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What is the Voltage at Floating node ??

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MammPp

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Dear All,

I have some question about floating node. As the following picture,

If I set V1 to be Vdd all the time.
and V2 is Vdd at the first period, then set V2 is 0 (Which is close NMOS)

My question is, what is the value of the voltage at node C ?

and if V2 is zero, the other transistor, which node C is gate of them, is on/off ?

Thank you so much for your help

MammPp

38_1288233081.jpg
 

Hi
Voltage at node C would be V(V1)+q/C, where
q= charge injected by transistor V2;
C= capacitance at node C;
If transistor is of NMOS type than q would be negative - so resultant voltage would be lower than V(V1).
 
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    MammPp

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Hi MammPq,

I think that: voltage at node C will be a little bit less than VDD due to the charge injection but enough to keep transistor in the right side on (i. e. VGS higher VT)
 
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    MammPp

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Hi all,

Thank you for your answer,

I used HSPICE to simulate this circuit, by consider at V1 and V2 is Vdd.
Let Vdd = 0.9V

The strange result this voltage at node C is equal to 0.13 something, which I concerned that,

From this circuit, if V1 and V2 is Vdd, the other transistor is on/off ?

Moreover, when V2 is off suddenly, voltage at node C is still there or gone ?

and why ???
 

Hi MammPq

1) If V1 and V2 is VDD, so both transistors are ON, and node_C is ~ (VDD - VTH_n). Where VTH_n is the threshold voltage of NMOS.
2) If V1 is VDD and V2 is gnd, so, node_C is a little bit less than (VDD - VTH_n) due to the charge loss. Therefore, transistor in the rigth is still on.

You can check this in my simulations using 0.18 um technology and VDD = 1.8.



I hope the explanation is clear now. If you have other concern, you can post it again.
 
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    MammPp

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The figuring of q, is always tricky. Up until the point
that the channel "goes away", the gate charge will
be returned to the (presumably stiff) V1 with the G-S
portion of charge bleeding back at an RC limited rate.

It is not Q=cox*(VH-VL) in the end, because of this.
A charge-conserving model and one well fitted to
(this wafer & lot's) reality is wanted.
 
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    MammPp

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thank you for your answer.
It's clear for me now
 

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