darkk
Member level 3
I have a question while reading the Razavi's analog design book. If you have the book at hand, you can go to the page 425 (in the Chapter 12 "intro. to switched capacitor ckts). Otherwise, I still like to redraw the schematic below:
---------P---X------------------
vin--~----||----[(-) >----vout
---------GND --[(+)
---------------------------------
"~" a CMOS switch
"||" a capacitor
"[(-) " the inverting input of an amplifier
"[(+)" the noninverting input of the amplifier (connected 2 GND)
">" the output of the amplifier
"P" and "X" stand for the voltage at the left and right plate of the capacitor
The book saids that "If vin is zero, turning off the CMOS switch will injects a charge Q1 to the left plate of the capacitor. If there's no capacitance bewteen X (right plate of the capacitor) and GND, VP and VX jumps to infinity. I think that P and X float as swtich turned off and meanwhile a charge Q1 injects to the left plate of the capacitor. Anyone can give me a clue why P and Q will go to infinity? Thx a lot![/img]
---------P---X------------------
vin--~----||----[(-) >----vout
---------GND --[(+)
---------------------------------
"~" a CMOS switch
"||" a capacitor
"[(-) " the inverting input of an amplifier
"[(+)" the noninverting input of the amplifier (connected 2 GND)
">" the output of the amplifier
"P" and "X" stand for the voltage at the left and right plate of the capacitor
The book saids that "If vin is zero, turning off the CMOS switch will injects a charge Q1 to the left plate of the capacitor. If there's no capacitance bewteen X (right plate of the capacitor) and GND, VP and VX jumps to infinity. I think that P and X float as swtich turned off and meanwhile a charge Q1 injects to the left plate of the capacitor. Anyone can give me a clue why P and Q will go to infinity? Thx a lot![/img]