Chip with simple program for Toy

[goose]

Keith R. Williams <krw@attglobal.net> wrote in message news:<MPG.1a0ba68ef518a1d798a818@enews.newsguy.com>...

In article <vq2i8b3uc0jibd@corp.supernews.com>,

<snipped>

for those who would like to play it on a guitar.

Holly came from Miami, Fla
C F
Hitchhiked her way across the USA.
C F
Plucked her eyebrows on the way
C D
Shaved her leg and then he was she - she said:
F D

Hey Babe, take a walk on the wild side,
C F
Said hey honey, take a walk on the wild side.
C F

hth
goose,

 

[Keith R. Williams]

In article <vqbu81juqgoa9@corp.supernews.com>, hmurray@suespammers.org
says...

I am working on some project that works like the Jeopardy Game. What happens
is, there are three teams, with buttons infront of them, and on each team's
desk is a light bulb.

Anyone who can give me an idea of how to do this? I know how to work if it
were only two teams ...

Two suggestions...

If you can do it for two, then start with AxB, BxC, and AxC. Then feed
the results of that to another layer of the same logic. That should
give you a simple result.

I'd do it with silicon rather than relays. I's use a PIC or AVR, and
do it in software. First run each input signal through a pair of FFs
to avoid metastability. (That's probably included in the PIC/AVR front
end.) Then in software, spin until at least one bit comes on. If
there is only one, your job is done. If two or three, then you have
more work to do.

One approach is to change the rules of the game. Say everybody gets
an extra $100 in the case of a tie. (Or only the teams that tied.)

Or you could run a counter in the background, and use the counter to
decide ties. Say divide the counter by 2 or 3 and use the remainder
to decide.

Yep. Just pick one at "random". If the quantization period is short
enough and your "random" choice is fair enough no one will notice.

--
Keith

 

[Keith R. Williams]

In article <vq97c7ipfpvvf9@corp.supernews.com>,
fandaDEATH2SPAMMERS@catskill.net says...

"Keith R. Williams" <krw@attglobal.net> wrote in message
news:MPG.1a0ba68ef518a1d798a818@enews.newsguy.com...
In article <vq2i8b3uc0jibd@corp.supernews.com>,
fandaDEATH2SPAMMERS@catskill.net says...

"John Larkin" <jjlarkin@highlandSNIPtechTHISnologyPLEASE.com> wrote in
message news:kkh2qvojevvlovg58m4e3p2vb5ip1unttc@4ax.com...
On 30 Oct 2003 16:47:26 GMT, dbowey@aol.com (Dbowey) wrote:



And, as I quickly learned after being an idiot just once, when a
young
lady
asks " do you have the time?" she ain't asking where the clocks hands
are
pointing.


She must have been a horologist.

John

And what makes you so sure that "she" was a she?

No! Tell me they wouldn't lie! EEww!


--
Holly came from Miami, Fla
Hitchhiked her way across the USA.
Plucked her eyebrows on the way
Shaved her leg and then he was she - she said:

Hey Babe, take a walk on the wild side,
Said hey honey, take a walk on the wild side.
...

;-)

--
Keith

Requires RealOne Player and fast internet connection and speakers cranked to
the max!
http://www.globecorp.org/sci.electronics.basics/WalkOnTheWildSide.rm ;-)

That be it! Eeww! ;-)

--
Keith

 

[Watson A.Name - "Watt Sun]

In article <3fa19b6f_2@news.arcor-ip.de>, dklingb@cs.tu-berlin.de
mentioned...

"Chris Roberts" <ihatespam@spamsucks.com> wrote in message
news:8dgob.9346$3o3.8044@news01.roc.ny...

traces on the back to really know whats goin on but I figured u gurus
could
recognize certain things right off the bat.

Don't rely on that. Prom the pic, it seeme that the diodes that are most
likely to form the PSU - D3 and D4 are identical or at least very similar.
It is not a full-wave rectifier as such one would need 4 diodes, but it is

Full wave rectifiers need two diodes. Full wave bridge rectifiers use
four.

also very unlikely that a half-wave rectifier would need 2 diodes (unless
one is a zener). As both D3 and D4 do not look like being zeners, but the
diode partially behind the big cap does, it could well be that the device
employs more than one low voltage. Could you please also post a photo of the
other side of the device, a view of the traces might be a great help.

