Chip with simple program for Toy

[Richard]

I take it, the 10 amp jump is the initial in rush current?

Yes. You are charging the bulk bypass caps with it.

then the 7.5 amps is a very breif spike from the SBC kinking off?

Hard to say (even if I wished to guess what "kinking off"
is, but I don't want to know ;-).

Sorry, typo, I meant that was the CPU kicking off or Starting up. The SBC
has it's on DC Converter inside so it could be it starting up.

That's a little much for a 14 inch wire run. At those
currents, heating is not really an issue, but drop may
be depending on your supply tolerance and how the
SBC is specified. 16 guage should be plenty big.

Any thoughts,

Is there a reason to go for the smaller wire?

No, we just had a 100 premade harness kits made because I looked at the
original motherboard specs and then looked up the 18 ga. and it seemed to be
sized ok, and our other harnesses were 18 ga.

I made up a 16 ga harness to see if it helps. Besides the fact that the
voltage drop will be much less, will I see differences with the inrush
currents as well? I think the wire being too small and the high inrush is
what keeps the computer from booting sometimes.

Richard



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[Larry Brasfield]

"Richard" <REMOVE AT-DOTrwskinnerAT@awesomenetDOTnet>
wrote in message news:4266e6de$1_1@127.0.0.1...

I take it, the 10 amp jump is the initial in rush current?

Yes. You are charging the bulk bypass caps with it.

then the 7.5 amps is a very breif spike from the SBC kinking off?

Hard to say (even if I wished to guess what "kinking off"
is, but I don't want to know ;-).

Sorry, typo, I meant that was the CPU kicking off or Starting up. The SBC has it's on DC Converter inside so it could be it
starting up.

Whew!

That's a little much for a 14 inch wire run. At those
currents, heating is not really an issue, but drop may
be depending on your supply tolerance and how the
SBC is specified. 16 guage should be plenty big.

Any thoughts,

Is there a reason to go for the smaller wire?

No, we just had a 100 premade harness kits made because I looked at the original motherboard specs and then looked up the 18 ga.
and it seemed to be sized ok, and our other harnesses were 18 ga.

If the spec on your supply output voltage is enough tighter
than the spec on your SBC supply requirement so that
the 90 mV drop is not a problem, then I would use those
100 harnesses. And have more made up with bigger wire.

I made up a 16 ga harness to see if it helps. Besides the fact that the voltage drop will be much less, will I see differences
with the inrush currents as well?

The inrush current is most likely limited by the supply
rather than the wire. I would expect no real change.

I think the wire being too small and the high inrush is what keeps the computer from booting sometimes.

That seems unlikely to me. Your SBC probably has
a spec on how fast the supply should come up. Be
sure you meet it. Also, be sure there is no overshoot
just after that initial current spike driving the supply
voltage above the SBC voltage spec.

You might want to look carefully at the supply voltage
waveform as it comes up. Maybe the DC-DC converter
is initially putting the supply into current limit and causing
a wobbly turn-on waveform rather than what the SBC
designers assumed (and should have specified). You
might try a higher current limit supply, experimentally,
to see if that reduces the misbooting incidence.

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

 

[Lord Garth]

"Richard" <REMOVE AT-DOTrwskinnerAT@awesomenetDOTnet> wrote in message
news:4266d96f$1_2@127.0.0.1...
<snip>

Any thoughts,
Richard

It occurs to me that the modern ATX power supply has extra power supply
lines. These
are intended as a parallel path to supply the CPU with the extra power it
needs. You
may have noted the (usually) two odd power connectors coming from the power
supply.

Are you using these?

 

[John Smith]

Impedance mismatch will mean that the antenna will not radiate 260mW, a
significant amount will be reflected back to the op-amp. That is why the
traditional approach exists.

"BobG" <bobgardner@aol.com> wrote in message
news:1114052307.475407.266430@o13g2000cwo.googlegroups.com...

