Chip with simple program for Toy

[Rich Grise]

On Thu, 14 Oct 2004 22:13:56 -0400, Rodney Kelp wrote:

If you are targeted what on earth have you done? or are planning to do?
On the other hand if you have nothing to hide why would they waste their
time and resources to watch you?

"He MUST be guilty! He's under arrest, isn't he?"

Maybe you'd better go back and re-take civics 101.

Good Luck!
Rich

 

[Rich Grise]

On Thu, 14 Oct 2004 10:35:58 -0700, jm wrote:

BarryNL <nospam@nospam.es> wrote in message news:<416e31df$0$34762$e4fe514c@news.xs4all.nl>...
jm wrote:
Should I buy a breadboard with a 5 volt regulated power supply or get
a regular breadboard without one?

I'd get one without (it's more flexible) and use some of the money saved
to buy a few 7805's. Personally, I mostly use a 9v battery with a 7805
and capacitor soldered directly to the battery clip - effectively a
regulated 5v battery


Thank you.

So I don't have to have one plugged into the wall?

Well, obviously, _something_ has to be plugged into the wall, unless
you're just using batteries. But that doesn't have to be the breadboard.
I have a "bare" breadboard like that, and I have a couple of power
supplies lying around that I use when I need them. All home-built,
of course.

Have Fun!
Rich

 

[Jonathan Kirwan]

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

 

[CFoley1064]

Subject: where does all the extra current go?
From: "anonymous" anonymous@catfarm.com
Date: 10/11/2004 10:40 PM Central Daylight Time
Message-id: <pmIad.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?

Not a dumb question -- just a newbie one.

First, I hope your transformer has a center tap on the secondary, because if it
doesn't and you're using a 9V linear regulator and are drawing 100mA, you're
talking about

(37V - 9V) * 0.1A = 2.8 Watts

Need a heat sink here. It might be somewhat easier to do it this way if your
25VAC secondary transformer has a center tap (view in fixed font or M$
Notepad):


25.2 VCT Sec ____
| | +
o-----. ,------>|--o---o---|7809|--o-----o
)|( | C1|+ |____| +|C2
)|( | --- | ---
) ,---. | --- | ---
)|( | | | | |
)|( | | | | |
o-----' '------>|--o---o-----o-----o-----o
| | -
| ===
=== GND
GND
created by Andy´s ASCII-Circuit v1.24.140803 Beta www.tech-chat.de

Use a 1000uF electrolytic for C1 and a 10uF electrolytic for C2.

This will give you about 18VDC peak on C1, which should make your 7809 a lot
more comfortable. With total power dissipation less than a watt, you should be
able to get away without a heat sink as long as the 7809 is in room temperature
ambient.

Oh, yes -- transformers are rated for maximum current with resistive load. By
definition, a transformer is usually rated so that the output voltage will be
within 20% of the rated voltage for the rated current. Usually, if you're
loading it lightly, the voltage will be higher than rated, and if you're
loading it toward its limit, the voltage will be lower. The transformer is an
on-demand machine -- it will produce the voltage as long as the current
requirement is within limits., There isn't any extra to take care of.

Good luck
Chris

 

[Don Kelly]

"D-Zyl" <d_zyl38@hotmail.com> wrote in message
news:ef7e39ee.0410151042.1caa6945@posting.google.com...

Hi,

I need to rotate a mass of 10kg around a vertical tube.
rotation will be of about 90 degrees.
the motor must be the smallest as possible.
how can I do to choose the right motor (power, technology, w/ or w/o
gearbox...)
?
where can I find useful infos ???

--------
How fast? That can make a big difference. This is only 1/4 revolution so a
rotary actuator of some sort may be better than a motor intended to rotate
continuously.
--
Don Kelly
dhky@peeshaw.ca
remove the urine to answer

 

[anonymous]

"tempus fugit" <toccata@no.spam.ciaccess.com> wrote in message
news:OhGad.591$57.51@fe51.usenetserver.com...

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?

