v GATE voltage , the saturation drain current is about 1.25A. now I want to switch a device (say a tiny motor) which demands 1.25 A at 2volts , Indicating the resistance 1.6 ohms. now I am supplying the drain source voltage 4 volts.(as usual the motor and the MOSFET is in series) which means the mosfet is saturated , and capable of supplying 1.25A , now the question is , usually we are advised to use ohmic region to switch, but how about this? won't this switch work?

the schematic is here :enter image description here

  • \$\begingroup\$ Show the circuit please. \$\endgroup\$ – Andy aka Mar 17 at 18:25
  • \$\begingroup\$ @Andyaka ADDED. \$\endgroup\$ – Sayan Mar 17 at 18:31

but how about this? won't this switch work?

It will work (for a short period of time) but, the MOSFET will be dissipating about 3.2 volts x 1.25 amps = 4 watts. You'll need a heatsink and this is quite wasteful of power. However, the MOSFET you linked is a 2N7000 and, a heatsink is out of the question hence it will burn in a few seconds: -

enter image description here

What you should be doing is choosing a MOSFET with much much lower on resistance and providing a supply of 2 volts.

  • \$\begingroup\$ it is clear as you said here power dissipation is the main issue, for a while to understand the theory let me use such a high power MOSFET which can withstand such a power, how about that time? or even if it works will it be economic? \$\endgroup\$ – Sayan Mar 17 at 18:42
  • \$\begingroup\$ @Sayan I'm unsure what you are asking me. Are you saying what will happen if the MOSFET can survive the burn> \$\endgroup\$ – Andy aka Mar 17 at 18:47
  • \$\begingroup\$ so in a nutshell if somebody drives a MOSFET in saturation as a switch, unnecessary power loss is the main issue right? \$\endgroup\$ – Sayan Mar 17 at 18:50
  • \$\begingroup\$ @Sayan correct. \$\endgroup\$ – Andy aka Mar 17 at 18:52

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