Based on an answerable but closed question:

If a given shunt regulator does not provide enough power at the desired voltage, how can the output power available be increased?

This relates to the shunt element and series input resistor, but how?

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    \$\begingroup\$ We shouldn't have to translate. \$\endgroup\$ Apr 26, 2012 at 13:52
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    \$\begingroup\$ @OlinLathrop - in an ideal world, yes. But in this imprfec reality, allowing just a small amount of slack and a few iterations seems a reasonable thing to do. If no progress is made in say a few days (about 1/10,000 th of a lifetime) then action can be taken. \$\endgroup\$
    – Russell McMahon
    Apr 26, 2012 at 17:16

1 Answer 1


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.

enter image description here

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.


Given Vin, Vout, Power wanted.
Imax = Power_wanted/Vout Rin <= Vdrop/Imax or Rin <= (Vin_min-Vout) / Imax
or Rin <= *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.


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