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'[EE:] Shunt regulator? 317/337 solution'
2001\01\25@023728 by Vasile Surducan

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lm317 solution: input current to Vi pin, Vout pin connected to A
pin of a resistor, ADJ (adjustment) pin connected to B resistor pin, this
is also output current.

Iout = Vref/R + Iadj where Vref = 1.2V, Iadj = 50uA

Vasile

On Wed, 24 Jan 2001, Roman Black wrote:

{Quote hidden}

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2001\01\25@060520 by Roman Black

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Vasile Surducan wrote:
>
> lm317 solution: input current to Vi pin, Vout pin connected to A
> pin of a resistor, ADJ (adjustment) pin connected to B resistor pin, this
> is also output current.
>
> Iout = Vref/R + Iadj where Vref = 1.2V, Iadj = 50uA
>
> Vasile


Hi Vasile, that is a series regulator? I think the
person wanted a shunt regulator, and I don't think you
can do it with a 317 as it uses negative feedback,
not positive. :o)
-Roman

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2001\01\25@071708 by Vasile Surducan

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On Thu, 25 Jan 2001, Roman Black wrote:

> Vasile Surducan wrote:
> >
> > lm317 solution: input current to Vi pin, Vout pin connected to A
> > pin of a resistor, ADJ (adjustment) pin connected to B resistor pin, this
> > is also output current.
> >
> > Iout = Vref/R + Iadj where Vref = 1.2V, Iadj = 50uA
> >
> > Vasile
>
>
> Hi Vasile, that is a series regulator? I think the
> person wanted a shunt regulator, and I don't think you
> can do it with a 317 as it uses negative feedback,
> not positive. :o)
> -Roman
>
 No, it's not. It's a current controlled source and can be used as shunt
on positive supply. On the negative supply just the LM337.
If on the negative or positive supply  in series with this device ( 317 or 337)
is placed a power resistor, the whole sistem becomes a shunt regulator.
 BTW, ANY and ALL linear stabiliser circuits use ONLY negative feedback
ALSO becomes oscillators !
 Cheers, Vasile

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2001\01\25@073437 by Michael Rigby-Jones

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{Quote hidden}

It's actually a constant current source in that configuration.  Been
thinking about this and I can't see a simple solution.  I reckon you could
make it work with the addition of an opamp (to invert the sense of the
feedback), but then you might as well stick with a 723 etc.

Mike

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2001\01\25@102116 by Russell McMahon

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> > Hi Vasile, that is a series regulator? I think the
> > person wanted a shunt regulator, and I don't think you
> > can do it with a 317 as it uses negative feedback,
> > not positive. :o)
> > -Roman
> >
> It's actually a constant current source in that configuration.  Been
> thinking about this and I can't see a simple solution.  I reckon you could
> make it work with the addition of an opamp (to invert the sense of the
> feedback), but then you might as well stick with a 723 etc.


If you MUST make a shunt regukator with an LM317 try this.
The 317 acts as the controller and the transistor as the dissipative shunt
element.
I haven't tried this circuit (but it will work :-) )

R1 feeds supply voltage to Vout line.
NPN TO220 (or whatever) transistor
   Collector to Vout
   Emitter to ground
   Base via R1 to LM317 input.

LM317
   Input to Vout via R3
   Input to transistor base via R2 as above
   Gnd to gnd
   Output to gnd via R4
   Output to supply via R5

R4/R3 form divide.
LM317 attempts to keep V across R4 at its standard value (1.25V I think it
is)

If Vout is too high LM317 shuts down and draws little.current
Transistor is driven on via R2/R3 and shunts Vout to lower it.

If Vout is too low LM317 turns on.
This draws current through R3 thereby lowereing R2/R3 centre point and
turning off transistor which deshunts supply as required.

The advantage of this over a zener is that it would have a MUCH sharper
control knee.
A zener is a truly pathetic regulator by almost any standard.

R2/R3 would need to be fiddled slightly to have transistor biased about
right with the current drawn by R4/R5 but this should not be hard to
achieve.

I would try and use the LM317 as a series pass regulator as it was designed
to be unless there was some really good reason to use a shunt regulator.


regards

       Russell McMahon

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2001\01\25@123554 by Harold M Hallikainen

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       This is a current regulator, right? I'm looking for an "adjustable power
zener", a shunt regulator.

Thanks though!

Harold


On Thu, 25 Jan 2001 09:13:50 +0200 Vasile Surducan
<KILLspamvasileKILLspamspamL30.ITIM-CJ.RO> writes:
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2001\01\26@021804 by Vasile Surducan

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This is an ajustable power zenner like you ask for. R may be a
potentiometer but must dissipate an important power if you need large
current as 0.5A. It's not a tipical shunt voltage regulator.
If you need to keep at precise value a potential and to use it is not too
good. But if you want only to minimise dissipation power on to a load
( like a serial power supply transistor or something else ) is perfect.
Vasile

On Thu, 25 Jan 2001, Harold M Hallikainen wrote:

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2001\01\26@050828 by Roman Black

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Michael Rigby-Jones wrote:
{Quote hidden}

As I said, a shunt regulator needs positive feedback,
ie; if voltage is too high, must turn on more.
The actual inversion is provided by the nature
of it being a shunt reg.

I really don't think you can do it with a 317,
but I already posted a one transistor solution that
I have used for years that is cheaper and just
as simple. You are not going to get short circuit
protection from a shunt reg anyway so the extra
cost of a regulator chip is not justified. :o)

-Roman

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2001\01\26@052103 by Alan B. Pearce

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>You are not going to get short circuit
>protection from a shunt reg anyway so the extra
>cost of a regulator chip is not justified. :o)

I always understood that shunt regulators were inherently short circuit
protected by the series impedance they need to do the regulation. If you S/C the
output the current flow is limited by the impedance, and if open circuit the
shunt regulator needs to be able to sink the total design load current plus what
it needs to maintain regulation.

