Chip with simple program for Toy

[Rich Grise]

On Monday 11 October 2004 08:00 am, jm did deign to grace us with the
following:

It is an older pentium mobo. It is useless. Not even ebay will do
good for me. Should I just throw it away or save it for parts? I
don't know what parts on it I should get. Thanks for opinions.

If you're a hobbyist and like used parts with short leads, you can
strip it with a torch and a puller of some kind - wear safety goggles
and a breath mask. Take the processor, get the chip out, and make a
brooch for your lady. Be sure and get those .025 square pin headers -
you never know when you'll need one. And connectors are always handy.
Any socketed chips, of course, you just pull. At this age, I'd toss
the battery.

Have Fun!
Rich

 

[Jag Man]

Chris,

Thanks a bunch! It may take me a while to implement since I'm
a bit of a novice, but I understand what you are doing and think
I can follow your diagram. Looks like an elegant solutuion.

Ed

 

[Peter Bennett]

On 11 Oct 2004 05:20:50 -0700, pascaldamian@icqmail.com (Pascal
Damian) wrote:

The sales rep that sold me AC told me that my AC unit will consume
about 1400W when starting up, then 700W *constantly* regardless of
temperature setting (e.g. 20C vs 25C). Is that true? I'm guessing not,
but I'm not sure.

I would expect it to consume 700W _while running_, regardless of the
thermostat setting. If it produces enough cooling to bring the room
temperature below the thermostat setting, then it will turn off, and
consume no power, until the temperature rises enough for it to
re-start.

The higher you set the thermostat, the less cooling it will have to
do, so it will run for a smaller percentage of the time, and so the
average power consumption will be lower.

--
Peter Bennett, VE7CEI
peterbb4 (at) interchange.ubc.ca
new newsgroup users info : http://vancouver-webpages.com/nnq
GPS and NMEA info: http://vancouver-webpages.com/peter
Vancouver Power Squadron: http://vancouver.powersquadron.ca

 

[Brian]

Hi Ed,
The 74LS123 is much like the 555 timer, if you understand one, the other
will be easy to learn. To show you how easy it would be, I drew the
schematic for you. You can see it at http://www.fncwired.com/SignalEx

Brian

"Jag Man" <Jag_Man653R-E-MOVE@hotmail.com> wrote in message
news:Ylxad.1934$6q2.406@newssvr14.news.prodigy.com...

Thanks, Brian!

Guess I'll have to do some reading up on the 74LS123
to see how to tell it what side to trigger on.

Ed

 

[Steve Evans]

On Mon, 11 Oct 2004 21:01:46 GMT, "Joe Rocci" <joe@roccis.com> wrote:

Steve,
I'm not sure, but I think the original post said this stage was a frequency
multiplier with an OUTPUT frequency of about 145 MHz. If that's the case,
then the INPUT frequency would be 72 MHz or less.

No Joe! The frequency is 145Mhz throughout.

An aside to Steve... I'm pouring over your expansive explanation.
It'll take a while to sink in, though!

Thanks,

Steve.
--

Fat, sugar, salt, beer: the four essentials for a healthy diet.

 

[Uncle Al]

Alan Horowitz wrote:

when a current just starts flowing into a RL or RC circuit, how does
the voltage "know" that it should be increasing exactly 63% during
each time-constant period?

And whence the number 63%?

1-(1/e). Crack your textbook.

<http://www.iop.org/EJ/S/UNREG/aS0aOQcupwxj4CNeiM5vaQ/article/-featured=jnl/0143-0807/23/1/304/ej2104.pdf>
"Demonstration of the exponential decay law using beer froth"

Google
"exponential decay" 63 20,500 hits

Uncle Al gotta think of everything.

--
Uncle Al
http://www.mazepath.com/uncleal/
(Toxic URL! Unsafe for children and most mammals)
http://www.mazepath.com/uncleal/qz.pdf

 

[John T Lowry]

"Alan Horowitz" <alanh_27@yahoo.com> wrote in message
news:1e3670a7.0410111511.47223a3b@posting.google.com...

when a current just starts flowing into a RL or RC circuit, how does
the voltage "know" that it should be increasing exactly 63% during
each time-constant period?

Definition of time-constant period.

John Lowry
Flight Physics

And whence the number 63%?

 

[tempus fugit]

One other question: Why do you have 4 7809s at the output? You could get
away with just one and run parallel lines to each effect from the one
ouptut. I wonder if you've got some goofy ground loop with all these
regulators.
FWIW I power 5 effects and a tuner with one wall wart connected to a single
7809 with about the same filtering you have here with no problems. Your plan
should work.
Also, have you tried the obvious culprits (bad cables, connectors, etc)?
Have you tried plugging in one effect at a time to see if any single effect
is the problem? Have you double checked your solder connections and wiring?





