Chip with simple program for Toy

[Dmitri(Cabling-Design.com]

Steve Evans wrote:

Do yourself a favor and buy a few dozen SMD capacitors or resistors at
the same time. They're great to practice the technique with and it
doesn't matter if you damage them since they're *so* cheap. When your
confidence is gained, go on to the LEDs.

Good point!

I knew quick soldering will be a problem at first, so I got myself ceramic
body 1206-type LEDs for starters.

Where do you stock up on discrete SMD components, anyways? They are rather
chap on eBay but the problem is variety: I don't really need 10,000 of the
same type on a tape, I may hardly use 100 in a year or so. What's the best
strategy to procure the small amounts?

--
Dmitri Abaimov, RCDD
http://www.cabling-design.com
Cabling Forum, color codes, pinouts and other useful resources for
premises cabling users and pros
http://www.cabling-design.com/homecabling
Residential Cabling Guide
-------------------------------------



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[Roger Johansson]

John Popelish <jpopelish@rica.net> wrote:

Saying that the batteries are 3 amps each is a bit nonsensical.

Maybe a fire extinguisher could be a good analogy for a battery.

It can deliver a certain amount of liquid and the pressure is high if it
newly filled.

If it contains 10 liter of water it can deliver a current of 1 liter per
second for 10 seconds. But the flow is not constant through these 10
seconds, the pressure and the current is a little higher in the beginning
and slows down as it becomes emptier.

There is a certain flow resistans in the output tubes and stuff, and that
limits the flow somewhat.

A battery has a capacity rating, like the amount of water in the fire
extinguisher, it is measured in AmpereHours (Ah) or mAh for smaller
batteries.

A battery with a capacity of 3000mAh can deliver 3000mA for one hour, or
1000mA for 3 hours, or 100mA for 30 hours.

Batteries also have a voltage rating, like the water pressure in the fire
extinguisher, it is measured in Volt.

The amount of work a battery can deliver is the capacity times the
voltage. A 3 Volt battery with 3000mAh capacity can do a work of 3*3 VAh,
or 9 Watthours.

That is what you pay for when you pay the electricity bill, the work you
have bought, measured in VoltAmperehours, or Watthours, or kiloWatthours.

The effekt is how much work is performed per time unit, like Watthours
per hour, or simply Watt.



--
Roger J.

 

[Robert Monsen]

Jean-Marie Vaneskahian wrote:

I Need To Trigger a Relay When an LED Lights Up On a Smoke Detector

My goal is to use simple battery operated combination Smoke / Carbon
Monoxide detectors in various parts of my house. I want these battery
operated detectors to trigger a simple contact closure when they
alarm.

I noticed that the detectors have a Red LED that lights up when the
Smoke / Carbon Monoxide detector is triggered. I soldered two small
wires to the PCB on the back of the detector in "parallel" with the
Red LED.

When I connect a voltmeter to the two wires I soldered in "parallel"
with the Red LED and now run from the back of the PCB of the Smoke /
Carbon Monoxide detector I get about +0.001V when nothing happens and
+1.78V when I hit the test button on the Smoke / Carbon Monoxide
detector. By the way, the Smoke / Carbon Monoxide detector runs on a
total of 3 AAA 1.5V batteries.

I have a very basic electronics understanding. One concept that I do
NOT understand is that of "Ground". I do understand how to "Wire"
components though.

My question is this:

How do I take the 2 wires running from the back of the PCB on the
Smoke / Carbon Monoxide detector and generate a basic dry "Contact
Closure" when the detector is triggered?

I am sure this requires transistors, diodes, resistors, reed switches
and a separate battery source, but I have no clue how to connect them
and what types and values to purchase. I buy most of these components
from Radio Shack (Part Numbers Would Be Great!).

Please help me put this together. I really appreciate any wisdom on
this topic. My goal is to protect my family by wiring these battery
operated Smoke / Carbon Monoxide detectors to my home alarm system
that uses contact closures.

This sounds like a fun project.

One thing to consider is that you probably won't be able to power the
relay open using the current through the LED. If you are willing to use
a separate battery, however, then there is an easy way to do this.

