Chip with simple program for Toy

[R. Steve Walz]

me wrote:

jackbruce9999@yahoo.com wrote in news:1118461679.394136.299150
@g14g2000cwa.googlegroups.com:

Again, is the term "DC Sine Wave" problematic because it is
fundametnally wrong OR is it problematic because it is at odds with
conventional terminology and nomenclature.....if it is fundamentally
wrong, then please show how.....

Look jackass, AC means it is alternating in time. If it is DC it is
constant. A sine wave alternates in time, thus is AC. Now shut up.
--------------------

Listen, you impolite little shit. Everyone in PHYSICS knows that any
changing current has an AC COMPONENT, and that the unchanging part
is the DC component, so while a sine will always be the AC component,
it does NOT have to be other than UNIPOLAR VOLTAGE-WISE!!

-Steve
--
-Steve Walz rstevew@armory.com ftp://ftp.armory.com/pub/user/rstevew
Electronics Site!! 1000's of Files and Dirs!! With Schematics Galore!!
http://www.armory.com/~rstevew or http://www.armory.com/~rstevew/Public

 

[R. Steve Walz]

soolun@webtv.net wrote:

I have a 12V battery pack. When the voltage drops to 9 volts, I need to
shut off the power. Is there a device to do this? A wiring diagram
would help as I am weak in electronics.
-------------

Wire an open-collector op-amp comparator with 100 ohm/5.1V zener-diode
defined reference voltage to turn on and off a latch that runs a
transistor. Use a 10K pot to set the 9V set-point, and then replace
with two resistors if you like.

If you want a diagram send me $10. Or you can look it all up using
those words and phrases in Google. Clues: LM339, 74LS74, 2N3055.

-Steve
--
-Steve Walz rstevew@armory.com ftp://ftp.armory.com/pub/user/rstevew
Electronics Site!! 1000's of Files and Dirs!! With Schematics Galore!!
http://www.armory.com/~rstevew or http://www.armory.com/~rstevew/Public

 

[John Larkin]

On 11 Jun 2005 19:37:25 -0700, "Bret Cahill" <BretCahill@aol.com>
wrote:

Spread a lot of cathodes around the outside of a vacuum chamber and
target a deterium anode in the center.

If the cold fusionists can use electrons, under equal peotection, so
can you.


Bret Cahill

Where do you want that Nobel prize sent?

John

 

[Lord Garth]

"Thaqalain" <saqlain92110@yahoo.com> wrote in message
news:1118541776.730800.56060@o13g2000cwo.googlegroups.com...

Truck is having 6 cells of 1.5 v general size.I connected all cell
positions by fixing wire pieces in cell slots.Later I tried to charge
by choosing required voltage of 9v,by connecting two end terminals of
opposite polarity,but device is not producing sounds.
What remedy I can do?

You need only concern yourself with the anode terminal of the first battery
chamber and the cathode terminal of the last battery chamber. You do not
need the connect a wire where each battery used to be. Apply the proper
polarity at 9 volts and with sufficient current and the toy will run.

 

[Don Kelly]

"BobG" <bobgardner@aol.com> wrote in message
news:1118284135.294538.233840@f14g2000cwb.googlegroups.com...

Is there a 'rule of thumb' for figuring out much torque it takes to put
out xx volts and yy amps from a PMA given the gauss of the magnet and
the N turns of the coils? Should spin free open circuit, and really
stall out short circuit, and be somewhere in the middle during
operation. I guess the 'max power' point is when the load drags the
volts down to half the open circuit value? I guess the current here
would be half the short circuit current too? I want to rig the pwm
controller to not load down the pma too much under low power/low
torque/ slow spinning conditions, like from a stirling or micro hydro.
In general volts (and current) is proportional to rpm... someone got
some formulas that are a little more specific?

First of all. the torque to produce a voltage xx is 0. Torque is not related

to voltage per se. Torque is related to current.
If you can find the voltage -easiest done by spinning it and measuring it.
then you will get a relationship E=Kw where E is in volts and w is in
radians/ sec (2*Pi*60 rad/sec =1 rpm)
Then a rough guide is Torque =KI torque in Newton meters and I in amps.
same K as above.