"anonymous" <anonymous@catfarm.com> wrote in message
news:lCcad.215994$MQ5.49226@attbi_s52...

"tempus fugit" <toccata@no.spam.ciaccess.com> wrote in message
news:fU%9d.336$57.170@fe51.usenetserver.com...
Is this to power guitar effects by any chance? If your components are
working properly, you might want to look into the possibility of a
ground
loop somewhere. For instance, if you have your amp plugged in one
outlet,
and your effects power supply in another one on the other side of the
room,
you're almost guaranteed to create a ground loop. Try plugging all the
gear
your plugging into into one outlet or power bar and see if the hum goes
away. Alternately, try unplugging (from the wall) each piece of
equipment
one at a time to see if the hum goes away. If it does, you've found the
problem.



yes these are guitar effects. they are all plugged into the same power
strip. they work properly (normally).

 

[colin]

"anonymous" <anonymous@catfarm.com> wrote in message
newsmIad.350483$mD.296451@attbi_s02...

WARNING - dumb question follows - WARNING

consider yourself warned.

So I have a nice transformer that puts out 25V and 1.5A. I want to step it
down to 9V with some voltage regulators. The regulators say they are rated
for 1A and 37V max (given proper heat sinking).

If the output of the regulator is drawing < 100mA - do I have to worry
about
the remaining output from the transformer?

No.

Colin =^.^=

 

[Roger Johansson]

bagnaninchi@tele2.fr (BPO) wrote:

I'm a biologist and I would like to try to culture some cells on a
Gold (or other biocompatible metal) rod/needle. I would like to try
some basic electronics measurements, as capacitance, conductance,
etc... and try to relate them to cell growth.

As I'm really bad in electronics, I come up with several basic
questions for you.

1)How can I measure the capacitance of a metal needle ?

In relation to what? You need to define another body and seek the
capacitance between these two bodies.
Between your needle and the planet earth, for example.

2) What sort of instrument (cheap one)do I need ? I expect changes in
the 0.1 pF range.

Hardly cheap. Such small capacitances are difficult to measure, because
the capacitances of bodies around it will interfere.

One of the common ways to measure small capacitances is to involve it in
an oscillator, so the frequency of the oscillator depends on the
capacitance.

3) Could I measure these electrical properties by accessing only one
end of the metal needle ?

Yes. A piece of metal is to the electrical current like water is to
sound. You can explore a water body with the help of sound, measure
echoes and timings and determine the size of the water body, its length,
etc..

In a similar way can we explore a piece of metal through measuring
reflected waves, capacitance, etc.., even though we only have access to
one end of the metal body.

When measuring small capacitances remember that two inch-long pieces of
isolated wire tightly twisted together have a capacitance of something
like 7pF. It is a trick often used to create a small capacitor out of
wires. I hope you understand that an instrument measuring sub-pF
capacitances would have a giant problem with its own terminals, unless
they were placed at least a meter apart. And the connecting leads will
have even bigger problems as their capacitance will be dependent on their
position.



--
Roger J.

 

[CFoley1064]

Subject: getting the right part # from mouser
From: "anonymous" anonymous@catfarm.com
Date: 10/11/2004 11:13 PM Central Daylight Time
Message-id: <oRIad.245546$3l3.214465@attbi_s03

Im trying to find .047uF and 47pF non surface mount caps.. and I can't seem
to get the site to cough up the correct part #. Any suggestions on how to
phrase my search query?

Also - whats the diff between linear and audio pots? lin vs log I can
understand.

Go to mouser.com, and work from the product guide on the left side of the main
page. Follow the string from passives to caps to the type you want.

Mouser has so many through-hole caps that, even working down to general
purpose, you've got hundreds of choices. It's not the ideal way to search.
Sometimes, nothing beats having the catalog, and the Mouser catalog will be
sent for free if you just get to the request page, available from the main
page. I think having Mouser and Digi-Key catalogs on the desk is a good idea.

If this is a hobbyist application, you're bogged down with too many choices,
and you're not particular about specs and need to get this done on the web, try
Jameco. They're fast and inexpensive.

Audio volume is logarithmic -- an audio pot is a log pot. See:

http://sound.westhost.com/pots.htm

For the graph.

