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There are many factors that affect efficiency and Load Regulation in the Dickson Voltage Doubler, inverting or not.

Impedance ratios at fundamental frequency of the the square wave and highest harmonic are the primary factors.  

This includes impedance of;


supply, (battery ESR)
diodes in conducting (ESR) and reverse mode (Cjcn)
Zout of the Emitter Follower which is Rb/hFE. or RdsOn of CMOS
switching frequency of driver
clock driver Zout ( 25~50 typ. for 74ALC@3.3V to 74HC at 5V)


The criteria of concern here was % Load Regulation. 


i.e. excessive % voltage drop with load R.
10% max Load error is consider max.while 2% is good.




So how to choose the optimal parameters.

What not to do


Drive the Q's too hard results in I²Rce losses
Drive the base with too low a Rb and excessive current spikes &amp; EMI.
Use Caps with excessive ESR and too large ESR*C value compared to 1/f . 


Caps should be ESR*C &lt;&lt;1us  
load R * C should be 10* 1/f for low ripple.
thus ESR or Zout of Dickson Doubler should be &lt; 10% of R load



So what is the Zout of the Doubler?


first if you know the load R or current then R you want a Zout = &lt;10% or load R


many factors but a simple approximation is the impedance of the two output shunt Caps \$Zc= 2* { \frac{1}{2\pi fC}} \$
so if Zc &lt;=10% of Load R then 
\$Zc =\frac{1}{\pi*f*C} \ &lt;=10R_L\$
the best Dickson IC CMOS chips can supply 150mA typ. using 0.1uF
1st make Rb for bipolars 3% to 30% of Load R affects Zout and load regulation and efficiency. 


  
  I suggest Rb=10% of load using hFe of 100 min.
  

then \$f = \frac{10\pi \ R_L }{C}\$
e.g. 10V  with no load expect 1V drop to -9V
thus 10% drop with max load current drop Vout to -8.1V out
Load = 600 Ω, f = 100 kHz , C = 1 uF  ESR &lt;70 Ohms or ...
Load = 600 Ω, f = 1.0 Mhz , C = 0.1uF ESR &lt;70 Ohms




  Due to all the losses dont' expect more than 10% efficiency and compare with commercial IC's. These are intended for low current because of charge transfer losses.
  
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