Dimitrij

--
@@F@r@o@m@@O@r@a@n@g@e@@C@o@u@n@t@y@,@@C@a@l@,@@w@h@e@r@e@@
###Got a Question about ELECTRONICS? Check HERE First:###
http://users.pandora.be/educypedia/electronics/databank.htm
My email address is whitelisted. *All* email sent to it
goes directly to the trash unless you add NOSPAM in the
Subject: line with other stuff. alondra101 <at> hotmail.com
Don't be ripped off by the big book dealers. Go to the URL
that will give you a choice and save you money(up to half).
http://www.everybookstore.com You'll be glad you did!
Just when you thought you had all this figured out, the gov't
changed it: http://physics.nist.gov/cuu/Units/binary.html
@@t@h@e@@a@f@f@l@u@e@n@t@@m@e@e@t@@t@h@e@@E@f@f@l@u@e@n@t@@

 

[Chris Roberts]

Wow thank you Dimitrij for taking the time to look over the circuit! I been
busy drawing it out (more difficult than I thought) and was wondering about
the power supply part after drawing that out.

Anyway, I measured the voltage across the cap and it runs about 22.5 volts.
Won't be able to play with it more for a while but I figured I'd let you
know I read your post and really appreciate your assistance.


Chris

Sorry for the misinformation in my first response, mea culpa. I have just
looked at the 2 pictures after half-transparently layering the traces
picture over the top side one to see both at once. The rectifier is a
full-bridge one, the diodes D1, D3, D4 and D(2?, number not visible on
picture) are used in a non-standard full-bridge configuration where D1 and
D(2?) seem to be Zeners. This rectifier powers the big black cap to which
everything else is connected. It is the only low voltage source. You
should
measure the voltage across the cap and use a low voltage supply (battery?)
with this spec. When using such a voltage source, remove the 4 diodes (R1,
R(2?), R3, R4), the long red capacitor, the resistor in parallel to it
(small, almost below the cap) and the large resistor standing on its side
between P3 and the PSU cap before connecting the low voltage source. The
connection with the white wire will not be used any longer, the other two
connections (P1, P3) go to the relay contacts and can be used for whatever
you need the relay for. The low voltage source should be connected in
parallel to (or instead of, your choice) the PSU cap (big black one).
Respect polarity! The schematic below illustrates how the power supply
part
works:

P1 P3 P2?
| | |
| R__ C | |
x----|__|--||---x |
| | |
| relay_contact| |
x-----o/ o------x |
| |
| R__ R__ D2? _ _ D1 |
x--|__|--x--|__|--x--|>-|--x---|-<|--x
| C | | | |
'--||----' '--|<|----x---|>|--'
D4 || D3
.---''---.
| +C - |
x---||---x
| |
|+ -|
to LV electronics

I hope this helps,

Dimitrij

 

[Dbowey]

BFD. Sometimes it makes lot's of sense, but why should anyone have tight
pucker strings over it?

Fried ones posted:

On Thu, 06 Nov 2003 16:09:03 GMT, "Bob Stephens"
stephensdigital@earthlink.net> wrote:

Why is top posting so unpopular? I think that especially in long threads
with dozens of posts, that it is a pain in the ass to have to scroll through
the same initial post and all previous responses
for each message in the thread. Or am I missing something?

A: Because top posting doesn't make sense.
Q: Why bottom post?

 

[Keith R. Williams]

In article <slrnbr973m.q8.ericlada@eric-ladas-computer.local>,
ericlada@eric-ladas-computer.local says...

Hi! Sorry for the basic nature of this post, but I figured this is the best
h place for it. I understand volage is Potential energy with respect to some
reference point. This is why you must always measure across a component, or
measure whith respect to ground. How then do scopy probes work. You are
determining a time varing voltaage with only one point. What is the voltage
referenced to?

Ground. The closer to the device, the better. If you're using a
scope on mains circuits you'd better know what you're doing!

--
Keith

 

[Charlie+]

Passive probes - By minimising the errors of attaching the probe to a
working circuit - usually at the expense of dividing downwards the
voltage sensitivity of the oscilloscope.
Active Probes - As above except without the loss of sensitivity but
substituting a heavy increase in cost.
Charlie+

 

[Keith R. Williams]

In article <slrnbrc9br.r8k.ericlada@eric-ladas-computer.local>,
ericlada@eric-ladas-computer.local says...

