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'100vdc to 5vdc conversion'
1999\02\19@040636 by Rod Phillips

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<DIV><FONT color=#000000 size=2>Hi to all,</FONT></DIV>
<DIV><FONT color=#000000 size=2></FONT>&nbsp;</DIV>
<DIV><FONT color=#000000 size=2>Could I pick some brains about the best method
to drive a Pic based control module from a very noisy 100vdc supply?&nbsp; Power
from the source is not a concern so a voltage divider circuit with its output
feeding a voltage regulator came to mind.&nbsp; I know a switching supply is
much more efficient, but aren't they very intolerant of dirty inputs.&nbsp; The
current requirements would be &lt; 500ma.&nbsp;&nbsp; Thanks in advance for any
ideas!</FONT></DIV>
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size=2>&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;
Rod</FONT></DIV>
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1999\02\19@130038 by Gerhard Fiedler

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At 03:06 02/19/99 -0800, Rod Phillips wrote:
>
> Could I pick some brains about the best method to drive a Pic based control
> module from a very noisy 100vdc supply?  Power from the source is not a
> concern so a voltage divider circuit with its output feeding a voltage
> regulator came to mind.  I know a switching supply is much more efficient,
> but aren't they very intolerant of dirty inputs.  The current requirements
> would be < 500ma.   Thanks in advance for any ideas!


what do you mean by "dirty"? with 500mA, i'd go with a switcher... otherwise,
you have 48W to dissipate.

ge

1999\02\19@130910 by D. Schouten

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>Could I pick some brains about the best method to drive a Pic based
control >module from a very noisy 100vdc supply?  Power from the
source is not a >concern so a voltage divider circuit with its output
feeding a voltage >regulator came to mind.  I know a switching supply
is much more efficient, >but aren't they very intolerant of dirty
inputs.  The current requirements >would be < 500ma.   Thanks in
advance for any ideas!

A step-down smps shouldn't be such a problem when powered from a dirty
source, but this depends ofcourse on how dirty and noisy it really is.
Just some minor extra filtering at the smps input should be sufficient
in most cases.
There's also the question whether you need input to output isolation
or not. Isolated or non isolated step down, there are lots of
possibilities nowadays. You can check out the Power Integrations
website at http://www.powerint.com or ST at http://www.st.com for example.

Using a linear voltage regulator approach for stepping down from
100VDC to 5VDC at currents up to 500mA isn't the way to go IMHO.

Daniel...

1999\02\19@234107 by dave vanhorn

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At 03:06 AM 2/19/99 -0800, Rod Phillips wrote:
>
> Hi to all,
>
> Could I pick some brains about the best method to drive a Pic based control
> module from a very noisy 100vdc supply?  Power from the source is not a
> concern so a voltage divider circuit with its output feeding a voltage
> regulator came to mind.  I know a switching supply is much more efficient,
> but aren't they very intolerant of dirty inputs.  The current requirements
> would be < 500ma.   Thanks in advance for any ideas!
>                                                                 Rod
>


Dirty supplies aren't a problem for switchers, at least not anything I ever
worked with.
Any decent buck design would work, although if a switcher failure applying full
input scares you, then you might look at a flyback.

1999\02\20@132445 by Mike Keitz

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On Fri, 19 Feb 1999 23:39:06 -0500 dave vanhorn <spam_OUTdvanhornTakeThisOuTspamCEDAR.NET>
writes:

>Dirty supplies aren't a problem for switchers, at least not anything I
>ever
>worked with.
>Any decent buck design would work, although if a switcher failure
>applying full
>input scares you, then you might look at a flyback.

Consider dropping the voltage to 30V or so with a resistor and use that
as the input to a switching buck regulator (30V -> 5V).  Have a fairly
large capacitor on the 30V node.  The resistor/capacitor will absorb
spikes, surges etc. and deliver decent power for the switcher to use.
Also there will be a lot more choices of how to build the switcher since
the input voltage is lower.  The resistor will only dissipate about 7 W
because a switching regulator only needs about 100 mA at 30V to deliver
500 mA at 5V.  You need a shunt regulator to keep the 30V node voltage
from rising when the load is light.  A 5W Zener diode could work, though
it may be cheaper to use a low-power Zener and a transistor.

Beware though that as the "30V" line drops, the switcher will demand more
current, which the resistor won't be able to supply.  The result is the
circuit may latch off.  One solution may be to inhibit the switching
regulator unless the voltage is more than 25V or so.  That doesn't give
as much immunity to dropouts on the 100V bus.  Another way would be to
have a linear regulator going from 100V to 30V.  This regulator will
dissipate a lot less power (7 W vs. 48 W) than one going from 100V to 5V.

