Marriage is under fire!!

[John Woodgate]

I read in alt.binaries.schematics.electronic that Gary Richardson
<garyr@fidalgo.net> wrote (in <10jrl2uc19v435a@corp.supernews.com&gt
about 'Ascii Circuit Art', on Tue, 7 Sep 2004:

Just click on the filename to start the download.

I bet you say that to all the girls.
--
Regards, John Woodgate, OOO - Own Opinions Only.
The good news is that nothing is compulsory.
The bad news is that everything is prohibited.
http://www.jmwa.demon.co.uk Also see http://www.isce.org.uk

 

[Jamie]

all it means really is the register
sizes are bits over 8 bits.
this could lead from the cpu
regs, Harward down to the memory
level..
you have to do more research because
even if the cpu was 4 bits multiple
regs can be used to form larger registers!


billy_evil wrote:

> What are the differences between 4-bit and 8-bit MCU?

 

[CFoley1064]

Subject: changing my VMeter from 0.25A to 25A
From: abcd@uk2.net (Hadry LittleWood)
Date: 9/8/2004 12:15 PM Central Daylight Time
Message-id: <2bf0330d.0409080915.3bf1ce11@posting.google.com

hi

I have a volt meter which can reads DC current up to 0.250A
How do I make it able to read up to 25A

maybe just adding some resistors?

thanks

jif

Hi, Jif. Most multimeters have a shunt resistor built-in, and use the internal
voltmeter to read voltage across that shunt resistor. Technically, if you know
that resistor value, all you would have to do is find a resistor 1/99th that
value, and place it in parallel with the ammeter. Then 99% of the current will
flow through your shunt, and 1% will flow through your ammeter. A reading of
0.25A = 25A.

Practically speaking, there's a bit of a problem, though. It's very hard to
get precision oddball value low ohm resistors at reasonable price/delivery.
Also, depending on the value, your R shunt value may be high enough to result
in an uncomfortable amount of power dissipation.

As a practical matter, it's a lot easier to just buy a 50A shunt and read the
voltage with any standard voltmeter. The shunt has a 1 milliohm resistor,
which results in 1mV per Amp up to 50A. This harkens back to the time when
analog meters ruled the earth, and they had 50mV full scale deflection burden.
Easy and straightforward.

And, lo and behold, All Electronics, one of the better US electronics surplus
outfits, has a 50A shunt available for $12 USD in stock. They claim 0.5%
accuracy:

http://www.allelectronics.com/cgi-bin/category.cgi?category=385&item=SNT-5
0&type=store

Remember to use the big screw terminals for the current, and the small ones for
your voltmeter, if you need that accuracy.

I guess today's your day.

Good luck
Chris

 

[Jonathan Kirwan]

On 8 Sep 2004 11:33:49 -0700, mike_uk31415@yahoo.co.uk (mike) wrote:

Could anyone tell me the difference between a LED and a laser diode.

Hehe. I'm sure you'll get various, more serious answers. But from a practical
optics standpoint I sometimes imagine they aren't so much a laser as they are a
somewhat narrower band LED. Some spatial and temporal coherence, but only
enough for a meter or two. Polarization is something like 100:1, I think.

Jon

 

[Jonathan Kirwan]

On Thu, 9 Sep 2004 01:13:58 +0000 (UTC), don@manx.misty.com (Don Klipstein)
wrote:

Another thing: The light emitting surface of a usual laser diode is of
roughly bacterium size - a few to several micrometers by a micrometer or
two. The light emitting surface of most regular LEDs is 200-300
micrometers wide.
The small size of the light emitting surface of most laser diodes
results in a beam that has significant divergence despite the coherence.

Excellent point. I had forgotten to add it!

Also, do you know if, by addition of appropriate external cavities, laser diodes
can maintain their coherence over longer ranges than just a few meters?

Jon

 

[Don Klipstein]

In article <8ccca585.0409081033.6804bbcf@posting.google.com>, mike wrote:

Hi

Could anyone tell me the difference between a LED and a laser diode.

I know that LEDs produce light via spontaneous emission that occurs
when electrons and holes recombine in a PN junction and that laser
diodes work via spontaneous emission but what is the difference in
construction?

In a laser diode, excited bits in the semiconductor matrix ("excited" by
having their electrons "boosted" from the "valence band" to the
"conduction band") mostly do not release their stored energy as photons
spontaneously, but do so in response to stimulation by similar photons
(this is "stimulated emission").

