Mosquito Sound

[Byron A Jeff]

In article <xn7jsmlpnx.fsf@delorie.com>, DJ Delorie <dj@delorie.com> wrote:
-
-"Mark (UK)" <jumbos.bazzar@btopenworld.com> writes:
-> http://www.cadsoft.de/
->
-> The trial version is totally free, and does 10 x 8 PCBs - useful for
-> small projects.
->
-> The difference with Eagle is - it's actually GOOD!
-
-gEDA and PCB (unix/cygwin hosts) are also $0.00 and fully featured
-(i.e. not crippleware).

D.J.

gEDA/PCB is on my list of things to look at. I've played with the pair a bit,
but I'm still looking for a blow by blow tutorial from circuit to printout.

The last time I Goggle researched was about 6 months ago with no real
paydirt.

Any pointers welcome.

BAJ

 

[Mark (UK)]

Hi!

Nope, it's fully functional in every way - the ONLY limitation is the
size of the PCB, and one page schematics - but that page can be any size.

The advantage of it is that it's a bang up to date product, no worrying
about older software with no support - and if you really get into it,
then at least if you want to pay for an upgrade to do bigger boards,
you'll know your way round it.

JM2C.

Yours, Mark.

Spehro Pefhany wrote:

On Thu, 29 Jul 2004 13:40:59 +0000 (UTC), the renowned "Mark (UK)"
jumbos.bazzar@btopenworld.com> wrote:


http://www.cadsoft.de/

The trial version is totally free, and does 10 x 8 PCBs - useful for
small projects.


10 x 8 being cm (100mm x 80mm), which isn't too bad.


The difference with Eagle is - it's actually GOOD!


Hmm.. isn't it a crippleware version of the same thing?

Best regards,
Spehro Pefhany

 

[John Miles]

In article <33f83f1d.0407291103.6c0dced0@posting.google.com>,
Wolfgangweinmann@aol.com says...

Hello,

i have a HP 1652B Logic-Analyzer. I am searching a printer for it. In
the Menue there are following printers selectable: Thinkjet, Quietjet,
Laserjet and Alternate.

I am searching a ink-printer.

Who knows a newer printer which is compatible and can be used which
this logic analyzer?

Best regards

Wolfgang

Does it support HP-GL/2 plotters like the HP7470A via a GPIB port? If
so, you may be able to use the 7470.exe app from
http://www.speakeasy.net/~jmiles1/7470.zip to capture screenshots on a
PC and generate hardcopy on any desired printer.

If the 7470 emulator doesn't work, you might also try the evaluation
version of PrintCapture (www.printcapture.com). Their HP-GL/2 renderer
is somewhat more robust than mine.

-- jm

------------------------------------------------------
http://www.qsl.net/ke5fx
Note: My E-mail address has been altered to avoid spam
------------------------------------------------------

 

[Activ8]

On Thu, 29 Jul 2004 09:55:08 -0500, Dave VanHorn wrote:

Couple months back I purchased Circuit Creator Pro PCB Design Software
for just $195. By now, have designed 7 PCBs and got the boards made.
Software has worked very well. It has some features available in
expensive package like Orcad. Demo can be downloaded from following
link.

When you say it has features like orcad, I assume you mean the current
version.
This alone is enough to send me running away, very fast.

Anyone tried this yet? www.pad2pad.com

It only works for that vendor.
--
Best Regards,
Mike

 

[Dan Fraser]

Yes there is a modification that will work.

Put the ballast back.

And fix the auto-repeat on your keyboard.

 

[Ken Taylor]

"Mike" <mjb@posie.local.dom> wrote in message
news:ce6l7q$pab$1@posie.local.dom...

In article <U9$AJ4AXcf$AFwyR@jmwa.demon.co.uk>,
John Woodgate <noone@yuk.yuk> wrote:

Do the names KIM, AIM-65, SYM-1, PET, Apple II, VIC-20, C-64, Atari &
Nintendo remind you of anything? I believe the 6502 was the most-sold
microprocessor ever.

You can add 'BBC Micro' to that list.

... And Oric-1 and Oric Atmos!
--
I have a chess computer here at home that uses a 6502. Plays a crap game,

and mighty slowly, but a chess computer it is!

