Let us take the example of the MOSFET with open drain.
There are 2 cases:
- When VGS << VTH, the drain terminal is indeed a high impedance floating node
- When VGS >> VTH,, the drain terminal is not a floating node. In this case, there is a low impedance between the drain and source.
Assume initially VGS << VTH and the drain is sitting at 0V according to your simulation without pullup. Basically, you can visualize this situation as a capacitance across drain-bulk junction as shown which is holding a charge of 5V (drain voltage = 0V and bulk/source voltage = -5V). Subsequently, when VGS >> VTH the drain-source becomes a low impedance and hence discharges the cap fully. Thus, the drain voltage becomes equal to the source voltage i.e. -5V.
Essentially, the pullup resistor is only needed to pull the drain high. To pull the drain low, we just need VGS >> VTH to be satisfied. If there is no pullup in this case, the drain capacitance will get discharged and eventually drain voltage = source voltage
When Vgs>Vth, channel inversion happens and hence there is a path available for the current. If there is a path available for current, it is low impedance. Drain to source is hence low impedance but drain to supply is high impedance because it has no path to supply.