Weeeell... Ill try removing 3 of them. I didn't want to mess with a daisy
chain.

When I plug in my Boss DS-1 (which should draw 5mA max) I get tons of hum
and not much sound. Plugging in the echo box doesn't work at all. Both work
fine from the boss wall wart.

This circuit shouldn't be giving me fits like this. I think its a rite of
passage.

 

[anonymous]

"tempus fugit" <toccata@no.spam.ciaccess.com> wrote in message
news:OhGad.591$57.51@fe51.usenetserver.com...

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?

FWIW - the regulators share the ground line. I was hoping to spread out any
heat load by having multiple regulators. Im a long shot from an EE - which
is probably why Im having so many issues with this.

 

[R. Steve Walz]

jm wrote:

BarryNL <nospam@nospam.es> wrote in message news:<416e31df$0$34762$e4fe514c@news.xs4all.nl>...
jm wrote:
Should I buy a breadboard with a 5 volt regulated power supply or get
a regular breadboard without one?

I'd get one without (it's more flexible) and use some of the money saved
to buy a few 7805's. Personally, I mostly use a 9v battery with a 7805
and capacitor soldered directly to the battery clip - effectively a
regulated 5v battery


Thank you.

So I don't have to have one plugged into the wall?
------------

No, any battery pack greater than about 7V can be used with a 7805
voltage regulator to provide regulated +5.0V +/- 5% .

-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

 

[Roy Lewallen]

Steve Evans wrote:

Sure did, Steve! I ran it through a spice program and you're right in
every detail. So there's obviously some basic flaw in my understanding
of caps. The textbooks say to treat a cap as a short circuit at AC
(assuiming its reactance isn't too high at the frequency of interest).
That would appear to be grossly misleading as there's a huge
difference between the mean voltage levels on each side. . .

The mean voltage on each side is the DC component. The fact that it's
different on the two sides illustrates the fact that the capacitor is an
open circuit to DC. The shape of the waveforms on the two sides of the
capacitor are the same. That illustrates that the AC component is the
same on both sides -- the capacitor is a short circuit to AC.

It's gonna

take me a while to get this trough my thick skull. :-(
So what happens when you have a small ac ripple riding on a DC bias?

The bias (DC) is removed, and the ripple (AC) is passed through.

I'll spice it but won't understand it, I guess. :-(

Roy Lewallen, W7EL

 

[Robert Monsen]

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%?

Suppose you are trying to fill up a box with balls. However, for some
strange reason, you've decided that each time you throw in balls, you'll
throw in 1/2 of the balls that will fit in the remaining space.

At the first second, you have 1/2 the balls. Next second, you'll have
that plus 1/2 of the remaining space, which is 1/2 + 1/4 = 3/4. The
third second, you'll have that plus 1/2 the remaining space, ie, 1/2 +
1/4 + 1/8 = 7/8...

So, the number of balls at any time t will be:

B(t) = 1 - (1/2)^t

Thus, after 3 seconds, there will be B(3) = 1 - (1/2)^3 = 1 - 1/8 = 7/8,
just like above.

Now, apply that same reasoning, only instead of using the ratio 1/2, use
the ratio 1/e (since we are applying arbitrary rules)

Then

B(t) = 1 - (1/e)^t

After the first second, you'll have

B(1) = 1 - (1/e)^1 = 1 - 1/e = 0.632 (that is, 63%)

Strange coincidence, isn't it? It happens because when you are charging
a capacitor through a resistor, you are throwing balls, in the form of
charges, into a box (the capacitor), and the number of charges you throw
at any given time (the current) depends on how many charges are already
on the capacitor (the voltage).

Each step of the formula above is one time constant, RC. By dividing out
the RC, you can get the answer given seconds, ie

B(t) = 1 - (1/e)^(t/RC) = 1 - e^(-t/RC)

Where B is the percentage 'filled' the capacitor is (ie, what percentage
it is of the input voltage).

Why is 1/e used instead of 1/2? That has to do with the fact that we
must have a continuous solution, not a solution based on ratios of
existing values; the rate of change of the current (ie, how many balls
we throw in per unit time) is proportional to the voltage remaining,
which is continuously changing. Using 1/e instead of 1/2 allows us to
generalize to this, in the same way as the compound interest formula
allows us to compute 'continuously compounding' interest.

--
Regards,
Robert Monsen

"Your Highness, I have no need of this hypothesis."
- Pierre Laplace (1749-1827), to Napoleon,
on why his works on celestial mechanics make no mention of God.

 

[Ban]

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%?

Let's try another way. You can actually experiment yourself.

Take a 1000uF electrolytic cap and charge it up to +5V, then disconnect the
power supply. Put your voltmeter on the cap terminals and read +5V. Now take
a 10k resistor and put it across the terminals as well. The cap discharges
slowly. In the first moment the discharge current was 5V/10k = 0.5mA. With
this current the cap would be discharged in 10s, this is the time constant
"tau" = RC

But since the voltage is dropping also the discharge current drops. Now you
can use a stopwatch and read the voltage after 10s and you find it to be
1.84V, which is (1-0.624)5V. So this is where your 63% come from. Since we
are discharging, the value is 37% of the initial voltage. You can also note
down the values for 20s, 30s etc. until your meter has no more resolution
and find the corresponding values for multiple time constants.
BTW you do not need to do this experiment yourself but use a simulator or
solve the equations others have already written in their answers.

--
ciao Ban
Bordighera, Italy

 

[John Popelish]

Jack wrote:

hi all

For a 1n540x diode average forward current is given 3amps, read in the data sheet.

Is this average current the same what i measure in my dmm.

Thanks for reply

Almost certainly. But I never use a component at its absolute maximum
rating, voltage, current, or power.

--
John Popelish

 

[Paul Burridge]

On Fri, 15 Oct 2004 23:06:38 +0000 (UTC), "Reg Edwards"
<g4fgq.regp@ZZZbtinternet.com> wrote:

Sure did, Steve! I ran it through a spice program and you're right in
every detail.
=======================

What makes you think Spice is correct? Its only a buggy computer program.
Rubbish in - rubbish out!

Because it agreed exactly with Steve's prediction. I think therefore
I'm entitled to rely on it in this instance at least.

--

"What is now proved was once only imagin'd." - William Blake, 1793.

 

[Paul Burridge]

On Fri, 15 Oct 2004 23:06:38 +0000 (UTC), "Reg Edwards"
<g4fgq.regp@ZZZbtinternet.com> wrote:

What makes you think Spice is correct? Its only a buggy computer program.
Rubbish in - rubbish out!

BTW, Reg. Is there a program on your site that can handle stripline
calculations for UHF circuit board inductors?
--

"What is now proved was once only imagin'd." - William Blake, 1793.

 

[Dr. Anton T. Squeegee]

In article <mfydnREG3Kn_r_LcRVn-2A@adelphia.com>, Rodneykelp605
@hotmail.com says...

Top-posting corrected. Please don't top-post. See this link for
why: http://www.html-faq.com/etiquette/?toppost

"Soundweapon" <soundweapon@cs.com> wrote in message
news:20041014194621.18589.00002580@mb-m29.news.cs.com...
Classified US Government Technology

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This US government has also incorporated this technology into targeting
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This technology was developed to protect the national security of the
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For more information, please visit the website listed below.

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http://ourworld.cs.com/soundweapon/

Rodney Kelp responded...

If you are targeted what on earth have you done? or are planning to do?
On the other hand if you have nothing to hide why would they waste their
time and resources to watch you?

The original post has 'Tinfoil Hat' sign all over it. Remember,
Usenet has a sufficient supply of nutcases and 'NetKooks' to keep us all
amused for at least a couple of centuries.

"Soundweapon" is obviously one of them. There's another fellow who
posts to the sci.electronics hierarchy on occasion who keeps insisting
that his neighbors are shooting "burning" laser beams and electric
shocks at him.

