mmitchell
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Hello,
I read the following from Wikipedia:
According to Wikipedia article:
To describe the ideal operation of the circuit, number the diodes D1, D2 etc. from left to right and the capacitors C1, C2 etc. When the clock Φ1 is low, D1 will charge C1 to Vin. When Φ1 goes high the top plate of C1 is pushed up to 2Vin. D1 is then turned off and D2 turned on and C2 begins to charge to 2Vin. On the next clock cycle Φ1 again goes low and now Φ2 goes high pushing the top plate of C2 to 3Vin. D2 switches off and D3 switches on, charging C3 to 3Vin and so on with charge passing up the chain, hence the name charge pump. The final diode-capacitor cell in the cascade is connected to ground rather than a clock phase and hence is not a multiplier; it is a peak detector which merely provides smoothing.[2]
A critical question is that when C1, with Q=C1Vin, is raised to 2Vin by Φ1, when charging C2 it will lose charge and the voltage will drop. The same problem extends to all subsequent chaining steps.
After carefully thinking I still feel it a legit question and don’t see why it is left unexplained. I actually suspect the charging process takes multiple iterations to reach equilibrium: in each iteration the charging capacitor (like C1) gives some charge to the next one (C2), and regains the lost capacity in the next iteration. Several iteration are needed to final charge output (Vo) to the desired value, but NOT in one iteration.
Could anyone explain this problem?
Matt
I read the following from Wikipedia:
According to Wikipedia article:
To describe the ideal operation of the circuit, number the diodes D1, D2 etc. from left to right and the capacitors C1, C2 etc. When the clock Φ1 is low, D1 will charge C1 to Vin. When Φ1 goes high the top plate of C1 is pushed up to 2Vin. D1 is then turned off and D2 turned on and C2 begins to charge to 2Vin. On the next clock cycle Φ1 again goes low and now Φ2 goes high pushing the top plate of C2 to 3Vin. D2 switches off and D3 switches on, charging C3 to 3Vin and so on with charge passing up the chain, hence the name charge pump. The final diode-capacitor cell in the cascade is connected to ground rather than a clock phase and hence is not a multiplier; it is a peak detector which merely provides smoothing.[2]
A critical question is that when C1, with Q=C1Vin, is raised to 2Vin by Φ1, when charging C2 it will lose charge and the voltage will drop. The same problem extends to all subsequent chaining steps.
After carefully thinking I still feel it a legit question and don’t see why it is left unexplained. I actually suspect the charging process takes multiple iterations to reach equilibrium: in each iteration the charging capacitor (like C1) gives some charge to the next one (C2), and regains the lost capacity in the next iteration. Several iteration are needed to final charge output (Vo) to the desired value, but NOT in one iteration.
Could anyone explain this problem?
Matt