Chip with simple program for Toy

[Rich Grise]

On Mon, 18 Oct 2004 05:36:31 -0700, mark.mcgee@csfb.com wrote:

Hi

Is there available a PCB header pin set, with matching socket which can
be assembled by an amateur without expensive specialist tools?

Sure. Try mouser or digikey, or http://www.google.com .

Good Luck!
Rich

 

[Rich Grise]

On Tue, 19 Oct 2004 23:48:21 +0000, Rich Webb wrote:

On 19 Oct 2004 03:26:02 -0700, "mark.mcgee@csfb.com"
mark.mcgee@csfb.com> wrote:

Thanks for the suggestions. I was put off by the cost of the tools,
but the actual items are cheap enough to have a play with, thanks.

You can (and should) use a proper crimper in a production environment.
For hobby or "just learning" pretty much anything that works, works.
You'll probably "explode" a few IDC connectors -- at least I've HEARD
that can happen, yeah that's it, just heard about it... ;-)

If you have access to a good machinists' vise, they're a breeze.

Cheers!
Rich

 

[Rich Grise]

On Mon, 18 Oct 2004 09:24:45 -0700, mark.mcgee@csfb.com wrote:

He he. Unfortunately, the wires only need to be added for 5 mins at a
time to re-program an on-board PIC microcontroller. The rest of the
time, the connection isn't required.

Oh, then just mount a header on the board, and use a small IDC
socket on a ribbon cable. I think you can get them down to six
pins, and I have used one-pin female contacts, which were called
"amp modu", but they seem to have disappeared.

Hm. Digikey has them down to 10 pins.

Good Luck!
RIch

 

[Robert Monsen]

Rich Grise wrote:

On Tue, 19 Oct 2004 23:50:20 -0700, mike wrote:


wow, thanks for the response. here is exactly what i want to do:

i live in a basement w/o windows, and i want to rig inside light so it
coincides with the outside light. i was figuring i could run an LDR
placed outside, to a dimmable ballast in my basement apt.


I seem to remember this from all those months ago, and it seems like
someone had suggested another LDR inside, with a comparator/difference
amp, to make the inside lights _really_ track the outside. Then, as
long as you get it scaled right, you won't have to worry about polarity,
or even nonlinearity.

Of course, on cloudy days, it could be a little dim. )-;

Have Fun!
Rich

But eclipses and airplanes would be fun.

--
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.

 

[Peter Fairbrother]

Michael Buchholz wrote:

"Nicholas O. Lindan" <see@sig.com> schrieb:
Side 2 looks like it was one (or more) of:

o under etched
o under developed resist
o bad artwork
o under exposed resist

My guess: under etched.

Sorry to disagree:

I don't agree either - but I don't know the cause.

Is this a positive or a negative resist (are the tracks on the artwork black
or clear)?

Are you applying the resist yourself, or are you buying ready-covered board?
Could the board be duff?



--
Peter Fairbrother

 

[Rich Webb]

On Thu, 21 Oct 2004 19:25:26 GMT, Rich Grise <rich@example.net> wrote:

On Tue, 19 Oct 2004 23:48:21 +0000, Rich Webb wrote:

On 19 Oct 2004 03:26:02 -0700, "mark.mcgee@csfb.com"
mark.mcgee@csfb.com> wrote:

Thanks for the suggestions. I was put off by the cost of the tools,
but the actual items are cheap enough to have a play with, thanks.

You can (and should) use a proper crimper in a production environment.
For hobby or "just learning" pretty much anything that works, works.
You'll probably "explode" a few IDC connectors -- at least I've HEARD
that can happen, yeah that's it, just heard about it... ;-)

If you have access to a good machinists' vise, they're a breeze.

Ya - I usually use a Panavise or one of these gadgets (long URL)
http://www.jameco.com/webapp/wcs/stores/servlet/ProductDisplay?langId=-1&storeId=10001&catalogId=10001&productId=118859
when I'm prototyping. But there's always the occasion when neither is
handy and the question arises whether it's better to try with the needle
nose pliers or to walk across the building to the lab with the right
tools... ;-)


--
Rich Webb Norfolk, VA

 

[John Popelish]

Robert Monsen wrote:

John Popelish wrote:
"Kelvin@!!!" wrote:

Hi:
how dose bypass capacitor filter out the noise on the signal?
say i have a 74HC74 chip... with a bypass capacitor connecting the Vcc and
Gnd, i get a really nice wave form. but w/o one, i got noice all over the
place! can't even get a waveform on the scope...
it's like magic!
i know how a cap can filter out high freq. signal. but just can't figure out
how a bypass cap work....

thank you for any answers...
--
{ Kelvin@!!! }


Wires have inductance. Circuit board traces have inductance. It
takes time for current to make its way from a power supply to a chip
that has a sudden drop in resistance across its supply lines (and
modern chips can do this extremely quickly, compared to the time it
takes light to make its way to the supply and back). Inductance takes
time for current to change by the differential equation:
V=L*(di/dt). A capacitor is defined by the differential equation:
I=C*(dv/dt). This means that the current through a capacitor is
proportional to the time rate of change of the voltage across it.
(Whew!)

So taking all that into account at the same time (imagination is
faster than the speed of light) when a chip passes a sudden pulse of
current, instead of that pulse having to make its way through the
circuit board trace inductance and supply wiring inductance (including
the time it takes for the current to change through those inductance
and not even worrying about the speed of light over that distance) and
having to put up with all the voltage sag it takes to drive that
current pulse through all that inductance, if you put a capacitor very
close to the chip, and if the capacitor is big enough, a small sag in
the capacitor voltage allows it to supply a very quick pulse of
current to the chip.

Then the inductive supply distribution system charges the cap back up
before the next pulse.

Isn't that easy?

;-)


How much inductance does a trace generally have?

That is difficult to nail down without modeling the surroundings, but
here is an article about the inductance of loops (and the supply
connections to a bypass capacitor form a loop):
http://members.aol.com/marctt/CV/Abstracts/inductance.htm

If you pick the wrong

capacitor, can it resonate with this inductance, given a particular
frequency for a chip's pulses of current?

Yes. Somewhat lossy capacitors are not necessarily bad for bypassing
for this reason. I have connected an ohm in series with a .1 uf on a
board near the connection points for the supply wiring to damp such
ringing, in addition to the distributed bypass capacitors.

I recall a T1 card that was designed at a former company (not by me, I'm
a software guy) that had issues with the 32 bit address/data bus having
too much inductance.

Bus drivers are especially bad actors, because they have high current
capability per bit line, and are switched on and off in large groups.
Big knocks.

There were data corruption issues, and when they
clipped on logic analyzer probes, the corruption went away. They fixed
it by hacking on tiny smt capacitors. Is that kind of thing easy to
predict? (It was between a QED MIPS core and a Galileo memory/pci
controller.)

Not for me, maybe for others. I try to select the adjacent bypass
values based on peak current drawn and duration of pulse, to limit the
sag during the pulses, neglecting the current supplied by the rest of
the system. If no part of the system can generate a significant
instantaneous voltage swing across the supply rails, there is little
likelihood that any large resonance will occur. But I always check
and have provision for ringing dampers as I described above, as cheap
insurance.

--
John Popelish

 

[CBarn24050]

Subject: Re: pcb development
From: Rich Grise rich@example.net
Date: 21/10/2004 19:28 GMT Standard Time
Message-id: <pan.2004.10.21.18.32.00.855961@example.net

On Thu, 21 Oct 2004 14:02:51 +0100, James Varga wrote:

Okay - I've started trying to make some d/s boards and have run into
another
funny. It seem that the image after developing is 'smudged' - its not
actually smudged but it looks like that. It looks like the image is
missing
an area like it has been rubbed off or something.


This is as good as I've gotten it so far - as you can see there is still a
problem with the second side, although the one side is perfect. Each step
is
done vertically so I'm really struggling to figure out whats going wrong.

Several possibilities come to mind. 1. Original artwork not dense enough, laser
and inkjet are not usually good enough. 2. Artwork on the wrong side of the
film. 3. wrong exposure time/development time/temp, make some test strips and
examine under a magnifier for best combination. 4. Uneven UV lighting, try
using sunlight.

 

[Kevin Aylward]

Jack// ani wrote:

For analog design, you might find someting but I dont know it.

And i am pretty sure that op is taking about analog synthesis in terms
of transfer function.

Specifically, I think the poster means non-linear time domain synthesis,
as in a real circuit topology. AC "transfer functions" of linear
circuits is pretty much a solved problem. Its called filter design!

I would also love to see such software.

I dont. I make my living by designing analogue circuits

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.