While we're talking about transmitters, let me ask this question: Lets
say I have a video opamp that will put out +-10V at 1 MHz into 377
ohms... can I hook this to an antenna? P=E^2/R=100/377= 260mw In
other words, the 'traditional' approach of making everything 50 ohms
and using tuned class c amps is just how it was done with tubes in the
40s right? If the impedance of the antenna is 377 ohms, and I can drive
it direct, thats all it takes, right?

 

[Larry Brasfield]

"mjohnson" <crvmp3@hotmail.com> wrote in message
news:1114085855.831868.141050@z14g2000cwz.googlegroups.com...

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?

The voltage applied to the buzzer is undoubtedly
higher than 395 mV. That might be its DC value,
but if you measure AC, you will find quite a bit
more RMS.

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.

The LED can be driven with a small fraction of the
power consumed by the buzzer, so your concern is
not one that should take much more your time.

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?

Put a 1k limiting resistor in series with the LED,
put the combination across the buzzer, then
measure DC voltage across the resistor when the
buzzer goes off. You'll see more than you might
have expected.

thanks for your time and help!
You're welcome.

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

 

[John Fields]

On 21 Apr 2005 07:06:35 -0700, Tommi-Vogel@gmx.de (Thomas Vogel)
wrote:

John Fields <jfields@austininstruments.com> wrote in message news:<agta611dbh5utnvos47ggtfdve6ufglr48@4ax.com>...
On 19 Apr 2005 11:03:51 -0700, Tommi-Vogel@gmx.de (Thomas Vogel)
wrote:

Hi,

no, it´s not for a bomb! I just need it for my music station to woke up in
the morning. My new music station has no timer and I can´t switch it with a
timer. So I want to construct a circuit, which switch by an alarm clock over
a relay a remote control. Pleas email me the schematic+parts list.

---
Here ya go:


+---+---------------+------+------+---+---+-------+
| | | | | | | |
| [10k] [1M] [100K] [10K] | |K | O------>C
| | | | | |[1N4148] [COIL]- -|
| | +-|+\ | | | | | O--> |
| | | | >-+------|-|+\ | | |
[BAT] +-[0.1ľF]-+-----|-|-/ | | | >--+-------+ +----------->NO
| | | | | +-|-/
| | | | |+ | |
| [MIC] [100k] [10K] [10ľF] [1M] |
| | | | | | |
+---+---------+-----+------+------+---+


The microphone is a Panasonic WM-61A, the comparators are part of an
LMV393, the battery is 3X 1.5VAA, and the relay is a COTO 9007-05-40.

Here's how it works:

After you finish building and testing it, put it in a soundproof box
with your alarm clock and connect the relay contacs (C and NO, above)
to the remote control. Voila!, when the alarm clock goes off you
won't hear it, bit it'll turn on your music station. (Radio???)

Or, you could forget the whole thing and just let the alarm clock wake
you up.




hi,

thank you for this schematic, but please send it at my e-mail adress
in a better format as a picture, that I can better realize it.

No. I prefer to keep discussions like this here, so that everyone can
participate in arriving at a solution for you, if they choose.

Writing schematics in ASCII is a way to use very little bandwidth to
convey a lot of information, and also to make sure that when the
article is archived the information stays with it essentially forever.
If you want to play, you need to learn how to use the toys. If the
schematic is garbled, view it using a non-proportional font like
Courier, and if you don't understand the operation of the circuit
post your questions here and one or some of us will be happy to help
you.

--
John Fields
Professional Circuit Designer

 

[Larry Brasfield]

"Up in Canada" <aleve@canada.com> wrote in message
news:1114101919.953870.42840@o13g2000cwo.googlegroups.com...

Hello-
Hi.
I am using a 4013 as a toggle to control a LED.
Simple school project.

I have Qbar connected to Data, Q is to a small fet that turns on a LED.

The clk input is connected to a MOM push button switch that grounds it
out when pushed.

You probably should debounce that switch to
derive one active clock edge per push.

I have RESET and SET at ground.