Define the ground to be the more negative side of the LED. (Note: you
can define ground to be anything. It's just the point you measure other
voltages against)

Then, attach one side of a 4.7k resistor to the positive side of the
LED, and the other side to the base of an NPN transistor. Attach the
emitter of the NPN to the negative side of the LED. Now, when the LED
turns on, it'll allow current to flow through the transistor.

Hook the negative terminal of a 9V battery to the emitter of the
transistor, and the positive side to one side of the coil for the relay.
Hook the other coil of the relay to the collector of the NPN transistor.

Now, when the LED is on, the transistor will allow current to flow, so
current will flow in a circuit through the relay coil, the transistor
and the 9V battery. This should cause the relay to open.

--
Regards,
Robert Monsen

"Your Highness, I have no need of this hypothesis."
- Pierre Laplace (1749-1827), to Napoleon,
on why his works on celestial mechanics make no mention of God.

 

[John Fields]

On 8 Nov 2004 13:23:42 -0800, jean@vaneskahian.com (Jean-Marie
Vaneskahian) wrote:

I Need To Trigger a Relay When an LED Lights Up On a Smoke Detector

My goal is to use simple battery operated combination Smoke / Carbon
Monoxide detectors in various parts of my house. I want these battery
operated detectors to trigger a simple contact closure when they
alarm.

I noticed that the detectors have a Red LED that lights up when the
Smoke / Carbon Monoxide detector is triggered. I soldered two small
wires to the PCB on the back of the detector in "parallel" with the
Red LED.

When I connect a voltmeter to the two wires I soldered in "parallel"
with the Red LED and now run from the back of the PCB of the Smoke /
Carbon Monoxide detector I get about +0.001V when nothing happens and
+1.78V when I hit the test button on the Smoke / Carbon Monoxide
detector. By the way, the Smoke / Carbon Monoxide detector runs on a
total of 3 AAA 1.5V batteries.

I have a very basic electronics understanding. One concept that I do
NOT understand is that of "Ground". I do understand how to "Wire"
components though.

My question is this:

How do I take the 2 wires running from the back of the PCB on the
Smoke / Carbon Monoxide detector and generate a basic dry "Contact
Closure" when the detector is triggered?

I am sure this requires transistors, diodes, resistors, reed switches
and a separate battery source, but I have no clue how to connect them
and what types and values to purchase. I buy most of these components
from Radio Shack (Part Numbers Would Be Great!).

Please help me put this together. I really appreciate any wisdom on
this topic. My goal is to protect my family by wiring these battery
operated Smoke / Carbon Monoxide detectors to my home alarm system
that uses contact closures.

---
I'd wind a reed relay and substitute it for the LED or, if I needed to
keep the LED for some reason, substitute it for at least _part_ of the
LED's current limiting resistor. Before I could proceed, though, I'd
need to know, as a bare minimum, the amount of current flowing through
the LED, the value of the current-limiting resistor in series with it,
and whether the alarm system is looking for a closing or an opening
contact closure as the triggering event.

If you can reverse-engineer all of the LED driving circuitry and post
what you find, so much the better!

--
John Fields

 

[Lord Garth]

"Gerhard v d Berg" <gvdberg@az.oc.risc> wrote in message
news:L_adnSFrkPlIGg7cRVn-sQ@is.co.za...

"Gerhard v d Berg" <gvdberg@az.oc.risc> wrote in message
news:Ff6dne1kJp_lVA3cRVn-gw@is.co.za...
Multi-rail DC supply from a Wall-wart

Snip ...

Take 2
I left out two diodes in the charge-pump circuits.
Nobody queried or corrected me - the hawks in the group is slipping up !
Maybe there is no interest :-/

I reposted the whole text and circuit with the corrections and add a
reference.
snip

The plural police have flagged your sentence above as incorrect. The
correction is as follows, "Nobody queried or corrected me - the hawks
in the group are slipping up !"