 

[John Larkin]

On Fri, 15 Oct 2004 23:52:03 GMT, Rich Grise <rich@example.net> wrote:

On Fri, 15 Oct 2004 02:12:31 -0700, Wong wrote:

Hi,

Is it possible to saturate the NPN BJT transistor in this biasing ?
Vcc
___
|
|
|
_| C
B |
---- Resistor ----|
|
-
| E
|---------------- Output
|
Resistor
|
|
|
---
- Ground


Since the transistor is OFF when '0' to base and hence output is '0'.
Then when '1' to base I would like to have 4.5V at the output, is that
possible to bias the transistor in saturation region (since Vce no
more <0.2V)?

Is "saturation" even meaningful in an emitter follower?

Thanks,
Rich

Yup, "inverted state saturation". This can be very cool; if one were
to pull the left side of the resistor up above Vcc, the transistor
drop will go very low, like normal saturation. Pull up more, and it
can go to zero, and then *below* zero with the right transistor. I did
a 16-bit DAC once using discrete transistors and an R-2R wirewound
resistor network, with the base currents of the first few stages
tweaked for exactly zero saturation voltage.

John

 

[CFoley1064]

Subject: Re: LDR / photo resistor - increasing resistance with increasing
light???
From: myothermailishotmail@gmail.com (mike)
Date: 10/22/2004 4:01 AM Central Daylight Time
Message-id: <f6382c10.0410220101.4a7f2c3d@posting.google.com



BUT... it there a way to do this not with voltage, but with
resistance in to the ballast control? from your suggestions maybe the
ballast 0-10v controls hooked up to one side of an op amp, a similar
but steady vcc hooked up to the other side, both sides with LDRs in
series and whatever necessary trimmers/resistors-- but how would this
raise or lower the voltage of the ballast control wires?

sorry, i'm a little dense. but i'm slowly learning

Again, take a look at the H11F1. Its output is a bilateral FET which is seen
by your ballast as a plain old resistor, as long as the voltage applied across
it is less than 30V or so. Depending on the amount of LED current, it runs
from 300 Megohms down to a couple of hundred ohms -- ideal for your
application. The op amp circuit I provided is definitely a fiirst cut -- it
can definitely be improved. With a little more effort and a trimmer with an RC
filter instead of a straight voltage divider to set the current servo point,
you should have a good solution. But that shows the way.

Good luck
Chris

 

[peterken]

as i said, channels are formed during construction

(see drawing with fixed font)

_________
/ gate /
/ /
------------------
\______/

channel
______
/ gate\
------------------

They all have PHYSICAL dimensions.

The ELECTRICAL (virtual) width of the channel is formed by a voltage applied
to gate, thereby forming an electric field in the channel, thereby
reducing(depletion) or allowing (enhancement) the current through the
channel itself, it has nothing to do with the given dimensions in the
datasheets.
It's all basic fet-stuff ya know.



"Kranthi Q" <kranthi_q@graffiti.net> wrote in message
news:ecef0ac4.0410151956.26855a15@posting.google.com...
But aren't channel and depletion layer formed because of difference
between gate and source?? I understand that for gate the length and
width are physical dimensions but what about channel L, W and D. I
want what effects the change of L, W and D.

Thanx.

Kranthi.

"peterken" <peter273@hotmail.com> wrote in message
news:<V2Ubd.279403$zW5.14405696@phobos.telenet-ops.be>...

All of your parameters refer to the physical dimensions of the fet at
silicon level, they are defined at the manufacturing process
All of them refer to what they stand for, seen in a 3-D manner.
Width = how wide is it
Length = how long is it
Depth = what's the thickness of the given layer
Look at the parameters as a lying down rectangle with a specific thickness

None is affected by threshold voltage and gate voltage, in fact it's the
other way around, the physical dimensions at silicon level affect these
parameters.

None of them you can have any influence on, whatever you try, since it's a
mechanical fact



"Kranthi Q" <kranthi_q@graffiti.net> wrote in message
news:ecef0ac4.0410150700.119d7b57@posting.google.com...
Hi

I need some help in understanding the difference between -

Gate Length

Gate width

Channel Length

Channel Width

Channel Depth

Depletion Layer Width

Depeltion Layer length

for a MOS transistor.
Which of these quantities are equal and how are they efected by threshold
voltage and gate voltage.
I am confused as to especially w.r.t depth and width.