It works great so far!!


Now, how do I control the Startup condition?

I need to have one that powers up with the Q high (led on)

I also will use the other half of the 4013 in the same exact way,
totally independent, but need this Q to startup LOW( that led off).

Build the following circuit with a 1N914 diode
(or equivalent), a 100k resistor, and a 100 nF
cap. Connec TrueBriefly to the SET or RESET
input of each FF, as dictated by your requirement.

VCC
|
---
---
|
o---------TrueBriefly
|
o---.
| |
.-. -
| | ^
| | |
'-' |
o---'
|
===
GND

(created by AACircuit v1.28.4 beta 13/12/04 www.tech-chat.de)
(View with a fixed-width font.)

Thanks.

Welcome.

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

 

[Larry Brasfield]

"Up in Canada" <aleve@canada.com> wrote in message
news:1114104659.063837.214040@f14g2000cwb.googlegroups.com...

You probably should debounce that switch to
derive one active clock edge per push.

Errrr.... how do you reccomend?

One way is to use an RC lowpass filter to take
out most of the bounce trash, and follow it with
a device having input hysteresis to preclude any
multiple transitions should the LPF output cross
a single stationary threshold more than once per
push or release of the switch.

Another way is to use a form C switch and hook
the NO and NC signals to the SET and RESET
inputs of a flip-flop whose output is bounce free.

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

 

[Larry Brasfield]

"Up in Canada" <aleve@canada.com> wrote in message
news:1114110289.492293.299960@o13g2000cwo.googlegroups.com...

one thing more.

I may need to add more momentary switches in parallel for remote
operation. Well not remote, but think of a light switch on each side of
a room.
It still has to be momentary toggles.

Uh oh, I think it just got more complicated. I have to use the least
amount of wires.

Anyway to have a debounce with just the momentary spst?

Here is a circuit which was once published in EDN, IIRC.
It relies on positive feedback thru the series RC pair to
effect a short duration threshold shift that suffices to avoid
bounce effects for normal switch bounce.


VCC
| ___ ||
.-. .-|___|--||---.
| | | || |
10k| | | 20k 100 nF |
'-' | |
| ___ | |\ |
o-|___|--o-| >---------o--Debounced
| |/
| 20k
| o
|=|>
| o
|
GND
(created by AACircuit v1.28.4 beta 13/12/04 www.tech-chat.de)

You can buy buffers with input hysteresis, so this approach
makes the most sense when you just want one or two of
these inputs and have leftover gates or PLD pins to use.

If I have to
use a remote box with cable conected to mulitple leds, the less wiring
in cables the better. If I have to do 30 leds you can see what I mean.

That is a disadvantage, alright. That's why I mentioned
the LPF and hysteresis (aka "Schmidt trigger") approach.
....

Thanks
Welcome.

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

 

[JeB]

On 21 Apr 2005 00:09:53 -0700, "twags6" <twags6@hotmail.com> wrote:

Is there any way to make a dc power supply capable of 50+ amps at
12-13.8 vdc? Basically, I'm looking for something to simulate the
power from a car without the whole battery and charger setup. Parts
express sells a nice rackmount one, but I'd rather not part with $190.

u might find something on eBay for about half that.

 

[Ariane]

Hi, hier sind meine geilen Bilder!
My nude Pics!!!
http://www.geile-tipps.info/go/

--
Posted by News Bulk Poster
Unregistered version

 

[Terry Pinnell]

Tommi-Vogel@gmx.de (Thomas Vogel) wrote (to John Fields):

thank you for this schematic, but please send it at my e-mail adress
in a better format as a picture, that I can better realize it.


Thomas

Did you not try the approach I suggested over two weeks ago to your
original post?
Subject: Re: a circuit to switch a relay by a alarm clock
Date: Wed, 06 Apr 2005 12:23:13 +0100
Message-ID: <6bh751pmrss0soi22e3ji1e6tt9qmgon6p@4ax.com>
Click this: news:6bh751pmrss0soi22e3ji1e6tt9qmgon6p@4ax.com

--
Terry Pinnell
Hobbyist, West Sussex, UK

 

[Hans-Bernhard Broeker]

[F'up2 c.a.e --- this would be even *more* off-topic in the
electronics newsgroup, I think.]