On 2003-11-15, Keith R Williams <krw@attglobal.net> wrote:
In article <slrnbr973m.q8.ericlada@eric-ladas-computer.local>,
ericlada@eric-ladas-computer.local says...
Hi! Sorry for the basic nature of this post, but I figured this is the best
h place for it. I understand volage is Potential energy with respect to some
reference point. This is why you must always measure across a component, or
measure whith respect to ground. How then do scopy probes work. You are
determining a time varing voltaage with only one point. What is the voltage
referenced to?

Ground. The closer to the device, the better. If you're using a
scope on mains circuits you'd better know what you're doing!



Thanks for answering. However, how does the scope still give me a reading,
even when the ground lead is not connected in the circuit, as is usually the
case.

Your grounding back through the case and safety grounds. This is
a pretty poor ground and will cause all sorts of screwy readings.
You should be grounding your scope probes as close to the signal
you're probing with as short of a ground lead as is physically
possible. The higher the frequency of interest the more
important this is. In some cases a ground lead of a few tenths
of an inch is too long.

--
Keith

 

[Robert Monsen]

"William" <kingtech@i-cable.com> wrote in message
news:3fb7b05d_2@storm.i-cable.com...

Help!!


I am now making a design in converting 230Vac into 12Vdc.
Before that I put a 1W resistor in series to a 240V ac mylar capacitor, a
full bridge rectifier. Then the rectifier output is connected to a 1W
12V
zenor diode. The circuit work great to give us a 12Vdc output.

But now, I have to design another circuit with higher output current. I
think about to replace the diode by 7812 regulator IC.

I concern whether or not the 7812 regulator IC can suffer high input
voltage
from bridge rectifier?


Any expert Please advise??

I'm hardly an expert, but here goes...

The voltage regulator won't give you more current. You need a bigger
capacitor or a smaller resistor to get more current. Given an incremental
increase in the cap, the 12V zener should continue to work properly.

Assuming you are running 230V RMS at 50Hz, and you need about 100mA, I'd say
you need to use a 2.2uF 400V non-polar cap, in series with a 2W 10 ohm
resistor. That should give you adequate current. If you need more current
and increase the size of the cap, be careful about the max power rating of
the resistor.

Also, I'm sure you know that this supply is not isolated from the mains, so
you should make sure to proceed with caution. Any point of the circuit (even
the areas that seem like they should be low voltage) may contain lethal
voltage differentials with ground, unless you are very careful. If you don't
know this, then I'd say you should use a transformer in order to
electrically isolate the supply from the mains.

Regards,
Bob Monsen

 

[Don Kelly]

From: "Fred Abse" <excretatauris@cerebrumconfus.it>
Subject: Re: 50-AMP remote meter needed
Date: Sunday, November 16, 2003 9:55 AM

On Sun, 16 Nov 2003 09:43:10 -0600, John Fields wrote:

There is no such thing as a "current to voltage converter" transformer.

What you're describing is using what's called a "current transformer" and
a "burden resistor" to generate an AC voltage which is proportional to the
current flowing in the primary, which is the single wire running through
or looped through the center of the toroidal core.

Sorry, John, beg to differ here. Any transformer having current in the
primary will induce a voltage in the secondary. There is no need for a
burden resistor. OK, I know there's no such thing as an infinite impedance
measuring device, but a 200 meg input DVM is reasonable approximation, and
I've used one of those with Rogowski coils before now .......
----------------
John is right. The problem is that something like a 50/5 current transformer
is that it is a step up voltage transformer. ignoring saturation, (which
will always occur on open circuit) , the result would be that for a 480V,
50A circuit , the open circuit voltage of the transformer secondary will be
4800V. That would not be good for the DVM. or the person holding it.
Put frankly, open circuit CT's are dangerous as many ex-technicians in the
power industry have found out.
Many CT's are made so that the secondary would short out if the secondary
circuit was opened and are designed for low burdens.
--
Don Kelly
dhky@peeshaw.ca
remove the urine to answer

 

[Baphomet]

"Keith R. Williams" <krw@attglobal.net> wrote in message
news:MPG.1a1f61e73cd990bc98a8aa@enews.newsguy.com...

