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I am unable to understand how the voltage at Node x in the below circuit is 5V/3 once A rises and M1 NMOS turns off.

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So A is initially 0 and once A rises to 5 V, the NMOS turns off.

Since NODE X is floating, I think the node X should be 5V, however simulation shows it's 1.66.

Could anyone help me figure this out?

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  • \$\begingroup\$ C{1,2,3} form a capacitive divider. For node X to return to 5V you also have to charge/discharge C1&2. \$\endgroup\$
    – sstobbe
    Commented Sep 24, 2017 at 0:54

1 Answer 1

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Since you're doing a DC operating point analysis from the perspective of the DC signal, the pulse source is a ground since alternating sources act as short circuits. If you add C1 and C2 together, you can simplify the circuit to just a 5V DC source connected across two capacitors: C12 = 2pF and C3 = 1pF.

For capacitors in series, each capacitor has the same charge on it. Therefore Q12 = Q3 = Q. If we further reduce the series capacitors C12 and C3 to

Ct = = (1/C12 + 1/C3)^-1 = (1/2pF + 1/1pF)^-1 = 2/3pF,

we know this simplified capacitor has a voltage of 5V and a charge of 2Q, since charges in series add. So, since CV=Q:

2/3pF * 5V = 2Q -> Q = 1.6667 pC

Now, if we go back to the C12 and C3 combination, Q is the charge across C3. And again from CV=Q, we have:

C3 * V = Q -> V = 1.6667pC/1pF = 1.6667 V

or V = 5/3V

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  • \$\begingroup\$ When node x = 1.67V the voltage across C3 is 3.33V, the charges are equal not additive. V(x) = 5 - 3.33pC/1pF I believe \$\endgroup\$
    – sstobbe
    Commented Sep 24, 2017 at 2:20
  • \$\begingroup\$ Since the charge across C12 and C3 are equal, the charge across both of them is the sum of them. Q12=Q3=Q, which is why the charge across both C12 and C3 is 2Q. And are you talking about the DC voltage? If the DC voltage at X is 1.67V, then the voltage across C3 would also be 1.67V since the pulse source acts as a short circuit, i.e. ground. \$\endgroup\$
    – royalt213
    Commented Sep 24, 2017 at 2:24
  • \$\begingroup\$ If the pulse acts as short in DC op, then the voltage at node x should be the voltage across C3, which should be twice of voltage across C1 and C2, so 3.33 V. However if I run a transient then voltage at node X is 1.67 V. \$\endgroup\$
    – Ash
    Commented Sep 24, 2017 at 5:04
  • \$\begingroup\$ How did you get that? I've looked at my analysis a few times and can't see anything wrong with it. Where are you getting 3.33V and why should the voltage across C3 be twice that of C1 and C2? \$\endgroup\$
    – royalt213
    Commented Sep 24, 2017 at 6:03
  • \$\begingroup\$ "We know this simplified capacitor has a voltage of 5V and a charge of 2Q, since charges in series add" The equivalent cap will have charge Q on it not 2Q. Think of charge as current flowing through circuit, the equivalent Cap will also have charge Q. \$\endgroup\$
    – Ash
    Commented Sep 25, 2017 at 19:31

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