In comp.arch.embedded bg <byju.ganga@gmail.com> wrote:

i have a dayton shaded pole single phase ac induction motor with
1/150HP, 3000 RPM, 230 V, 60 Hz frequency and Full Load Amps 0.24A.

230 V at 0.24 A is 55 VA, or, assuming a cos(\phi) of 1 for the
moment, 55 Watt. That doesn't quite figure with your HP figure, I
think --- it's off by one order of magnitude. Should this beast
really have a phase angle of acos(0.1)==84 degrees?

I would like to know whether there is any equation for
troque.

Torque times angular velocity, just like force times linear velocity,
equals power. So:

M = P / \omega
= P / (2*\pi*f)

Using the 1/150 HP (or roughly 5 Watt), that gives us a torque of
0.013 Nm. Using the 55 VA, you get 0.146 Nm. I'll leave it to the
metrically challenged to convert that into strange units ;-)

--
Hans-Bernhard Broeker (broeker@physik.rwth-aachen.de)
Even if all the snow were burnt, ashes would remain.

 

[Richard]

No Sir,

This is an Industrial, Embedded Motherboard running a Via 667mhz on a single
5 vdc supply. We are using a Datel 18-36vdc input to 5 vdc output rated at
5.0 amps. It's 14" away from the SBC.

Richard

"Lord Garth" <LGarth@Tantalus.net> wrote in message
news:mAG9e.997$l45.83@newssvr12.news.prodigy.com...

"Richard" <REMOVE AT-DOTrwskinnerAT@awesomenetDOTnet> wrote in message
news:4266d96f$1_2@127.0.0.1...
snip
Any thoughts,
Richard


It occurs to me that the modern ATX power supply has extra power supply
lines. These
are intended as a parallel path to supply the CPU with the extra power it
needs. You
may have noted the (usually) two odd power connectors coming from the
power
supply.

Are you using these?

----== Posted via Newsfeeds.Com - Unlimited-Uncensored-Secure Usenet News==----
http://www.newsfeeds.com The #1 Newsgroup Service in the World! 120,000+ Newsgroups
----= East and West-Coast Server Farms - Total Privacy via Encryption =----

 

[John Fields]

On 22 Apr 2005 05:57:14 -0700, "mjohnson" <crvmp3@hotmail.com> wrote:

Specifically, on your remote control are the OPEN and CLOSE functions
separate or does a single button toggle them?

It's a toggle switch, I need it "pressed" for a little less than a
second for it to activate the garage door then I have to let it go or
else the motor won't respond. So I would imagine I need a pluse that
last about 800 to 900 milliseconds.

The alarm clock runs off 3V DC. I measured the DC voltage with a
multimeter between the two solder points on the buzzer itself -- so it
was across the buzzer. Should I be measuring it someplace else?

---
Across the buzzer should be OK, but 395mV sounds awfully low, so I
suspect it's being driven by AC. Measure it with your meter set to AC
VOLTS and see what you get.
---

I'm assuming that you're planning on paralleling the contacts on the
remote's keypad with whatever will be actuating it. Am I right?

I wanted to extend two leads from the solder points on the buzzer to an
opto-isolator which would activate the "pulse" to the garage door
remote through an intermediate circuit.

