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'Direct Connection-Power Feed'
1998\05\05@200416 by Solon Caceres Moreno

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Engineers:

For a project based on PIC, How it can be feed from a source of 110 V AC 60
Hz without using a transformer?

Thanks in advance


_________________________________________
Sol—n C‡ceres Moreno
Ingeniero de Sistemas UIS
Gerente C.M. & C’a Ltda.
e-mail: spam_OUTsolcaTakeThisOuTspamusa.net
             .....cdmbKILLspamspam@spam@multinet.com.co
tel: (57-7)6347873
Colombia, South America

1998\05\05@203517 by David VanHorn

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Engineers:

For a project based on PIC, How it can be feed from a source of 110 V AC 60
Hz without using a transformer?


Will humans or other living things come into contact with it, EVER?  Will it
ever be connected to another system?

If the answer to either of those questions is yes, then you have some
serious safety issues to contend with.

You also get to take powerline glitches full force.

You didn't give much info here, so the answer is yes it's possible, but it's
a VERY narrow road.

1998\05\06@010323 by Calvin

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It depends on the current you need, but theres is a Microchip application
note describing this, I believe it is called "Transformerless Power Supply".
It is real simple, but works reasonable well.

Calvin

{Original Message removed}

1998\05\06@044532 by Frank A. Vorstenbosch

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Solon Caceres Moreno wrote:
>
> Engineers:
>
> For a project based on PIC, How it can be feed from a source of 110 V AC 60
> Hz without using a transformer?

There are several ways for doing this, the easiest involves using a
single-wave
rectifier in series with a resistor to charge up an electrolytic
capacitor.  A
zener diode parallel to the capacitor limits the output voltage to some
sensible
level.  Caveats: the resistor dissipates lots and lots of energy (to
give 50mA at 5V the resistor will dissipate 50mA*105V=5.3W) and the
output voltage will have a very noticable ripple.  To improve on the
ripple, replace the 5.1V zener with an 8V2 zener, and use a 78L05 to
regulate down to 5V.  Of course, your circuit may work reasonably well
without removing all the ripple.

             R
 AC in ---\/\/\/---->|----+---+------- DC out
                          |+  | +
                         ===  ZD 5V1
                          |   |
AC rtn -------------------+---+------- DC return

The second solution involves using one of the integrated
line-AC-to-5V-DC circuits -- these are available and require even fewer
components and will give you far less headaches.  These work by
switching the AC line to a holding capacitor whenever the AC line
voltage is less than a set voltage, and regulating the switch so that
the output voltage is close to 5V.  Someone on the list will remember a
part number, I'm sure :-)

Frank
------------------------------------------------------------------------
Frank A. Vorstenbosch     <SPAM_ACCEPT="NONE">    Phone: 0181 - 636 3000
Electronics and Software Engineer                 Mobile: 0976 - 430 569
Eidos Technologies Ltd., Wimbledon, London        Email: favspamKILLspameidos.co.uk

1998\05\06@071910 by Caisson

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> Van: Solon Caceres Moreno <.....cdmbKILLspamspam.....MULTINET.COM.CO>
> Aan: EraseMEPICLISTspam_OUTspamTakeThisOuTMITVMA.MIT.EDU
> Onderwerp: Direct Connection-Power Feed
> Datum: woensdag 6 mei 1998 1:29
>
> Engineers:
>
> For a project based on PIC, How it can be feed from a source of 110 V AC
60
> Hz without using a transformer?
>
> Thanks in advance

If the project uses a _very_ low, constant current use a resistor
(cheapest).
If the power-consumption is _very_ low, but not constant, add a zener-diode
over the project.

But remember, a current of 10 mili-Amp will generate 10mA * 110 V =
1.1 Watt's of heat in your resistor !

An other possebility is to use a condensator as a AC-Resistor.  You will
have
to do some math to find the right value though ...

Greetz,
 Rudy Wieser

1998\05\13@101214 by lilel

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Solon Wrote:


> Engineers:
>
> For a project based on PIC, How it can be feed from a source of 110
> V AC 60 Hz without using a transformer?
>
> Thanks in advance

Use a big, ugly, high power dropping resistor and a half wave diode.
As dumb as a post, this is the cheapest simplest power supply
available.  Also one of the worst performing.  PIC's work fine on it.


I do this all the time.  Several other folks have noted some problems
- high power dissipation, and danger of people coming in contact with
the AC line.  These things must be in an insulated case, or a
grounded enclosure.

I've used a half wave rectirfier - dropping resistor - capacitor
power supply in several appliances.  Someone suggested not using a
zener diode - that's a bad idea.  Zeners are neccesary.  The AC power
line is not very well regulated and fulla spikes.  A zener will do a
little toward mitigating spikes (but not much!).

You'll get lots better ripple and regulation if you drop in two steps
- first go from half wave AC to 24 volts, then to 5 volts.  I often
need 24 volts to power a relay anyway, so this works well.  In my
designs the 24 volt line goes to 40 or 50 volts when the relay is
off, so use a higher voltage capacitor.

A better design, still cheap, is to use a metallized polyester film
capacitor of about .68 to 1.5 mfd, followed by a full wave bridge and
a much smaller, lower power dropping resistor.  This avoids all that
heat and the expensive power resistor.  It won't work with a half
wave bridge.


Best Regards,

Lawrence Lile

1998\05\18@163902 by H.P. de Vries

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At 08:48 AM 5/13/98 +0000, Lawrence Lile wrote:
{Quote hidden}

Maybe try a condensor-divider (in stead of resistors) in comb. ith some
diodes???? Should dissipate a lot less. .

(for replies contact me private, i'm temporarily off the list)
Hans de Vries - H.P.d.Vriesspamspam_OUTstud.tue.nl
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