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if drain terminal of an nmos is open and source terminal is connected to 1v+ and the gate is also biased to a positive voltage, the Id=0 and Vds=0. can anyone explain? especially "Vds=0" I thought when a terminal is open it means we don't know its voltage, how does it say: Vds=0??

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If the gate and source are both at 1V the MOSFET is off, but there is still a bit of leakage that will quickly discharge any charge that might be on the drain. If the gate is at a high enough potential to turn the MOSFET on, of course the discharge will be much faster.

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  • \$\begingroup\$ unfortunately I didn't get your point @Sphehro Pefhany \$\endgroup\$ – Fateme Oct 10 '16 at 6:53
  • \$\begingroup\$ The terminal is not externally connected, but there is a small amount of leakage (nA, probably) inside the MOSFET to the terminals that are connected that will cause the drain terminal to be at the same potential as the source and gate. If the whole MOSFET was floating you would be correct, at least in theory. Imagine a large value resistor, say 10G\$\Omega\$ always internally connected between drain and source. \$\endgroup\$ – Spehro Pefhany Oct 10 '16 at 7:15
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The spice circuit would simulate, but drain current will be in fA range due to noise in the mosfet. And yes, you are right that Vds cannot be 0, because drain is open.

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