Chip with simple program for Toy

[Larry Brasfield]

"mjohnson" <crvmp3@hotmail.com> wrote in message
news:1114607172.412350.295340@g14g2000cwa.googlegroups.com...

Ok, I tried using the diode but I think my multimeter isn't fast enough
to sample the data but I did see values ranging from 800mV to 1V this
time (actually, I would see it ramp up from 80mV to generally topping
out at 800mV, 900mV, and 1V).

What do you think? Does ~1V sound like a reasonable voltage across the
buzzer?

I think you clearly have AC, probably in the form
of pulses, being applied to that buzzer transducer.
Given that your meter loads that diode detector
with a few hundred KOhms, ~1 V is believable.

Are there any other tests I should try?

A 0.1 uF cap across the meter would produce a
value closer to what I mentioned, and provide a
more accurate estimate of the peak voltage being
applied across the buzzer.

At this point, if stranded on a desert island without
more instruments, I would hook an optoisolator LED
with a 1K series resistor, anode side, to the buzzer.
Then measure the collector-to-emitter "resistance"
with the Ohms function while the buzzer sounds.
For one polarity of that measurement, you probably
will see significant conduction when the buzzer runs.
This is probably your best alarm pickoff circuit if you
are still happy with the buzzer's audibility.

I probably should invest in an analog multimeter.

An oscilliscope is a good idea for the serious hobbiest.

Is this where SWAG comes into play?

That would be a last resort.

thanks again!

You're welcome.

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

 

[mike rothe]

You are correct, if you connected your circuit in a three phase system you would not end up with 240 Volts, but with 208V instead.

Here is a link to a nice site explaining 3-Phase AC

http://www.du.edu/~jcalvert/tech/threeph.htm

Mike Rothe


--------------= Posted using GrabIt =----------------
------= Binary Usenet downloading made easy =---------
-= Get GrabIt for free from http://www.shemes.com/ =-

 

[Lord Garth]

"Tom Del Rosso" <ng01@att.net.invalid> wrote in message
news:58Rbe.657240$w62.141900@bgtnsc05-news.ops.worldnet.att.net...

"Lord Garth" <LGarth@Tantalus.net> wrote in message
news:gqNbe.1604$gd5.510@newssvr12.news.prodigy.com...

Assuming you've gone to classic menus within WinXP, dumping the
Fisher-Price look [......]

I always thought it looked like Candy Land.

I just got an idea for an original work of art. Take a dead monitor, chew
up various colors of bubble gum, and stick bits of it on the screen in a
mosaic that looks exactly like XP.

That sounds like the currently running commercial for Starburst candy.

The high school geek sculpts a bust of the high school 'hottie'. Upon
showing it to her he begins eating the face off to the horror of the
subject.
It's funny but I like the several featuring Clapton tunes.

Candy Land *shudder* from a great Uncle who hope to never play
that particular game again!

 

[John Fields]

On 27 Apr 2005 10:27:14 -0700, Tommi-Vogel@gmx.de (Thomas Vogel)
wrote:

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




how is the comparator conected?

* *
| |
|-|+\
| |
+-|-/
| |
* *


has the comparator four inputs? Or is that on the right a vertical
wire?

---
The comparator has two signal inputs. The vertical wire on the top is
connected to the comparator's positive supply input. The vertical
wore on the bottom is connected to the comparator's negative supply
input.
---

----+
|
| O-------> C
[COIL]- -|
| O--> |
| |
----+ +------------> NO

And with how many inputs is the relay connected? Are the + + also
connected, or the [COIL]- - ?

---
The relay coil has two terminals. C and NO are the Common and
Normally Open terminals, which go to the remote. + + is not a
connection.

----+
|
| O-------> C
[COIL]- - -|
| O--> |
| |
----+ +------------> NO

---

And please, answear me my last question to the LM393, too!

---
Use whichever one can supply 4mA into the relay with a Vcesat of less
than 0.75V. Also, if your alarm clock stays on for a long time, you
may want to think about a timer for the circuit.