Anything you can do to use less than 500 mA at 5V will help a lot.  If
the current is more like 50 mA, simply dropping from 100V to 5V linearly
through resistors and regulators is quite practical.

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1999\02\20@144508 by dave vanhorn

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At 01:21 PM 2/20/99 -0500, Mike Keitz wrote:
>On Fri, 19 Feb 1999 23:39:06 -0500 dave vanhorn <.....dvanhornKILLspamspam@spam@CEDAR.NET>
>writes:
>
>>Dirty supplies aren't a problem for switchers, at least not anything I
>>ever
>>worked with.
>>Any decent buck design would work, although if a switcher failure
>>applying full
>>input scares you, then you might look at a flyback.
>
>
>Beware though that as the "30V" line drops, the switcher will demand more
>current, which the resistor won't be able to supply.  The result is the
>circuit may latch off.  One solution may be to inhibit the switching
>regulator unless the voltage is more than 25V or so.  That doesn't give
>as much immunity to dropouts on the 100V bus.  Another way would be to
>have a linear regulator going from 100V to 30V.  This regulator will
>dissipate a lot less power (7 W vs. 48 W) than one going from 100V to 5V.

I have a hard time imagining what level of noise he's got that would cause
a problem with a switcher that's already designed to do this.. A diode and
capacitor on the input would handle reverse voltage, and dropouts to zero
of limited duration, and instead of blowing the energy as heat, it's saved
as available energy in the input cap at E=0.5*C*V^2  (note the V^2 term!)

1999\02\20@151719 by Mike Keitz

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On Sat, 20 Feb 1999 14:42:20 -0500 dave vanhorn <dvanhornspamKILLspamCEDAR.NET>
writes:

>I have a hard time imagining what level of noise he's got that would
>cause
>a problem with a switcher that's already designed to do this..

I think it may be useful to first drop to 30V or so because few (are
there any?) of the all-in-one switching ICs can handle more than 40 to 60
V.  A regulator suitable for 100 V input would be more complicated
involving a discrete transistor output and possibly a custom transformer
instead of just an inductor.

>A diode
>and
>capacitor on the input would handle reverse voltage, and dropouts to
>zero
>of limited duration,

Spikes in the positive direction will send a lot of current through the
diode, possibly burning it out.  That's easily dealt with by adding a
relatively small resistor in series with the diode.  However if the
positive spikes and surges are extensive enough to start charging up the
capacitor, it or the regulator may fail.

and instead of blowing the energy as heat, it's
>saved
>as available energy in the input cap at E=0.5*C*V^2  (note the V^2
>term!)

The term is noted.  But, if you look at capacitors having the same energy
storage capacity (0.5 * (voltage rating ^ 2) * capacitance), you'll find
that they are about the same physical size regardless of the voltage
rating.  Going to a higher voltage in the same size case means less
capacitance.  So it doesn't seem that you're gaining much running the
capacitor at a higher voltage.

Of course you may have less heat, but 7W isn't hard to deal with in a big
industrial device.

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1999\02\20@154847 by dave vanhorn

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>I think it may be useful to first drop to 30V or so because few (are
>there any?) of the all-in-one switching ICs can handle more than 40 to 60
>V.  A regulator suitable for 100 V input would be more complicated
>involving a discrete transistor output and possibly a custom transformer
>instead of just an inductor.

I wasn't thinking in terms of "simple switcher" chips.. You'd need an
external fet, but so what?
CS384x would work well here, they're cheap. 200V mosfets are available, and
cheap.
If he wants a flyback, he's almost certainly into custom magnetics anyway,
and that's much less of a deal than it looks. At 300kHz, the inductor could
be done on an RM-6 core easily.
If you have to do it with a simple inductor, then you'll probably want some
output clamping to prevent a catastrophic failure, and that will be as much
bulk and $ as the magnetics for the flyback.  If it's a onesie, sample a
core, bobbin, and clips, and wind your own. Takes about 30 mins. If it's
production, then talk to the chineese, they make inductors rather
inexpensively.

>Spikes in the positive direction will send a lot of current through the
>diode, possibly burning it out.  That's easily dealt with by adding a
>relatively small resistor in series with the diode.  However if the
>positive spikes and surges are extensive enough to start charging up the
>capacitor, it or the regulator may fail.

We really need to put some numbers on this. A design that takes a 110V line
with an electric drill plugged in next to it isn't hard, and I'd call that
a noisy line.  Are we talking noise, or nuclear weapons?


>The term is noted.  But, if you look at capacitors having the same energy
>storage capacity (0.5 * (voltage rating ^ 2) * capacitance), you'll find
>that they are about the same physical size regardless of the voltage
>rating.  Going to a higher voltage in the same size case means less
>capacitance.  So it doesn't seem that you're gaining much running the
>capacitor at a higher voltage.

Panasonic TS series, a 100V (surge 125) cap has 680uF in a 22 x 31.5 mm can
That same can at 50V is 2200uF.

100*100*0.5*0.000680 = 3.4j
50*50*0.5*0.002200 = 2.75j

Other lines are similar.
The reason is this. To double the voltage, you have to double the
dielectric thickness.
To double the C, (and maintain the same voltage) you have to double the
plate area.
There's two plates areas (and volumes) to every one dielectric.

The resistor approach assumes that the load will always be drawing at least
that current, otherwise it goes BOOM. Or are you going to include the
protection zener which would, of course, have to dissipate the rest of the
power? (and the heatsink and fan :)

>Of course you may have less heat, but 7W isn't hard to deal with in a big
>industrial device.

Cap lifetimes are halved for every 10c rise above their specified temperature.
Better keep all caps away from your nuclear reactor.

1999\02\21@140318 by Mark A Moss

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On Fri, 19 Feb 1999 03:06:32 -0800 Rod Phillips <.....rphillipKILLspamspam.....NETINS.NET>
writes:
>Hi to all,
>
>Could I pick some brains about the best method to drive a Pic based
>control module from a very noisy 100vdc supply?  Power from the source
>is not a concern so a voltage divider circuit with its output feeding
>a voltage regulator came to mind.  <snip>

That would waste a lot of power. 95V * 500mA = 47.5W.  There are some
DC-DC converter modules that would do it.  We use one to drop about 300V
to 40V at several amps (to power a brushless DC motor).  I can't think of
the manufacturer at the moment, but the modules are about 3" by 6" and
are designed to mount on a heatsink.  I would assume a 500mA converter
would be significantly smaller.

Mark Moss
Amateur Radio Operator, Technician, and General Tinkerer
Kalamazoo,  MI   U.S.A.

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1999\02\21@143050 by paulb

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Mark A Moss wrote, quoting Rod Phillips.

>> ... the best method to drive a Pic based control module from a very
>> noisy 100vdc supply?  Power from the source is not a concern so a
>> voltage divider circuit with its output feeding a voltage regulator
>> came to mind.  <snip>

> That would waste a lot of power. 95V * 500mA = 47.5W.

 And my suggestion of using a lamp ballast and a shunt regulator picked
up on the comment "Power from the source is not a concern", so I do
think it's quite valid.

>  There are some DC-DC converter modules that would do it.

 They're an absolutely stock-standard, common-as-dirt part.  100VDC is
well within tolerance of many if not most "mini" SMPSUs.  Just look them
up in the catalogue and check for guaranteed operation down to 50VAC
(corresponding to 72V peak) and you're home.

 The fact that they are specified for AC is of no consequence since
they simply contain a bridge rectifier.  (OK, *do* check that they are
in fact the design with a bridge rectifier and *one* electrolytic
capacitor but a good proportion of the "minis" are such.)
--
 Cheers,
       Paul B.

1999\02\21@150804 by D. Schouten

picon face
>>I think it may be useful to first drop to 30V or so because few (are
>>there any?) of the all-in-one switching ICs can handle more than 40
to 60
>>V.  A regulator suitable for 100 V input would be more complicated
>>involving a discrete transistor output and possibly a custom
transformer
>>instead of just an inductor.
>
>I wasn't thinking in terms of "simple switcher" chips.. You'd need an
>external fet, but so what?
>CS384x would work well here, they're cheap. 200V mosfets are
available, and
>cheap.

Or use a TOPswitch. They have build-in MOSFET and modulator, need very
few external components and have very good design support. Reading
their design notes makes winding a flyback transformer a piece of
cake. You can even use the TOPswitch in BUCK mode requiring an
inductor instead of a transformer.

Just skip the linear step down and use the switcher directly on your
100V DC bus with some proper filtering on the input side.
I have such small switchers working as an auxiliary power supply on
the same DC bus as a 25kW Full Bridge power converter. Now That's
noisy!

Daniel...

1999\02\21@152036 by Gerhard Fiedler

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At 21:02 02/21/99 +0100, D. Schouten wrote:
>Or use a TOPswitch.

interesting parts. the link is http://www.powerint.com/datasheets.htm

ge

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