Working principle of a mosfet

Thread Starter

jsnx

Joined Mar 17, 2022
21
Hello, I have a dumb question about the working principle of a normal mosfet.

When it is used as a switch, when we apply a gate voltage higher than the threshold it will output the voltage present at the source. It happens because it is working on its linear region.

When we use a mosfet in a buffer, we’ll sample the gate at the output because it is working in saturation.

is this reasoning correct?

thank you in advance.
 

ci139

Joined Jul 11, 2016
1,864
the 1-st one somewhat applies

the second one is a bit rushed - as the (presumably) non-inverting , non-digital buffer (in simplified interpretation) still takes (an easy to achieve) a threshold offset to the output voltage . . . while the word "saturation" would apply to the digital buffer only - but then (at the least case) it won't be a voltage follower - but more likely a double inverter (with several mosfets and amplifier stages)

the "normal MOS-FET" may be misleading - especially with the today´s fields of application and the variety available
https://www.google.com/search?q=mosfet+types+mode+of+operation
 

Ramussons

Joined May 3, 2013
1,279
An easier way to understand the working of a mosfet is to consider the Drain-Source as a "Voltage Controlled Resistor". The controlling voltage is the Gate-Source voltage. Depending on whether the mosfet is an Enhancement or Depletion type, we Reduce or Increase the D-S resistance by varying the G-S voltage.
 

crutschow

Joined Mar 14, 2008
30,151
When we use a MOSFET as a switch in the linear (ohmic) mode (left of the red line below), then a gate-source voltage well above the gate-threshold voltage is applied (typically several volts above) so that the MOSFET is fully on, and the source-drain voltage is then determined by the current flowing from drain-to-source times the MOSFET on-resistance.

When the MOSFET is used as amplifier or buffer, then it is operating in the saturation (active) region to the right of the red line below, where the gate-source voltage is just a little above Vgs(th).
In that mode the output is not ohmic but essentially a high-resistance constant-current.

1648045343716.png
 
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