The relationship between E and I depens on the load.

Finding the max power is another problem and it will not be at the half
voltage level. Do you really want to be operating there? the alternator
won't want to be operating there. It is a little different than what is
implied by the maximum power theorem which implies matching load impedance
to the alternator impedance., not the other way around (it also implies <50%
efficiency)


--
Don Kelly
dhky@peeshaw.ca
remove the urine to answer

 

[Nigel Heather]

You also need to consider bolting the 7812 to a large heat sink.

You haven't what current you what to supply but for the sake of argument
assume that it is 1A.

You have a 24V source and you are delivering 12V. This means that the
difference of 12V will be across the 7812. With 1A being supplied this
means that 7812 will be disipating 12W (12V x 1A) which will mostly be
turned into heat. This may not seem much compared with everyday domestic
appliances but for an electronics component it is a lot. Without an
effective way to get rid of this unwanted power (heat) the 7812 will cook
and probably fail.

Cheers,

Nigel

 

[John Fields]

On 12 Jun 2005 09:01:11 -0700, mabelmapleleaf@yahoo.com wrote:

It's a shame you have to weed thru all the crap from some
of the posters here who have a lot of time on their hands and
have no tolerance for those who are just learning their craft....

---
It's a shame that those of us who give of our time in an effort to
edify the ignorant are often abused by imbeciles who can't take
correction gracefully.

--
John Fields
Professional Circuit Designer

 

[Tom Biasi]

"Nigel Heather" <No-Spam@No-Spam.com> wrote in message
news:aO2dnRYDrawkqjHfRVnyuQ@pipex.net...

You also need to consider bolting the 7812 to a large heat sink.

You haven't what current you what to supply but for the sake of argument
assume that it is 1A.

You have a 24V source and you are delivering 12V. This means that the
difference of 12V will be across the 7812. With 1A being supplied this
means that 7812 will be disipating 12W (12V x 1A) which will mostly be
turned into heat. This may not seem much compared with everyday domestic
appliances but for an electronics component it is a lot. Without an
effective way to get rid of this unwanted power (heat) the 7812 will cook
and probably fail.

Cheers,

Nigel
Yes, That occurred to me after I posted. The 78xx series are thermal
protected ( IIRC) and also sometimes go into thermal oscillation.
A heat sink is definitely called for.

Tom

 

[Kitchen Man]

On Sun, 12 Jun 2005 04:55:45 -0800, floyd@barrow.com (Floyd L.
Davidson) wrote:

But rather than look at DC as current going all going in one
direction, and AC as anything else, it is *far* easier to view
it as AC is any current that is changing, and DC is anything
else (i.e., the current is steady).

Technically those definitions are exactly the same, but one
leads to a lot of confusion.

Thanks, Floyd, for that excellent and understandable post.

--
Al Brennan

"If you only knew the magnificence of the 3, 6 and 9,
then you would have a key to the universe." Nicola Tesla

 

[Roger Johansson]

Floyd L. Davidson wrote:

Except, polarity reversals are not significant to the definition
of AC.

You are right. AC and DC terms are about rate of change.

DC is when rate of change is zero, AC is everything else.


The problem started when the OP thought that the word "fully" would
give the meaning "never changing direction" to the term "fully DC".

But people who work with electronics will read the word "fully" as
meaning a perfect DC signal, with no change of voltage at all.

Because rate of change is important to us, absolute direction of
current is not important.


--
Roger J.

 

[keith]

On Mon, 13 Jun 2005 02:23:55 +0000, Ian Stirling wrote:

In sci.electronics.design keith <krw@att.bizzzz> wrote:
On Sat, 11 Jun 2005 20:37:00 +0000, Joerg wrote:

Hello Jak,

They need to save their bandwidth for pictures & video....

Let's see how that goes. Maybe it's like with pool heaters. People love
it until that first bill arrives in the mail.

What amazes me is the cost of text-messaging. Why do I want to do that
again?

Time?

Typing a text-message in on my phone is going to save me time? OTOH,
their time is certainly cheaper.

I've ditched my land-line and gone cell, but there are many things that
make no sense. Land-lines are expensive because the government decides
it is so, but that doesn't excuse the cell companies from sanity. Sure,
I know, it's all the teeny-boppers that think text-messaging is cool.