In a laser diode, the "stimulated emission" results in the photons hving
coherence. For one thing, the bandwidth of a usual laser diode is around
or less than 1/10 naometer, while that of regular LEDs is a few to a few
10's of nanometers.

The light of a laser diode is polarized while that of a regular LED is
not.

Another thing: The light emitting surface of a usual laser diode is of
roughly bacterium size - a few to several micrometers by a micrometer or
two. The light emitting surface of most regular LEDs is 200-300
micrometers wide.
The small size of the light emitting surface of most laser diodes
results in a beam that has significant divergence despite the coherence.
As a result, laser pointers have lenses that collimate the beam into one
with low divergence. The same lens cannot do the same with ordinary LEDs,
due to the larger light source size with lack of coherence.

- Don Klipstein (don@misty.com, http://www.misty.com/~don/laserdon.html)

 

[mark krawczuk]

hi, laser light is a coherent beam of light, led light is spread out : non
coherent.
mark k

--


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"mike" <mike_uk31415@yahoo.co.uk> wrote in message
news:8ccca585.0409081033.6804bbcf@posting.google.com...

Hi

Could anyone tell me the difference between a LED and a laser diode.

I know that LEDs produce light via spontaneous emission that occurs
when electrons and holes recombine in a PN junction and that laser
diodes work via spontaneous emission but what is the difference in
construction?

Thanks

Mike

 

[dan]

What's that Lassie? You say that BruceW..1 fell down the old
sci.electronics.basics mine and will die if we don't mount a rescue by
Mon, 06 Sep 2004 00:33:27 GMT:

I'd like to power a 1-watt Luxeon Star
http://www.lumileds.com/products/family.cfm?familyId=2
with two AA batteries.

And I'd like to be able to dim the light. I've found that the Luxeon
Star puts out useable light down to 50mA, or up to about 350 mA.

This is for a headlamp to be used for camping and backpacking. Most
manufacturers of these headlamps use 3 batteries and a voltage dropping
resistor. IMHO, this is pretty cheesy and besides I'd like to mod one
of my existing AA headlamps.

Actually I'd like to make it run off of a single AA battery because I'd
rarely use full brightness.

Has Maxim come out with any new IC'c in the last year that might be
right for the job? I just ran across this, don't know what's inside,
but it doesn't dim:
http://www.luxeonstar.com/item.php?id=415&link_str=89&partno=02009A

I'm no electronics expert but I'm smarter than the average bear in this
respect and I can solder.

Can anyone recommend a circuit?

Ask here:
http://www.candlepowerforums.com/ubbthreads/ubbthreads.php?Cat=

Search in the electronics forum.
http://www.candlepowerforums.com/ubbthreads/postlist.php?Cat=&Board=UBB6

--

Dan

 

[Jamie]

it varies among sound cards a bit but 5k ohm on some Mic inputs
and 50K on line inputs. these are just some.
the levels for the line in are usually around .3 to .6 p-p
and mic input levels can be very sensitive and very alot among
sound cards.
if you work around .050 for the max average i think you should
be ok.



Animesh Maurya wrote:

Hi all,

I'm using my sound card for data acquisition. It is inbuilt on my
Intel 845GL motherboard, named Avance AC97 Audio.

I need to know the input impendence(line in port) for calculating
coupling capacitor. Also interested to know max p-p voltage?

Any help in highly appreciated.

Thanks

AM

 

[Robert C Monsen]

"the yeti" <returnoftheyeti@aol.com> wrote in message
news:d22f77dc.0409110821.52f13d4c@posting.google.com...

I have two of these R100D1205-12TS
(http://www.aconinc.com/pdf/R.pdf)and I want to use it to run a
computer on a car battery. Its a 100 Watt, so I think I can do it.
I'll use the first one to power the motherboard and the second one
to
power the hard drive.

My 2 questins are:

1) Can I just use a resistor to step down the voltage?

Going from 5V to 3.3V using a resistor will waste about 35% of the
available power. At 5A, the resistor will dissipate something like
8.5W, so it'll get hot. A little SMPS would be better, although more
complex.

2) Do you think I could get away with just using one to power the
hard
drive and the motherboard?

This depends on the current draw of the mainboard and drive. If you
are using an EPIA nano-itx mainboard, you might be able to get away
with it.

I realize I have some reading ahead of me to learn som of these
basic
electronics. Thanks for any advice you can give me.

