Chip with simple program for Toy

[John Bokma]

Anthony Fremont wrote:

"John Fields" wrote
On Sat, 23 Apr 2005 05:13:57 GMT, "Anthony Fremont"

I came up with the same calculation as you. If LEDs are dieing, then
20mA may be a bit too much current. Since there are two LEDs in
series,
one may hog more current than the other resulting in its demise.

---
So, you're saying that because the LEDs may not be identical one may
be drawing more current than the other?

Perhaps "dissipate more power" would have been more appropriate than
"hog more current".

Since they are in series, yes. And this is possible if the voltage *over*
the LED differs (which it very likely does). However the current should
stay 20 mA, and due to the resistor, one LED can only have a higher voltage
if another LED has less. The current will stay the same though.

It's like you can hog water in an open tube, it has to go somewhere.

--
John MexIT: http://johnbokma.com/mexit/
personal page: http://johnbokma.com/
Experienced programmer available: http://castleamber.com/
Happy Customers: http://castleamber.com/testimonials.html

 

[John Fields]

On Sat, 23 Apr 2005 10:51:19 -0700, Peter Bennett
<peterbb@somewhere.invalid> wrote:

On 23 Apr 2005 04:41:48 -0700, Tommi-Vogel@gmx.de (Thomas Vogel)
wrote:


You´re right, but for me it had been a big puzzle work, to know how I
must construct the elements, if I don`t have seen the schematic with
the elements in the right order, here while I have answear you.

What are comparators? I don´t have find it in my dictornary. Can you
explain it?

A comparator is an integrated circuit which has two inputs (called +
and -), and one output. When the + input is more positive than the -
input, the output will be high. If the - input is more positive, the
output will be low.

And what is (COIL) and (1N4148)? Is the last one a diode?

(COIL) is the coil of the relay, and (1N4148) is a diode. The "K"
indicates the cathode end of the diode.



And who are these conected? * *

| | +-|+\ | | | | | O--> |
| | | | >-+------|-|+\ | | |
[BAT] +-[0.1ľF]-+-----|-|-/ | | | >--+-------+ +----------->NO
| | | | | +-|-/


And what did this mean:

--|+\
| >---
--|-/

That is a comparator

You will have to go to the National Semiconductor website (or google
for "LMV393 datasheet") to get the datasheet for the comparator that
John suggested - The LMV393 is a dual comparator - there are two
separate comparators in the package. I would suggest using an LM393
instead, as the LMV393 is in a tiny surface mount package which will
be hard to handle.

---
I chose the LMV393 because (at 5V) its supply current is 200ľA max,
yet it can sink 20mA, so it beats the LM393 on both counts. But
you're right, it is surface mount, which the OP might not be able to
deal with. So, he gets to trade one off against the other or, maybe,
some kind soul will find him a DIP alternative.

--
John Fields
Professional Circuit Designer

 

[Larry Brasfield]

"aman" <aman.bindra@gmail.com> wrote in message
news:1114287466.891003.249680@f14g2000cwb.googlegroups.com...

So doesnt this imply that if there are free ions in water there will
be a different K from a nuetral sample of water with no free ions ?

I don't think so. The free ions will drift under the
influence of an E field, contributing in-phase current,
not lagged current such as displacement produces.
The only way free ions could change the K would be
if there were enough of them to act as a dielectric
and they had a K different than the H2O they displace.

I
am actually adding chemicals(ionic) to water which nuetralises harmful
ions in water to form a nuetral particle floc. So I need to detect zero
crossing. As i am adding ions into water i need to detect a zero
crossing in going from postive charge to negetive.

Sorry, but I cannot make sense of that. Maybe it
would help to show your circuit or block diagram.

--
--Larry Brasfield
email: donotspam_larry_brasfield@hotmail.com
Above views may belong only to me.

 

[Larry Brasfield]

"Larry Brasfield" <donotspam_larry_brasfield@hotmail.com> wrote
in message newsiyae.45$QL5.2154@news.uswest.net...

"aman" <aman.bindra@gmail.com> wrote in message
news:1114287466.891003.249680@f14g2000cwb.googlegroups.com...
So doesnt this imply that if there are free ions in water there will
be a different K from a nuetral sample of water with no free ions ?