I will say this much: Soundweapon's site may be a good source of
material if you're out to write some bad science fiction.

Keep the peace(es).


--
Dr. Anton T. Squeegee, Director, Dutch Surrealist Plumbing Institute.
(Known to some as Bruce Lane, ARS KC7GR,
kyrrin (a/t) bluefeathertech[d=o=t]calm -- www.bluefeathertech.com
"If Salvador Dali had owned a computer, would it have been equipped
with surreal ports?"

 

[Steve Evans]

On Fri, 15 Oct 2004 14:23:24 -0500, "Steve Nosko"
<suteuve.nosukowicuz@moutouroula.com> wrote:

While not necessary for understanding this circuit, I'll fill-in this bit
here. A "clamper" is a diode and cap circuit which will clamp a particular
point on a waveform to a specific voltage (kinda like being clamped in a
vice on the bench), but NOT change the wave's wiggling *shape*. Sometimes
this is needed. In the coupling circuit in question, it is the POSITIVE
PEAK of the signal that gets clamped to +0.7 volts. The wave's SHAPE is
un-changed, but the whole thing is shifted in its DC component.

What you are thinking of is called a "CLIPPER" because it CLIPS *off*
part of the waveform like barber's scissors.

Gotcha, Steve. I won't forget that distinction in a hurrq! It\s
amazing after nearly 2 years of studying this subject that I didn't
tumble the true meahing of clamping. Duh!

you state (and the spice progs agree) that what *actually* happens in
this case is that the whole AC waveform gets shifted south into
negative territory. It's still a full wave, but it's way down into
the negative and only the highest peaks just creep above zero volts.
Is this effect *solely* attributable to the steady build-up of
negative charge on the cap's RHS?



YUP !




I think what's really freaking me
out here is the fact that the signal source is grounded on the neg.
side and yet we have that same signal that after going through a cap
can end up going fully negative *below* ground. It just seems like any
such voltage beneath zero/ground potential is breaking the laws of
physics. Ground should be the 'absolute zero' of the potentials in any
circuit and here it is being violated. I need some help to get my
thick head around the concept! :-(



Oooooo. BEWARE! GROUND IS NOT ABSOLUTE ! ! ! Nope, nope, nope. This is
going to take some time to explain and more experience/study will be needed
for you to really _get it_.

The bad news for you may be that ground, I must sadly inform you, is
relative. There is no one, solid, never varying, absolute thing which is
ground, except in our imaginations. Many hams believe there is, but there
isn't.



That being said, let's start out simply and build.



Here is a very applicable analogy:

Voltage, also called "potential difference", is a lot like altitude --
height. We can talk about the height above the street level. We might
consider the street level to be "ground". In Physics, moving some object to
a higher level gives it the "Potential" to do damage if it falls on your
head, so the "potential energy" of it is greater. Holding it three feet
above your head gives it a certain potential, right? Voltage is just like
this.

HOWEVER, what about standing on the roof of a building THEN moving the same
object three feet above your head. You must agree that it has the SAME
potential to do damage TO YOUR HEAD that it did in the first example, right?
In this case, the roof of the building is our ground. So ground is relative
and *we* get to pick it. It is usually a known point in our circuit and we
use special symbols to show it. Note that this is why voltage is also is
called potential *DIFFERENCE*.

Unfortunately, this analogy will fall apart when trying to use it for
negative voltage, if we put this negative voltage "Below" our ground, but
for the "relative" concept, I hope it worked.



Now I'll try *negative*.

Okay, Steve, I've snipped your explanation about negative voltages as
that wasn't quite what I was getting at. I'm familiar with the
analogos you used (which I'm sure will help ohters in future via
Google). I've done quite a lot of experimentng with opamps using split
supplies, hence I don't have a problem envisaging below ground
voltages where the supply is say +15V - 0 -15V. The problem I was
having was with below ground voltages in a circuit with only one
ground and one V+ supply! I'm gojng to have a good think about this
before posting back for further clarification. I'm not completly sure
that its all down to the cap alone. Mebe the diode has an effect on t
producing his sub-ground signal as well?

