The discussion centers on the misunderstanding of drain voltage in a MOSFET circuit, specifically questioning why the drain voltage is not equal to 1.8V. It is established that the drain voltage would only equal 1.8V if the MOSFET is in a cut-off state, meaning it is completely non-conducting. The conversation emphasizes the importance of considering the gate terminal's influence on the MOSFET's operation and the necessity of using provided parameters to determine the FET's operating mode and DC operating point.
PREREQUISITESElectrical engineering students, circuit designers, and anyone involved in analyzing or designing MOSFET-based circuits will benefit from this discussion.
It would be 1.8 V if the FET were cut off (completely non-conducting). Do you have reason to believe that this is the case?CoolDude420 said:
no. Ah, I keep considering the branches of the terminals of the MOSFET independent due to the gap in the circuit diagram of it. So basically because of the gate terminal I can't just assume its 1.8Vgneill said:
Schematic diagram symbols often offer a stylized hint as to the underlying construction of the components they represent, but really they can't do more than jog your memory about how they actually function.CoolDude420 said:
You're contradicting yourself here.CoolDude420 said: