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'[EE]: 5V from 40V'
2003\05\07@223141 by Andy Shaw

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Hi Folks,
Help!
Can anyone suggest a good way of creating a 5V supply (max 250mA) from a 40V
supply. Most of the simple regulators don't seem to like this high an input
voltage. Oh and if this is too easy how about generating a +5V supply again
250mA from a -40V supply, needs to be able to share the same ground as the
+40v and -40v supply and must be +5V not -5V.

What's it for well I'm building a controller for my 3 Axis milling machine.
It uses chop mode bipolar stepper motors. I've found a supply from an old
HiFi amp that provides -40V 0 +40V I would like to use this as the power for
the steppers. I would also like to drive the logic (PIC based) from the same
supply. Ideally I'll use the +40v for the stepper and the -40V for the logic
to reduce noise, hence the questions... See http://www.gloomy-place.com/cnc.htm for
pictures of the milling machine...

Thanks

Andy

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2003\05\07@225222 by Picdude

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Three things I can think of for +40V to +5V...

(1) Since you're considering a linear reg, perhaps try to do this in two steps ... drop to say 15V then again to 5V with a second reg.  BUT that assumes that you can find a higher-voltage-output regulator that will allow 40V in.

(2) Use a series of diodes (like 7 or 8 of them) on the input to drop the 40V to 35V?  Silly IMO, but will work.

(3) Better option ... I know that you can do this with one of the National Semi dc-dc converters.  Look at the datasheets for the LM2677-5.0, LM2671-5.0, LM2674-5.0, LM2670-5.0.  If you go to http://www.national.com, there is an online tool that will specify workable parts based on your requirements and design the circuit for you.

They may also have something for -40 to +5, but if not, you may be able to do +40V to -5V *isolated* and connect to the other parts of your circuit accordingly.  This last one is a guess, so don't hold me to it.

One last thing, and actually a question for the experts ... the 35V max input voltage specification for the linear regs should be with reference to it's ground pin, right?  So shouldn't it be possible to raise the ground reference on a regulator to >+5V, so that the 40V could be dropped to an intermediary voltage (between 35V and (5V+dropout)), then use another linear regulator with a "proper" ground reference to drop it to an accurate +5V?

Cheers,
-Neil.



On Wednesday 07 May 2003 21:31, Andy Shaw scribbled:
{Quote hidden}

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2003\05\07@230859 by hard Prosser

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Look at Roman Blacks "Black Converter".

With a suitable output transistor you should be able to get 250mA.
If the output noise is too high - drop to 8V or so & then use a linear reg
for 5V.

You may need something more elaborate to generate it from the -40V supply -
but again, you may be able to adapt Roman's converter to drive a charge
pump type circuit - probably about as simple as you'll get.

Richard P




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Hi Folks,
Help!
Can anyone suggest a good way of creating a 5V supply (max 250mA) from a
40V
supply. Most of the simple regulators don't seem to like this high an input
voltage. Oh and if this is too easy how about generating a +5V supply again
250mA from a -40V supply, needs to be able to share the same ground as the
+40v and -40v supply and must be +5V not -5V.

What's it for well I'm building a controller for my 3 Axis milling machine.
It uses chop mode bipolar stepper motors. I've found a supply from an old
HiFi amp that provides -40V 0 +40V I would like to use this as the power
for
the steppers. I would also like to drive the logic (PIC based) from the
same
supply. Ideally I'll use the +40v for the stepper and the -40V for the
logic
to reduce noise, hence the questions... See http://www.gloomy-place.com/cnc.htm
for
pictures of the milling machine...

Thanks

Andy

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2003\05\08@050528 by William Chops Westfield

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   Three things I can think of for +40V to +5V...

   (1) Since you're considering a linear reg, perhaps try to do this in
   two steps ... drop to say 15V then again to 5V with a second reg.  BUT
   that assumes that you can find a higher-voltage-output regulator that
   will allow 40V in.

Hmm.  The popular and cheap LM317 (variable 3-terminal voltage regulator)
doesn't require that the reference voltage be ground.  There are examples
in the the datasheets/appnotes of using them as regulators for quite high
voltages...

BillW

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2003\05\08@060644 by Mike Harrison

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you do not want a linear reg. - it will dissipate nearly 9 watts at 250mA
A switcher will cost less than the heatsink!

On Thu, 8 May 2003 02:05:05 PDT, you wrote:

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2003\05\08@083927 by Russell McMahon

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>you do not want a linear reg. - it will dissipate nearly 9 watts at 250mA
>A switcher will cost less than the heatsink!

Better still, try to use a more sensible input voltage. From what you say it
appears the supply isbeing used "because its there". Dissipating most of the
voltage or using a switcher are both introducing complications you don't
need. Suitable transformers or plug packs (wall warts?) cost very little
compared to the value of the system to you and other costs.