I don't think so. The free ions will drift under the
influence of an E field, contributing in-phase current,
not lagged current such as displacement produces.
Woops, that should be:

not leading current such as displacement produces.

--
--Larry Brasfield
email: donotspam_larry_brasfield@hotmail.com
Above views may belong only to me.

 

[John Popelish]

aman wrote:

I just need 2.5V to be at one of the inputs of the opamp. I will be
using the virtual ground property of opamp getting 2.5V at the other
input of the opamp also. I dont need practically any current to be
drawn by the load because after opamp theren will be only a schmitt
trigger(comparator). So that you know this is part of an water overflow
detector. Still you think the regulator or voltage divider are good
ideas for this.

A simple two resistor divider may be all you need.
If only an opamp input is connected to it (no current load except the
opamp bias current) a pair of 100k resistors may be stiff enough.

 

[Lord Garth]

<cornytheclown@hotmail.com> wrote in message
news:1114294794.337813.297580@l41g2000cwc.googlegroups.com...

Ive seen a lot of industrial controllers such as plcs and they all seem
to use a signal of 4-20 ma in proportion to relay sensor information to
the controller.

My question is .....why 4-20 ma.....why not 1-10 ma or 1-50 ma... what
is the reasoning between using 4-20 milliamps

thanks

It probably comes from the old standard for a Teletype serial current loop.
I would guess that alarm systems also maintain a loop current in this range
in order to detect cut sensor lines.

 

[Larry Brasfield]

"aman" <aman.bindra@gmail.com> wrote in message
news:1114294879.893388.38930@l41g2000cwc.googlegroups.com...

I am adding salt to water. I read something in one of the forums below
which confuses me. It also says "adding virtually any contaminant to
water will alter its conductivity. That also ruins the diaelectric
constant.

I don't know what a ruined dielectric constant is.
That is gibberish as far as I am concerned.

Only pure water will have a measurable diaelectric constant."

http://forum.allaboutcircuits.com/lofiversion/index.php/t1636.html

It says that pure water is insulator and adding salt changes
conductivity. I am adding salts to contaminated water to which makes it
an insulator. But if more salts are added water crosses the point at
which it is an insulator(pure) and again starts conducting. So I need
to detect that zero crossing.

I don't understand the zero crossing. Pure water
conducts. There are always H+ and OH- ions
present, and they can carry current. What you
will see, if you can reduce free ions by adding
salts to react with already present ions so as to
produce a non-ionic product (a precipitate, I
presume), is a decrease in conductivity which
will approach the conductivity of pure water,
possibly followed by an increase if you add too
much. There will be no zero crossing.

To me, it looks like you should almost forget
about capacitance and measure conductance.
If you are using AC for that measurement, it
may be a good idea to measure only in-phase
current so as to ignore the capacitance.

But according to the forum in the link above, conductors cant be
dielectric.

For good conductors, the dielectric properties do
not matter and are difficult or impossible to measure.
A counter-example to that statement can be found
in any electrolytic capacitor. They all leak current,
so, because the material between their plates is a
conductor (albeit a poor one), and taking the above
as true, they cannot also be capacitors. Countless
devices say otherwise.

But what if I use conducting water as dielectric. What
happens. I know dielectric is supposed to support electric field but
oppose current. But if there is leakage current the capacitor can't
hold charge or what ?

The question should not be "Can it hold charge?" but
"How long can it hold charge?" Leakage limits the
time that a capacitor can hold charge.

Also what is meant by "Only pure water will have a measurable
diaelectric constant."

That overstates the case. If we are to believe that,
then adding one ion molecule to a vat of pure water
will turn it from being a dielectric to being a poor
conductor.

--
--Larry Brasfield
email: donotspam_larry_brasfield@hotmail.com
Above views may belong only to me.

 

[John Popelish]

cornytheclown@hotmail.com wrote:

Ive seen a lot of industrial controllers such as plcs and they all seem
to use a signal of 4-20 ma in proportion to relay sensor information to
the controller.

My question is .....why 4-20 ma.....why not 1-10 ma or 1-50 ma... what
is the reasoning between using 4-20 milliamps

thanks

There have been lots of competing signal standards over the years, but

this one has enough advantages to have become very common. First of
all, using current instead of a voltage eliminates errors from wire
resistance. This also allows several devices to be wired in series on
the same current loop, as long as the total supply voltage can account
for the sum of all their drops at maximum current.

Having an elevated (live zero) distinguishes between a zero signal and
a broken circuit, and allows off scale below zero readings, also. The
minimum power delivered to the field device at 4 milliamps minimum
current and a few volts drop is enough to power lots of micro power
devices from sensors to microprocessors to local LCD displays. And a
short circuit across the signal lines causes no particular damage.

There is a similar 10 to 50 ma standard, but the extra power it
consumes does not help with anything, since micro power instruments
have become available.

 

[John Popelish]

aman wrote:

I am adding salt to water. I read something in one of the forums below
which confuses me. It also says "adding virtually any contaminant to
water will alter its conductivity. That also ruins the diaelectric
constant . Only pure water will have a measurable diaelectric
constant."

A more accurate statement might have been that adding things to water
that increase its conductivity interferes with many dielectric
measurement methods.

Imagine measuring the value of a capacitor with a capacitance bridge.
The null is nice and sharp and determining the value with high
precision is easy. Now, connect various resistances in parallel with
that capacitor. The bridge null gets shallower and wider as more and
more of the instrument's bridge current detours through the resistance
and less and less of it passes through the capacitance. But
paralleling those resistors did not alter the dielectric constant of
the insulating material in the capacitor.

http://forum.allaboutcircuits.com/lofiversion/index.php/t1636.html

It says that pure water is insulator and adding salt changes
conductivity. I am adding salts to contaminated water to which makes it
an insulator. But if more salts are added water crosses the point at
which it is an insulator(pure) and again starts conducting. So I need
to detect that zero crossing.

But according to the forum in the link above, conductors cant be
dielectric. But what if I use conducting water as dielectric. What
happens. I know dielectric is supposed to support electric field but
oppose current. But if there is leakage current the capacitor can't
hold charge or what ?

Also what is meant by "Only pure water will have a measurable
diaelectric constant."

The key word, here, is "measurable". Using a higher AC excitation
frequency lowers the capacitive impedance while the Resistive
component is essentially unchanged. So measuring conductive samples
is more practical with a higher excitation frequency. The caution is
that the actual value of high dielectric materials, especially liquids
can vary dramatically as the excitation frequency passes through
motional resonances of the polar molecules. So you will need
calibration standards that you can use at the frequency you choose.

 

[Gareth]

Larry Brasfield wrote:

"aman" <aman.bindra@gmail.com> wrote in message
news:1114294879.893388.38930@l41g2000cwc.googlegroups.com...

I am adding salt to water. I read something in one of the forums below
which confuses me. It also says "adding virtually any contaminant to
water will alter its conductivity. That also ruins the diaelectric
constant.


I don't know what a ruined dielectric constant is.
That is gibberish as far as I am concerned.


Only pure water will have a measurable diaelectric constant."

http://forum.allaboutcircuits.com/lofiversion/index.php/t1636.html

It says that pure water is insulator and adding salt changes
conductivity. I am adding salts to contaminated water to which makes it
an insulator. But if more salts are added water crosses the point at
which it is an insulator(pure) and again starts conducting. So I need
to detect that zero crossing.


I don't understand the zero crossing. Pure water
conducts. There are always H+ and OH- ions
present, and they can carry current.

There will be some ions, but not many. Only about 1 in 500 million
water molecules in pure water break up into H+ and OH- ions.

Pure water is actually a good insulator (conductivity < 0.06 uS/cm)


--
-----------------------------------------------------------------------
To reply to me directly:

Replace privacy.net with: totalise DOT co DOT uk and replace me with
gareth.harris

 

[mike]

mjohnson wrote:

I want to build an automatic garage door closer with an alarm clock and
my garage door remote. I realize that I could just buy something but I
want to build it so I can learn something and have some fun (and
frustration).

Here is a block diagram of what I am imagining:
http://img98.echo.cx/img98/3411/phase12nt.jpg

My question is what do I need to do to take the output voltage at the
clock's buzzer to activate the interface circuit. The voltage I read
on the buzzer when it's going off is 395mV (.395V). If for example, I
just want to turn on an LED (baby steps right) what would I need to do
to couple the alarm clock to the LED circuit?

I'm assuming that the actual coupling of the LED circuit to the buzzer
will represent a new load to the alarm clock which it wasn't designed
to take. So my guess is that I would need an opto-isolator and run the
LED circuit on it's own power supply? But is 395mV is enough to drive
the opto-isolator?

thanks for your time and help!

Why do you need the remote? If you're close enough to sense the door
state, you're close enough to use the wired actuation.

Only you and your insurance company can determine what should be in the
door sensor block.

I can tell you from personal experience that unattended operation of a
garage door is a BAD idea. Once had the door hang up on a gasoline can.
It hung up on the (sharp)edge of the door brace, so a sensor along the
bottom of the door wouldn't have helped.
Once had a broom handle hang up in the door guides. Kids, pets,
newspaper, bicycle...
It's no fun to have the door close as you're driving in from that late
night on the town.
Yes, if you put enough safety interlocks and timeouts, you can make it
as safe as you think you need.

A much easier/safer way would be to use the door state sensor to inhibit
the alarm. If the alarm goes off, get off your butt and go close the door.
More technology/automation is not always a good thing.

If you're determined to do this, stick a scope on the buzzer and see
what's actually there. Some clocks use DC on a buzzer module. Others
drive the piezo directly with AC. YMMV.

Also do a LOT of testing on what happens to the clock with various power
line glitches/outages, battery failure...Interesting things can happen
to a battery powered clock as temperature variations, as might be
encountered in a garage, change the battery voltage slightly near the
cutoff point.

This is one of those situations with a very low probability of having a
very BIG problem. Multiply the numbers and you'll feel very
safe...until it fails.
mike
--
Return address is VALID but some sites block emails
with links. Delete this sig when replying.
..
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Wanted 12" LCD for Compaq Armada 7770MT.
Bunch of stuff For Sale and Wanted at the link below.
MAKE THE OBVIOUS CHANGES TO THE LINK
ht<removethis>tp://www.geocities.com/SiliconValley/Monitor/4710/

 

[Watson A.Name - \"Watt Su]

"John Fields" <jfields@austininstruments.com> wrote in message
news:vcuk611q8nljblh2q4bd4tta7uvimhn3pl@4ax.com...

On Sat, 23 Apr 2005 05:13:57 GMT, "Anthony Fremont"
spam@anywhere.com> wrote:


"John Bokma" wrote:

snipped exactly what I was going to post

Note that I did this calculation, later checked it with a voltage
meter,
and yet already 2 exotic LEDs died on me. I think it's a bad badge.
(Unless someone can point out my errors).

I came up with the same calculation as you. If LEDs are dieing, then
20mA may be a bit too much current. Since there are two LEDs in
series,
one may hog more current than the other resulting in its demise.

---
So, you're saying that because the LEDs may not be identical one may
be drawing more current than the other?
---

Heh. Some advice handed out in the NG is really atrocious, ain't it?

You
might consider not driving them so hard. There is probably a
relatively
insignificant brightness difference between 10mA and 20mA anyway.

--
John Fields
Professional Circuit Designer

 

[Lord Garth]

"Watson A.Name - "Watt Sun, the Dark Remover"" <NOSPAM@dslextreme.com> wrote
in message news:116m36rp23in359@corp.supernews.com...

"John Fields" <jfields@austininstruments.com> wrote in message
news:vcuk611q8nljblh2q4bd4tta7uvimhn3pl@4ax.com...
On Sat, 23 Apr 2005 05:13:57 GMT, "Anthony Fremont"
spam@anywhere.com> wrote:


"John Bokma" wrote:

snipped exactly what I was going to post

Note that I did this calculation, later checked it with a voltage
meter,
and yet already 2 exotic LEDs died on me. I think it's a bad badge.
(Unless someone can point out my errors).

I came up with the same calculation as you. If LEDs are dieing, then
20mA may be a bit too much current. Since there are two LEDs in
series,
one may hog more current than the other resulting in its demise.

---
So, you're saying that because the LEDs may not be identical one may
be drawing more current than the other?
---

Heh. Some advice handed out in the NG is really atrocious, ain't it?

I found an author of an article published in Electronics Now commenting
about the voltage through the circuit...

 

[Don Kelly]

--
Don Kelly
dhky@peeshaw.ca
remove the urine to answer
"Larry Brasfield" <donotspam_larry_brasfield@hotmail.com> wrote in message
news:U5wae.22$QL5.1414@news.uswest.net...

"CNM" <charles.macleod@gmail.com> wrote in message
news:19123424.0504230929.efc95c8@posting.google.com...
Why does the effciency of my low voltage transformer very much
decrease (to eg 10%) when increasing the frequency to around 40000hz.
Why does this increase in frequency affect the efficiency value so
much?

The most likely reason is that eddy current losses
in the core go up and magnetizing current goes up.
In effect, the shunt inductance of the transformer
goes down as frequency goes up when eddy
currents become signficant. Those currents are
(nearly) in phase with the applied voltage and
represent a loss.
--------

Actually the "magnetising" or inductive component of the current decreases
because of lower flux density for a given voltage, at higher frequencies
(The inductance won't decrease much if at all) .but the total exciting
current increases due to hysteresis and eddy current loss increases. This
increased loss current will coupled with skin effects will also lead to
higher I^R loss.
---------------

You probably have a transformer designed for
line frequency with the lamination thickness set
accordingly. At 40 KHz, that thickness is way
too large.
-----------

Right on- design a 40KHz transformer for 40KHz- don't use a 60Hz transformer
and expect good operation. It is more than just lamination thickness, it is
excessive iron.volume as well as capacitive effects. It is a wonder if a
60Hz transformer is not completely useless at 40KHz.>

--
--Larry Brasfield
email: donotspam_larry_brasfield@hotmail.com
Above views may belong only to me.

 

[John Bokma]

Lord Garth wrote:

I found an author of an article published in Electronics Now
commenting about the voltage through the circuit...

Batteries in toy racing cars?

--
John MexIT: http://johnbokma.com/mexit/
personal page: http://johnbokma.com/
Experienced programmer available: http://castleamber.com/
Happy Customers: http://castleamber.com/testimonials.html

 

[Larry Brasfield]

"Don Kelly" <dhky@peeshaw.ca> wrote in message
news:a7Eae.1121148$8l.65597@pd7tw1no...

"Larry Brasfield" <donotspam_larry_brasfield@hotmail.com> wrote in message
news:U5wae.22$QL5.1414@news.uswest.net...
"CNM" <charles.macleod@gmail.com> wrote in message
news:19123424.0504230929.efc95c8@posting.google.com...
Why does the effciency of my low voltage transformer very much
decrease (to eg 10%) when increasing the frequency to around 40000hz.
Why does this increase in frequency affect the efficiency value so
much?

The most likely reason is that eddy current losses
in the core go up and magnetizing current goes up.
In effect, the shunt inductance of the transformer
goes down as frequency goes up when eddy
currents become signficant. Those currents are
(nearly) in phase with the applied voltage and
represent a loss.
--------
Actually the "magnetising" or inductive component of the current decreases
because of lower flux density for a given voltage, at higher frequencies
(The inductance won't decrease much if at all) .but the total exciting
current increases due to hysteresis and eddy current loss increases.

I agree with your correction on terminology. I was
incorrectly referring to the shunt current term as
"magnetising current". As for inductance not being
reduced, I have to disagree. If you consider that
eddy current prevents flux from penetrating into
the interior of the laminations, you should be able to
see that the effective area of the core is reduced.

This effect shows up as a non-polynomial relation
between impedance and frequency, such that the
impedance rises approximately as sqrt(frequency).
If you take the imaginary part of the impedance in
any part of that curve, you will find that inductance
(defined as that component divided by radians/S)
is indeed decreasing with frequency.

This
increased loss current will coupled with skin effects will also lead to
higher I^R loss.

Yes.

You probably have a transformer designed for
line frequency with the lamination thickness set
accordingly. At 40 KHz, that thickness is way
too large.
-----------
Right on- design a 40KHz transformer for 40KHz- don't use a 60Hz transformer
and expect good operation. It is more than just lamination thickness, it is
excessive iron.volume as well as capacitive effects. It is a wonder if a
60Hz transformer is not completely useless at 40KHz.

Agreed.

--
--Larry Brasfield
email: donotspam_larry_brasfield@hotmail.com
Above views may belong only to me.

 

[Bob Eldred]

<cornytheclown@hotmail.com> wrote in message
news:1114294794.337813.297580@l41g2000cwc.googlegroups.com...

Ive seen a lot of industrial controllers such as plcs and they all seem
to use a signal of 4-20 ma in proportion to relay sensor information to
the controller.

My question is .....why 4-20 ma.....why not 1-10 ma or 1-50 ma... what
is the reasoning between using 4-20 milliamps

thanks

Several reasons: One, The current to the device powers the device and it
cannot to go below some minimum or there would be insufficient power for the
device. That's why it's not zero to 20mA. Four mA was chosen as a
convenient minimum still allowing enough to run the device. Having the
current power the device allows the device to operate on only two wires. It
would take a minimum of three wires for a voltage output device and
sometimes four wires. You need power, ground and signal and maybe signal
ground. A current 4 to 20mA device has +V usually 24 volts and signal and
that's it, two wires.

Second: The current in a line is always the same, barring leakage, Therefore
there are no losses regardless of the length of the line. The resistance of
the line does not matter.

Third: The signal is loaded at the receiving end of the line with a sense
resistor, often 100 ohms or 250 ohms. These are low impedance loads and help
keep down noise and interference. Voltage lines, by contrast, are usually
high impedance loads and are more susceptable to noise.

Fourth: One could go to higher current but that would be wasteful especially
in a system with hundreds of channels. Twenty milliamps is sufficient for
most purposes. Four to twenty milliamps gives one to five volts on a 250 ohm
load, a good range for data acquisition hardware. That still leaves 19 volts
to run the device on the transmitting end of the line with a 24 volt supply.
Bob

 

[Roger Johansson]

dB wrote:

The resistor limits the current, it has no direct effect on the
voltage developed across each l.e.d. The actual value across each
l.e.d. varies from device to device at any current. The data sheets
give a "typical" Vf and sometimes a max figure.

In this case we should not trust the value, 3.6V, which was
given earlier. That should be checked with a voltmeter in reality.

If these LEDs have a lower voltage that would explain why some suddenly
die.

Send 10mA through a LED and measure the voltage over it, and the
voltage over the resistor.
Do the math and find out how much current is passing the resistor, and
the LED.



--
Roger J.

 

[Anthony Fremont]

"Watson A.Name - "Watt Sun, the Dark Remover"" <NOSPAM@dslextreme.com>
wrote in message

Heh. Some advice handed out in the NG is really atrocious, ain't it?

I guess the restatement I made is not good enough. Should I start
another thread and offer a formal apology to the world for making such a
heinous mis-statement about current vs. dissipation? Maybe I could help
make amends by belittling others, nit-picking posts and posting a bunch
of OT crap?

Lets see if we can't get on to the road to recovery now. Speaking of
good advice, why are you trying to get a poster to use non-rechargeable
alkaline batteries when he clearly expressed a preference for
rechargeable? Hey Fields, are you ever going to acknowledge/correct
your mistake in S.E.D about max collector current on the 2N4401?
sheez....

 

[Andrew Holme]

eeh wrote:

Hi,

I saw the datasheet of pulse H1102 ethernet magnetic but cannot
understand the theory.

I want to know what is the function of the transformers interfacing to
the RJ45 ethernet network. Is it used to reject common mode noise?

Yes; and for isolation - to protect sensitive inputs from different mains
earth voltages around the building.

Besides, why do I need to use the magnetic to interface to ethernet?

Well, if you're just experimenting - with short CAT5 cable runs - you don't.
This guy connected the pins of an FPGA directly to his hub for transmitting,
and just used a simple discrete-transistor differential-amplifier for
receiving:

http://www.fpga4fun.com/10BASE-T.html

I wouldn't do it without magnetics myself. It's not that hard. I used the
SI-10021 RJ45 connector with integral magnetics from RS Components in this
project:

http://www.holmea.demon.co.uk/Ethernet/EthernetRx.htm