--
John Fields
Professional Circuit Designer

 

[Jamie]

aman wrote:

I dont understand the real meaning of DC and AC readings on the Digital
voltmeter. Is it correct to say that if the signal = DC component + AC
component then DC reading gives the DC component and AC reading gives
the AC component ?

no.

the DMM (digital Multimeter) will give you DC (direct current)
from a steady source of energy , like a Battery for example.
AC voltages are alternating types of energy meaning, take the battery
and keep reversing the leads on the meter at 60 times per second or
50 times depending from what country your coming from. this Voltage
is changing polarity..
lets look at it from a more simpler view.
a common cell(1 part of a battery) (carbon) puts out aprox 1.56 volts
new. imagine putting this cell on a rotating plate on it's side.
now imagine the probes are on the edge of the plate equally spaced
apart.
as the plate turns with the cell lying on it, the - + terminals are
getting closer to the probe tips, lets say the meter is reading the
approaching terminals of the cell giving you a stronger reading until
they are perfectly lined up with the 2 probes! at this point the reading
is at its strongest. note that the DMM may have the - or + sign on the
left of the reading indicating the polarity. now the cell is still
turning and as it does the voltage is now going down. when the 2 cell
tips are at the far points of each probe tip, the voltage reading will
become 0. the cell keeps turning and you then will see voltage start
increasing again ! but this time the polarity is now the opposite as it
was before.
this continues until we get back to the starting point. this is
call alternating current because the voltage is raising and falling
back and forth alternating its polarity as it does so.
Hence the need for a meter to have a different mode to display to you
the Average or RMS. (root mean Square) reading which is really not the
highest voltage of the sine wave (PEAK).
for example.
10 Volts RMS. would be something like 10.4 if this AC wave was
converted to a clean DC to represent the peak..
so when measuring that 120 Volts of house current , the actual peak
voltage is around 160 volts.
any ways..
enough of this.

 

[Rheilly Phoull]

One day Randy2747@msn.com got dressed and committed to text

Thank You all for writing to help me out,sorry I didnt get to check
back and reply but my dad caught me surfing porn on his computer and
he grounded me LOL

I already have my leds here are the spec that came with them

DATA SPECIFICATION
5mm
Reverse Voltage (VR)5V
Reverse Current ( at VR=5V)<=10uA
Operating Temperature-40C ~ 85C
Storage Temperature-40C ~ 100C
Lead Soldering Temperature 260C for 5 seconds (Max)
Power Dissipation 5.5 mW
Emitted Color White
Chromaticity Coordinates (20mA) X : 0.32Y : 0.31
Lens Appearance Water Clear
Absolute max Rating Pd : 120 mW IF : 20 mA
Peak IF : 120 mA
Electro-optical Date (20mA)
VF : 3.2V (Max 3.4V)
IV : 15,000 mcd
(14,000/16,000 mcd Min/Max)
Viewing Angle 15 +- 5
Life (normal usage) 80,000 - 100,000 hours
___________________________
my other leds I ordered are:

20 x 18000mcd SUPER WHITE 10mm LED Lamp
Product Description :
Emitted Colour : WHITE
Size (mm) : 10mm
Lens Colour : Water Clear
Peak Wave Length (nm) : N/A
Forward Voltage (V) : 3.2 ~ 3.8
Reverse Current (uA) : <=30
Luminous Intensity Typ Iv (mcd) : Average in 18000
Life Rating : 100,000 Hours
Viewing Angle : ?10?
Absolute Maximum Ratings (Ta=25?C)
Max Power Dissipation : 80mw
Max Continuous Forward Current : 30mA
Max Peak Forward Current : 75mA
Reverse Voltage : 5~6V
Lead Soldering Temperature : 240?C (<5Sec)
Operating Temperature Range : -25?C ~ +85?C
Preservative Temperature Range : -30?C ~ +100?C
_________________________________________

I would like to make two differant circuits,one using the 5mm leds and
another using the 10mm leds to see which leds are whiter. I prefer to
use a transformer instead of 120 volt.