Help is appreciated.

Kranthi.

 

[Don Kelly]

"Bill Bowden" <wrongaddress@att.net> wrote in message
news:ad025737.0410221306.39ed4816@posting.google.com...

kianmeng.tey@gmail.com (KM) wrote in message
news:<ecdf7787.0410200555.2e97365d@posting.google.com>...
Use Delta to STAR conversion,


----- A -----
| | |
| Ra |
R1 | R2
| / \ |
| Rb Rc |
--/----R3----\-
| |
| |
R4 R5
| |
| |
----- B -----

the result became, and it is series with parallel series.

A > | | |
Ra
|
/ \
Rb Rc
--/ \-
| |
| |
R4 R5
| |
| |
----- B -----


Ra = (R1+R2)/(R1+R2+R3)
Rb = (R1+R3)/(R1+R2+R3)
Ra = (R2+R3)/(R1+R2+R3)


I'm not following these formulas.
If all the resistors R1-R5 are 1 ohm, then the
total resistance from a to b will be 1 ohm.
But, according to the formulas, Ra, Rb ,Rc will
be 2/3 or 0.67 ohms. So, we have 1.67 in parallel
with 1.67 which is 0.83 which is in series with
Ra (0.67). That all adds up to 1.5 ohms, but it
should be 1 ohm.

What am I missing?


-Bill
---------

Nothing:

The formula is wrong for delta to star conversion- try Ra=(R1*R2)/(R1
+R2+R3) etc
This will give you 1 ohm.

In the case where R1/R4 =R2/R5, you can simply remove R3 as it won't have
any effect. (figure out why).
--
Don Kelly
dhky@peeshaw.ca
remove the urine to answer

 

[w_tom]

What does a surge seek? Destructive transients are
typically longitudinal mode. IOW they seek earth ground. If
not earthed at the building entrance, then they will seek
earth ground destructively through household appliances. Too
many fail to learn of multiple types of transients. The
destructive transient is not stopped, blocked, or absorbed.
And yet that is what a plug-in protector must do. Is that
plug-in protector going to stop what miles of sky could not?

Introduction to protection principles in "Pull the wall plug
or not?" in nz.comp on 7 Sept 2004 at
http://tinyurl.com/5ttwl

For more technical sources, then three consecutive posts: "
Belkin Surgemaster worth the money?" in the newsgroup
uk.comp.homebuilt on 29 Sept 2004 at
http://tinyurl.com/6zfps

Surge protection demonstrated by a simple example:
"Whole house surge suppressors" in alt.home.repair on 12 Jul
2004 at
http://tinyurl.com/6gl67

A classic case of this is proven at the School District where
I work. Every electrical storm would blow the phone system
cards in the switch after a major remodel to the tune of
$2000 - $5000. All kinds of surge protection blah blah.
The "brains" came in scratched their heads. I have been
reading your posts for a long time on this. I said lets
find and install a GOOD ground.

Problem solved. Has NOT happened since.

Transients first form a complete electrical path from cloud
to earth. Only after current is passing through everything,
then something in that path fails. Destructive transients seek
earth ground. Earthing - and not some protector - defines a
protection 'system'. A surge protector is only as effective
as its earth ground. Too many forget about essential earthing
and instead hope what they see on retail shelves is actually
protection. No earth ground means no effective protection.


Sean wrote:

Here's another source:
http://www.eeel.nist.gov/817/817g/spd-anthology/files/Text%20Part%204.doc

From "Coordination: 1980" - F.D. Martzloff

Fact 3.
Without substantial connected loads in the system, the open-circuit
surges appearing at the service entrance propagate along the branch
circuits with very little attenuation.

Conclusion 4.
Coordination of surge suppressors requires a finite impedance to
separate the two devices, enabling the lower voltage device to perform
its voltage-clamping function while the higher voltage device performs
the energy-diverting function.

Conclusion 5.
The concept that surge voltages decrease from the service entrance to
the outlets is misleading for a lightly loaded System. Rather, the
protection scheme must be based on the propagation of unattenuated
voltages

Conclusion 6.
Indiscriminate application of surge protectors may, at best, fail to
provide the intended protection and, at worst, cause disruptive
operation of the suppressors. What is needed is a coordinated
approach based on the recognition of the essential factors governing
devices and surge propagation.

There is a wealth of papers concerning surge suppression available tn
the directory of the example above, that I am trying to sort through.

