voltage bootstrapping question

[020170]

this is voltage bootstrapping circuit.

**broken link removed**

I understand why I_Cboot current flow through Cboot capacitor.

because Vx is fixed voltage, but Vout voltage is not fixed, when Vin voltage switches from its logic-high level to 0 v.

But voltage cross the Cs capacitor is fixed, not changed.

nevertheless, why Ics = Icboot ?

in my opinion, voltage cross the Cs is fixed, therefore Ics = 0.

however it is wrong. why?

 

[Btrend]

I understand why I_Cboot current flow through Cboot capacitor.

because Vx is fixed voltage, but Vout voltage is not fixed, when Vin voltage switches from its logic-high level to 0 v.

But voltage cross the Cs capacitor is fixed, not changed.

nevertheless, why Ics = Icboot ?

in my opinion, voltage cross the Cs is fixed, therefore Ics = 0.

however it is wrong. why?

the voltage at Vx is not fixed.

1. Vin=high, Vout=0, Vx=Vdd-Vt3,
the charge storing in Cs and Cboot is Q1=(Vdd-Vt3)*(Cs+Cboot)
2. Vin=0, Q2=V*Cs+(Vx-Vout)*Cboot
3. cause charge conservation, so Q1=Q2
===> Vx=(Vdd-Vt3)+Vdd*Cboot/(Cs+Cboot)
===> if Cboot >> Cs
===> Vx=2Vdd-Vt3 ................. this is the name of bootstrapping.

4. Ics=ΔQcs/Δt = Cs*ΔVcs/Δt=Cs*[(Vx2-0)-(Vx1-0)]/Δt ≈Cs*Vdd/Δt
5.Icboot=ΔQboot/Δt =Cboot*ΔVboot/Δt=Cboot*[(Vx2-Vx1)-(Vx1-0)]/Δt =Cboot*Vt3/Δt
6. Ics≠Icboot ... different from ur result !??
am I wrong ?

 

[020170]

Mm...I think that your method is right. don't worry.

in my opinion, Ics = Icboot means just approximation.

thanks.