Now my questions are:

1. Is the 9V/300mA AC-to-DC Power Adapter going to put out enough
voltage for 8 leds in each of the two circuits I want to make?

2. Should I still use the 330 ohm 1/4W 5% Carbon Film Resistor for
all leds in each circuit?

Thank you all again for helping a rookie like myself,Randy

The power adaptor should be OK if you run the LED's at 20mA. Since the
forward voltage of the LED's id 3.2v 3 in series would not be good for the
9v supply, so you will have to have 4 series sets of 2 LED's connected in
parallel (if that makes sense ?). Each group would need its own resistor (=
4 resistors).
You should work out the values of the resistors and the current dissipation
yourself for yourself as a start in your experience. Folk in this NG will
help if you show some indication of effort.


--
Regards ..... Rheilly Phoull

 

[Rich Grise]

On Thu, 28 Apr 2005 10:48:47 -0700, jreim wrote:
[motor diagram snipped]

It says on the label this is for cw direction and if switch the black
and the red wire I get ccw direction. (or vice versa I forget which)

I bought a Double Pole Double Throw three position toggle switch
(on-off-on) but I just cannot figure out the proper wiring. I want it
so that I have cw on one side of the toggle - center being off and ccw
on the other side of the toggle. Can this be done? - If so - cna
someone tell me how to wire it. If not - is it the switch or is it that
I cannot do what I want with this motor. I guess I can always just use
one direction but it would be nice if I can reverse it.

Crossover Switch:

former Black ----+---------o on 1A
| \
| off o o------- New Black
| 1C
Former Red --- | ---+----o on 1B
| |
| |
| `----o on 2A
| \
| off o o------- New Red
| 2C
`---------o on 2B

Cheers!
Rich

 

[Lord Garth]

<Randy2747@msn.com> wrote in message
news:1114709876.547704.100990@l41g2000cwc.googlegroups.com...

Like I said I know very little about electronics but let me see if this
is correct:
R=E/I

9 voltage
-6.4 (3.2 volt bulb x 2 since its in series)
=2.6 volts

2.6 volts/0.040 amps (.020 amps x 2 in series=.040)
=65 ohms

In a series circuit, the current is the same through both lamps
so do not multiply by 2.

130 ohms is your target value.

The power dissipated by the resistor is P=I*E
..02*2.6=.052 Watts so a 1/4 watt resistor is fine.

 

[Jonathan Westhues]

<Randy2747@msn.com> wrote in message
news:1114709876.547704.100990@l41g2000cwc.googlegroups.com...

Like I said I know very little about electronics but let me see if this
is correct:
R=E/I

9 voltage
-6.4 (3.2 volt bulb x 2 since its in series)
=2.6 volts

Yes.

2.6 volts/0.040 amps (.020 amps x 2 in series=.040)

No. When you wire LEDs in series, the voltage drops add up but the current
is the same for all of them. 2.6/0.020 = 130 ohms.

see that the Max Continuous Forward Current is 30mA. Would this make
the leds dimmer?

Yes, just dimmer.

Notice that your LEDs actually specify a range of forward voltage drops,
from 3.2 to 3.8 V for the 10mm LEDs. That means that the voltage drop might
be anywhere in that range. If it is on the low end, 3.2 V, then the current
will be perfect, because you used 3.2 V in the calculations. If it is on the
high end then the current is (9 - 3.8 - 3.8)/130 = 1.4/130 = 0.011 A or 11
mA, which is maybe a bit low. You probably don't care though.

Jonathan
http://cq.cx/

 

[Rich Grise]

On Thu, 28 Apr 2005 10:37:56 -0700, Randy2747 top-posted:

Like I said I know very little about electronics but let me see if this
is correct:
R=E/I

9 voltage
-6.4 (3.2 volt bulb x 2 since its in series)
=2.6 volts

2.6 volts/0.040 amps (.020 amps x 2 in series=.040)
=65 ohms

NO. They're in series, so the current through the whole series string
is still 20 mA. The current is the same through every point in a
series circuit. Remember, volts are pressure, amps are rate of flow.

so the colors on my resistor would be Blue,Green,Black,Gold

Is this correct and am I on the right path?