Ken

 

[Mike Harrison]

On Thu, 29 Jul 2004 18:50:13 -0700, John Miles <jmiles@pop.removethistomailme.net> wrote:

In article <33f83f1d.0407291103.6c0dced0@posting.google.com>,
Wolfgangweinmann@aol.com says...
Hello,

i have a HP 1652B Logic-Analyzer. I am searching a printer for it. In
the Menue there are following printers selectable: Thinkjet, Quietjet,
Laserjet and Alternate.

I am searching a ink-printer.

Who knows a newer printer which is compatible and can be used which
this logic analyzer?

Best regards

I'm pretty sure the older deskjet inkjets (e.g. 500) are laserjet compatible, but I;m not sure if
any of them have RS232

 

[Tom Woodrow]

Available in the US from http://www.saelig.com/tracker.htm for about $899

Tom Woodrow

Chris Stephens wrote:

The Tracker 110 is a USB bus analyser which can record low-speed (1.5Mbps),
full-speed (12Mbps) and mixed transactions. All bus data is available but,
to avoid total confusion, only the data most often referenced is displayed
by default. Transactions can be expanded to show the individual packets
whose contents can be reviewed in detail. Comprehensive performance
statistics are displayed during data collection and the data, once saved,
can be searched for specific features or displayed with unwanted packets
filtered out. Transactions belonging to individual peripherals can be colour
coded for clarity and standard requests and descriptors are clearly flagged.

Best of all it is so affordable that each engineer on the project can have
one for just 520 UK Pounds, 780 Euro or $940 .

We also have the Explorer 200 for those who need the High Speed 480Mbps
transfer rate of full USB2

To see more details of these products and to see how easy the they are to
use visit

http://www.computer-solutions.co.uk/gendev/usb.htm

To learn more about USB , download the trial software (which includes a
number of annotated USB captures)

The Tracker 110 is now available for world wide sales from our web shop -
next day delivery ( most of EU ).

----------------------------------------------------------------------

Chris Stephens E-mail: sales@computer-solutions.co.uk
Computer Solutions Ltd. Phone: +44 (0)1 932 829 460
1a New Haw Road, Fax: +44 (0)1 932 840 603
Addlestone, Surrey,
KT15 2BZ England http://www.computer-solutions.co.uk

For the largest range of embedded microprocessor development tools
now available for next day delivery - from our web shop

 

[L.]

You sure they're not "BNC" connectors? Those are common on many scopes and
probes are available from many sources.

L.

"Too_Many_Tools" <too_many_tools@yahoo.com> wrote in message
news:2dc6c2d9.0407301127.52fa1854@posting.google.com...

I recently acquired a Gould 2608 oscilloscope that I have several
questions about.

First is the fact that the imputs use a connector that I have never
seen before. They almost look like a push on F-connector but are not.
Does anyone have any ideas where to go to get probes that would fit
this apparently Gould specific connector?

Second, does anyone know where the USA contact for Gould is located?

I have done the usual Google searches with no results so any
information would be appreciated.

Thanks in advance.

TMT

 

[Tom Del Rosso]

"Ken Taylor" <ken@home.nz> wrote in message
news:4JoOc.7106$N77.369727@news.xtra.co.nz

--
I have a chess computer here at home that uses a 6502. Plays a crap
game, and mighty slowly, but a chess computer it is!

In '78-'79 Popular Electronics had a computer chess project with a
"2650", which I remember noticing was a 6502 with the digits rearranged.
Apparently it was a completely different CPU.

 

[Mark (UK)]

Hi!

Their factory used to be on the industrial estate behind my house! I
remember going there for a tour as a kid.

It's now an Oxfam charity warehouse.

Yours, Mark.

Too_Many_Tools wrote:

I recently acquired a Gould 2608 oscilloscope that I have several
questions about.

First is the fact that the imputs use a connector that I have never
seen before. They almost look like a push on F-connector but are not.
Does anyone have any ideas where to go to get probes that would fit
this apparently Gould specific connector?

Second, does anyone know where the USA contact for Gould is located?

I have done the usual Google searches with no results so any
information would be appreciated.

Thanks in advance.

TMT

 

[Jerry G.]