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2001\01\26@062838 by Michael Rigby-Jones

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{Quote hidden}

Agreed, even the safe operating area protection that most 3 terminal
regulators empliy is wasted, if the device overheats it will stop conducting
and the voltage to your device will soar, probably toasting it.

Mike

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2001\01\26@073535 by Vasile Surducan

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On Fri, 26 Jan 2001, Michael Rigby-Jones wrote:

> > -----Original Message-----
> > From: Roman Black [SMTP:KILLspamfastvidspamBeGonespamEZY.NET.AU]
> > Sent: Friday, January 26, 2001 10:04 AM
> > To:   EraseMEPICLISTspamEraseMEMITVMA.MIT.EDU
> > Subject:      Re: [EE:] Shunt regulator? 317/337 solution
> >
> >
> > As I said, a shunt regulator needs positive feedback,
> > ie; if voltage is too high, must turn on more.

   I'm agree with this only if "regulator" means for you the final
transistor of a shunt regulator. But the effect of regulation comes
because of the global negative feedback. If you have a NPN driven in base
from an operational amplifier you must connect +in to feedback and -in to
reference to achieve a shunt regulator. Colector of the npn conected to
load and balast resistor to V+. But this is not a positive
feedback. Increasing output voltage level will determine the amplifier to
turn on more the transistor and the final effect is the output voltage
will decrease at initial value ( assuming you have a good reference )
In any language we are talking about this is negative feedback !

> > The actual inversion is provided by the nature
> > of it being a shunt reg.
> >
> > I really don't think you can do it with a 317,

   Well, the ideea was to do it with a 317. Of course is neither the
cheapest nor the best methode. But can be done in some circumstances.
Don't shut the piano player just because you don't like how it sing !
And if you don't think make first a try.
Cheers, Vasile

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2001\01\26@082814 by Roman Black

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Alan B. Pearce wrote:
>
> >You are not going to get short circuit
> >protection from a shunt reg anyway so the extra
> >cost of a regulator chip is not justified. :o)
>
> I always understood that shunt regulators were inherently short circuit
> protected by the series impedance they need to do the regulation. If you S/C the
> output the current flow is limited by the impedance, and if open circuit the
> shunt regulator needs to be able to sink the total design load current plus what
> it needs to maintain regulation.

Hi Alan, you may be thinking of current regulation?
A shunt regulator fixes the output voltage by
drawing off excess current that the load does not
require, with no series resistor etc needed.
If the load becomes faulty, max current is limited
only by the impedance of the psu, which probably
has big caps!! Normally the regulator ICs like
317 etc are nice because of the current limiting,
but using one as a shunt reg will not utilise this
benefit, so you may as well use a transistor. :o)
-Roman

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2001\01\26@085951 by Roman Black

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Vasile Surducan wrote:
>
> On Fri, 26 Jan 2001, Michael Rigby-Jones wrote:
>
> > > {Original Message removed}

2001\01\26@092031 by Alan B. Pearce

face picon face
>Hi Alan, you may be thinking of current regulation?
>A shunt regulator fixes the output voltage by
>drawing off excess current that the load does not
>require, with no series resistor etc needed.
>If the load becomes faulty, max current is limited
>only by the impedance of the psu, which probably
>has big caps!! Normally the regulator ICs like
>317 etc are nice because of the current limiting,
>but using one as a shunt reg will not utilise this
>benefit, so you may as well use a transistor. :o)
>-Roman

sheesh - can I have your contract for heat sinks? I would not want to do this. I
always understood you used a shunt regulator zener diode style with a current
limiting source, at minimum a resistor which can supply the full current that is
needed in the load plus a tiny bit for the regulator. Then if part of the load
is switched off the shunt regulator draws more current. I have never heard of
people having current limiting within the shunt regulator before.

If I was doing the sort of thing you describe above I would be looking at having
some form of OVP to drop the input to the supply and shut it down.

the only other situation I could see where one might do it is if a minimum
current draw is required to keep the voltage from going too high, like what
might happen with a wall wart supply, where the transformer is being pushed to
its limits and has no low current regulation. But I would not class such a
device as a real shunt regulator because I would not be designing it to have
high stability - again it would be a OVP current sink, but not to the full
current capability of the supply, just enough to maintain the peak voltage
within limits.

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2001\01\26@101108 by Roman Black

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>
> sheesh - can I have your contract for heat sinks? I would not want to do this. I
> always understood you used a shunt regulator zener diode style with a current
> limiting source, at minimum a resistor which can supply the full current that is
> needed in the load plus a tiny bit for the regulator. Then if part of the load
> is switched off the shunt regulator draws more current. I have never heard of
> people having current limiting within the shunt regulator before.
>
> If I was doing the sort of thing you describe above I would be looking at having
> some form of OVP to drop the input to the supply and shut it down.
>
> the only other situation I could see where one might do it is if a minimum
> current draw is required to keep the voltage from going too high, like what
> might happen with a wall wart supply, where the transformer is being pushed to
> its limits and has no low current regulation. But I would not class such a
> device as a real shunt regulator because I would not be designing it to have
> high stability - again it would be a OVP current sink, but not to the full
> current capability of the supply, just enough to maintain the peak voltage
> within limits.

Hi Alan, no, with high current supplies the main
regulation concern is the transformer, and to keep
efficiency high it is often better to use a linear
shunt reg than a linear series reg. Especially for
very high currents with a definite high-low range,
that is not too far apart. Like a load that always
draws between 15 and 20 amps, but needs a regulated
voltage. :o)
-Roman

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