Steve


--

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

 

[Roy Lewallen]

Reg Edwards wrote:

. . .
The only way of accumulating confidence in a computer program is to use it
and compare results with what you are already aware of as being true. . .
. . .

SPICE has been used for decades in the design of countless products that
you undoubtedly use daily. It's an extremely useful and valuable tool,
without which many modern designs simply wouldn't be possible.

Roy Lewallen, W7EL

 

[Steve Evans]

On Sat, 16 Oct 2004 16:35:00 -0700, Roy Lewallen <w7el@eznec.com>
wrote:

Reg Edwards wrote:
. . .
The only way of accumulating confidence in a computer program is to use it
and compare results with what you are already aware of as being true. . .
. . .

SPICE has been used for decades in the design of countless products that
you undoubtedly use daily. It's an extremely useful and valuable tool,
without which many modern designs simply wouldn't be possible.

Here again, Roy I"m confused. You say its indespensible; Reg Edwards
says its' unreliable. Who am I to beleive? When the experts disagree,
its imposible to form a reliable conculsion.
--

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

 

[Paul Burridge]

On Sat, 16 Oct 2004 22:43:50 +0000 (UTC), "Reg Edwards"
<g4fgq.regp@ZZZbtinternet.com> wrote:

Without any intended offence, your Multisim program is worthless to me for
the purpose of checking or confirming anything. Why? Because I've never
even heard of it! Therefore it carries no weight in attempting to convince
me of anything.

I can't believe you haven't heard of Multisim, Reg! It used to be
Electronics Work Bench and is with little doubt the worst heap of sh*t
simulator on the market. One can but hope the OP's got the demo
version for nowt or he's spunked a load of cash on SFA. They wanted 4
grand for this hopless crock of sh*t last time I looked!! :-D
--

"What is now proved was once only imagin'd." - William Blake, 1793.

 

[Roy Lewallen]

It's only impossible to form a reliable conclusion when "experts"
disagree if your sole source of knowledge and information is from those
"experts". That's a bad spot to be in. There are many sources of
information available to help you learn about the topic and come to a
more informed opinion. That's the solution to your dilemma -- develop a
wider range of sources of information and decrease your dependence on
the "experts".

Originally developed at Berkeley, SPICE has been commercialized by a
number of companies, one of the most popular being PSpice by MicroSim.
It's in very wide use, and has been for decades. Do you think that a
couple of generations of engineers would have paid several thousand
dollars each for software with a reputation of unreliability?

My own experience includes about 30 years designing a variety of test
and measurement equipment for several companies, followed by several
years doing electronics design as a consultant. In that time, I and my
colleages at Tektronix and other companies used SPICE very often. (In
fact, recognizing it value, Tek spent a large amount of money and
devoted resources to development of its own internal version of SPICE,
which included schematic entry and other features before they were
available in outside commercial versions.) It's virtually impossible to
design an analog or mixed analog-digital integrated circuit without it,
and I and my colleagues found it indispensible for many other projects.

You might compare this with Reg's experience with SPICE, if he'll tell
you what it is, and see what brought him to make the unequivocal
statement about it which he did. I don't personally think he really
believes that it's unreliable, though, but was just making one of his
characteristic trolls in order to relieve boredom.

Roy Lewallen, W7EL

Steve Evans wrote:

On Sat, 16 Oct 2004 16:35:00 -0700, Roy Lewallen <w7el@eznec.com
wrote:


Reg Edwards wrote:

. . .
The only way of accumulating confidence in a computer program is to use it
and compare results with what you are already aware of as being true. . .
. . .

SPICE has been used for decades in the design of countless products that
you undoubtedly use daily. It's an extremely useful and valuable tool,
without which many modern designs simply wouldn't be possible.


Here again, Roy I"m confused. You say its indespensible; Reg Edwards
says its' unreliable. Who am I to beleive? When the experts disagree,
its imposible to form a reliable conculsion.