Chip with simple program for Toy

[Don Bruder]

In article <2HM4e.198$6_4.2466@nnrp1.ozemail.com.au>,
"John G" <Greentest@ozemail.com.au> wrote:

Don,
I am glad you said all that because I often get bagged for saying things
like "You do not know enough to do what you want" when I am trying to
save the OP from killing himself or others.

Yeah, I've seen that syndrome many times. I consider it for exactly what
it's worth: The babblings of fools more interested in hearing the sound
of their own imaginary "authority" on a topic they themselves don't know
enough about to be saying "shit" about if they were standing there with
a mouthful of it. The "politically correct" climate today is really
conducive to "Oh, don't listen to that jerk saying you don't know
enough!"

Never mind that the querant has posed a question about a fundamental
task that even the rankest qualified beginner knows either can't be done
safely, or can't be done at all using the proposed methods.

As I made a point of distinguishing, there's two classes of people:
There's "stupid" (AKA "Just plain dumb"), and there's "ignorant" (AKA
"Unknowing"). Stupid is incurable. You don't know, you can't know, you
won't know, you make no effort to know - You're stupid, with every
negative connotation the word can be loaded down with, and you deserve
to die in agony from the results of a stupid deed. Ignorance is cured
with incredible ease. You don't know, but you acknowledge that you
don't, and seek assistance. You gain knowledge. You're no longer
ignorant. Cured! Holy of holies, I'm cured!

OP on this one struck me as ignorant, and was handled accordingly.
Sounds like he might have actually learned something, too

--
Don Bruder - dakidd@sonic.net - New Email policy in effect as of Feb. 21, 2004.
Short form: I'm trashing EVERY E-mail that doesn't contain a password in the
subject unless it comes from a "whitelisted" (pre-approved by me) address.
See <http://www.sonic.net/~dakidd/main/contact.html> for full details.

 

[Kevin Aylward]

PeteS wrote:

You are correct that the cap charged with the higher voltage has more
electrons on one plate (and a corresponding lack of them on the other
plate). The basic equations for capacitors (useful in both the
electrical and physics domains) are:

Q=CV, where Q = charge in coulombs, C the capacitance, and V the
voltage across the plates, which when integrated to get energy, gives

E = 1/2 [C (V^2)]

To know the current, or the rate of change of voltage, then take the
derivative of the first equation wrt time (Q/t = current, steady
state)

I = C (dV/dt).

Giving the useful piece of information that charging a cap with a
current source gives a linear ramp of voltage (used extensively in
timing circuits).

A coulomb, incidentally, is roughly equal to 6.241506×10^18 electron
charges (One might say electrons for a generality).

From the physical aspect, a symmetric [each plate has equal area]
plate
style capacitor is given by
C = (k(o) * k(r) * A) / d
where k(o) is the permittivity of free space (8.854 * 10^-12 F/m),
k(r) the relative permittivity of the dielectric, A the area of each
plate (one plate - remember they should be symmetric) and d the
distance between the plates.

That ignores leakage at the edges, but it's a starting point.

This comes in very handy when using power planes next to ground planes
in multilayer boards to figure out the effective capacitance (which I
have used as part of the local high speed filtering).

For an interesting experiment, charge two capacitors in parallel,
disconnect the supply and reconnect them in series. if the caps were
equal, you have a voltage doubler (because you now have half the
capacitance).

Nope. Its because one cap has V and so does the other, the sum is 2V.

Then do it with one cap 10 times the capacitance of the
other. Measure the new voltage. You'll be quite surprised.

I would be, if it were anything other than V1 + V2. If this were not so
connecting different sized batteries in series would not produce V1 +
V2.

Charged caps look like voltage sources. Voltage sources in series add
voltages. I am at a complete loss as to why you think one can generate
high voltages in this manner. Maybe I have missed something in the last
25 years...

This trick
is used in Parametric amplifiers (common in Radars, for example).

Oh?

Parametric amplifies typically use a non-linear capacitor, a pump
oscillator and a signal. The mixing produces gain at the signal
frequency due to the 3rd harmonic term non-linearity, much the same as
in http://www.anasoft.co.uk/EE/tapebias/tapebias.html.

Kevin Aylward
informationEXTRACT@anasoft.co.uk
http://www.anasoft.co.uk
SuperSpice, a very affordable Mixed-Mode
Windows Simulator with Schematic Capture,
Waveform Display, FFT's and Filter Design.

 

[Kevin Aylward]

PeteS wrote:

Let's do the actual mathematics

Oh?