More Later [YMMV]

 

[Rich Grise]

On Sat, 23 Oct 2004 02:11:41 +0000, Don Kelly wrote:

"Bill Bowden" <wrongaddress@att.net> wrote in message
news:ad025737.0410221306.39ed4816@posting.google.com...
kianmeng.tey@gmail.com (KM) wrote in message
news:<ecdf7787.0410200555.2e97365d@posting.google.com>...
Use Delta to STAR conversion,


----- A -----
| | |
| Ra |
R1 | R2
| / \ |
| Rb Rc |
--/----R3----\-
| |
| |
R4 R5
| |
| |
----- B -----

the result became, and it is series with parallel series.

A > | | |
Ra
|
/ \
Rb Rc
--/ \-
| |
| |
R4 R5
| |
| |
----- B -----


Ra = (R1+R2)/(R1+R2+R3)
Rb = (R1+R3)/(R1+R2+R3)
Ra = (R2+R3)/(R1+R2+R3)


I'm not following these formulas.
If all the resistors R1-R5 are 1 ohm, then the total resistance from a
to b will be 1 ohm. But, according to the formulas, Ra, Rb ,Rc will be
2/3 or 0.67 ohms. So, we have 1.67 in parallel with 1.67 which is 0.83
which is in series with Ra (0.67). That all adds up to 1.5 ohms, but it
should be 1 ohm.

What am I missing?


-Bill
---------
Nothing:

The formula is wrong for delta to star conversion- try Ra=(R1*R2)/(R1
+R2+R3) etc
This will give you 1 ohm.

In the case where R1/R4 =R2/R5, you can simply remove R3 as it won't have
any effect. (figure out why).

Because the current's flowing both ways simultaneously, and so, from the
POV of our perceptive systems, cancel each other out and so don't exist.

Cheers!
Rich

 

[Reg Edwards]

All programs have bugs. I only asked why people think it IS reliable.

I am aware of its existence only from the frequent mentions made on
newsgroups.

Why do the arguments continue after Spice has arbitrated?

I have no reason to think Roy's opinion of Spice is anything other than
true.

But whatever it is, it is not a device intended to be used as a means of
instructing learners on the theory of electrical circuits. All programs
have many limitations which eventually always become serious and which are
UNKNOWN to the user. Very often they are unknown even to the programmer.
Limitations should not be allowed to cross over the borders of knowledge.
Programs should not be worshipped for always telling the gospel truth. They
don't.

For example, a sensible circuit designer invariably checks the output of a
program by making a hardware prototype - or several. Why? Because he
trusts neither himself nor the program!
----
Reg

"Steve Evans" <smevans@jif-lemon.co.mars> wrote in message
news:b5h3n0d7uj15gg1a50t2l083suoel29hj8@4ax.com...

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.

 

[Roger Johansson]

"Reg Edwards" <g4fgq.regp@ZZZbtinternet.com> wrote:

Some years back, after retirement, I bought out of curiosity a copy of
Electronics Work Bench. It was and still is the only such program I
have ever had my hands on. I think it arrived on a collection of
floppies.

After a few days curiosity was satisfied. Then I junked it.
---
Reg

Just an explanation to the person who asked:

Reg is one of the few people in the world who have no use for a spice
simulator. He writes lots of very good calculation programs and knows
electronics like he had a built-in electronics simulator in his brain. He
has studied electronics during a long life and knows what he talks about.

For other people though, a spice simulator can be very useful.

Multisim has a very bad reputation because is has a lot of bugs.
The best version of EWB is 5.c
After that version multisim replaced the central parts of the program and
the new program had a lot of problems.

Nobody uses later versions of multisim, so we do not know if they have
solved the bugs yet.

EWB 5 is used by beginners, because it has a very good user interface.
But even EWB 5 is not regarded as a good spice simulator.

Professionals often like the freeware spice simulator Switchercad3 from
Linear Technology. http://www.linear.com/company/software.jsp

I use EWB 5.c myself, because it is good enough for my purposes, and it
is a lot easier to work with than Switchercad3. But I use Switchercad3
sometimes too, because people in newsgroups often give a circuit in
switchercad's text file format.

Another advantage for EWB is that it has been around a while, so there
are many add-ons to the program, like a converter from EWB circuit to an
Eagle layout (Eagle is a freeware layout program for circuit boards).
Translation programs from ewb to standard netlist is also available.