If you feel comfortable talking to people by just reciting a message,
then you can probably get it done faster/cheaper.

Huh?

But, how often is that the case?
Many people on the other end will digress into weather/how their dog is/...

Huh^2?

--
Keith

 

[steamer]

--Here's a better design for ya...
http://www.hunkinsexperiments.com/pages/nuclearfission.htm

--
"Steamboat Ed" Haas : Just another fart in
Hacking the Trailing Edge! : the Elevator of Life...
http://www.nmpproducts.com/intro.htm
---Decks a-wash in a sea of words---

 

[Don Kelly]

"Fred Abse" <excretatauris@cerebrumconfus.it> wrote in message
newsan.2005.06.12.09.23.18.98450@cerebrumconfus.it...

On Sat, 11 Jun 2005 23:02:05 +0000, Rich The Newsgroup Wacko wrote:

They are not "out of phase" - they are simply opposite
polarity! It's a significant difference, in the realm of phasors and
imaginary power and stuff.

Well. They are not "in phase". they are "antiphase" with reference to the
center tap.

One is (162 sin(wt))
The other is (162 sin(wt+PI))

That's out of phase enough for me

--
Actually, if you were in the UK where the 3 wire single phase 120/240V

(Edison) system is not used, it is called 2 phase. Where it is used it is
called 3 wire single phase. Technically the Brits are right (balanced
polyphase = N equal voltages , 360/N degrees apart) but the terminology
single phase 3 wire is a historical rather than a strict technical
description dating back to Edison where it was used for DC.
--
Don Kelly
dhky@peeshaw.ca
remove the urine to answer


P.S love the Leacock quote
-----------

"Electricity is of two kinds, positive and negative. The difference
is, I presume, that one comes a little more expensive, but is more
durable; the other is a cheaper thing, but the moths get into it."
(Stephen Leacock)

 

[ehsjr]

soolun@webtv.net wrote:

I have a 12V battery pack. When the voltage drops to 9 volts, I need to
shut off the power. Is there a device to do this? A wiring diagram
would help as I am weak in electronics.

+----/ o-----+-------------------- +12 output
| |
+ ---+---o o-----+-----RELAY-----+
PB | |
R1 \ -----
/ / \ TL431
|<------------/ \ or
/ /_____\ LM431
R2 \ |
| |
- ----------------+---------------+

You need 2 resistors, a relay and a TL431, plus a momentary
pushbutton switch. MPB - 1B for the switch, and RLY-428 for
the relay are good choices, available from Allelectronics
http://www.allelectronics.com/
The specified relay will handle 8 amps and can be driven
directly from the TL431. If you use a different relay,
you need to be sure that the TL431 can drive it directly.
Make sure that the relay's coil resistance is over 120
ohms. The higher the resistance of the relay coil, the
better. It will be tough to find a better relay choice than
the one specified. R1 is 2200 ohms. R2 is 352 ohms and can
be made with a 330 ohm and a 22 ohm resistor in series.

When you press the pushbutton, the relay will energize,
provided the battery voltage is over about 9.1 volts.
The relay contact, shown above the PB (pushbutton)
symbol will close conducting the 12 volts to the output
and to the TL431 (or LM431 - same device) through the
relay coil. The 431 will conduct as long as the voltage
from the battery is over about 9.1 volts.

The circuit shown will draw about .025 amps from your
battery pack. When the voltage drops to close to below
about 9.1 volts the relay will de-energize, and no
current will flow either to the load or to the circuit.

Ed

 

[John Larkin]

On 12 Jun 2005 13:25:30 -0700, "Bret Cahill" <BretCahill@aol.com>
wrote:

It was probably the first thing they tried.

They have all kinds of deals where they convert eletron beams to
radiation and vice versa for fusion.

Electrons won't work, but it's not hard to whack duterium with
duterium or tritium ions to get fusion... it just takes a couple
hundred KEV or something. The net energy yield will be some insanely
small fraction of a per cent.

John

 

[Larry Brasfield]

<htewam@yahoo.com> wrote in message
news:1118691011.414013.202840@f14g2000cwb.googlegroups.com...

