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'AC Adapter Substitute'
1996\05\22@135014 by rdmiller

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Edwin wrote:
[...]
>      (2) If I have A/C anyway, is there a inexpensive way (i.e. less than
>      the cost of a AC/DC adapter) to convert A/C to a DC in the PIC
>      operating voltage?
>
>      -Edwin

You could do it with a resistive voltage divider, diode, and capacitor
to form a half-wave rectifier which would feed a voltage regulator.  The
single power resistor you would need (it'd have to be able to handle at
least 2 or 3 Watts) may be expensive, but it would eliminate the need
for a transformer.

[The design of such a supply is left as an excercise for the reader.  :-]

Overall I don't know if it would turn out to be cheaper than an average
wall adapter but it should be smaller and certainly lighter.

Hmmm...  Now I'm thinking of all the uses *I* might have for such an
adapter.  As if I don't already have enough to do...

Rick Miller

1996\05\22@141306 by Brent Miller

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Rick Miller wrote:

> You could do it with a resistive voltage divider, diode, and capacitor
> to form a half-wave rectifier which would feed a voltage regulator.  The
> single power resistor you would need (it'd have to be able to handle at
> least 2 or 3 Watts) may be expensive, but it would eliminate the need
> for a transformer.

You might also consider a capacitive voltage divider instead of resistive.
This would not dissipate the 2 or 3 Watts (energy is stored, not dissipated
in the capacitors).

Brent Miller

1996\05\22@150403 by Jonathan King

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At 01:50 PM 5/22/96 -0400, you wrote:
>
>Edwin wrote:
>[...]
>>      (2) If I have A/C anyway, is there a inexpensive way (i.e. less than
>>      the cost of a AC/DC adapter) to convert A/C to a DC in the PIC
>>      operating voltage?
>>
>>      -Edwin
>
>You could do it with a resistive voltage divider, diode, and capacitor
>to form a half-wave rectifier which would feed a voltage regulator.  The
>single power resistor you would need (it'd have to be able to handle at
>least 2 or 3 Watts) may be expensive, but it would eliminate the need
>for a transformer.


For safety reasons, you might want to use multiple lower power resistors instead
of a single large one.  That way, failure of one resistor will not result in
a high voltage into the regulator/PIC.  Also, the PIC will NOT be isolated from
the AC line, which can be a hazard as well as a grounding problem.

There are also small DC-Dc converter chips that can handle this with less power
dissipation.  Email me if you are interested in more info.


Good luck and be careful!


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1996\05\25@020840 by nigelg

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In message  <.....96May22.150102edt.5423-1KILLspamspam@spam@firewall.uicc.com> PICLISTspamKILLspammitvma.mit.edu
writes:

> For safety reasons, you might want to use multiple lower power resistors
instead
> of a single large one.  That way, failure of one resistor will not result in
> a high voltage into the regulator/PIC.  Also, the PIC will NOT be isolated
from
> the AC line, which can be a hazard as well as a grounding problem.

From a practical point of view, using resistors in series like this is
VERY unreliable. Most TV & Satellite receiver manufacturers use two resistors
in series for starting the switchmode PSU (values vary from 15K to 120K).
As the two resistors are never exactly the same value the slightly higher
one will dissipate more power. This will tend to make the resistor go
higher in value which then dissipates even more. Eventually the resistor
will go open circuit resulting in the unit no longer starting.

I would estimate the failure rate on some units to be greater than 70-80%
over the first two or three years. I usually replace the two small resistors
with two 1 watt ones, this greatly improves reliability.

BTW, the reason for using two small resistors in production is down to cost,
it's cheaper to use two small resistors than one large one.

Nigel.

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1996\05\26@111706 by Jonathan King

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At 03:08 AM 5/23/96 -0400, you wrote:
>
>In message  <EraseME96May22.150102edt.5423-1spam_OUTspamTakeThisOuTfirewall.uicc.com> PICLISTspamspam_OUTmitvma.mit.edu
> writes:
>
>> For safety reasons, you might want to use multiple lower power resistors
> instead
>> of a single large one.  That way, failure of one resistor will not result in
>> a high voltage into the regulator/PIC.  Also, the PIC will NOT be isolated
> from
>> the AC line, which can be a hazard as well as a grounding problem.
>
>>From a practical point of view, using resistors in series like this is
>VERY unreliable.