After multisim changed the program there was a big wave of protests among
the EWB users and buyers. New buyers demanded to get a copy of the old
working program when they bought the new version, and Multisim accepted
that and delivered a EWB version 5c for free, on demand, to all buyers of
the new program.

I don't know if they still do.


--
Roger J.

 

[peterken]

Ray

Correct, but as I read the question
- a light bulb and a heater are both resistive loads
- a radio is a more complex load

As I see the question I read "all these apparatus consuming the same
wattage..."
so I in fact read "do all these apparatus turn the applied (identical) power
into heat"

Answer is no, given the fact all of them are designed for different purposes
so some of the applied power is turned into something else as intended per
design, and the rest into heat

Proof for this can be established empirically

Or just take next situation :
I'd like to stand next to my 1kW heater on a cold day, it gets me warm
I wouldn't like to stand next to my 1kW amplifier on a cold day, it mostly
creates noise but gives me ampel heat




"rayjking" <rayjking@bellsouth.net> wrote in message
news:d5Bdd.38181$pi7.10280@bignews4.bellsouth.net...
Rene,

You are correct. If you measure watts to be equal then you are correct
except for the power factor which is ( Power = E x I x Cos of the angle
between the voltage and current ) ( theta ) which is different for loads
other than resistive. Just measuring the voltage and just measuring the
current ( E x I ) gives you apparent power which is always more than that
which is creating heat-watts.

Ray


..
"peterken" <peter273@hotmail.com> wrote in message
news:j3ydd.284682$FW5.14407816@phobos.telenet-ops.be...

yes, if they all converted exactly the same amount of energy into
something else (movement, light, sound)
no, if they all convert different amounts of energy into something else

the bulb will convert an amount into light, the non-light-energy will be
turned mostly into heat
the heater will have losses in light-radiation (usually IR), the rest will
be dissipated as heat
the radio will convert energy into sound, the rest will be dissipated as
heat

so it all depends on the efficiency of every aparatus that will be
compared


"Rene" <nospam@nospam.nospam> wrote in message
news:UYidd.394645$mD.336927@attbi_s02...

Suppose that I have a light bulb consuming 100 watts, a radio also
consuming
100 watts and an electric heater also consuming 100 watts.

If I were to measure the heat generated by all these 3 devices, would I
get the same amount of heat? In other words, would all devices consuming
the same wattage (solid state, no moving parts) generate the same amount
of heat?

Thank you.

 

[TW]

Joe-

I am so glad I came across your post. I have PICBasic (not the pro
version), and I needed a better way to get started on a project
involving an LCD readout. I demoed and then bought Vladimir's
simulator program, and it is worth much more than the $19. Not only
is it a great simulation, but the built in BASIC compiler includes LCD
instructions.

Cheers,
Ted

On Thu, 30 Sep 2004 06:30:43 +0000 (UTC), Joe McElvenney
<ximac@btinternet.com> wrote:

Hi,

...................... Everything costs money, or is a crappy
trial and all I want to do is program a damn Pic in Picbasic.

Here's a definitely 'non-crappy' trial PIC simulation and programming
package which will cost you a mere USD-19 if you can spare it.


http://www.oshonsoft.com/


IMHO, they don't come much better than this at the price. FREE is
nice but you can spend half your life looking for it. Just like a
motorist driving from petrol station to petrol station looking for the
cheapest fuel.


Cheers - Joe

 

[John Fields]

On Sat, 23 Oct 2004 04:03:04 GMT, Rich Grise <rich@example.net> wrote:

On Sat, 23 Oct 2004 02:11:41 +0000, Don Kelly wrote:

The formula is wrong for delta to star conversion- try Ra=(R1*R2)/(R1
+R2+R3) etc
This will give you 1 ohm.

In the case where R1/R4 =R2/R5, you can simply remove R3 as it won't have
any effect. (figure out why).

Because the current's flowing both ways simultaneously, and so, from the
POV of our perceptive systems, cancel each other out and so don't exist.

---
Puh-leeze...

If R1/R4 = R2/R5, then the voltages at the junctions of R1R4 and R2R5
will be equal, and with no potential difference across R3 current
can't flow through it, whether it's an open, a short, or anything in
between.

As a matter of fact that's why, at null, the detector across a
Wheatstone bridge vanishes.

--
John Fields