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When the switch turns off, this charge is injected to the either direction and part of the charge (ΔQ) is dumped into the capacitor CS. The charge ΔQ results in an error voltage of ΔQ/CS on CS. In a fast switching-off condition when the falling time of the control clock is very fast, the transistor channel disappears quickly. In practice, the charge ΔQ is hardly can be predicted precisely.
budzz said:When the switch turns off, this charge is injected to the either direction and part of the charge (ΔQ) is dumped into the capacitor CS. The charge ΔQ results in an error voltage of ΔQ/CS on CS. In a fast switching-off condition when the falling time of the control clock is very fast, the transistor channel disappears quickly. In practice, the charge ΔQ is hardly can be predicted precisely.
I have also read something like this too.
I have known that, there are charge that must be dumped when clock is turned to '0'.
but, a question comes out that I don't know the fraction of charge that dumped to Drain and that dumped to Source. and how the impedance in Drain and Source will interference.(the impedance can be same/different between Drain and Source)
thanks
leo_o2 said:clock feedthrough and charge injection are different. Clock feedthrough means clock signal change at the gate node lead to a change in drain/source voltage through Cgd and Cgs coupling.
And charge injection means the charge is squeezed into/absorbed
from drain/source during channel forming or disappearing.
Though the distribution of chrge between source anmd drain maynot be equal, but this does reduce the error.
Well... it reduces the error, but we don't know and can't quantified the amount of this reduction..??
Indeed yes you can quantify the amount of charge injected on either terminal of the switch by calculating the dynamic impedance seen on each of these terminals.
Hi Fabien,
I'm not sure if charge injection is caused by clock edges.