Why not use a laptop, and charge the battery directly?

Regards,
Bob Monsen

 

[Tim Auton]

"Jeroen" <jayjay.1974@xs4all.nl> wrote:

"smpaladin" <smpaladin@yahoo.com> wrote in message
news:6753235c.0409111600.686eb556@posting.google.com...
I have a simple battery/electromagnet setup. Does anyone know how I
could set up a switch to reverse the polarity of the electromagnet?
(Besides switching the wires on the battery)

A simple DPDT switch will do the trick:
[snip]

If it's not centre-off, make sure it's a break-before-make switch!


Tim
--
Guns Don’t Kill People, Rappers Do.

 

[Joe McElvenney]

Hi,

If you have two system, one SN is a, another is b. If they are summed, what is SN?
Assuming they can be summed for same properties.

In the case of two uncorrelated noise sources mixed with
identical signals of the same amplitude, I think there should be
a 3dB improvement in S/N ratio. The voltage resulting from two
such noise signals would be equal to the square-root of their
sum but the signals alone would add normally.

That is, uncorrelated noise powers add normally but the sum of
their voltages do not.


Cheers - Joe

 

[Joe McElvenney]

Hi,

Oops! the last sentence should read -

That is, uncorrelated noise powers add normally but
their voltages do not.


Cheers - Joe

 

[Claude Sylvain]

"Jeroen" <jayjay.1974@xs4all.nl> wrote in message
news:414398c1$0$568$e4fe514c@news.xs4all.nl...

"smpaladin" <smpaladin@yahoo.com> wrote in message
news:6753235c.0409111600.686eb556@posting.google.com...
I have a simple battery/electromagnet setup. Does anyone know how I
could set up a switch to reverse the polarity of the electromagnet?
(Besides switching the wires on the battery)

A simple DPDT switch will do the trick: (view with a fixed font)

+ --+--o
| \----o magnet +
| -o
| |
---o
\----o magnet -
- ----+o

If you take a switch with 3 positions the middle position will turn the
magnet off

Sorry, but your schematics do not work.
I think this one will be better...

+ --+---o
| \----o magnet +
| +-o
| |
- ----+-o
| \----o magnet -
+---o


Never forget the make a design review ;-)
Claude

 

[Tzortzakakis Dimitrios]

You are in the 19th century.The germans have used this since 1990.The ICE
train (InterCityExpress) has 13,000 HP, probably 8 traction motors and 2
"locomotives" (see www.db.de if you can read german).The motors are
asynchronous, three-phase squirrel cage and of course they must have
sufficient torque to pull the train at stop and enough rpm to reach the
maximum speed of 200 km/h.The cetenary system is 15 kV 16 2/3 Hz.Of course,
there's such a drive as in the DVD.In normal electric locomotives there's a
motor with brushes and excitation in series (like the one your drill has or
your mixer, the one you make the ice-cream) and a transformer with 18
taps.To start the train, you need high current and low voltage.For maximum
speed, you need high voltage and sufficient current.These motors are
directly coupled on the wheels.

--
Dimitris Tzortzakakis,Iraklion Crete,Greece
major in electrical engineering
freelance electrician
dimtzort AT otenet DOT gr
? "~Dude17~" <dude17@sacbeemail.com> ?????? ??? ??????
news:b959931f.0409102228.5a14eff2@posting.google.com...

X-No-Archive: Yes

I took apart a DVD-ROM drive for the heck of it and the control LSI is
amazing stuff.

The driver chip has everything to directly drive all the mechanical
parts in a CD-ROM drive from loading tray, focus, tracking, sled and
the spindle motor. I find the spindle motor control the most
fascinating.

http://www.rohm.com/products/databook/optdisc/pdf/bd7902cfs.pdf

The chip controls the spindle motor using three phase PWM and reading
the controller documentation leads me to believe the thing can be
controlled somehow with pin 24.

The range is rather wide 230RPM while playing back audio CD at outer
diameter and about 10,000RPM at 48x CAV mode. The chip can also apply
reverse torque to quickly bring the disc to stop.

Is it difficult to make a variable speed drive using the spindle motor
and the LSI pulled from a DVD-ROm drive to let me run the motor
anywhere from 280 to 10,000RPM outside of the original drive? It
would surely make a cool project part.


If this sophisticated control can be built into a $20 DVD-ROM drive,
how expensive would it be to integrate a similar controller with
beefier drive circuit to drive a motor in few hundred watt to a few
kilowatt range?