Good luck
Chris

 

[Kevin Aylward]

Steve Evans wrote:

On Mon, 11 Oct 2004 12:29:46 GMT, "peterken" <peter273@hotmail.com
wrote:

In every configuration, a transistor needs to be set using resistors
for having a known amplification to avoid unknown behaviour.
Starting point is using the *lowest* possible Hfe to begin
calculations, since *all* transistors of the same type will work
then, despite of their (most likely higher) Hfe.

Beginning in a reverse way, and assuming a minimum Hfe of say 100

I don't see how designing at the low-end range of Betas is going to
help much.

It does. Thats the way its done.

If you do that and you end up in practice with a device
with a 300 Beta (not too unlikely) then either that stage or a
subsequent one is going to have so much current gain that the output
voltage across Rc is going to swing into the supply rail on one peak
and the base voltage on the other, leading to hideous distortion.

It wont if you design it *correctly*. One designs the circuit so that
the circuit is essentially insensitive to hfe as far as bias current
goes. To do this one might have a bleed current that is calculated
knowing the worst case base current and making it a good bit larger.
Secondly, for example, if the transistor is driving an output load it
will need a minimum base drive drive. If the hfe is larger, it wont
matter, it only takes the current it needs, e.g. emitter follower. Only
if the drive runs out of steam, will there be a problem.

I
don't see how it's possible to overcome the problem without measuring
each transistor's hfe individually and biasing accordingly.

The issue is designing bias circuits that are insensitive to hfe. This
usually require designing based on the minimum hfe expected. One makes a
base potential that is essentially indepednant of base current. If the
bleed current (resister from base to ground) is say 10 times the minimum
base current, the base voltage wont change much. Assuming there is a
resister in the emitter (class A amp), then this voltage will also be
independent of base current. The voltage across this resister will fix
the emitter current.

http://www.anasoft.co.uk/EE/bipolardesign3/bipolardesign3.html
http://www.anasoft.co.uk/EE/index.html

Kevin Aylward
salesEXTRACT@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.

 

[Linden]

Thanks Miguel, I have something similar to your ringdetect but my
ringdetect signal doesnt stay low for the duration of the ring it
oscillates low/high in sync with the ring.

Does your circuit give a steady output ? Maybe I have to integrate the
input or the output first ?

Cheers,
Linden.

 

[Roy McCammon]

Archer wrote:

Roy McCammon <barkupine-news@yahoo.com> wrote in message news:<415DAD43.1030306@yahoo.com>...

you the impedence (complex) with the far end
shorted and with it open and the characteristic
impedence is the (complex) geometric mean of those
two measurements.

Archer wrote:

how to measure the characteristic impedance of a coax?
I'm starting to study the rf circuit design.


may i measure the impedence using meter?

I suppose you might do with a vector volt meter. Not sure.
Been too long since I used one. I'd use a scope.

and how would it be done that measure the open far end coax?

you don't measure at the far end. you just short the far end
or open the far end.




--
local optimization seldom leads to global optimization

my e-mail address is: <my first name> <my last name> AT mmm DOT com

 

[Roy McCammon]

Brian Raab wrote:

And liked it.

Is an electro-shock dangerous for me

If you mean electro-shock as administered in a hospital,
ask an MD.

If you mean use the household electricity, then
yes, it is dangerous and it can cause harm that
shows up later.




--
local optimization seldom leads to global optimization

my e-mail address is: <my first name> <my last name> AT mmm DOT com

 

[peterken]

simple
as far as I see the output of the regulator is totally under-buffered
try putting an extra elco at the output of the reg, say 2200u something



"anonymous" <anonymous@catfarm.com> wrote in message
news:hkZ9d.212267$MQ5.19587@attbi_s52...
Im working on a power supply for some musical bits. Unfortunately its
producing a great deal of "hum" in the circuit. Is there any way to filter
out residual AC in the DC line?

What I have:

+18V (wall wart) ---->3500uF
(electrolytic) ---->.22uF--->7809--->.1uF--->out
|
| |
ground
ground ground

the ground is common between the wall wart and all components. There are 4
7809s in parallel at the output.

 

[Steve Nosko]

**** Y I K E S !!!



--
Steve N, K,9;d, c. i My email has no u's.


"Jonathan Kirwan" <jkirwan@easystreet.com> wrote in message
news:kpemm0d5379mui0buc2v8v78g3d806gq6l@4ax.com...

On 11 Oct 2004 16:11:16 -0700, alanh_27@yahoo.com (Alan Horowitz) wrote:

when a current just starts flowing into a RL or RC circuit, how does
the voltage "know" that it should be increasing exactly 63% during
each time-constant period?

And whence the number 63%?

