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Re: gEDA-user: Dropping a few volts



On Friday 12 October 2007, Randall Nortman wrote:
> So I have a 42VDC supply that I want to feed into a linear
> regulator. But almost all linear regulators want 40V absolute
> maximum, and the ones that have higher maximums are not
> reliably stocked anywhere.  So I need to drop that down by at
> least 5V or so.  The good news is that I'm only pulling
> around 70mA on this, so whatever drops it doesn't have to
> dissipate much power.  It occured to me to use a normal shunt
> zener, but that will end up dissipating extra power through
> the zener, which is just wasteful.  It also occured to me to
> put a bunch of diodes in series and use their Vf drops, but
> that would require quite a few diodes.  So why not use a 5V
> zener (reverse biased) in *series* with the load?  It will
> never dissipate extra power, only what is required by the
> load.

The responses to this have been entertaining.

My circuit would be a series resistor, shunt zener and 
capacitor, with values chosen so the zener does not conduct in 
normal use.

Don't rely on any series element alone.  With zero current, the 
voltage across any passive element will be zero.  (until the 
chip you are trying to protect breaks down and draws some 
current.)

The series resistor is chosen for appropriate voltage drop, 
considering the variations in input voltage and load, and the 
characteristics of the device you want to protect.  If 
possible, choose it so the zener does not conduct.  Since you 
didn't give any info, I can't tell you more than this.  By 
making calculations, you will get a range that works.  Try to 
put the input voltage of the regulator in the middle of the 
range that works.  You might want it other than in the middle 
to get more favorable distribution of power dissipation.

The zener is for protection, just in case something goes wrong.  
In case of low load current, it might conduct.

The capacitor is to keep the chip from oscillating.  You may not 
need it, but it is a good idea.


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