In article <slrnbr973m.q8.ericlada@eric-ladas-computer.local>,
ericlada@eric-ladas-computer.local says...
Hi! Sorry for the basic nature of this post, but I figured this is the
best
h place for it. I understand volage is Potential energy with respect to
some
reference point. This is why you must always measure across a
component, or
measure whith respect to ground. How then do scopy probes work. You
are
determining a time varing voltaage with only one point. What is the
voltage
referenced to?

Ground. The closer to the device, the better. If you're using a
scope on mains circuits you'd better know what you're doing!

--
Keith

You could always plug the scope into a cheater (lifts chassis ground). A
better way is to use an isolation transformer, if possible, on the equipment
to be measured by the scope.

 

[Fred Abse]

On Tue, 18 Nov 2003 00:19:18 +0000, Don Kelly wrote:

John is right. The problem is that something like a 50/5 current
transformer is that it is a step up voltage transformer. ignoring
saturation, (which will always occur on open circuit) , the result would
be that for a 480V, 50A circuit , the open circuit voltage of the
transformer secondary will be 4800V.

So you're telling me that a current transformer with an o/c secondary will
have 480V across its (single turn) primary? Come on.......

A typical current transformer using a single turn primary has a primary
inductance of approximately 1 microhenry (I went and measured one). This
will drop 18.8 millivolts at 50 amps, 60Hz. Assuming a ratio of 50:5
, the voltage at the secondary will be 188mV on open circuit.

Using John's example, a 5000:1 CT will give 94 V on open circuit.

Still not infinite

BTW, how do you come to the
conclusion that saturation will always occur on open circuit?



--
Then there's duct tape ...
(Garrison Keillor)
nofr@sbhevre.pbzchyvax.pb.hx

 

[Robert Monsen]

"Fred Abse" <excretatauris@cerebrumconfus.it> wrote in message
newsan.2003.11.18.22.37.14.954617@cerebrumconfus.it...

On Tue, 18 Nov 2003 00:19:18 +0000, Don Kelly wrote:

John is right. The problem is that something like a 50/5 current
transformer is that it is a step up voltage transformer. ignoring
saturation, (which will always occur on open circuit) , the result would
be that for a 480V, 50A circuit , the open circuit voltage of the
transformer secondary will be 4800V.

So you're telling me that a current transformer with an o/c secondary will
have 480V across its (single turn) primary? Come on.......

A typical current transformer using a single turn primary has a primary
inductance of approximately 1 microhenry (I went and measured one). This
will drop 18.8 millivolts at 50 amps, 60Hz. Assuming a ratio of 50:5
, the voltage at the secondary will be 188mV on open circuit.

Using John's example, a 5000:1 CT will give 94 V on open circuit.

Still not infinite

BTW, how do you come to the
conclusion that saturation will always occur on open circuit?

The current transformers I've used have much higher number of secondary
windings, and the manufacturer warns to always have a load across the
secondary. I have one here that has a 3:2000 winding ratio. Its only rated
for 30A, however.

The nice thing about larger numbers of turns is that the current goes way
down. For a 3:2000 ratio using the recommended 100 ohm sense resistor, given
a 50A input, you get a 7.5V RMS differential across the resistor. At the 30A
limit for this device, its 4.5V.

http://www.manutech.us/_Catalog/MN204.htm

Regards,
Bob Monsen

 

[Don Kelly]

"Fred Abse" <excretatauris@cerebrumconfus.it> wrote in message
newsan.2003.11.18.22.37.14.954617@cerebrumconfus.it...

On Tue, 18 Nov 2003 00:19:18 +0000, Don Kelly wrote:

John is right. The problem is that something like a 50/5 current
transformer is that it is a step up voltage transformer. ignoring
saturation, (which will always occur on open circuit) , the result would
be that for a 480V, 50A circuit , the open circuit voltage of the
transformer secondary will be 4800V.

So you're telling me that a current transformer with an o/c secondary will
have 480V across its (single turn) primary? Come on.......