Assume we charge a 10uF and 1uF to 10V. The 10uF will have 100uC
(Q=CV) and the 1uF 10uC of charge. The total charge is 110uC.
Now lets stick them in series.
The first thing is to calculate the new effective capacitance =
C1*C2/(C1 + C2) = 0.909uF. A thing about series caps is the total
capacitance is always less than the smallest cap.

As Q = CV, then V = Q/C so 110uC/0.909uF = 121V (near enough).

Amazing...Simply connecting two charged caps, of different capacitance
values produces, a higher voltage then either. Wow. Wonders will never
cease. Like, dude, you haven't ever tried this have you...You out to
lunch on this one.

So what you get is V(init) + [V(init] * C2/C1].
In Paramps, we use varactors (reverse biased diodes specifically
designed to be used as votage variable caps) to vary the effective
capacitance, and thence vary the plate voltage.

What are you actually trying to say here?

A voltage varies a capacitor and hence another voltage varies means
what, exactly?

Kevin Aylward
informationEXTRACT@anasoft.co.uk
http://www.anasoft.co.uk
SuperSpice, a very affordable Mixed-Mode
Windows Simulator with Schematic Capture,
Waveform Display, FFT's and Filter Design.

 

[Kevin Aylward]

Bob wrote:

"Michael Redmann" <redmann@despammed.com> wrote in message
news:d30a6c$7m9$04$1@news.t-online.com...
PeteS schrieb:

Let's do the actual mathematics

Assume we charge a 10uF and 1uF to 10V. The 10uF will have 100uC
(Q=CV) and the 1uF 10uC of charge. The total charge is 110uC.
Now lets stick them in series.
The first thing is to calculate the new effective capacitance =
C1*C2/(C1 + C2) = 0.909uF. A thing about series caps is the total
capacitance is always less than the smallest cap.

As Q = CV, then V = Q/C so 110uC/0.909uF = 121V (near enough).

Hi Pete

I did it by calculating the total energy stored in both capacitors
which is 5.5 mJ. The corresponding voltage with both caps in series
is V = V0 * (C1+C2)/sqrt(C1*C2) after charging each cap to V0.

So, what's correct? Conservation of charge or energy?

Let's test your calculation with two equal capacitors, say 10 uF at
10 volts. If you were right voltage would be 200 uC/5 uF = 40 V !
That's obviously wrong. Of course the correct answer is 20 V. That's
the same voltage you get using my formula.

Regards
--
Michael Redmann
"It's life, Jim, but not as we know it." (Spock)


If the method is:

1) Place two paralled caps, of different values, across a voltage
source (of magnitude V)
2) Remove the voltage source
3) Separate the caps and reconnect them in series
4) Measure the voltage across the series combination

It doesn't make sense that the series voltage would be anything other
than 2*V.

Indeed.

How can the voltage across either of the caps suddenly
increase when you separate them?

They cant.

Also, I tried it.

Yep. No getting around emperical evidence.

I used a 220uF cap and a 2200uF cap. The supply was
5V. After I separated them, and then place them in series, I measured
10V across the series combination.

Was this the method described for the so-called parametric amplifier?

Couldn't be. Parametric amplifiers use a different principle, more or
less

I did a little searching on this subject, but never found this method
described for parametric amplifiers.

It works, essentially, the same as
http://www.anasoft.co.uk/EE/tapebias/tapebias.html. Change magnetic
non-linearity to capacitor non-linearity.

Kevin Aylward
informationEXTRACT@anasoft.co.uk
http://www.anasoft.co.uk
SuperSpice, a very affordable Mixed-Mode
Windows Simulator with Schematic Capture,
Waveform Display, FFT's and Filter Design.

 

[Lord Garth]

"Thomas Vogel" <Tommi-Vogel@gmx.de> wrote in message
news:7f88d87.0504060612.538a39d@posting.google.com...

"Lord Garth" <LGarth@Tantalus.net> wrote in message
news:<orz4e.1859$3z3.1270@newssvr12.news.prodigy.com>...
"Thomas Vogel" <Tommi-Vogel@gmx.de> wrote in message
news:7f88d87.0504050839.7cbf63a1@posting.google.com...
I want to construct a circuit which switch a relay by a alarm clock.
Know anyone where I can find a plan in the internet or in a book or in
a magazine?


You must define what signal the alarm clock produces before you can
get any useful responses. What is the output voltage? Is it a pulse or
a level? What are the specs of the relay you wish to use? Is the relay
supposed to stay energized until it is manually reset? Is the relay
circuit
powered by a battery or a line source?