Just do the math again for 20 mA.

The total current through the 4 series strings (of two LEDS and R)
will be the sum of the currents through the legs, or 80 mA.

And would the 65 ohm resistor work for the 10mm leds as well because I
see that the Max Continuous Forward Current is 30mA. Would this make
the leds dimmer?

Other than the resistor value, (which you should calculate as above,
using the forward V of the other LEDs), your setup should give equal
currents through the LEDs, which is actually what you're looking for:
The dimmer LEDs are dimmer LEDs.

Thanks again for your help

My Pleasure!
Rich

P.S. Many people prefer bottom-posting, for continuity of the conversation
for newcomers to the thread.

Rheilly Phoull wrote:
The power adaptor should be OK if you run the LED's at 20mA. Since
the
forward voltage of the LED's id 3.2v 3 in series would not be good
for the
9v supply, so you will have to have 4 series sets of 2 LED's
connected in
parallel (if that makes sense ?). Each group would need its own
resistor (=
4 resistors).
You should work out the values of the resistors and the current
dissipation
yourself for yourself as a start in your experience. Folk in this NG
will
help if you show some indication of effort.


--
Regards ..... Rheilly Phoull

 

[cpemma]

deepaa@gmail.com wrote:

Wondering if I need to buy a NiMH charger for NiMH batteries or a
normal AA charger will work with the NiMH batteries. Can anyone tell
me if this is possible?

Thanks in advance!

 

[Larry Brasfield]

"Thinking123" <jpb193@yahoo.com> wrote in message
news:1114718917.124295.297310@l41g2000cwc.googlegroups.com...

Can anyone suggest the best way to shape a single wire carring a
current so that it's magnetic field cancels itself out? Completely if
possible. I'm trying to optimize the idea for light weight and varying
current values.

The cancellation will be nearly complete if half of the
wire runs back along the same path as the other half.
Twist the halves together and it will be better yet. Put
a hole down the middle of one half, enlarge it, and run
the other half back thru the hole, (coaxially), and the
cancellation is as perfect as the coaxiality.

If the single wire has to carry current from one point
to another in a different place, then you are chasing
a rainbow. There will be a magnetic field created
by that current, no matter what path it takes between
the two separated points.

All I can think of would be to bunch the wire into a
tight zig-zag plane. (From:---------- To:||||||||) Would that work? Or
would it produce the same field...and just be heavy?

It would produce very nearly the same field, and just
dissipate more heat to carry the same current.

Any help would be great, thanks.

You could get better help by stating more of your
real problem and constraints.

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

 

[cpemma]

Not according to
http://www.starbatteries.com/chargerfaqs1.html#Can%20I%20use%20my%20older%20NiCD

"Can I use my older NiCD battery charger to charge my NiMH batteries?
No, you can use a newer NiMH charger to charge your old NiCD batteries but
you should not try to charge NiMH batteries with a NiCD charger. Here's why:
NiMH and NiCD batteries are chemically very similar and when both are fully
charged the battery voltage will start to drop. The NiMH voltage drop is
much more subtle and more difficult to detect. NiMH chargers usually have an
overtemperature sensor and shutoff to prevent overcharge as well as a more
sensitive voltage drop detector where NiCD chargers usually only use the
voltage drop to determine when to stop charging. If you charge NiMH
batteries in a NiCD charger it will probably miss the voltage drop and keep
right on charging causing the batteries to overheat. Overcharge and
overheating NiMH batteries can damage them internally and reduce battery
life."

You'll also find a new charger will be better geared to the higher
capacities usual in NiMH.

deepaa@gmail.com wrote:

Wondering if I need to buy a NiMH charger for NiMH batteries or a
normal AA charger will work with the NiMH batteries. Can anyone tell
me if this is possible?

Thanks in advance!