A typical current transformer using a single turn primary has a primary
inductance of approximately 1 microhenry (I went and measured one). This
will drop 18.8 millivolts at 50 amps, 60Hz. Assuming a ratio of 50:5
, the voltage at the secondary will be 188mV on open circuit.
---------------

Did you do this with a secondary open circuit?. Did you do this measurement
using a low current inductance (probably at a high frequency) measuring
device as used in the electronics industry, or did you do this with 5 to 50A
at 60 Hz flowing and measurement of voltage current and power to find the
parameters? It makes a difference as for many core materials the
permeability at very low NI is near that of free space so the measurement
may be giving you the inductance of what is effectively an air cored
transformer. Much depends on the CT but a good one must have good coupling
between primary and secondary or its accuracy would be lousy. It also would
need a low magnetising current. Such a low primary inductance doesn't appear
to fit these conditions.
What I have said is a very real safety concern in the power industry.
Possibly they might know something you don't. Even a Weston CT used for
accurate measurements in the lab (in 100-500V circuits) has a shorting
switch across its terminals and the gap when the switch is open, is of the
order of 1 mm) and, in utilities, transfomers used for relaying are
automatically shorted when a meter is removed.
The following is from: http://www.kele.com/Tech/Monitor/Power/TRefPM2.html

"CAUTION: Never open-circuit a CT secondary while the primary is energized.
High crest voltages may occur across the open secondary circuit. To avoid
personal injury or equipment damage, the secondary must always be
short-circuited or connected to a burden. NOTE: A buzzing transformer is an
indication of an open secondary."

Other manufacturers indicate the same.

Using John's example, a 5000:1 CT will give 94 V on open circuit.

Still not infinite

BTW, how do you come to the
conclusion that saturation will always occur on open circuit?

---------

On the basis that I have postulated, a 480V, 50A circuit, the normal voltage
drop across the primary should be negligable. Suppose the meter used has an
impedance of the order of 0.2 ohms ( a bit high for the typical meter) so
that at 5A the secondary voltage is 1V (5VA typical) and the primary
voltage is about 0.1V ( the impedance seen is about about 0.02 ohms. This
is negligable as far as the load circuit is concerned. Since meter accuracy
is to be maintained up to 2 to 4 times the rating, and the transformer
should have some reasonable accuracy under short circuit conditions where
the current may be of the order of 500 to 1000 A. This puts the voltage at
the knee of the magnetising current somewhere in the order of 20V for this
particular hypothetical transformer.
If the transformer burden is 100 ohms then the primary impedance will be of
the order of 25 ohm as compared to the nominal load impedance of
480/50=9.6ohms. Not a problem except that the transformer primary voltage is
now about 350V and the transformer is saturated. In saturation, the accuracy
and coupling go out of whack so saturation does limit voltages for part of
the cycle. However, the transformer will be saturated for only part of the
cycle so the primary current and the voltage across the transformer will
have a decidedly spiky waveform. There will still be voltage spikes seen on
the secondary and these can be high. N.B. This is an off the cuff analysis
but I have seen oscillograms of secondary currents of a saturated CT. Peak
voltages occur at the time when d(phi)/dt is highest so peak voltages on the
secondary will still be dangerously high.
--
Don Kelly
dhky@peeshaw.ca
remove the urine to answer

 

[John Popelish]

Fred Abse wrote:

So you're telling me that a current transformer with an o/c secondary will
have 480V across its (single turn) primary? Come on.......

It doesn't happen. You would need a core with infinite permeability
for that to occur. You will see a sharp pulse of voltage every time
the current goes through zero, then nothing the rest of the half
cycle.

A typical current transformer using a single turn primary has a primary
inductance of approximately 1 microhenry (I went and measured one). This
will drop 18.8 millivolts at 50 amps, 60Hz. Assuming a ratio of 50:5
, the voltage at the secondary will be 188mV on open circuit.

Using John's example, a 5000:1 CT will give 94 V on open circuit.

Still not infinite

BTW, how do you come to the
conclusion that saturation will always occur on open circuit?

The better the CT, the higher its permeability, and the the fewer amp
turns it takes to saturate it. In normal operation (with burden
lower than the maximum specification) the secondary current cancels
almost all the amp turns of the primary.

