How to increase the power output of a shunt regulator?
I had this answer ready to send.
 The system would not accept it.
 I found that the question had been closed.
 So - 



In a shunt regulator Iload flows fron Vin to Vout via a series resistor Rseries.
 A "shunt element" connects from Vout to ground.
 The shunt element dissipates power when the load does not need it.  



Above circuit from here - good shunt regulator explanation.



Answer is straight forward


To get more power out reduce Rseries as required - see below
Shunt element MUST be able to dissipate required maximum power out (for when there is NO load)
Series dropper resistor MUST be able to handle maximum load current
Power in series resistor  = Imax^2 x Rseries


ALL shunt regulator circuits have a designed voltage rating.
 If you exceed the maximum design level the voltage output will drop below design level.

Shunt regulators are in some ways easier to design for max possible power level than most other types.
When there is NO load the shunt element is taking ALL the power output hat is possible.
 When you add a load some of this power is stakn by the load instead.
 When all the power is going to the load and none to the "shunt" the output voltage drops.
As long as the voltage is OK at no load then you know the maximum possible power out - it is te power being taken by the shunt element.
 In practice a very small amount of power will be required at "dropout" for the  shunt but this is typically 1% or less of the total power available.

So:

Given Vin, Vout, Power wanted.
 Imax = Power_wanted/Vout
 Rin &lt;= Vdrop/Imax 
 or Rin &lt;= (Vin_min-Vout) / Imax
 or Rin &lt;= *Vin_min - Vout) x Vout / Power_wanted

Select required Rin.
 If Vout = correct then shunt regulator WILL provide desired power - as long as shunt element does not burn up.



For the original question that this is based on:

You MUST tell us what the regulator is (IC name at least).
 You MUST tell us what the load is.
 A circuit is a very very very good idea.