 

[John Popelish]

Thinking123 wrote:

Can anyone suggest the best way to shape a single wire carring a
current so that it's magnetic field cancels itself out? Completely if
possible. I'm trying to optimize the idea for light weight and varying
current values. All I can think of would be to bunch the wire into a
tight zig-zag plane. (From:---------- To:||||||||) Would that work? Or
would it produce the same field...and just be heavy? Any help would be
great, thanks.

The only way to cancel the magnetic field of a current is to have that
current return by the same path. The best you can do is to use a
coaxial wire, with the current going out in one conductor and coming
back through the other. A twisted pair is pretty good. A twisted
quad (two wires at the corners of the cross section carrying one way,
the wires at the other two corners of the cross section carrying the
other way), is better, etc.

 

[Larry Brasfield]

"Thinking123" <jpb193@yahoo.com> wrote in message
news:1114722588.885280.248360@l41g2000cwc.googlegroups.com...

You're right, I meant to mention that it had to be between two points.
Chasing a rainbow then. Are you sure?

Well, I am sure that *if* you have a current not following
its own return path back coaxially, *then* there will be a
magetic field produced by that current.

What if I wrapped it in a coil
for half the distance, then wrapped it in the opposite direction for
the other half? Like magnet/solinoid wraps. I just thought of it, but
that would change the direction of the field and cancel out right?

Almost, but not quite. And the not quite is equivalent
to what you get if the current does no spiraling around
and just travels in a straight path along the axis of what
was the spiral.

Or would that be the same as the zigzag?

The same in the sense of no net improvement.
Your zigzag is probably easiest to think about.
Consider that each zig or zag can be broken
into two vector components. One is in line
with the net direction of the zig-zag, the other
is transverse to it. The transverse components
of the zigs cancel the transverse components of
the zags, but the in line components all add in
the same direction, as much as would the effect
from a plain straight current flow.

(This is for a math model that I
gave up on a while ago when I realized that I would have to create and
destroy electrons to have a single current between 2 points in a closed
system. Now this is another idea how to make it realistic...unless it's
impossible. Just trying to think and get knowledgable feedback. )

Now you're on to the practical issue here. In the
world we inhabit, there is no such thing as a current
not flowing in a loop in appreciable amounts. You
can imagine single or small numbers of charges being
made to movie that way, but either they have to get
back to their source, (an ever mounting positive or
negative) region, or you have to have an unreal device
to keep moving them indefinitely.

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

 

[Larry Brasfield]

"Thinking123" <jpb193@yahoo.com> wrote in message
news:1114729836.951714.191690@z14g2000cwz.googlegroups.com...

Alright, if you're still following this, maybe you can think of a way
to make this work: Just think of an x-y plane where the magnetic field
is in the z direction. There are two battteries, DC generators,
whatever at two points. Say (0,0) and (10,0). If two wires are
connected between the two sources, there would be a current in both
wires. Both would generate a magnetic field and cancel out.

They would cancel out, if the current flows were
coaxial, at points outside the outer conductor.
If the currents are not coaxial, there will be a
net magnetic field at most points. There could
be some cancellation, along some surfaces or
lines, depending on the flow geometry.

Can you
think of a way to shape one of the wires so that it would not effect
the forces on the whole system? So that it would not produce a net
force in any direction and only the one wire (along the x-axis) would
be in the math model? (eg. I think a tall triangle would make the
effects in the y direction negligable, but then it would produce a
large effect in the +- x direction...)

Your question is too vague for me to understand.
Only moving charges are subject to a force due
to a DC magnetic field. And the closed system
does not exert net forces upon itself, (at least not
in the DC case. If it emits photons, it could.)

Any ideas? Impossible?

Sorry, I don't know at this point.

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

 

[John Fields]

On 28 Apr 2005 11:01:01 -0700, Tommi-Vogel@gmx.de (Thomas Vogel)
wrote:

If I construct it with a timer, where in the circuit should I built it
in and which type of should I take?

---
Build what you've got first. If it works and you need a timer, come
back.