It depends what the alternative is.  Resistors have a max voltage rating as
well as a max power rating.  Leaded 1/4 W resistors are typically 200-250 Volts.
Surface mount devices are as low as 50 Volts.

For a 120V input, you may get by with a single resitor, but for a 220V input
you are
asking for trouble.

{Quote hidden}

It sounds like the resistors are overstressed by design. One "gotcha" I have
found
with resistors, especially surface mount ones, is the resistor temperature
at maximum power
dissipation.  I've seen cases where the resistor will be ~200C above ambient at
'rated' power.  I'm not sure I want a component that hot on my board!  It also
could be a voltage overstress issue.  Higher power components also have a
higher max
voltage rating.

You can also look at the failure modes of carbon comp vs film/wire wound
resistors and
try to make sure that dangerous voltages aren't applied to a circuit even
after a component
failure.

>
>BTW, the reason for using two small resistors in production is down to cost,
>it's cheaper to use two small resistors than one large one.

Very true.

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1996\05\26@130209 by Mark K Sullivan

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Have you considered using a capacitive reactance instead of a resistor to drop
line voltage?  I think I remember an RCA temperature controller chip, CA30??
maybe, that worked this way.  You can use a capacitive divider or a capacitor
sourcing current across a resistor to get low voltage AC and then rectify and
filter it after the divider.

Also, Supertex makes some high voltage depletion-mode FETs that are good if you
only need small current.  The advantage is wide input range.  I built a PIC
power supply that drew constant current from 12 to 240 volts this way.

- Mark Sullivan -

1996\05\26@151705 by reginald neale

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>
> Have you considered using a capacitive reactance instead of a resistor to drop
> line voltage?  I think I remember an RCA temperature controller chip, CA30??
> maybe, that worked this way.  You can use a capacitive divider or a capacitor
> sourcing current across a resistor to get low voltage AC and then rectify and
> filter it after the divider.
>
>
Yes, capacitive reactance is a more elegant solution from a conceptual
standpoint. But you need a voltage rating equal to the max peak line
voltage plus safety factor. The capacitance needed for practical values
of output current, combined with the required voltage rating, result in
a capacitor that's physically pretty large. A lot larger than the
power resistor it would replace. But if you have plenty of space and are
more worried about heat, it could be a good design approach.


....Reg Neale.............standard disclaimer applies.......
"Ignorance is a renewable resource."    P. J. O'Rourke......

1996\05\26@193611 by Mike Keitz

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>>
>> Have you considered using a capacitive reactance instead of a resistor to
drop
>> line voltage?  I think I remember an RCA temperature controller chip, CA30??
>> maybe, that worked this way.  You can use a capacitive divider or a capacitor
>> sourcing current across a resistor to get low voltage AC and then rectify and
>> filter it after the divider.
>>
>>
>Yes, capacitive reactance is a more elegant solution from a conceptual
>standpoint. But you need a voltage rating equal to the max peak line
>voltage plus safety factor. The capacitance needed for practical values
>of output current, combined with the required voltage rating, result in
>a capacitor that's physically pretty large. A lot larger than the
>power resistor it would replace. But if you have plenty of space and are
>more worried about heat, it could be a good design approach.
>

When using a capacitor, be sure to include a small (10-100 ohms) resistor in
series with it.  Since voltage spikes (by definition) have fast risetimes,
they will couple straight through the capacitor with little attenuation and
possibly damage the rectifier diodes or other parts of the regulator.  The
capacitor, being large, will not experience a voltage transient, but a heavy
current transient will couple through it.  The protection resistor and other
resistances in the circuit are the only limiting factor.   Special resistors
intended to function as fuses are available that can combine the fuse and
resistor functions in one part.  Lacking those, a conventional "flameproof"
resistor would be suitable for ordinary hacker use.

-Mike

1996\05\28@024610 by adrian

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Edwin wrote:
[...]
>      (2) If I have A/C anyway, is there a inexpensive way (i.e. less than
>      the cost of a AC/DC adapter) to convert A/C to a DC in the PIC
>      operating voltage?

As I understand it, there is an 8 pin IC designed to connect directly to
mains and provide 100mA at 5V. It switches the mains at the right point to
get the voltage you want on to a capacitor (or something like that).

I don't have details, but I thought Maplin (UK) do them. They obviously don't
provide any/much isolation, so are not necessarily safe in all failure
modes.

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