 

[John Popelish]

Gary Helfert wrote:

As good as lithium & Ni-mH batteries are, alkalines still are the champs as
far as energy density is concerned.
So, why is it so hard to find the amp-hour rating for these batteries?
Could somebody please give me amp-hour ratings for alkalines:
AAA cells:
AA cells:
C cells:
D cells:

http://www.zbattery.com/zbattery/batteryinfo.html
--
John Popelish

 

[John Fields]

On Fri, 12 Nov 2004 10:55:05 -0500, John Popelish <jpopelish@rica.net>
wrote:

Gary Helfert wrote:

As good as lithium & Ni-mH batteries are, alkalines still are the champs as
far as energy density is concerned.
So, why is it so hard to find the amp-hour rating for these batteries?
Could somebody please give me amp-hour ratings for alkalines:
AAA cells:
AA cells:
C cells:
D cells:

http://www.zbattery.com/zbattery/batteryinfo.html

---
WOW!!!

Great find!

--
John Fields

 

[CFoley1064]

Subject: Re: Alkaline batterys; why is it so hard to find the amp-hour
ratings
From: John Popelish jpopelish@rica.net
Date: 11/12/2004 9:55 AM Central Standard Time
Message-id: <4194DCD9.711CC35E@rica.net

<snip>

http://www.zbattery.com/zbattery/batteryinfo.html
--
John Popelish

Thank you, sir. Truly useful.

Chris

 

[Watson A.Name - \"Watt Su]

"John Larkin" <jjlarkin@highlandSNIPtechTHISnologyPLEASE.com> wrote in
message news:nn32o0t5skfdo7ao712vhlh894i0vmklne@4ax.com...

But I'm impressed that you're actually verifying your simulations by
experiment; some people just quit when the sim works.

Some people just don't bother with the sims, and do just the
experiments. ;-)

Move the 10k resistors from the cathodes to the anodes of the zeners.

> John

 

[John Fields]

On 8 Nov 2004 13:23:42 -0800, jean@vaneskahian.com (Jean-Marie
Vaneskahian) wrote:

I Need To Trigger a Relay When an LED Lights Up On a Smoke Detector

My goal is to use simple battery operated combination Smoke / Carbon
Monoxide detectors in various parts of my house. I want these battery
operated detectors to trigger a simple contact closure when they
alarm.

I noticed that the detectors have a Red LED that lights up when the
Smoke / Carbon Monoxide detector is triggered. I soldered two small
wires to the PCB on the back of the detector in "parallel" with the
Red LED.

When I connect a voltmeter to the two wires I soldered in "parallel"
with the Red LED and now run from the back of the PCB of the Smoke /
Carbon Monoxide detector I get about +0.001V when nothing happens and
+1.78V when I hit the test button on the Smoke / Carbon Monoxide
detector. By the way, the Smoke / Carbon Monoxide detector runs on a
total of 3 AAA 1.5V batteries.

I have a very basic electronics understanding. One concept that I do
NOT understand is that of "Ground". I do understand how to "Wire"
components though.

My question is this:

How do I take the 2 wires running from the back of the PCB on the
Smoke / Carbon Monoxide detector and generate a basic dry "Contact
Closure" when the detector is triggered?

I am sure this requires transistors, diodes, resistors, reed switches
and a separate battery source, but I have no clue how to connect them
and what types and values to purchase. I buy most of these components
from Radio Shack (Part Numbers Would Be Great!).

Please help me put this together. I really appreciate any wisdom on
this topic. My goal is to protect my family by wiring these battery
operated Smoke / Carbon Monoxide detectors to my home alarm system
that uses contact closures.

---
If the voltage across the LED is 1.78V and the drop across what's
driving it is 0.3V, then if the LED current is 20 mA and the supply
voltage is 4.5V, the current limiting resistor for the LED should be
something like

E 4.5 - (1.8 + 0.3)
R = --- = ------------------ = 120 ohms
I 0.02

Now, if you wanted to use a reed relay to do the switching you could
choose something like a Hamlin MITI-3V1 reed switch and wind it with
enough turns to make sure it closed.