Perhaps you mean you wish to build a digital alarm clock with a relay
output?

the alarm clock produce a signal which changes all the time. It goes
with a 1,5V battery, so the output is by ~1V.
The relay is powered by a battery.

You will need another power supply because that 1.5V battery is insufficient
for the new circuitry.

You say the signal changes all the time, I presume this means you are seeing
an oscillating alarm output which is probably meant to be buffered and then
drive a piezo speaker. There is the possibility that you are saying the
alarm
output changes as the clock battery ages however.

With such a low output voltage, you will need to amplify the output before
you can do anything useful.

The likely relay driver would first amplify the alarm output then use this
to trigger a one shot, to rid your alarm signal of the oscillations. The
last
step is to use the one shot output to activate the relay.

There are several approaches to a solution, however, I would suggest you
move this discussion to alt.binaries.schematics.electronic so that actual
pictures and diagrams can be attached. ASCII art is rather crude.

 

[John Fields]

On 6 Apr 2005 10:57:35 -0700, "Jack// ani" <nospam4u_jack@yahoo.com>
wrote:

Hi all,

I just took out a dead Panasonic lead acid battery form my UPS and
trying to understand the specifications written on it

Voltage Regulation
Cycle use : 14.5 - 14.9V
Initial current : less than 2.8A
Standby use : 13.6 - 13.8V

Will anybody please explain it?

---
How about Panasonic?

First, go to:

http://www.panasonic.com/industrial/battery/oem/chem/seal/index.html


and then click on:

VRLA Charge Methods

--
John Fields
Professional Circuit Designer

 

[Byron A Jeff]

In article <PeW4e.8421$c76.6375@newssvr11.news.prodigy.com>,
Lord Garth <LGarth@Tantalus.net> wrote:

Hi all,

I just took out a dead Panasonic lead acid battery form my UPS and
trying to understand the specifications written on it

Voltage Regulation
Cycle use : 14.5 - 14.9V
Initial current : less than 2.8A
Standby use : 13.6 - 13.8V

When you charge the battery, limit the applied voltage to between 14.5 and
14.9 volts DC

That's a bit high for standard gel cels. They usually want 14.4V.

and limit the current to no more than 2.8 Amperes.

Correct. A fully depleated battery can draw a lot of current if you allow
it to.

As the battery charges, the current will fall.

Correct. The basic 3 stage charging algorithm for lead acid batteries are:

1) Bulk charge with highest allowable voltage and current until battery
reaches 2.4V/cell (14.4V for a 12V battery).

2) You then fix the voltage to 14.4V and start watching the current. Continue
in this phase until the current draw drops to C/100, where C is the Amp-Hour
capacity of the batter. So for example with my 33 AHr battery, C/100 would be
330 mA.

3) Then go into float/trickle/top off voltage discussed below.

If you wish to 'top off' the battery or trickle charge it, keep the applied
voltage between 13.6 and 13.8 volts DC.

This voltage is safe indefinitely without venting.

Beware that hydrogen gas forms during charging so vent the area.

If a sealed lead acid gel cell is venting, then you have big problems.
It's one of the reasons why the specifications listed on the battery are
in place.

BAJ

 

[Lord Garth]

"Jack// ani" <nospam4u_jack@yahoo.com> wrote in message
news:1112815470.076422.68500@g14g2000cwa.googlegroups.com...

Thanks Lord. BTW this battery is sealed form every where( as I can
see), so where does the Hydrogen escapes??

There is likely a hidden vent that opens if the pressure gets too high.
If that happens, I'd guess the electrolyte would either leak or evaporate.
That's pretty standard for a *sealed* battery.

 

[John Fields]

On 6 Apr 2005 13:56:06 -0700, "SklettTheNewb" <SteveKlett@gmail.com>
wrote:

I'm trying to understand the basic LM239 comparator. I have read
several sites about them, looked at samples, read datasheet, etc, etc
and I still am not getting logical results from my simulations.

Here is the test circuit that I made using CircuitMaker, there are 2
version where I switched the reference/input legs
http://www.pmddirect.com/temp/comparator_a.gif
http://www.pmddirect.com/temp/comparator_b.gif

You will see in those images that the output voltages are either 15.8m
(millivolt?) or 21p (I have no idea what that means)

The reference and input voltages are 6v and 3v and I have tried
flipping them, I never get output voltage of +9v (my supply to the
comparator)

I would really appreciate a tip what I've done wrong. I'm sure it's
simple, but 2 hours now and I can't find it.

---
The LM239 has open collector outputs, so R4 needs to go to +V and it
needs to be about 10k.