--
John Popelish

 

[Keith R. Williams]

In article <vrl5lrh612a347@corp.supernews.com>,
fandaDEATH2SPAMMERS@catskill.net says...

"Keith R. Williams" <krw@attglobal.net> wrote in message
news:MPG.1a1f61e73cd990bc98a8aa@enews.newsguy.com...
In article <slrnbr973m.q8.ericlada@eric-ladas-computer.local>,
ericlada@eric-ladas-computer.local says...
Hi! Sorry for the basic nature of this post, but I figured this is the
best
h place for it. I understand volage is Potential energy with respect to
some
reference point. This is why you must always measure across a
component, or
measure whith respect to ground. How then do scopy probes work. You
are
determining a time varing voltaage with only one point. What is the
voltage
referenced to?

Ground. The closer to the device, the better. If you're using a
scope on mains circuits you'd better know what you're doing!

--
Keith

You could always plug the scope into a cheater (lifts chassis ground). A
better way is to use an isolation transformer, if possible, on the equipment
to be measured by the scope.

Yeah, and you'd still better know what you're doing! Your scope may be
at mains potential.

--
Keith

 

[Fred Abse]

On Wed, 19 Nov 2003 01:44:55 +0000, Don Kelly wrote:

---------------
Did you do this with a secondary open circuit?. Did you do this
measurement using a low current inductance (probably at a high
frequency) measuring device as used in the electronics industry, or did
you do this with 5 to 50A at 60 Hz flowing and measurement of voltage
current and power to find the parameters?

Rather crudely done, I admit, Just measurements of current and voltage on
the same conductor with and without the transformer, and assuming the
added impedance was due entirely to inductance (which it really wasn't
all) Current standardized at 10A, 60Hz on a 100:5 CT secondary O/C
Measurement done using 5.5 digit, true RMS DVM. There's a limit to what
can be done in odd spare moments and lunch breaks.

It makes a difference as for many core

materials the permeability at very low NI is near that of free space so
the measurement may be giving you the inductance of what is effectively
an air cored transformer.

i did realize that.

Much depends on the CT but a good one must
have good coupling between primary and secondary or its accuracy would
be lousy. It also would need a low magnetising current. Such a low
primary inductance doesn't appear to fit these conditions.

I thought it a bit low, too, but that's what I got. I'll try and find time
to do some proper measurements. I think I can lay my hands on a vector
voltmeter, or maybe a dynamometer instrument.

What you appeared to be saying was that *under normal operating
conditions*, in a 480V circuit, a CT would have the whole supply voltage
across its primary. That's what I was questioning.

What I have said is a very real safety concern in the power industry.
Possibly they might know something you don't.

I don't doubt that it is, and they do.

On the basis that I have postulated, a 480V, 50A circuit, the normal
voltage drop across the primary should be negligable. Suppose the meter
used has an impedance of the order of 0.2 ohms ( a bit high for the
typical meter) so that at 5A the secondary voltage is 1V (5VA typical)
and the primary voltage is about 0.1V ( the impedance seen is about
about 0.02 ohms. This is negligable as far as the load circuit is
concerned. Since meter accuracy is to be maintained up to 2 to 4 times
the rating, and the transformer should have some reasonable accuracy
under short circuit conditions where the current may be of the order of
500 to 1000 A. This puts the voltage at the knee of the magnetising
current somewhere in the order of 20V for this particular hypothetical
transformer. If the transformer burden is 100 ohms then the primary
impedance will be of the order of 25 ohm as compared to the nominal load
impedance of 480/50=9.6ohms. Not a problem except that the transformer
primary voltage is now about 350V and the transformer is saturated. In
saturation, the accuracy and coupling go out of whack so saturation does
limit voltages for part of the cycle. However, the transformer will be
saturated for only part of the cycle so the primary current and the
voltage across the transformer will have a decidedly spiky waveform.
There will still be voltage spikes seen on the secondary and these can
be high. N.B. This is an off the cuff analysis but I have seen
oscillograms of secondary currents of a saturated CT. Peak voltages
occur at the time when d(phi)/dt is highest so peak voltages on the
secondary will still be dangerously high.