Assuming a worst case of 15AT and, assuming end-of-life battery
voltage of 3V, then you'd have about 7.5mA to play with and you'd need

AT 15
n = ---- = -------- = 2000 turns
A 0.0075

The switch has a length of 0.275" and a diameter of 0.071", so you'd
probably need a bobbin to hold all the wire; say with an inside length
of 0.3" and a diameter 0.1".

If you used #40 wire for the coil, with a diameter of 0.003145" (and
you rounded up to 0.004", just because...) you could get

0.275
n = --------- ~ 68 turns
0.004

per layer. Since you need 2000 turns, that would be 29.4 layers and,
rounding up again that would give us 30 layers.

If we started with a bobbin diameter of 0.1" and added 0.004" for the
wire, the first layer would have a diameter of 0.104". Adding 0.004"
for each additional layer would result in a diameter of 0.220 for the
30th layer, resulting in a mean length of turn of 0.509" for the
winding, and with 2000 turns that would be 1018" or, about 85 feet of
wire. # 40 wire has a resistance of 1.05 ohms per foot, so that would
be about 90 ohms.

If the LED's current limiting resistor was 120 ohms, then, you could
replace it with a 30 ohm resistor and wire the coil in series with it
giving you 20 mA into the LED and a definite metallic contact closure
when the LED went on with no additional circuitry. If you wanted to,
you could wind the bobbin with enough wire to get you to 120 ohms,
which would eliminate the resistor altogether. If you decide to go
this route, it wouldn't hurt to put a diode (something like a 1N4148
in parallel with the coil, with the diode's cathode connected to the
most positive end of the coil.

Thinking about it, since you probably won't be able to level-wind the
coil you're going to wind up putting more than 85' on it for 2000
turns. Maybe, even, 114.28'


--
John Fields

 

[John Larkin]

On Fri, 12 Nov 2004 09:03:05 -0800, "Watson A.Name - \"Watt Sun, the
Dark Remover\"" <NOSPAM@dslextreme.com> wrote:

"John Larkin" <jjlarkin@highlandSNIPtechTHISnologyPLEASE.com> wrote in
message news:nn32o0t5skfdo7ao712vhlh894i0vmklne@4ax.com...



But I'm impressed that you're actually verifying your simulations by
experiment; some people just quit when the sim works.

Some people just don't bother with the sims, and do just the
experiments. ;-)

And some people just do the math, lay out boards, and go into
production.

John

 

[Jonathan Kirwan]

On Sat, 13 Nov 2004 23:06:19 +0800, "Rheilly Phoull" <Rheilly@bigpong.com>
wrote:

More like the finger has a resistance of a few K's. For real saturation just
moisten the digit.

I think the OP was saying that ONLY one finger was used. He's acting kind of
like a radio antenna, I suspect.

Jon

 

[Jack// ani]

Jonathan Kirwan <jkirwan@easystreet.com> wrote in message news:<44scp0938886tmqngurqeddts23bic66ev@4ax.com>...

I think the OP was saying that ONLY one finger was used.

Oh yeah, exactly.

 

[Ban]

Jack// ani wrote:

Hi all,

Placing a finger on the base pin of this darlington configuration
glows the LED (in fact reaches very close to saturation)!

I learned that you need a threshold voltage of about 0.7volts across
the base-emitter junction in order to bring transistor into
conduction.

So does this imply that my finger having a potential of 0.7volts??

..

In the first place you will need more than +1.2V, because the darlington has
2 junctions in series.
And your finger will have an AC-potential, which you can measure with an
oscilloscope. This is mainly because of 60Hz mains field radiating all over
the place. The transistor conducts only half of the time, so the intensity
will be less than half of the maximum. Since the current gain of a
darlington is very high (maybe 10000 or even more) you will need only a uA
to bias the base into conduction.

--
ciao Ban
Bordighera, Italy

 

[Jack// ani]

"Ban" <bansuri@web.de> wrote in message news:<gBBld.166734$b5.8315447@news3.tin.it>...

In the first place you will need more than +1.2V, because the darlington has
2 junctions in series.

whoops, I forget that fact.

This is mainly because of 60Hz mains field radiating all over
the place.

