Double CMOS inverter time constant

The problem involves finding the time constant for the circuit that arises when the switch is flipped to the U position. What excactly is happening here? I am having trouble visualizing what is going on with the capacitors once the flip happens.

(And what about flipping it to the D position? Will the time constant change?)
Thank you

2 Answers

The 5V source on the right side can be ignored. The capacitor Cp will be charged to a voltage that equals the difference between vc and the 5V source. Therefore its easier to set this voltage to zero.

For this reason Cn and Cp are effectively in parallel. When the switch is moved to the U position the time constant is Rp (Cp+Cn). When the switch is flipped to the D position the time constant is Rn (Cp+Cn).

Including the 5V source we can visualize the behavior as follows. Assuming vc is initially zero, the charge on Cn is zero and Cp is charged to 5V. Now the switch is moved into to U position. Cn gets charged to 5V, a current is flowing into Cn. Cp needs to be discharged to zero volts, a current is flowing into Cp. The capacitors are in parallel, since the current is flowing into both capacitors.

Now----what if the ON time (the pullup time) is much shorter than the R*C time constant: