Chip with simple program for Toy

[billcalley]

Thanks for all of your help guys!

Cheers,

Bill

 

[Dr1]

but I only get about 3 charges before the 12V has dropped too far.
You need to use a power supply. Using a battery to supply voltage to a
battery charger is a no win situation. It works with field chargers
only because the supply battery is MUCH larger than the battery being
charged, and can be kept peaked by the car's engine.

(about 70Ah).
Good luck finding a car battery that small. They usually come in
ratings of 600Ah or more.

The question is; will I be able to charge the 12V car battery with the
same charger that I use to charge the 12V sealed lead acid?
Yup <giggle> but you may be there for days.

Dr.1

 

[obliquez]

Chris wrote:

obliquez wrote:
Thank you guys so much for your help so far!

I'm sorry that i seem to be ignorant (which i am) and slow (that too)
and i don't quite understand all of which you guys are trying to
explain to me. I don't really describe my problems well either. so
thanks alot for your patience.

The reason why i laid out my problems in parts, is because i thought it
would be easier for me to solve one step at a time. I apologise for the
trouble i caused.

So i shall now try to summarize everything.

For my project, I am supposed to embed an LLE sensor which will sound a
buzzer when the liquid reaches that certain level in the mug. A temp
sensor, or thermistor, which will sound the buzzer (I think it's
another buzzer, my teacher did not clarify this with me) if the liquid
is too hot. And lastly, a tilt sensor, which will either 1. turn off
the LLE and temp sensor or 2. turn off the buzzer when the mug is in a
tilted position.

My 1st problem was not knowing how to get a suitable buffer/driver
circuit to connect the LLE sensor and buzzer. But from all your help, i
have decided to use the PNP with the 2 10k resistors. If that works
best.

Now i have to figure out how to connect everything together and make it
work as a single unit. I have yet to find a suitable temp
sensor/thermistor or tilt sensor.

Btw, I'm sorry not to have mentioned this, but i only noticed it
yesterday when i was testing the LLE sensor. The LLE sensor has 4
wires, not 3. REd, blue, green and black. What the heck is the black
one for? I hope this doesn't create another problem.

erm.. is this laying out of my problem better?

Lastly, what is an OP? Does it refer to me? If it does.. I'm a her not
a he. -smiles-

Thank you so so much!!

Hello again. First off, you don't seem to have a Honeywell LLE sensor
after all. I believe they make six different models, all of which have
identical electrical circuits and only three wires. The differences
are solely in housing types for different applications.

I think you've got a Honeywell LL-type sensor, which accounts for the
four wires. They make nine different kinds of LL-type sensors, and I
believe all of them are electrically the same and have 4 wires. The
Honeywell datasheets are typically inscrutable, but the installation
guides are usually a bit better. I've never been happy with Honeywell
industrial documentation on anything they make, I'm afraid. Usually
that would be a killer, but they happen to make really good stuff, for
the most part. That makes up for it, and their distributors can
usually answer any questions eventually. Look at this:

http://content.honeywell.com/sensing/prodinfo/liquidlevel/installation/xp4025-1.pdf

It shows a device that operates on a 5VDC to 16VDC supply, and has an
internal voltage regulator. The functions of the four wires on the
LL-type sensors are as follows:

* Red Vcc (+5VDC to +16VDC)
* Blue 0VDC
* Green Schmitt Trigger Output (internal open-collector transistor
with an internal 10K pullup to Vcc)
* Black To internal LED (use series resistor to Vcc such that I(LED)
= 30mA nominal -- do not exceed 40 mA. I(f)(LED)(typ) = 1.2V

This is kind of a critical problem here. You have to know what you're
working with, unless you want to let the smoke out. However, if you do
have this sensor, your job is now easy:

` VCC VCC VCC
` + + +
` | | |
` .-. | |
` R1| | | .-. VCC
` | | Red| R2| | +
` '-' .----o-----. | | |
` | | | '-' |
` | Black| LL |Green ___ | |
` '-------o Sensor o-----|___|--o----|2N3906
` | | R2 |\
` | | |
` '----o-----' |
` Blue| |
` | |
` === / \
` GND (BZ1)
` \_/
` |
` |
` ===
` GND
(created by AACircuit v1.28.5 beta 02/06/05 www.tech-chat.de)

By the way, I believe but can't determine for sure if the output is
active-high or active-low. By that I mean whether the logic level is 1
(Vcc) or 0 (0V) when water is present in front of the sensor. In most
all sensors of this type, it's active-low, and the driver above works
for that (turns the buzzer on when the sensor output goes low).
However, if you want the reverse effect for some reason (buzzer ON when
sensor output is HIGH), or if I'm wrong in my assumption that the
output is active-low, the internal 10K pullup makes your driver circuit
the easiest yet:

` VCC VCC VCC
` + + +
` | | |
` .-. | / \
` R1| | | (BZ1)
` | | Red| \_/
` '-' .----o-----. |
` | | | |
` | Black| LL |Green |/
` '-------o Sensor o------| 2N3904
` | | |
` | | |
` '----o-----' |
` Blue| ===
` | GND
` ===
` GND
(created by AACircuit v1.28.5 beta 02/06/05 www.tech-chat.de)

The internal 10K pullup resistor means that when the sensor output is
high, it will source enough current into the base of the NPN transistor
so that it will turn on for a 10mA load quite well.

Since you're supposed to be pulling 30mA for the LEDs, I wouldn't think
a few coin cells would last that long. They're basically intended for
a load of a mA or so at best. You might do better with a 9V
"transistor" battery -- that's more along the lines of what's
considered a normal load for them. The 9V battery will also be cheaper
in the long run -- you can get a lot more power from one of these than
a couple of coin cells.

R1 is used to regulate LED current. The installation sheet suggests
120 ohms, 1/4 watt for a 5V supply. If you're using a 9V battery, you
should have a 270 ohm resistor (1/4 watt is acceptable, 1/2 watt is
preferred). If you've got a 12VDC supply, you should use a 390 ohm,
1/2 watt resistor. R2 (2 ea.) can be 10K, 1/4 watt for all values of
Vcc.

Now, most tilt switches are just normally open mechanical switches,
where either a metal ball or a blob of mercury will close a connection
if the device tilts. You can just connect the normally open switch
from the base to the emitter of the transistor for either of the
diagrams above. That way, when the tilt switch turns on, it will
prevent the transistor (and buzzer) from turning on. Simple. If your
tilt switch is normall closed (the contacts open when the device tilts)
you can either use it to interrupt the power supply to the LL sensor,
or interrupt the flow of current to the base of the transistor. That
works for either of the above diagrams, too. Again, simple.

From your control logic, I'm not sure you actually want the hot alarm
to be disabled when someone's about to scald their lips. So, all we
have to do to complete this is determine what you're going to use as a
temperature sensor. Please post back and let us know what the teacher
suggests, or if you've already bought something.

I can say that sensing temp with a thermistor will require an EXACT
part number. Thermistor resistances are usually specified at room temp
(25C), and have an almost infinite number of different resistance value
vs. temp. graphs. If you have to use a thermistor, you will need to
make a current source to create a voltage across the thermistor
(usually just another transistor and a couple of resistors will
suffice), and an IC comparator to compare the voltage across the
thermistor with a contrived voltage which would represent overtemp. If
the thermistor voltage exceeds the setpoint, the comparator can turn on
and drive another small beeper directly.

As I said in an earlier post, it might be easier for a newbie to spring
for an LM34 IC, which works on 5V to 30VDC as a power supply, and
outputs 10mV per degree. You can then feed that voltage directly into
the comparator without worrying about thermistor curves. You may have
a bit of a problem in that the LM34 looks like a TO-92 transistor and
is not immersible (it will short out your power supply), so you might
have to glue the sensor to the bottom or outside of the cup. Actually,
you usually need to purchase special thermistors which are made with a
watertight sheath, allowing them to be immersed, too. Anyway, you need
to find this stuff out, and get back to us.

You're almost there. Get back to us with more information about
whether you've got an LL-type level sensor, and the type and specific
part number of the temp sensor you need.

Good luck
Chris

hi, just checking in over the weekend. I'm having a migraine now, so i
can't absorb a single thing Chris just posted. But, i will read it
again tommorrow. Just wanna tell ya all that, my name is Jean. -smiles-

OP = Original Poster, lol.. -slaps forehead- Now i know.. thanks John
Popelish for the heads up.

Jon, the tilt sensor is supposed to prevent the buzzer from going off
when the mug is in a tilted position. Which ever way is easier. But i
think it's either by, turning out the LLE and temp sensor when the mug
is tilted. or turning off the buzzer.

The tilt sensor was not specified. The LLE sensor was approved by my
teacher. He asked me to find a suitable sensor by using search engines.
Then i will print out the specifications of the component, and he will
decide whether or not it can be used.

I will read the all the post again tommorrow. And try to answer and
clarify all the fuzzy details. Thanks for bearing with me guys.

 

[Grant Edwards]

On 2005-06-28, Steve Banks <none@here.isfine> wrote:

| weight on the pedal for 1200/200 seconds or 6 SECONDS!

Your understanding of the idea of work (in the physics sense) is very
flawed bla bla bla.

I guess you missed the relationship between thrust and time.
Hint: pretend the airplane is just a fancy rocket. In this
example it is putting out 2 lbs. of thrust, for 600 seconds,
we are disregarding the weight of the rocket.

Yes, you actually have to be moving the pedal with your entire
weight for those 6 seconds, but with a spring and proper
gearing or leverage, you can reproduce the same two pound push
for 600 seconds.

Wrong.

Gears and levers do not provide a ratio between thrust/time and
thrust/time. The ratio provided by gearing/leverage provides
ratios of thrust/distance -- which is what was explained to you
using the formula for work.

Reaching for the work formula will not help you see this relationship

Because there is no such relationship.

without a bit of algebra, and that woud be contrary to my
stated goal of a quick guess .

Quick, but not founded in real physics.

--
Grant Edwards grante Yow! Boys, you have ALL
at been selected to LEAVE th'
visi.com PLANET in 15 minutes!!

 

junk1@davidbevan.co.uk wrote:

This may sound a bit of a silly question, but currently I use one of
the small 12V 7Ah sealed lead acid batteries to charge my lipos with
(via a lipo charger) - but I only get about 3 charges before the 12V
has dropped too far.

...so the obvious solution is to buy a 12v car battery (about 70Ah).

The question is; will I be able to charge the 12V car battery with the
same charger that I use to charge the 12V sealed lead acid?

As I understand it a car battery is just a lead acid battery? So same
technology and same voltage just higher Ah. Will I blow myself up if I
use the small 12v lead acid charger on the big 12v car batt?


Thanks

David Bevan
http://www.davidbevan.co.uk

I hope you can forgive me. This way off topic. I don't know what
happened, but I have been unable to log onto this usenet for the
longest time and am just trying to see if I could do it. Strangely, now
I can whereas I could not for almost a year.

However I do know people have been using car batteries for charging Li
Polies sucessfully. They just use a 12 volt charger to recharge the
batery. No problems.

Wan

 

Andrew

I have been away from this discussion group for a long. Now I am on a
roll. Good to be back. I thought everyone knows what a Lipo (Lithium
Polymer) is by now. This is the type of battery that is all the rage
now. Beats NiMh, Nicad. Lipos are lighter, smaller, longer flight
times.

I am by far not an expert, been using them for about 3 years. Loved
them.

Wan

 

Andrew

I have been away from this discussion group for a long. Now I am on a
roll. Good to be back. I thought everyone knows what a Lipo (Lithium
Polymer) is by now. This is the type of battery that is all the rage
now. Beats NiMh, Nicad. Lipos are lighter, smaller, longer flight
times.

I am by far from being an expert, been using them for about 3 years.
Loved them.

Wan

 

So thanks for lending validity
to my theory, which BTW didn't take me years of >specialized college
coursework to realize.

You need years of specialized college coursework, or actual
experimentation with human-powered generators, to realize just how far
off you are. Any kid who has visited a science center and pedaled one
of those little generators to power a small television (maybe 75 or 100
watts) quickly understands that it takes a LOT of work to generate any
significant amount of juice. Makes one appreciate cheap electricity...

Dan

 

[redbelly]

Dr Engelbert Buxbaum wrote:

Shannons theorem says that in order to see a signal with frequency n,
the sampling rate needs to be at least 2*n.

Fred Abse wrote:
That's Nyquist's sampling theorem, surely.

Yep. And if the 2*f sampling is 90 degrees out of phase with the
signal (namely, sampling at the zero-crossings), then the signal will
not appear even then.

Mark

 

[Rich Grise]

On Wed, 06 Jul 2005 12:55:51 -0500, John Fields wrote:

On Wed, 06 Jul 2005 17:18:31 GMT, "Mook Johnson" <mook@mook.net
"John Fields" <jfields@austininstruments.com> wrote in message
news:64hnc15fnbahrmlf8lop7j0g7cvf2glqv4@4ax.com...
On Wed, 06 Jul 2005 02:17:26 GMT, "dimention11"
dimention11@bigpond.com> wrote:
"John Popelish" <jpopelish@rica.net> wrote in message
news:j72dnfJ7ZIMqqVbfRVn-uw@adelphia.com...
dimention11 wrote:
hello i have an analog 0-5v i wish to change to a digital (ttl) 0-5v.
I assume it is as simple as hooking up a transistor.
am i right...

Might be. At exactly at what input voltage do you want the TTL output
to
change from low to high. Then, at exactly what input voltage do you
want
the output to transition back from high to low?

I will want this to switch from approx 2v to on(5v)
if this can be adjusted it would be even better.

Try this: (View in a non-proportional font like Courier)

+5V>-------+---------+--------+
| | |
| | [1k]
| | |
| +---|-[100K]-+
| | | |
VIN>--[1K]-------+--|+\ |
| | >------+-->VOUT
[10k]<--+--|-/ LM393
| *| |
| *| |
GND>-------+-----+---+----------->GND


Adjust the switching voltage with the 10k pot and be sure to ground
the inputs and the output of the other comparator.


looks good but omit the connection from vin- to ground.

What are you talking about?

Probably the one I've marked with the '*'s.

And please bottom post. It makes it much easier to follow a
discussion.

I've also crossposted this to sci.electronics.basics, because
others could benefit from the discussion so far, and it kinda
belongs there anyway.

Incidentally, John - I noticed your hysteresis feedback, and
I concur, albeit one might have to do some calculations to
get a specific amount. To my techie eye, it looks like
about 1%, which is definitely enough to keep it from
osculating[sic]. ;-)

Cheers!
Rich

 

[John Fields]

On Wed, 06 Jul 2005 23:38:06 GMT, Rich Grise <richgrise@example.net>
wrote:

On Wed, 06 Jul 2005 12:55:51 -0500, John Fields wrote:
On Wed, 06 Jul 2005 17:18:31 GMT, "Mook Johnson" <mook@mook.net
"John Fields" <jfields@austininstruments.com> wrote in message
news:64hnc15fnbahrmlf8lop7j0g7cvf2glqv4@4ax.com...
On Wed, 06 Jul 2005 02:17:26 GMT, "dimention11"
dimention11@bigpond.com> wrote:
"John Popelish" <jpopelish@rica.net> wrote in message
news:j72dnfJ7ZIMqqVbfRVn-uw@adelphia.com...
dimention11 wrote:
hello i have an analog 0-5v i wish to change to a digital (ttl) 0-5v.
I assume it is as simple as hooking up a transistor.
am i right...

Might be. At exactly at what input voltage do you want the TTL output
to
change from low to high. Then, at exactly what input voltage do you
want
the output to transition back from high to low?

I will want this to switch from approx 2v to on(5v)
if this can be adjusted it would be even better.

Try this: (View in a non-proportional font like Courier)

+5V>-------+---------+--------+
| | |
| | [1k]
| | |
| +---|-[100K]-+
| | | |
VIN>--[1K]-------+--|+\ |
| | >------+-->VOUT
[10k]<--+--|-/ LM393
| *| |
| *| |
GND>-------+-----+---+----------->GND


Adjust the switching voltage with the 10k pot and be sure to ground
the inputs and the output of the other comparator.


looks good but omit the connection from vin- to ground.

What are you talking about?

Probably the one I've marked with the '*'s.

---
Geez, I completely missed that! Thanks.
---



--
John Fields
Professional Circuit Designer

 

[Ed Price]

"Rudolf Wiesendanger" <ruediw@lycos.com> wrote in message
news:dat31c$vcq$1@news.hispeed.ch...

Hi folks,
i have some questions abaut international standards.

If a company produces electronic products for the international market,
to which standards have this product to complie ?

Who controls this ?

Thank You all for your responses
R.Wiesendanger

There is no such thing as "the international market." Your product must
comply with the regulatory standards of the country in which you wish to
sell the product. If the market is the USA, then the FCC (and possibly other
groups, depending on what your electronic product is and does) would have
requirements. If the market is the EU, then there are a number of Directives
that would be applicable. China is another game, so is Australia.

Identify the market and identify the product's function and users to get
more specific advice.


--
Ed
WB6WSN
El Cajon, CA USA

 

[Pooh Bear]

Ed Price wrote:

"Rudolf Wiesendanger" <ruediw@lycos.com> wrote in message
news:dat31c$vcq$1@news.hispeed.ch...
Hi folks,
i have some questions abaut international standards.

If a company produces electronic products for the international market,
to which standards have this product to complie ?

Who controls this ?

Thank You all for your responses
R.Wiesendanger

There is no such thing as "the international market."

No ? Getting pretty damn close !

Your product must
comply with the regulatory standards of the country in which you wish to
sell the product. If the market is the USA, then the FCC (and possibly other
groups, depending on what your electronic product is and does) would have
requirements. If the market is the EU, then there are a number of Directives
that would be applicable. China is another game, so is Australia.

No ! You're wrong.

Identify the market and identify the product's function and users to get
more specific advice.

With the exception of the USA. IEC norms apply. End of story. The USA is slowly
applying IEC norms in UL etc standards too.

Graham

 

[OBones]

Pooh Bear wrote:

With the exception of the USA. IEC norms apply. End of story. The USA is slowly
applying IEC norms in UL etc standards too.

Uho, the EEC has it's own standards with respect to safety. Sure, most
of them are based on IEC, but some of them are much more restrictive.
For instance, the plugs in the US or Australia could not even make it to
market in Europe, they are not safe enough. I mean, it's way too easy to
touch the metal part if the plug is not fully inserted.
And that's only one example.

 

[Pooh Bear]

OBones wrote:

Pooh Bear wrote:

With the exception of the USA. IEC norms apply. End of story. The USA is slowly
applying IEC norms in UL etc standards too.

Uho, the EEC has it's own standards with respect to safety. Sure, most
of them are based on IEC, but some of them are much more restrictive.
For instance, the plugs in the US or Australia could not even make it to
market in Europe, they are not safe enough. I mean, it's way too easy to
touch the metal part if the plug is not fully inserted.
And that's only one example.

National plugs and sockets aren't set by IEC norms. Just the preferred 'equipment
end' as per IEC 320 etc....

Graham

 

[Joe McElvenney]

Hi,

For instance, the plugs in the US or Australia could not even make it to
market in Europe, they are not safe enough. I mean, it's way too easy to
touch the metal part if the plug is not fully inserted.

I notice that most 3-pin sockets in the U.S. have the ground pin
at the bottom which is opposite to the U.K. norm. From a safety angle
I would have thought this to be a 'no-no'. Turning it round would
put the ground pin where 'little fingers' and dropped objects would
be most likely to make contact instead of the live pin as now.

Am I missing something then, apart from numerous brain cells?


Cheers - Joe

 

[Chris]

"OBones" <obones_gfd_@_gfd_altern.org> wrote in message
news:datcc5$i38$1@reader1.imaginet.fr...

Pooh Bear wrote:

With the exception of the USA. IEC norms apply. End of story. The USA is
slowly
applying IEC norms in UL etc standards too.

Uho, the EEC has it's own standards with respect to safety. Sure, most
of them are based on IEC, but some of them are much more restrictive.
For instance, the plugs in the US or Australia could not even make it to
market in Europe, they are not safe enough. I mean, it's way too easy to
touch the metal part if the plug is not fully inserted.
And that's only one example.

Not true.

In Australia since March it is mandatory for insulated plugs. Beside that
the configuration is different and they wouldn't fit anyway.

Cheers.

 

[OBones]

Chris wrote:

Not true.

In Australia since March it is mandatory for insulated plugs. Beside that
the configuration is different and they wouldn't fit anyway.

Oh, I didn't know it had changed. That's good news for it was very
common to see the pins accessible by a few millimeters between the plug
and the outlet.
But is it only for new devices, or do everyone have to change their
hardware?

 

[Rudolf Wiesendanger]

Ed Price wrote:

"Rudolf Wiesendanger" <ruediw@lycos.com> wrote in message
news:dat31c$vcq$1@news.hispeed.ch...
Hi folks,
i have some questions abaut international standards.

If a company produces electronic products for the international
market, to which standards have this product to complie ?

Who controls this ?

Thank You all for your responses
R.Wiesendanger


There is no such thing as "the international market." Your product
must comply with the regulatory standards of the country in which you
wish to sell the product. If the market is the USA, then the FCC (and
possibly other groups, depending on what your electronic product is
and does) would have requirements. If the market is the EU, then
there are a number of Directives that would be applicable. China is
another game, so is Australia.

Identify the market and identify the product's function and users to
get more specific advice.

So i will be more specific. The products are for water analysing
(conductivity, oxygen, ...) and used in power plants.

What requirement would this product have in the USA ?
Does the FCC control if the product comply with the
regulatory standard used un the USA ?

What requirement would this product have in the EU ?
Who controls if the product comply with the
regulatory standard used in EU ?

Thank you for your answers
Rudolf

 

On Mon, 11 Jul 2005 06:19:24 +0000 (UTC), "Rudolf Wiesendanger"
<ruediw@lycos.com> wrote:

Hi folks,
i have some questions abaut international standards.

If a company produces electronic products for the international market,
to which standards have this product to complie ?

Who controls this ?

Thank You all for your responses
R.Wiesendanger

Approximately, the relevant standards that are current in every
country in which the company wishes to sell the product. That can get
quite complicated!
Regards
JonH