Application of charge conservation
When in doubt go back to first principles.



The circuit above shows two capacitors, C1 and C2 connected in a basic voltage doubler circuit (charge pump)

C1 is connected to a changeover switch (equivalent to the MOSFETS in your more complex circuit). With the switch in the position shown C1 will be charged through R1 from a voltage source V. (R1 represents a simplified representation of the total series resistance of the charging circuit)

The 'normal' RC (exponential) charging of C1 takes place through R1 from the voltage source, V. Let's not worry too much about the precise maths at this stage - you can fill that in later. We know that after about 5 time constants (t=CR) C1 will be just about fully charged. So let's take it we've operated the switch for a long enough period and the voltage across C1 is equal to V. 

To start with the voltage across C2 is 0 (completely discharged). When the switch is operated C1 and C2 become connected.

Basic principle 1 - you cannot instantly change the voltage across a capacitor.



This means that C1 will have voltage V across it and this will be ADDED to the supply voltage (V) giving us a voltage of 2*V at the node between C1 and R2. This voltage (2*V) will appear at each switch cycle after C1 has be re-charged. 

C2 initially has no voltage across it and so will begin to charge up. In doing so its voltage will increase and at the same time the voltage across C1 will decrease as the total charge is re-distributed.

Basic principle 2 - charge is conserved.

The total charge on C1 is Q = C1 * V  (Q1 in Coulombs, C in Farads)

Let the final voltage across C2 be Vx after the first 'dump' of charge.

Then     Vx * C2 (charge on C2) + (V - Vx) * C1 (charge on C1) = C1 * V = Q (initial charge)

On the next switch cycle C1 is charged again (C1 * V) but C2 already is partially charged to Vx (a voltage lower than 2 *V) so the total charge is Q + (C2 * Vx). This charge is then shared and raises the voltage across C2. After several cycles the voltage across C2 will be close to 2*V. (pumped up). Once C2 is at 2*V there is no voltage difference across R2 so no more charging current will flow.

(In reality current taken from C2 at the output will reduce this voltage so the charge pump 'tops up' C2 with each cycle)

What about R2? This will have the effect of limiting the initial charging current and of course dissipating some energy. (This is a source of inefficiency in the circuit and one of the reasons why (low resistance) MOSFETS are used as series switches.)  It doesn't, however, have an effect on the amount of charge transferred - just the time it takes for the circuit to settle.