The LED's will not normally be accurate enough to each other. You will most
likely have a situation where one would light normally, and the rest would
be dim, or not lit at all.

You need a separate resistor for each one, or you should look for LED's with
built in resistors. If they have built in ones, then you will be restricted
to the voltage range indicated for the LED's.

--

Jerry G.
==========================


"wylbur37" <wylbur37nospam@yahoo.com> wrote in message
news:8028c236.0407310039.45737ad0@posting.google.com...
When using an LED from a power source whose voltage is higher than the
rating of the LED, a resistor is typically used.
If more than one LED is to be used (let's say three),
each one would have its own respective resistor,
and the schematic would look something like this ...


+----------+-----------+-----------+--------------
| | | |
| | | |
| LED-1 LED-2 LED-3
power | | |
source | | |
| resistor-1 resistor-2 resistor-3
| | | |
| | | |
+----------+-----------+-----------+--------------


But if all the LEDs were of the same voltage and amperage
(e.g., if they were all of the typical 3.6V 20mA),
isn't there a simpler way to do this by using only ONE resistor?
In other words, can't it be done like the following ?


+--------------+-----------+-----------+---
| | | |
| | | |
power LED-1 LED-2 LED-3
source | | |
| | | |
| | | |
+---resistor---+-----------+-----------+---


And what would be the proper value of the resistor?

(I'm guessing it would be one third of what it was in the first diagram
because the required voltage reduction would be the same but the
effective current draw of the 3-LED assembly would be three times what it
was before.
So if the power source were 6V and the LEDs were 3.6V and 20mA each,
then the required resistance would be 120 ohms in the first case
and 40 ohms in the second case. Correct?)

 

[Michael A. Covington]

Put the LEDs in series, not parallel. That is:

+ -----/\/\/\/-----LED-----LED----LED----- -

They will all carry the same current but need not have the same voltage
(e.g., you can have a 3.2-volt blue LED, a 2.1-volt green, and a 1.8-volt
red, all in series).

The power supply must be higher than the sum of the LED voltages, with the
resistor calculated to make up the difference.

Example: With the 3 LEDs I just mentioned, 3.2 + 2.1 + 1.8 = 7.1 volts.

Suppose the supply is 12 volts and you want 20 mA.

Then the resistor is (12 - 7.1)/0.020 = 245 ohms (use 270, which is a
standard value).

As someone pointed out, if you put the LEDs in parallel, they will not share
current equally unless their voltages are *exactly* equal, which depends on
temperature, manufacturing lot, etc., even if they're all the same color.

 

[Michael A. Covington]

"Nigel Heather" <nigel@NOSPAM.the-heathers.nildram.co.uk> wrote in message
news:Z9CdnVEFk40J7pbcRVn-hg@nildram.net...

I agree that there is no guarantee about the internal characteristics of
similar diodes. But then this argument equally holds true for the first
diagram - if they all used 300 ohm resistors you couldn't gurantee that
they
all glowed identically, just the same as if you powered all three through
a
single 100 ohm resistor.

My answer to the original question is -- Yes that should work and yes the
resitor will need to be a third the value of the individual ones in the
first diagram. Yes the diodes might differ internally but in my opinion
the
differences will be small and I doubt that you will notice them. Give it
a
try.

Yes; contrary to what I just said, they probably *do* have enough internal
resistance to even things out if you make a good faith attempt to use the
same kind. Obviously one red and one green will not do... even one bright
red and one dark red might not do.

 

[si]

There's no guarantee the LEDs will share the current equally due to internal
differences (even between LEDs of a similar batch). You will at best get one
or two at different brightness, at worse one or more LEDs will be totally
off. Hence the preference to use the seperate resistors.

Si.


"wylbur37" <wylbur37nospam@yahoo.com> wrote in message
news:8028c236.0407310039.45737ad0@posting.google.com...

When using an LED from a power source whose voltage is higher than the
rating of the LED, a resistor is typically used.
If more than one LED is to be used (let's say three),
each one would have its own respective resistor,
and the schematic would look something like this ...