--
John Fields
Professional Circuit Designer

 

[Thomas A. Horsley]

Light - simulating the sun beaming down onto her face. A bright
spotlamp perhaps, or an infra-red lamp of the sort sometimes fitted in
bathrooms. Quickly dismissed as I didn't know enough to be sure of its
safety over prolonged exposure.

Actually this works great for me. A floor lamp with a timer that turns
on about 15 or 20 minutes before I want to get up convinces my
subconcious that the sun has been up for a while .
--

==>> The *Best* political site <URL:http://www.vote-smart.org/> >>==+
email: Tom.Horsley@worldnet.att.net icbm: Delray Beach, FL |

<URL:http://home.att.net/~Tom.Horsley> Free Software and Politics <<==+

 

[Lord Garth]

"Thomas A. Horsley" <tom.horsley@att.net> wrote in message
news:upsx7bj9r.fsf@att.net...

Light - simulating the sun beaming down onto her face. A bright
spotlamp perhaps, or an infra-red lamp of the sort sometimes fitted in
bathrooms. Quickly dismissed as I didn't know enough to be sure of its
safety over prolonged exposure.

Actually this works great for me. A floor lamp with a timer that turns
on about 15 or 20 minutes before I want to get up convinces my
subconcious that the sun has been up for a while .

....and if you're a gecko, so much the better!!!

 

[Larry Brasfield]

"Active8" <reply2group@ndbbm.net> wrote in
message news:d2ira0.2qc.1@active8.fqdn.th-h.de...

On Fri, 1 Apr 2005 03:12:12 -0800, Larry Brasfield wrote:

"Active8" <reply2group@ndbbm.net> wrote in message
snip

Volume in drive C is unlabeled Serial number is CCCE:6A5D
Directory of C:\Play\*.url

3/31/2005 9:37 44 x.url
44 bytes in 1 file and 0 dirs 4,096 bytes allocated
^^^^^^^^^^ WTF?

That means the directory listing specified did not include
any subdirectories. (Nothing that strange, really.)

That's what "0 dirs" means.

Woops, I forgot about the alignment issue with
proportional fonts. Seems obvious, now!

"^^^^^^^^^^^ WTF?" means that I'd like to know what's up with the
"4,096 bytes allocated" part. That's why I highlighted those
particular characters.

As far as I can tell, it is summarizing the same statistic reported on
a per-file or per-directory basis by explorer.exe as "Size on disk".
At least that's what a few comparisons show, and given the 4NT
author's discipline, it must be using the available API for that.

It's been fun, Mike, but I'm on a tight schedule this
morning so will have to neglect this for awhile.

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

 

[Michael Redmann]

Kevin Aylward schrieb:

Bob wrote:
It doesn't make sense that the series voltage would be anything other
than 2*V.

Indeed.

Bob and Kevin: you're absolutely right! After replying on Pete's post I
was in doubt if this strange effect could really be true. It can't.

But what's wrong with Pete's and my computation. Neither conservation of
charge nor of energy lead to the correct answer. (Okay, if nothing
really happens there's nothing to compute ;-)

regards
--
Michael Redmann
"It's life, Jim, but not as we know it." (Spock)

 

[John Smith]

"Roger Johansson" <no-email@no.invalid> wrote in message
news:4253679a$0$43988$14726298@news.sunsite.dk...

Lord Garth wrote:

I'm studying electrical, what's the best way to remember the
P.I.R.E. wheel.

Learn just one formula such as V=IR and rearrange terms mathematically
as needed. No escaping the math if you want this field.

Pedantic mode on.
You have to learn _two_ equations, actually.

V=I*R (Volt=Amp*Ohm)
P=U*I (Watt=Volt*Amp)

Then learn how to re-arrange these equations as needed for the problem
at hand, and use a calculator to get the result.

The art of re-arranging equations is called algebra, and you need some
basic knowledge and experience in this.


An alternative is to use a visual diagram like the ones I have put on a
web site

http://humanist.250free.com/

Click on the two .jpg files at the bottom of the list, save them to
hard disk. Can be distributed freely.


--
Roger J.

If you going to learn P=VI rather learn P=VIcos(Phi) where Phi is the phase
between V and I.

 

[Tom Biasi]

"js5895" <JoshTmp@nycap.rr.com> wrote in message
news:1112760564.180299.140220@g14g2000cwa.googlegroups.com...

Hi,

I'm studying electrical, what's the best way to remember the P.I.R.E.
wheel.

Thanks.