---
Before anything can happen we need to find out definitively what's
happening at the buzzer when the alarm goes off, or we need to find a
signal somewhere in the clock which changes state when the alarm goes
off. Either that or couple to the clock acoustically in order to
detect the sound of the buzzer and use that to trigger the chain of
events leading to the activation of the remote.
---

Here's a new diagram, as well as front/back images of the remote.

http://img195.echo.cx/img195/4791/diagram6yf.jpg

Front side of the remote:
http://img195.echo.cx/img195/1594/remotefront5nd.jpg

Back side of the remote (red lines are show the traces)
http://img195.echo.cx/img195/6156/remoteback8tt.jpg

I'm having a hard time getting a for the voltage accross the toggle
switch on the remote. I read 0 no matter if the switch is toggled or
not. I am probably not laying my probes across the switch correctly.
Do I need to touch both pairs of connectors at the same time?

---
No, it looks like you only have to get across the terminals with trace
connected to them. But, since we don't know how the switch is being
used in the circuit, it would be best to use relay contacts (instead
of the out of an optocoupler) across the switch terminals in order to
activate the remote. Test it by shorting the terminals momentarily
and see if it works the door. If it does, then relay contacts will be
fine.
---

I'm assuming that I would use an opto-isolator to interface the pluse
circuit to the remote and that the output of the opto-isolator will be
enough to activiate the remote? I guess I'll just have to try it an
see.

---
I wouldn't use an opto because of the current required for its LED and
the uncertainty of being able to use its transistor output to trigger
the remote.

Here's what I see as a much simpler solution, with only the clock
output needing to be defined in order to make it work:

INTERFACE
BATTERY
CLOCK 4.5V REMOTE
BATTERY | BATTERY
3V +---------+---------+ 3V
| | | | |
+--+--+ +--+--+ +--+--+ +--+--+ +--+---+
|CLOCK|---|ALARM|---|1SEC |---|REED |---|REMOTE|
+-----+ | DET | |MONO | |RELAY|---|SWITCH|
+-----+ +-----+ +-----+ +------+


--
John Fields
Professional Circuit Designer

 

[Watson A.Name - \"Watt Su]

"Bill Bailley" <JustMe@Home> wrote in message
news:4268e16f$0$5180$afc38c87@news.optusnet.com.au...

The chairman of Citroen once said that the safest car in the world
would be made of glass, with a six inch spike sticking out of the
centre of the steering wheel.

d

Is that the gentleman who was killed by his briefcase when he made
an
emergency stop?

Dammit! Wrong again. It was the administrator of Reault, Pierre
Lefaucheux.

http://tinyurl.com/9emkv

I had a friend who was a HiFi stereo nut who was fairly tall, maybe
6'3", as were his mom and dad, both over 6'. But his fraternal twin
brother was even taller, about 6'8". And what puzzled me was why his
brother bought a little Renault Dauphine 'compact' when he could've got
a used American car, which in the '60s were full-sized autos with plenty
of legroom.

 

[Larry Brasfield]

"John Fields" <jfields@austininstruments.com> wrote in message
news:560i61tfmof8mhjis826fj5q3dcghb53of@4ax.com...

On 22 Apr 2005 05:57:14 -0700, "mjohnson" <crvmp3@hotmail.com> wrote:
....
Across the buzzer should be OK, but 395mV sounds awfully low, so I
suspect it's being driven by AC. Measure it with your meter set to AC
VOLTS and see what you get.

(To the OP I concur with that good advice.

....

Before anything can happen we need to find out definitively what's
happening at the buzzer when the alarm goes off, or we need to find a
signal somewhere in the clock which changes state when the alarm goes
off.

The power needed to drive a buzzer will be many
times larger than what needs to be picked off to
activate another circuit, (many mW versus uW).

....

I'm having a hard time getting a for the voltage accross the toggle
switch on the remote. I read 0 no matter if the switch is toggled or
not. I am probably not laying my probes across the switch correctly.
Do I need to touch both pairs of connectors at the same time?

---
No, it looks like you only have to get across the terminals with trace
connected to them. But, since we don't know how the switch is being
used in the circuit, it would be best to use relay contacts (instead
of the out of an optocoupler) across the switch terminals in order to
activate the remote. Test it by shorting the terminals momentarily
and see if it works the door. If it does, then relay contacts will be
fine.
---

I'm assuming that I would use an opto-isolator to interface the pluse
circuit to the remote and that the output of the opto-isolator will be
enough to activiate the remote? I guess I'll just have to try it an
see.