I haven't had a chance to read other responses, but here's mine:

Take the case of an RC:

,---,
| |
V| \
--- / R
- \
--- |
- +-----
| |
o --- C
/ ---
o |
'---+-----

Assume C is discharged and V has just been applied by closing the
switch...

The current through R is based on V, less the voltage on C (which counters
V),
so:

I(R) = ( V - V(C) ) / R

The above is a function of time, because V(C) is a function of time. So,
what's
V(C)? Well, that needs to be arrived at more slowly.

First, we know that this is true:

Q = C*V

Well, actually, that's an average statement. More exactly, it's:

dQ = C * dV

In other words, the instantaneous change in Coulombs is equal to the
capacitance
times the instantaneous change in voltage. Both sides can now be divided
by an
instant of time to give:

dQ / dt = C * dV / dt

Since dQ/dt is just current (I), for the above capacitor this becomes:

I(C) = C * dV(C) / dt

So how does this help? Well, we know that the current from R must
accumulate on
C. So, we know that:

I(C) = I(R) = ( V - V(C) ) / R

so, combining, we get:

C * dV(C) / dt = ( V - V(C) ) / R

Rearrangement of this gives:

dV(C) / dt + V(C) / (R*C) = V / ( R*C )

Which is the standard form for ordinary differential equations of this
type.

The standard form with general terms looks like: dy/dx + P(x)*y = Q(x).
In our
case, though, y = V(C), x = t, P(x) = 1 / (R*C), and Q(x) = V / (R*C).

The solution to this includes multiplying by what is called "the
integrating
factor", which is:

u(x) = e^(integral (P(x)*dx)) = e^(integral (dt/RC)) = e^(t/(R*C))

(This is a VERY POWERFUL method to learn, by the way, and it is probably
covered
in the first few chapters of any ordinary differential equations book.)
So,
going back to look at the general form and multiplying both sides:

u(x)*dy/dx + u(x)*P(x)*y = u(x)*Q(x)

But the left hand side is just d(u(x)*y)/dx, so:

d(u(x)*y)/dx = u(x)*Q(x)

or,

d(u(x)*y) = u(x)*Q(x) * dx

In our case, this means:

d( e^(t/(R*C)) * V(C) ) = V / ( R*C ) * e^(t/(R*C)) * dt

Taking the integral of both sides, we are left with:

e^(t/(R*C)) * V(C) = integral [ V / ( R*C ) * e^(t/(R*C)) * dt ]
= V / ( R*C ) * integral [ e^(t/(R*C)) * dt ]

setting z = t/(R*C), we have dz = dt/(R*C) or dt = R*C*dz, thus:

e^(t/(R*C)) * V(C) = V / ( R*C ) * R*C * [e^(t/(R*C)) + k1]
= V * [e^(t/(R*C)) + k1]
= V * e^(t/(R*C)) + V * k1
V(C) = V + V * k1 / e^(t/(R*C))
= V + V * k1 * e^(-t/(R*C))
= V * [ 1 + k1*e^(-t/(R*C)) ]

From initial conditions, where V(C) = 0V at t=0, we know that k1=-1, so:

V(C) = V * [ 1 - e^(-t/(R*C)) ]

Time constants are usually taken to be:

e^(-t/k)

with (k) being the constant. In our V(C) case, this means that k=R*C. So
that's the basic constant and it's in units of seconds.

So, what's the voltage after one such constant of time? Well:

V(C) = V * [ 1 - e^(R*C/(R*C)) ] = V * [ 1 - 1/e ] = V * .63212

Ah! There's that 63% figure. Actually, more like 63.212%.

Two time constants would be:

1 - 1/e^2 = .864665

and so on....

Oh... and there are other methods you can use to solve the simple RC
formula,
but the method I chose is a very general and powerful one worth learning
well.

Jon

 

[Steve Nosko]

Alan,

John Popelish got a good start with "e is a natural constant that has
some very sweet
properties in many applications of mathematics, and simplifying..."



Then, it looked as thought John Jardine was going to steal my thunder with
"the voltage knows nothing about how it's "supposed" to behave. "



This could resolve to a mater of faith Alan.



Indeed, the voltage/current "knows" nothing.



After observing what happens in such circuits, "we" (those who must
understand all things) very carefully examined what was going on and
"discovered" that there were mathematical expressions or equations which
would model what happens in nature. "We" came up with theories about what
was going on and what was causing it to happen. "We" then found ways to
make the math fit reality. In the case of time constants, we have a natural
phenomena which is very nicely described by the equations stated elsewhere
in this thread (the 1/e thingy). It is just like the F=MA equation. "We"
discovered that the force applied to a mass is equal to the mass times the
acceleration. The Mass knows nothing about force, acceleration or
mathematics. We found that this math describes nature.