That seems to answer my question rather well. I'm going off to look at
waveforms on an O/C secondary, and see if I can spot where saturation
happens. I just hope my 1500V probes will cope

--
Then there's duct tape ...
(Garrison Keillor)
nofr@sbhevre.pbzchyvax.pb.hx

 

[Ted Swift]

"InOverMyHead" <bbart@nospam.ix.netcom.com> wrote in message news:<fOktb.1124$sb4.175@newsread2.news.pas.earthlink.net>...

What am I looking for? I don't even know what to call it. But it would sense
temperature (or light or sound level or electrical current or ... ) and
input data (through serial or USB ports?) for capture into a database.
Presumably I would graph the results with Excel.

Going off-topic (OT), anyone got an inexpensive solution to this
related problem? Is there something like a Palm OS terminal emulator
that can be programmed with a macro to send out a character every,
say, 10 minutes, then log what comes back to a text file? I don't want
to tie up a full-size computer with this, but it needs enough smarts
to send and receive short strings, know roughly what time it is, etc.
A time stamp would be great, but not absolutely necessary. A used Palm
organizer seems like a great platform for this sort of thing (maybe
rigged with larger batteries to run for long periods). This could be
used for a bunch of applications, but what I have in mind at the
moment is this:
Quantum Research Group makes a nifty capacitive linear position
sensor chip (www.qprox.com/products/qt300_301.php), available for a
few bucks from DigiKey. You can configure it to chat on a one wire
serial line: You ping it with a short pulse train (a single character
should do, if you choose the right character), and it responds with a
16-bit position measurement (two 8-bit bytes with the usual start and
stop characters). Whatever terminal emulator I use would have to be
able to deal with *any* value from 0-255, which might be tricky, since
there be dragons such as ESC, RET, etc. that will cause weirdness in
the logged text file, but most of that could be handled later. I've
programmed in a handful of languages and systems, but haven't done any
Palm OS programming, and would like to avoid reinventing the wheel if
possible. There is a LOT of Palm shareware out there, but I haven't
located what I described above. Ideas? Pointers? Thanks.
-Ted

 

[Dennis Clark]

In sci.electronics.misc Ted Swift <tjswift@ucdavis.edu> wrote:

I've used the Palm m100 running hotpaw basic to talk to an OOPic data
logger. Talk about simple and very capable, and very, very cheap.

DLC

: "InOverMyHead" <bbart@nospam.ix.netcom.com> wrote in message news:<fOktb.1124$sb4.175@newsread2.news.pas.earthlink.net>...
:> What am I looking for? I don't even know what to call it. But it would sense
:> temperature (or light or sound level or electrical current or ... ) and
:> input data (through serial or USB ports?) for capture into a database.
:> Presumably I would graph the results with Excel.

: Going off-topic (OT), anyone got an inexpensive solution to this
: related problem? Is there something like a Palm OS terminal emulator
: that can be programmed with a macro to send out a character every,
: say, 10 minutes, then log what comes back to a text file? I don't want
: to tie up a full-size computer with this, but it needs enough smarts
: to send and receive short strings, know roughly what time it is, etc.
: A time stamp would be great, but not absolutely necessary. A used Palm
: organizer seems like a great platform for this sort of thing (maybe
: rigged with larger batteries to run for long periods). This could be
: used for a bunch of applications, but what I have in mind at the
: moment is this:
: Quantum Research Group makes a nifty capacitive linear position
: sensor chip (www.qprox.com/products/qt300_301.php), available for a
: few bucks from DigiKey. You can configure it to chat on a one wire
: serial line: You ping it with a short pulse train (a single character
: should do, if you choose the right character), and it responds with a
: 16-bit position measurement (two 8-bit bytes with the usual start and
: stop characters). Whatever terminal emulator I use would have to be
: able to deal with *any* value from 0-255, which might be tricky, since
: there be dragons such as ESC, RET, etc. that will cause weirdness in
: the logged text file, but most of that could be handled later. I've
: programmed in a handful of languages and systems, but haven't done any
: Palm OS programming, and would like to avoid reinventing the wheel if
: possible. There is a LOT of Palm shareware out there, but I haven't
: located what I described above. Ideas? Pointers? Thanks.
: -Ted

--
============================================================================
* Dennis Clark dlc@frii.com www.techtoystoday.com *
* "Programming and Customizing the OOPic Microcontroller" Mcgraw-Hill 2003 *
============================================================================