---
I wouldn't use an opto because of the current required for its LED and
the uncertainty of being able to use its transistor output to trigger
the remote.

Some such uncertainty is warranted, but I suggest that
there is reason to believe an opto-isolator will be fine.
The cheap (typically membrane) switches used in many
remotes are not asked to carry much current and, to
conserve battery power, large value pull-{up,down}
resistors are used. If a replacement for the contanct
had to carry more than 100 uA, I would be surprised.
The CTR (current transfer ratio) for opto-isolaters is
often guaranteed to be 100% or better, so a similar
current is all that the LED would need. Finally, the
signal sent thru the opto-isolator can be time limited
to just over what is needed for the remote in order to
conserve the battery.

Here's what I see as a much simpler solution, with only the clock
output needing to be defined in order to make it work:

INTERFACE
BATTERY
CLOCK 4.5V REMOTE
BATTERY | BATTERY
3V +---------+---------+ 3V
| | | | |
+--+--+ +--+--+ +--+--+ +--+--+ +--+---+
|CLOCK|---|ALARM|---|1SEC |---|REED |---|REMOTE|
+-----+ | DET | |MONO | |RELAY|---|SWITCH|
+-----+ +-----+ +-----+ +------+

The opto-isolator would plug into that with little
change except reduction of the 4.5V battery drain
(unless my power surmises are completely wrong).

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

 

[John Fields]

On Fri, 22 Apr 2005 09:42:24 -0700, "Larry Brasfield"
<donotspam_larry_brasfield@hotmail.com> wrote:

"John Fields" <jfields@austininstruments.com> wrote in message
news:560i61tfmof8mhjis826fj5q3dcghb53of@4ax.com...
On 22 Apr 2005 05:57:14 -0700, "mjohnson" <crvmp3@hotmail.com> wrote:
...
Across the buzzer should be OK, but 395mV sounds awfully low, so I
suspect it's being driven by AC. Measure it with your meter set to AC
VOLTS and see what you get.

(To the OP I concur with that good advice.

...
Before anything can happen we need to find out definitively what's
happening at the buzzer when the alarm goes off, or we need to find a
signal somewhere in the clock which changes state when the alarm goes
off.

The power needed to drive a buzzer will be many
times larger than what needs to be picked off to
activate another circuit, (many mW versus uW).

---
Depends. The OP's advocating using the signal driving the buzzer to
also drive the LED in an opto, which will be milliwatts VS milliwatts.
---

...
I'm having a hard time getting a for the voltage accross the toggle
switch on the remote. I read 0 no matter if the switch is toggled or
not. I am probably not laying my probes across the switch correctly.
Do I need to touch both pairs of connectors at the same time?

---
No, it looks like you only have to get across the terminals with trace
connected to them. But, since we don't know how the switch is being
used in the circuit, it would be best to use relay contacts (instead
of the out of an optocoupler) across the switch terminals in order to
activate the remote. Test it by shorting the terminals momentarily
and see if it works the door. If it does, then relay contacts will be
fine.
---

I'm assuming that I would use an opto-isolator to interface the pluse
circuit to the remote and that the output of the opto-isolator will be
enough to activiate the remote? I guess I'll just have to try it an
see.

---
I wouldn't use an opto because of the current required for its LED and
the uncertainty of being able to use its transistor output to trigger
the remote.

Some such uncertainty is warranted, but I suggest that
there is reason to believe an opto-isolator will be fine.
The cheap (typically membrane) switches used in many
remotes are not asked to carry much current and, to
conserve battery power, large value pull-{up,down}
resistors are used. If a replacement for the contanct
had to carry more than 100 uA, I would be surprised.
The CTR (current transfer ratio) for opto-isolaters is
often guaranteed to be 100% or better, so a similar
current is all that the LED would need.