If you MUST use a linear supply, place a series resistor from 40v to the
regulator input. Design this to drop most of the voltage drop needed at full
power. Rmax = (Vin-Vregulator_in-Vsafety)/Imax.
eg Vin = 40, Isafety = 3v, Vreg in = 7volt say.
   R = (40-7-3)/0.25 = 120r.
This will drop V=IR = 30v at max current and dissipate P = I^2R = 7.5 watts
at full current.
Use AT LEAST a 10 watt resistor.

There areany comercial IC switching regul;ators that will meet your
requirement, but introducing a switcher unnecessarily is seldom a good idea.



       Russell McMahon

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2003\05\08@090105 by Olin Lathrop

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> the 35V max
> input voltage specification for the linear regs should be with
> reference to it's ground pin, right?

Right.

> So shouldn't it be possible to
> raise the ground reference on a regulator to >+5V, so that the 40V
> could be dropped to an intermediary voltage (between 35V and
> (5V+dropout)), then use another linear regulator with a "proper" ground
> reference to drop it to an accurate +5V?

Yes, that would work, but as any linear regulator it will still dissipate
most of the power as heat.


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Embed Inc, embedded system specialists in Littleton Massachusetts
(978) 742-9014, http://www.embedinc.com

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2003\05\08@092454 by Mike Harrison

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On Fri, 9 May 2003 00:18:21 +1200, you wrote:

>>you do not want a linear reg. - it will dissipate nearly 9 watts at 250mA
>>A switcher will cost less than the heatsink!
>
>Better still, try to use a more sensible input voltage. From what you say it
>appears the supply isbeing used "because its there". Dissipating most of the
>voltage or using a switcher are both introducing complications you don't
>need. Suitable transformers or plug packs (wall warts?) cost very little
>compared to the value of the system to you and other costs.
>
>If you MUST use a linear supply, place a series resistor from 40v to the
>regulator input. Design this to drop most of the voltage drop needed at full
>power. Rmax = (Vin-Vregulator_in-Vsafety)/Imax.
>eg Vin = 40, Isafety = 3v, Vreg in = 7volt say.
>    R = (40-7-3)/0.25 = 120r.
>This will drop V=IR = 30v at max current and dissipate P = I^2R = 7.5 watts
>at full current.
>Use AT LEAST a 10 watt resistor
A 10 watt R dissipating 7.5 watts will get extremely hot unless airflow is very good. 7.5 watss is 7.5 watts, regardless of how big the resistor is.

If you really want a linear supply for quietness, use a switching pre-regulator, but a simple
switcher, maybe with an extra output LC filter will be plenty clean enough for most applications.
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2003\05\08@095139 by Andy Shaw

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Folks thanks a lot for the ideas. I've been doing a little more digging
around on the PCB of the old amp and it has a LM337 and a LM 317 to provide
a lower voltage (-12..+12) for the input stages of the amp. These regulators
are bolted to a huge heatsink that forms part of the case. Since I was
planning on using the case of the amp for this one off project anyway it
looks like that with change of the ref. voltage resistor I can get +5 from
this. Failing that I think I'll just invest in a wall wart! It's amazing how
much of this old amp I think I can re-use. Both parts of the power supply
(including smoothing caps), the case/heatsinks and I think I can make use of
most of the input sockets to provide outputs to the steppers and inputs from
limit switches! Oh and I get a nice safe mains switch and power setup thrown
in as well!

I've also learned a lot about switching supplies and the other non linear
regulators... I'm sure that will come in useful in the future. Once again
this list provides just the right amount of help....

Thanks


Andy

{Original Message removed}

2003\05\08@101549 by Dave VanHorn

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>
>A 10 watt R dissipating 7.5 watts will get extremely hot unless airflow is
>very good.
>7.5 watss is 7.5 watts, regardless of how big the resistor is.

I've worked with 5W solder irons.

A buck switcher would be my first cut at this application.
Done properly, they are pretty quiet.

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2003\05\09@095948 by Micro Eng

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hmmmm......you can BUY a 48 to X for around $15, and give you 5+ amps.  I've
got several samples of modules that I am evaluating.  You mght even be able
to get one sampled to you, but for the hassle of a single use thing, I'd go
that way.


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2003\05\09@121209 by William Chops Westfield

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   hmmmm......you can BUY a 48 to X for around $15, and give you 5+
   amps.  I've got several samples of modules that I am evaluating.
   You mght even be able to get one sampled to you, but for the
   hassle of a single use thing, I'd go that way.

That's an interesting idea.  Rip apart ay telco equipment and you find
these in large quantitities.  A "business" phone is likely to contain a
low-power variant as well, and they show up in surplus stores quite often.
(see for instance Jameco #206076)  Normal input range is 36-72V...

BillW

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