Oh yes, I can clearly see a 50Hz ac/noise. But why there exist only
50Hz, what about my cell and cordless phones which emits MHz, KHz
frequencies…

Any help is highly appreciated

 

[Ban]

Jack// ani wrote:

"Ban" <bansuri@web.de> wrote in message
news:<gBBld.166734$b5.8315447@news3.tin.it>...

In the first place you will need more than +1.2V, because the
darlington has 2 junctions in series.

whoops, I forget that fact.

This is mainly because of 60Hz mains field radiating all over
the place.

Oh yes, I can clearly see a 50Hz ac/noise. But why there exist only
50Hz, what about my cell and cordless phones which emits MHz, KHz
frequencies…

Any help is highly appreciated

Really? You remind me of my youngest daughter, one question answered, two
new ones are popping up immediately.
Check out the input impedance of the darlington, we have seen already
1.2V/1.2uA = 1Megohm DC-resistance and now calculate the input capacitance
of that darlington. Hint: it is high because of the Miller-capacitance.
And then you can calculate the 3dB frequency of that lowpass, and you will
see it is low.
And BTW, what kind of antenna is your finger? Must have some loss at 800MHz?
You need a tuned circuit for RF to establish some power transfer.
--
ciao Ban
Bordighera, Italy

 

[Roger Johansson]

"Mike" <mike@notemail.com> wrote:

It seems a zener diode is the way to go, but I can't find any that
will regulate 400V (only 200V).

Take two.

Two diodes in series, add the zener voltages.


--
Roger J.

 

[Jonathan Kirwan]

On 15 Nov 2004 12:14:58 GMT, Roger Johansson <no-email@home.se> wrote:

"Mike" <mike@notemail.com> wrote:

It seems a zener diode is the way to go, but I can't find any that
will regulate 400V (only 200V).

Take two.

Two diodes in series, add the zener voltages.

I'm not sure how good that will be. The 1N5281A is rated at 200V @ 650uA but
it's tolerance is 10% with some .11%/C temperature drift, on top. I suppose the
OP could hand select them to get the right voltage, though.

Also, since the Izt of the 1N5281A isn't anything like the 10mA desired, a BJT
emitter follower would be needed that only requires some percentage of that
650uA biasing, say 100uA (it's .16V change for 65uA change in bias, which is
probably fine.) This means beta of 100, which may be achievable so long as
there is at least a volt or two drop across V(CE). But how much the V(CE) will
have to stand off could be a problem. The MJ12005 is mentioned in the Art of
Electronics, 2nd ed., as an NPN with Vceo of 750V, though. Also, that same book
includes a couple of high voltage circuit examples on pages 369 and 370 that
could be used for something like this. The second example regulates the ground
return.

Jon

 

[Roger Johansson]

Jonathan Kirwan <jkirwan@easystreet.com> wrote:

Two diodes in series, add the zener voltages.

I'm not sure how good that will be. The 1N5281A is rated at 200V @
650uA but it's tolerance is 10% with some .11%/C temperature drift, on

Two of them in series will still have the same tolerance.

top. I suppose the OP could hand select them to get the right voltage,
though.

If his 200V zeners have the precision he needs then two of them will have
the same precision.



--
Roger J.

 

[Jonathan Kirwan]

On 15 Nov 2004 16:49:47 GMT, Roger Johansson <no-email@home.se> wrote:

Jonathan Kirwan <jkirwan@easystreet.com> wrote:

Two diodes in series, add the zener voltages.

I'm not sure how good that will be. The 1N5281A is rated at 200V @
650uA but it's tolerance is 10% with some .11%/C temperature drift, on

Two of them in series will still have the same tolerance.

10% tolerance, yes. At 400V, put in absolute terms, this is +/-40V, yes?

top. I suppose the OP could hand select them to get the right voltage,
though.

If his 200V zeners have the precision he needs then two of them will have
the same precision.

This still begs the question, though.

Is 10% okay?

And it still means that an emitter follower BJT is required as the 650uA Izt
rating simply cannot be used for a regulated 10mA supply without one.

Jon