It is exactly like a model airplane (or whatever). We make the model to
look like the real thing. The real thing knows not of the model that we
built, but if we did a good job, I or you can now look at the model and
"know" just how the real thing looks.



The math behind all of our sciences is just like this. *WE* found math
which models reality and because we did such a good job, we can now "do the
math" and "know" how the real thing should behave.



To be a little more specific, in the case of the time constant. we have
theories about current flow, charge, capacitance, inductance magnetism and
resistance which are borne out by countless experiments and then by
subsequent usage. These theories have all had mathematics fitted to them,
and by golly everything fits. We can now plug-in values to equations till
the cows come home and holy-cripes! The real thing does just what the math
predicted. Based upon the properties we have observed for each type of
component, this math works out such that this 1/e thingy fits just right.



In other words, the answer is: "It just does!"

73,
--
Steve N, K,9;d, c. i My email has no u's.




"Alan Horowitz" <alanh_27@yahoo.com> wrote in message
news:1e3670a7.0410111511.47223a3b@posting.google.com...

when a current just starts flowing into a RL or RC circuit, how does
the voltage "know" that it should be increasing exactly 63% during
each time-constant period?

And whence the number 63%?

 

[Steve Nosko]

"Steve Evans" <smevans@jif-lemon.co.mars> wrote in message
news:7u3mm0lun8fk658o0ibl23neqbf9vmlnt2@4ax.com...

On Mon, 11 Oct 2004 14:59:11 -0500, "Steve Nosko"
suteuve.nosukowicuz@moutouroula.com> wrote:

The Base-Emitter in a transistor is a semiconductor junction just like a
diode and in the (higher power) RF amplifiers behaves pretty much like a
diode. With RF applied to the base, there will be conduction on the
positive peaks only and this will constitute a DC current flow which must
have a DC path. The inductor provides such a path since the capacitor
can
not.

Okay, Steve, I'm gonna have to take your explanation in bite-size
chunks. Kindly indulge me...

I don't see that the inductor is necessary to provide such a path to
ground for the signal peaks, since they (the input signal pos. peaks)
turn on the transistor and complete the circuit to ground via the
base/emitter junction, which will be a low resistance path with
sufficient base drive level on the peaks. Can you tell me why this
path alone isn't good enough and there has to be an inductor across
B/E as well?
Thanks! Steve

Most certainly, Steve.... Indulge, I will.
This can get confusing... Steve Noskowicz here (K9DCI)

Joe Rocci tried, but I don't think he went through the proper step-by-step
explanation.

OK Here's what happens... Start with the capacitor completely discharged -
zero volts across it - right end same voltage as left side (whatever that
may be). On the first positive peak, some current flows through the base
emitter junction because the voltage on the right side of the cap is the
same as that on the left side. (this assumes there is at least 0.7 volts of
signal coming from the left. NPN transistors conduct current when the base
is about 0.7 volts positive.
The current is some quantity of electrons, right. Well these electrons
will start to "fill up" or charge the capacitor. Each time another positive
pulse happens, the right side of the capacitor collects more and more
electrons. This charges the cap more and more and makes the right side more
and more negative. There is no way to drain off this charge before the next
pulse comes along. The base-emitter junction will be reverse biased and not
conduct any current. I hope you see this because that's the key.
As the caps gets more charge from each pulse, the right side becomes more
negative. After each pulse, it will take more and more voltage on the caps
left side to get enough voltage on the right side (0.7 volts) to get to the
base-emitter conduction voltage and get any current to flow.
The end result is that the capacitor will charge to the peak input
voltage (less about 0.7 volts) and the base will never get to the 0.7 volt
level. At this point, the base voltage will be swinging (with the input
signal causing it) from about 0.7 volts plus to a value equal to
*negative* the peak-to-peak input voltage (less the 0.7 volts). That's
minus volts. The AC signal at the bacse will NOT be swinting equally around
zero volts.

This could be viewed as what we call a "clamper circuit" The AC voltage
AT THE BASE will have its positive peak "clamped" or moved to +0.7 volts and
the waveform will extend from there as negative as the waveform is tall.

IF we had 5 volts peak (10 volts peak-to-peak) the cap left side, the base
voltage will swing from +0.7 volts to -9.3 volts.

WHEW !
Did this work ???
--
Steve N, K,9;d, c. i My email has no u's.

 

[Jonathan Kirwan]

On Tue, 12 Oct 2004 16:20:32 -0500, "Steve Nosko"
<suteuve.nosukowicuz@moutouroula.com> wrote:

**** Y I K E S !!!

Jon