+----------+-----------+-----------+--------------
| | | |
| | | |
| LED-1 LED-2 LED-3
power | | |
source | | |
| resistor-1 resistor-2 resistor-3
| | | |
| | | |
+----------+-----------+-----------+--------------


But if all the LEDs were of the same voltage and amperage
(e.g., if they were all of the typical 3.6V 20mA),
isn't there a simpler way to do this by using only ONE resistor?
In other words, can't it be done like the following ?


+--------------+-----------+-----------+---
| | | |
| | | |
power LED-1 LED-2 LED-3
source | | |
| | | |
| | | |
+---resistor---+-----------+-----------+---


And what would be the proper value of the resistor?

(I'm guessing it would be one third of what it was in the first diagram
because the required voltage reduction would be the same but the
effective current draw of the 3-LED assembly would be three times what it
was before.
So if the power source were 6V and the LEDs were 3.6V and 20mA each,
then the required resistance would be 120 ohms in the first case
and 40 ohms in the second case. Correct?)

 

[Watson A.Name - \"Watt Su]

"Mark (UK)" <jumbos.bazzar@btopenworld.com> wrote in message
news:cefqqc$f0g$2@hercules.btinternet.com...

Hi!

Or you could use LEDs with built in resistors?

Yours, Mark.

si wrote:

Most LEDs don't come with a built-in resistor. Ones that do are more
expensive.

The LED driver chip makers get around this problem by connecting up to
maybe 6 LEDs in series. All the LEDs then get exactly the same current.

Also, if you are parelleling 3 or more LEDs to get more light, then
instead, use a single more powerful LED such as the Luxeon Star. There
are also other brighter LEDs such as the 'spider' LEDs.

With blue or white LEDs, if you do parallel LEDs without a separate
resistor for each, be prepared to have one or more LEDs that overheat
with a greatly shortened lifetime and/or much dimmer light output.
Also, one or more LEDs may overheat and its voltage drop will go up,
which then causes that LED to get dim, and the other LEDs will then end
up with more current. They then may overheat,, etc.

Murphy's Law always applies.

There's no guarantee the LEDs will share the current equally due to
internal
differences (even between LEDs of a similar batch). You will at best
get one
or two at different brightness, at worse one or more LEDs will be
totally
off. Hence the preference to use the seperate resistors.

Si.


"wylbur37" <wylbur37nospam@yahoo.com> wrote in message
news:8028c236.0407310039.45737ad0@posting.google.com...

When using an LED from a power source whose voltage is higher than
the
rating of the LED, a resistor is typically used.
If more than one LED is to be used (let's say three),
each one would have its own respective resistor,
and the schematic would look something like this ...


+----------+-----------+-----------+--------------
| | | |
| | | |
| LED-1 LED-2 LED-3
power | | |
source | | |
| resistor-1 resistor-2 resistor-3
| | | |
| | | |
+----------+-----------+-----------+--------------


But if all the LEDs were of the same voltage and amperage
(e.g., if they were all of the typical 3.6V 20mA),
isn't there a simpler way to do this by using only ONE resistor?
In other words, can't it be done like the following ?


+--------------+-----------+-----------+---
| | | |
| | | |
power LED-1 LED-2 LED-3
source | | |
| | | |
| | | |
+---resistor---+-----------+-----------+---


And what would be the proper value of the resistor?

(I'm guessing it would be one third of what it was in the first
diagram
because the required voltage reduction would be the same but the
effective current draw of the 3-LED assembly would be three times
what it
was before.
So if the power source were 6V and the LEDs were 3.6V and 20mA each,
then the required resistance would be 120 ohms in the first case
and 40 ohms in the second case. Correct?)

 

[Watson A.Name - \"Watt Su]

"Michael A. Covington" <look@ai.uga.edu.for.address> wrote in message
news:zIOdnUTC-5026JbcRVn-ug@speedfactory.net...

"Nigel Heather" <nigel@NOSPAM.the-heathers.nildram.co.uk> wrote in
message
news:Z9CdnVEFk40J7pbcRVn-hg@nildram.net...
I agree that there is no guarantee about the internal
characteristics of
similar diodes. But then this argument equally holds true for the
first
diagram - if they all used 300 ohm resistors you couldn't gurantee
that
they
all glowed identically, just the same as if you powered all three
through
a
single 100 ohm resistor.