--
John Fields
Professional Circuit Designer

 

[Rich Grise]

On Thu, 28 Apr 2005 11:44:49 -0700, jreim wrote:

Hi Rich,

Thanks for the info. I still do not get it though

There are six posts on the switch

3 wires on the cord (i know green goes to ground)

That leaves 7 wires to be attached to the switch. Where do I put those.

I realize that I probably have to keep

brown and orange together

yellow and white together

do I use the purple ungrounded wire at all?

If the diagram that you gave in your first post is accurate, and the
motor runs (in one direction) then leave them the way they are. According
to your original post, only the black and red needed to be swapped to
reverse the motor.

If I were to look at the posts


1----2 on cw
| |
| |
3 4 off
| |
| |
5----6 on ccw

I've replaced my designators with your numbers:

Crossover Switch:

Black ----+---------o 1
| \
| off o o------- to terminal black/red
| 3
Red --- | ---+----o 5
| |
| |
| `----o 2
| \
| off o o------- to terminal red/black
| 4
`---------o 6

I've done the switch diagram in the way a standard schematic
is done. For each switch, there are three positions:

On Off On
o 1,2 o 1,2 o 1,2
\
o o 3,4 o---o 3,4 o o 3,4
/
o 5,6 o 5,6 o 5,6

Interpose this between wherever the red and black are coming
from, and the point where they attach. Connect the black wire
to 1 and 6, and connect 3 to the terminal block on the motor.
Connect the red wire to 2 and 5, and connect 4 to the terminal
block on the motor.

But this is all assuming that the rest of the wires don't
need to be changed _at all_ - if that's not the case, then
we need more information.

Good Luck!
Rich

 

[John Smith]

"tlbs" <tlbs101@excite.com> wrote in message
news:1114703967.278251.279020@g14g2000cwa.googlegroups.com...

Disclaimer: you need to check your local building codes to be
absolutely sure.

What I have seen (and done) time-and-time again is this: Connect the 2
ground wires together along with a 3rd short pigtail, all in a
wire-nut. Connect the pigtail to the ground lug on the outlet or
switch (or to the box).

I believe it is safer to connect the ground pigtail directly to the
outlet or switch ground-lug.

One other suggestion: if you can, don't connect all the outlets and
the overhead light to the same circuit breaker. The load on the
circuit breaker is not the issue -- the problem is, if that particular
breaker goes bad you lose everything in that room simultaneously. If
one or two outlets are on a different breaker, and a desk lamp or
nightlight is plugged into those outlet, then you would still have some
light/power in the room when the light breaker failed.

This has happened to me (breaker failure -- overhead light only). I
have observed the practice I described in several homes.

As tibs says, put the plug outlets onto a separate breaker to the lights.

Also, create a ring for the plug outlets, not just daisy-chain. (I.e. loop
the end of the daisy-chain back to the breaker). Should give you (a) smaller
wire requirement, thus a bit cheaper, and (b) a second path if something
breaks in one leg.

 

[Genome]

"alpha" <newsgroupaddress@hotmail.com> wrote in message
news:1114775975.780112.14880@z14g2000cwz.googlegroups.com...

I need to shorten the end of my solder iron so i have more control of
direction and make it more like a pen.....can i cut it down so my
fingers are nearer the tip without causing a problem.....thought it
might cause a problem with heat distribution and overheat..........i
have no clue about anything electric related

was also wondering what material i can rap around the rod as a grip to
hold it closer to the end if cutting is not an option......

many thanks in advance if you can advise......A

If you are talking about shortening the tip then you will run into problems
because it is iron coated to stop the solder from dissolving the copper. If
you are talking about cutting the element then you will run into problems
because it will no longer work.

Soldering takes practice but one of the best things you can do is give
yourself a solid base from which to work. That means the piece you are
trying to solder should be stable on the bench and then, holding the solder
in one hand and the iron in the other, rest your forearms on the edge of the
bench and operate from that position. Don't try and do it with the whole of
your arms flapping about in the air, you introduce too many degrees a
freedom.

DNA