 

[Peter Bennett]

On 12 Sep 2004 17:18:20 -0700, smpaladin@yahoo.com (smpaladin) wrote:

"Claude Sylvain" <claudesylvain@sympatico.ca> wrote in message >> Sorry, but your schematics do not work.
I think this one will be better...

+ --+---o
| \----o magnet +
| +-o
| |
- ----+-o
| \----o magnet -
+---o


Never forget the make a design review ;-)
Claude

Now is there any kind of switch that will reverse the polarity of the
coil electrically like in this picture?
http://www.geocities.com/smpaladin/switch.jpg

Once a current is applied to the red switch, it reverses the polarity
of the coil as long as the current is applied. If this sounds
ridiculous, don't blame me since I may have no idea what I'm talking
about.

The "red switch" would be a double pole, double throw relay, with its
contacts wired as shown above (but you'd probably also want another
switch to turn the coil off...)



--
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

 

[Rich Grise]

On Saturday 11 September 2004 04:21 am, species8350 did deign to grace us
with the following:

I'd like to understand the circuitry that accompanies canon cartridges.

Has anyone found a good tutorial site.

The "circuitry" would be trivial - a trace from the pad to the ink gun.

Who the hell is going to pay somebody to come with a tutorial on a
throwaway part?

At the appropriate time, you send a current pulse to the correct ink
jet, and it squirts ink at the paper. I think it's little heating
elements, that boil the ink and shoot the remaining liquid ink by
the power of ink vapor.

Good luck!
Rich

 

[Rheilly Phoull]

"boki" <bokiteam@ms21.hinet.net> wrote in message
news:4c3c095a.0409130001.22fdde14@posting.google.com...

Dear All,

Could you please tell me more about DC jack?

What is that? connector?

Thank you very much.


Best regards,

Boki.

These are used for the DC power inputs on most equipment. They are round
with one connection on the exposed outer 'pin' on the plug which is simply a
tube there is an 'inner pin' which is 'female'.
Go here
http://www1.jaycar.com.au and then enter 'dc plug' to see one on a device.


--
Regards ........... Rheilly Phoull

 

[Tom Biasi]

"Rich Grise" <null@example.net> wrote in message
news:ONb1d.3994$ZP.894@trnddc05...

On Saturday 11 September 2004 04:21 am, species8350 did deign to grace us
with the following:

I'd like to understand the circuitry that accompanies canon cartridges.

Has anyone found a good tutorial site.

The "circuitry" would be trivial - a trace from the pad to the ink gun.
Who the hell is going to pay somebody to come with a tutorial on a
throwaway part?

At the appropriate time, you send a current pulse to the correct ink
jet, and it squirts ink at the paper. I think it's little heating
elements, that boil the ink and shoot the remaining liquid ink by
the power of ink vapor.

Good luck!
Rich

He probably wants to see the chip that tells the printer the cart has been

used before.
There are refill sites that show how to spoof the binary code sent to the
printer by the chip.

Tom

 

[Terry Pinnell]

Alessandro Mulloni <mullah@dontspam.loyalmail.co.uk> wrote:

Hi everyone,

I must state first that I'm a total beginner in electronics.

I have a problem with a simple Ni-Cd battery recharger circuit. The
diagram is the sequent

---------------------- 1N4004 diode
+ o----|----------|input (LM317) output|----|----------->|----o +
| | adj | > U
input | ---------------------- > 47ohm U
= 0.1uF | | battery
| |---------------- U
- o----|--------------------|---------------------------------o -

and the input is between 4V and 12V DC.

What I would like to get is 1.25V between the adj and output pins (as
should be from the LM317 specs)

What I get is a varying voltage depending on the input one (that is,
3.3V if the input is 4V, 4.8V if the input is 5.5V, and so on..)

Do you all know why this is possible? Shouldn't the LM317 in the above
circuit guarantee 1.25V _always_ between the adj and the output pins?

Thank you all in advance.

There are some aspects of your schematic that I'm unclear about. Is
the lower end of your 47R connected direct to ground, as apparently
drawn? Or to the wiper of a pot, as would be the case for variable
control? If the latter, what is its value? And, although not strictly
relevant to your question, what voltage battery are you charging?

Anyway, begging answers to those, here are a couple of simulations
that may help.

http://www.terrypin.dial.pipex.com/Images/317Source.gif

--
Terry Pinnell
Hobbyist, West Sussex, UK