---
Not necessarily, CTR falls off quickly as LED forward current
diminishes and there are temperature effects which need to be taken
into consideration which can largely be ignored with a comparator-reed
switch solution. Also, with the reed switch solution there is no
saturation voltage VS LED If problem since it's either on or off.
---

Finally, the signal sent thru the opto-isolator can be time limited
to just over what is needed for the remote in order to
conserve the battery.

---
Either solution will require the generation of a timed pulse to the
remote, so that's probably a wash.
---

Here's what I see as a much simpler solution, with only the clock
output needing to be defined in order to make it work:

INTERFACE
BATTERY
CLOCK 4.5V REMOTE
BATTERY | BATTERY
3V +---------+---------+ 3V
| | | | |
+--+--+ +--+--+ +--+--+ +--+--+ +--+---+
|CLOCK|---|ALARM|---|1SEC |---|REED |---|REMOTE|
+-----+ | DET | |MONO | |RELAY|---|SWITCH|
+-----+ +-----+ +-----+ +------+

The opto-isolator would plug into that with little
change except reduction of the 4.5V battery drain
(unless my power surmises are completely wrong).

---
Could be. I'll defer judgement and wait until the OP comes back with
something definitive on the buzzer signal to post my design. If he
doesn't, I can always fall back on the acoustic thing I've already
posted. You may want to ask him about the current being conducted by
the remote's switch switch to see whether you can use an opto in
there. An easy way to determine the current would be to jump the
switch contacts with a milli/microammeter...

--
John Fields
Professional Circuit Designer

 

[John Bokma]

Brilla wrote:

Hi, I want to make a simple LED circuit with 12 LEDs running off a
nine volt battery. I've managed to dig up enough information about
most things, so I know I'll have to wire them in parallel. But the
resistance I should be using still confuses me.
Should I have one (or more) resistors at the beginning of the circuit?
Or one before each LED in the circuit?

The LEDs I'm using have a 3.6v voltage drop, and they are supposed to
get 20mA I beleive.
So that's 12 LEDs off a 9v battery.

http://johnbokma.com/pet/scorpion/detection-using-uv-leds.html

9V, 3.6V drop ->

9V - ( 2 x 3.6 V ) = 1.8 V

so you want to have 1.8V over your resistor.

the 2 LEDs in series use 20mA ->

1.8V
----
20 mA = 90 Ohm


With 12 LEDs, you need 6 resistors. And the total current should be
around 6 x 20 mA = 120 mA.

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).

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

 

[Lord Garth]

"Lord Garth" <LGarth@Tantalus.net> wrote in message news:GOm8e.2305$zq4.245@newssvr11.news.prodigy.com...

"js5895" <JoshTmp@nycap.rr.com> wrote in message
news:1113713094.309142.67210@g14g2000cwa.googlegroups.com...
I find that strange, because, won't the load's resistance dim or
brighten the switches neon lamp. Or would if the load was a receptacle.

Neon lamps need a series resistance and the draw extremely little power.




O--o-----------------traveler------------------o---O
| | (NE2 on) |
Hot O --^^^^---neon-- --neon---^^^^--o O---O incandescent lamp O---O neutral
(NE2 on) | | | Off
O----------------o---traveler---o------------------O

OR

O--o-----------------traveler------------------o---O
| (NE2 on) | |
Hot O --^^^^---neon-- --neon---^^^^--o O---O incandescent lamp O---O neutral
| (NE2 on) | | Off
O----------------o---traveler---o------------------O





O--o-----------------traveler------------------o---O
| | (NE2 off) | |
Hot O --^^^^---neon-- --neon---^^^^--o O---O incandescent lamp O---O neutral
(NE2 off) | | On
O----------------o---traveler---o------------------O

OR

O--o-----------------traveler------------------o---O
| (NE2 off) |
Hot O --^^^^---neon-- --neon---^^^^--o O---O incandescent lamp O---O neutral
| (NE2 off) | | | On
O----------------o---traveler---o------------------O


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