Hi,
You are asking for a method to' remember' something and you are being given
answers on how to' learn' something.
I would suggest that you take the advice that says 'learn' algebra. If you
learn the relationship that is generically referred to as "Ohm's Law" and
learn the algebra to solve for all variables you will be in far better shape
when the formulae get more complicated.
Regards,
Tom

 

[Peter Bennett]

On 7 Apr 2005 06:32:29 -0700, "BobG" <bobgardner@aol.com> wrote:

Can someone sum up the top couple of rules of algebra for him? How
about something like: 'An equation has an expression on each side of
the equal sign. To solve the equation for any of the variables, you
need to get that variable over to the left side of the equal sign. To
eliminate a variable on one side, multiply both sides of the equation
by the inverse of that variable. This doesnt change the equality,
because you are multiplying both sides by the same number.' Is this the
necessary and sufficient information needed to solve ohms law for 3
variables?

I use three rules:

1. If you do something on one side of the equal sign, you must do the
same thing on the other.

2. Anything divided by itself equals 1.

3. Anything multiplied by 1 is unchanged, so the "1" can be discarded.

I've been working in electronics for some 40 years, and have no idea
what the "P.I.R.E. wheel" is - I just remember E = IR and P=EI, and
shuffle things around as needed. The same "shuffling" rules apply to
any simple equation.



--
Peter Bennett VE7CEI
email: peterbb4 (at) interchange.ubc.ca
GPS and NMEA info and programs: http://vancouver-webpages.com/peter/index.html
Newsgroup new user info: http://vancouver-webpages.com/nnq

 

[Lord Garth]

"GotCoffee" <Ray.Straub@gmail.com> wrote in message
news:1112893022.059056.138370@z14g2000cwz.googlegroups.com...

How about using the timer on an automatic coffee pot. The power
supplied to the heating element can be used. If hooked up properly,
you can get woke up a hot cup of coffee

YEOW!!!!!! Wouldn't you rather be awakened by the SMELL of
a freshly perked pot of coffee?

 

[Lord Garth]

"Peter Bennett" <peterbb@nowhere.invalid> wrote in message
news:l8pa515u0a1mgegm39m0d4f8mnornpftae@4ax.com...

On 7 Apr 2005 06:32:29 -0700, "BobG" <bobgardner@aol.com> wrote:

Can someone sum up the top couple of rules of algebra for him? How
about something like: 'An equation has an expression on each side of
the equal sign. To solve the equation for any of the variables, you
need to get that variable over to the left side of the equal sign. To
eliminate a variable on one side, multiply both sides of the equation
by the inverse of that variable. This doesnt change the equality,
because you are multiplying both sides by the same number.' Is this the
necessary and sufficient information needed to solve ohms law for 3
variables?

I use three rules:

1. If you do something on one side of the equal sign, you must do the
same thing on the other.

2. Anything divided by itself equals 1.

3. Anything multiplied by 1 is unchanged, so the "1" can be discarded.

I've been working in electronics for some 40 years, and have no idea
what the "P.I.R.E. wheel" is - I just remember E = IR and P=EI, and
shuffle things around as needed. The same "shuffling" rules apply to
any simple equation.

Exactly! And like most technical people, I don't give a rats ass about
being PC with the resistor color code mnemonic.

 

[Larry Brasfield]

"Jack// ani" <nospam4u_jack@yahoo.com> wrote in message
news:1112897029.222347.222440@o13g2000cwo.googlegroups.com...

Hi there,
Hi.
In 3-phase AC wiring, if phase to neutral voltage is 110V, then why is
phase to phase voltage 220? I know phase difference between any two
phases differ by 120 degree, so they should add up to give something
less than 220V!

The mathematical impossibility you ask about does not
happen. What makes you think it does?

It should sum up to give 220V if the phase difference
were 0 degree or 360degree!

Many people do not distinguish 0 and 360 degrees
for continuous sinusoids.

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

 

[Lord Garth]

"Jack// ani" <nospam4u_jack@yahoo.com> wrote in message
news:1112897029.222347.222440@o13g2000cwo.googlegroups.com...

Hi there,

In 3-phase AC wiring, if phase to neutral voltage is 110V, then why is
phase to phase voltage 220? I know phase difference between any two
phases differ by 120 degree, so they should add up to give something
less than 220V! It should sum up to give 220V if the phase difference
were 0 degree or 360degree!

Thanks

The phase to phase voltage in the USA is 207 VAC which is phase to ground x
1.732
120 x 1.732 = 207.84 You can find this voltage supply running the light in
many
buildings.

Houses get one 240 volt phase which is transformed to provide 2 outputs of
120 volts
that are 180 degrees apart, with respect to ground.

Three 240 volt phases implies the phase to phase voltage from this circuit
is 415.68 volts.