I need to find a transformer (converter?) that will allow me to
upconvert from 12VAC to 24VAC in order to drive a small motorized
device (50 watts) for continuous duty. I tried using a 110V/220V
'travel' transformer -- it gave me the correct voltage but not enough
output current (probably not enough turns on the windings). I need a
transformer rather than an electronic converter since I am driving an
AC motor... isolation isn't an issue so I could go with an
Autotransformer. All of the transformers I see out there are for high
voltage primaries (110V or more). Thanks.

The method I suggest below will require a slightly larger
transformer than one designed for your purpose, but it
would only leave about half the copper unused. For a
one-off application, that may be better than trying to
find an unusual transformer.

If you can find any transformer with a 24 VAC center-tapped
secondary rated for your load current (which appears to be a
bit above 2 A), then you can drive the center-tap and one end
with your 12 VAC and take 24 VAC from the ends. Be sure
to cover the primary lead ends as they will (likely) have a higher
voltage. This will function as an auto-transformer, with about
2 A flowing thru each half of the 24 V winding for a total of
4 A applied to the center-tap. If the center-tap is brought out
as a single wire, you need be sure it is good for 4 A. (If the
secondary is split, as is common, the wires should be large
enough for the rated current.)

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

 

[ehsjr]

ehsjr wrote:

soolun@webtv.net wrote:

I have a 12V battery pack. When the voltage drops to 9 volts, I need to
shut off the power. Is there a device to do this? A wiring diagram
would help as I am weak in electronics.


+----/ o-----+-------------------- +12 output
| |
+ ---+---o o-----+-----RELAY-----+
PB | |
R1 \ -----
/ / \ TL431
|<------------/ \ or
/ /_____\ LM431
R2 \ |
| |
- ----------------+---------------+

You need 2 resistors, a relay and a TL431, plus a momentary
pushbutton switch. MPB - 1B for the switch, and RLY-428 for
the relay are good choices, available from Allelectronics
http://www.allelectronics.com/
The specified relay will handle 8 amps and can be driven
directly from the TL431. If you use a different relay,
you need to be sure that the TL431 can drive it directly.
Make sure that the relay's coil resistance is over 120
ohms. The higher the resistance of the relay coil, the
better. It will be tough to find a better relay choice than
the one specified. R1 is 2200 ohms. R2 is 352 ohms and can
be made with a 330 ohm and a 22 ohm resistor in series.

When you press the pushbutton, the relay will energize,
provided the battery voltage is over about 9.1 volts.
The relay contact, shown above the PB (pushbutton)
symbol will close conducting the 12 volts to the output
and to the TL431 (or LM431 - same device) through the
relay coil. The 431 will conduct as long as the voltage
from the battery is over about 9.1 volts.

The circuit shown will draw about .025 amps from your
battery pack. When the voltage drops to close to below
about 9.1 volts the relay will de-energize, and no
current will flow either to the load or to the circuit.

Ed
Here's a corrected version - I forgot to show a diode:

+----/ o-----+-------------------- +12 output
| | D1
| | +---|<--+
| | | |
+ ---+---o o-----+---+-RELAY-+---+
PB | |
R1 \ -----
/ / \ TL431
|<------------/ \ or
/ /_____\ LM431
R2 \ |
| |
- ----------------+---------------+


Note the added (D1) 1N4001 diode


Ed

 

[ehsjr]

htewam@yahoo.com wrote:

I need to find a transformer (converter?) that will allow me to
upconvert from 12VAC to 24VAC in order to drive a small motorized
device (50 watts) for continuous duty. I tried using a 110V/220V
'travel' transformer -- it gave me the correct voltage but not enough
output current (probably not enough turns on the windings). I need a
transformer rather than an electronic converter since I am driving an
AC motor... isolation isn't an issue so I could go with an
Autotransformer. All of the transformers I see out there are for high
voltage primaries (110V or more). Thanks.

