Charge pumping system

Discussion in 'Homework Help' started by adashiu, May 1, 2011.

  1. adashiu

    Thread Starter New Member

    Nov 27, 2008
    Hi Guys !
    Again I need your help :p

    I have a circuit like this :

    I need to discover a principle of this circuit.
    I need to know if I think well.

    So... When the switches S1 and S3 are closed, C1 charges to the Vcc and then if S2 and S4 are closed (S1,S3 opened) C1 discharges and C2 charges to Vcc. With adjusting C1 and C2 we can obtain an amplified Vcc. (we omit all losses). Am I right ?
    How to obtain an expression of Vs in terms of Ve, C1, C2 ?

    Thanks in advance.
    Greets. Adam
  2. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    When S1 & S3 close then C1 will charge to Vcc.

    With S2 & S4 closed and S1 & S3 open then C1 is switched in parallel with C2.

    To analyze proceed as follows.....

    1. Calculate the charge supplied to C1 with S1 & S3 closed - call it Q1.
    2. Calculate the equivalent parallel capacitance of C1 & C2 with S2 & S4 closed - call it Ceq.
    3. The original charge supplied Q1 will be stored in the parallel capacitance combination Ceq.
    4. Vs=Q1/Ceq
    Last edited: May 1, 2011
    adashiu likes this.
  3. adashiu

    Thread Starter New Member

    Nov 27, 2008
    Thank you !
    Ive got something like :

    Vs = (C1*Vcc)/(C1+C2)

    So what is the purpouse of this circuit ?
  4. Wendy


    Mar 24, 2008
    How about to create more voltage than the battery is providing?

    MOSFETs require 10V between the gate and source, but if the MOSFET is connected to the high side of the power supply where will you get the voltage to turn it on? This is the answer.

    I have a simple LED flasher circuit in my experiments section where the LED requires more voltage than the power supply. Again, something like this was used.

    CMOS 555 Long Duration Blue LED Flasher

    There is a Theory of Operation at the end of the article that explains how this works.