A 110 to 220 transformer will work with 12 vac on its primary

to give you 24 vac output on its secondary.
Ed

 

[John G]

"ehsjr" <ehsjr@bellatlantic.net> wrote in message
newsmmre.7272$5s1.3158@trndny06...

htewam@yahoo.com wrote:
I need to find a transformer (converter?) that will allow me to
upconvert from 12VAC to 24VAC in order to drive a small motorized
device (50 watts) for continuous duty. I tried using a 110V/220V
'travel' transformer -- it gave me the correct voltage but not enough
output current (probably not enough turns on the windings). I need a
transformer rather than an electronic converter since I am driving an
AC motor... isolation isn't an issue so I could go with an
Autotransformer. All of the transformers I see out there are for
high
voltage primaries (110V or more). Thanks.

A 110 to 220 transformer will work with 12 vac on its primary
to give you 24 vac output on its secondary.
Ed

But to supply the OPs load the 110 to 220 volt transformer will need to
be rated at about 500 va to have heavy enough wire to cope with the 2
amps needed.

--
John G

Wot's Your Real Problem?

 

[Bob Monsen]

ehsjr wrote:

ehsjr wrote:

soolun@webtv.net wrote:

I have a 12V battery pack. When the voltage drops to 9 volts, I need to
shut off the power. Is there a device to do this? A wiring diagram
would help as I am weak in electronics.


+----/ o-----+-------------------- +12 output
| |
+ ---+---o o-----+-----RELAY-----+
PB | |
R1 \ -----
/ / \ TL431
|<------------/ \ or
/ /_____\ LM431
R2 \ |
| |
- ----------------+---------------+

You need 2 resistors, a relay and a TL431, plus a momentary
pushbutton switch. MPB - 1B for the switch, and RLY-428 for
the relay are good choices, available from Allelectronics
http://www.allelectronics.com/
The specified relay will handle 8 amps and can be driven
directly from the TL431. If you use a different relay,
you need to be sure that the TL431 can drive it directly.
Make sure that the relay's coil resistance is over 120
ohms. The higher the resistance of the relay coil, the
better. It will be tough to find a better relay choice than
the one specified. R1 is 2200 ohms. R2 is 352 ohms and can
be made with a 330 ohm and a 22 ohm resistor in series.

When you press the pushbutton, the relay will energize,
provided the battery voltage is over about 9.1 volts.
The relay contact, shown above the PB (pushbutton)
symbol will close conducting the 12 volts to the output
and to the TL431 (or LM431 - same device) through the
relay coil. The 431 will conduct as long as the voltage
from the battery is over about 9.1 volts.

The circuit shown will draw about .025 amps from your
battery pack. When the voltage drops to close to below
about 9.1 volts the relay will de-energize, and no
current will flow either to the load or to the circuit.

Ed

Here's a corrected version - I forgot to show a diode:

+----/ o-----+-------------------- +12 output
| | D1
| | +---|<--+
| | | |
+ ---+---o o-----+---+-RELAY-+---+
PB | |
R1 \ -----
/ / \ TL431
|<------------/ \ or
/ /_____\ LM431
R2 \ |
| |
- ----------------+---------------+


Note the added (D1) 1N4001 diode


Ed

You forgot the positive feedback. Also, using a P-MOSFET instead of a
relay will save alot of power.

The circuit is as follows:

P-MOSFET
Battery+ -------+------+---S+--. -D------.
| | | ^ | |
[620k] [22k] -- -- -- |
| | --------
| | G |
| +----' |
o------(---------[2.2MEG]--o
| | |
| --- |
o-----/ \ TL431 |
| ----- |
| | [Rload]
[100k] | |
| | |
Battery- --------o------o-------------------'

The TL431 will sink current when it's control lead is > 2.5V. That will
happen when the output is high, and battery gets down to 9V. It won't
turn back on until the battery gets back up to 10V. Thus, it may not
chatter as the battery gets close to 9V like the relay one could easily
do, depending on the load.

You actually need a suprisingly small MOSFET, because it'll always
either be off or completely on, so it won't need to deal with large
amounts of power. When the circuit is off, the simulator claims it
should only draw about 50uA.

I have not built this circuit, only simulated it, so you should
obviously experiment before committing to building 1000 of them.

This circuit has crappy temperature compensation, so it may not work
outside a reasonable temp range.

---
Regards,
Bob Monsen