My answer to the original question is -- Yes that should work and
yes the
resitor will need to be a third the value of the individual ones in
the
first diagram. Yes the diodes might differ internally but in my
opinion
the
differences will be small and I doubt that you will notice them.
Give it
a
try.

Yes; contrary to what I just said, they probably *do* have enough
internal
resistance to even things out if you make a good faith attempt to use
the
same kind. Obviously one red and one green will not do... even one
bright
red and one dark red might not do.

One factor needs further explanation. LEDs have a negative temperature
coefficient. So if one LED gets a bit more current, it gets hotter,
which then causes its voltage drop to be lower, which then causes it to
hog even more current. This vicious circle of current hogging usually
causes the LED to overheat and have a much shorter lifetime. This may
take hundreds or thousands of hours, but the closer the LEDs are run to
their maximum, the less time it will take.

One way to reduce this tendency is to have all the LEDs closely coupled
thermally. But then this usually means they will get hotter, because
there's a lot of heat in a small space. I put a dozen white LEDs on a
predrilled 0.1" hole spacing PCB spaced every other hole, and they got
too hot. So I spaced them out every third hole and they stayed cooler.
(View with Courier font)

Originally: 0 . 0 . 0 . 0 . 0 . 0
0 . 0 . 0 . 0 . 0 . 0

Much Better: 0 .. 0 .. 0 .. 0 .. 0 .. 0
0 .. 0 .. 0 .. 0 .. 0 .. 0

I've been leaving these cheap white LEDs from Hong Kong on 24/7 at their
maximum current, and they just get real dim after a few months and a
thousand or so hours. Just remember that whatever you do, heat is the
enemy of LEDs, and the hotter they get, the shorter they will live.
It's just a law of physics.

 

[si]

With different resistors LED characteristic differences are vastly swamped
by the series resistance. Under those circumstances the differences in
brightness will be minute and most probably not noticeable.
I only suggested NO GUARANTEE that the LEDs would share equally in direct
parallel it is not impossible, but I have found this so unlikely as to be
not worth what is saved in using the extra resistors! As you quite rightly
suggest, a R or 1/3rd original value will suffice.

Hey, but what about connecting them in series if you have sufficient supply
voltage? Current through the lot is the same and that only takes ONE
resistor too!

Si.


"Nigel Heather" <nigel@NOSPAM.the-heathers.nildram.co.uk> wrote in message
news:Z9CdnVEFk40J7pbcRVn-hg@nildram.net...

I agree that there is no guarantee about the internal characteristics of
similar diodes. But then this argument equally holds true for the first
diagram - if they all used 300 ohm resistors you couldn't gurantee that
they
all glowed identically, just the same as if you powered all three through
a
single 100 ohm resistor.

My answer to the original question is -- Yes that should work and yes the
resitor will need to be a third the value of the individual ones in the
first diagram. Yes the diodes might differ internally but in my opinion
the
differences will be small and I doubt that you will notice them. Give it
a
try.

Cheers,

Nigel


"si" <'blu_glo_uk@yahoo.com'> wrote in message
news:xOKOc.87$E25.0@newsfe3-gui.ntli.net...
There's no guarantee the LEDs will share the current equally due to
internal
differences (even between LEDs of a similar batch). You will at best get
one
or two at different brightness, at worse one or more LEDs will be
totally
off. Hence the preference to use the seperate resistors.

Si.


"wylbur37" <wylbur37nospam@yahoo.com> wrote in message
news:8028c236.0407310039.45737ad0@posting.google.com...
When using an LED from a power source whose voltage is higher than the
rating of the LED, a resistor is typically used.
If more than one LED is to be used (let's say three),
each one would have its own respective resistor,
and the schematic would look something like this ...


+----------+-----------+-----------+--------------
| | | |
| | | |
| LED-1 LED-2 LED-3
power | | |
source | | |
| resistor-1 resistor-2 resistor-3
| | | |
| | | |
+----------+-----------+-----------+--------------


But if all the LEDs were of the same voltage and amperage
(e.g., if they were all of the typical 3.6V 20mA),
isn't there a simpler way to do this by using only ONE resistor?
In other words, can't it be done like the following ?


+--------------+-----------+-----------+---
| | | |
| | | |
power LED-1 LED-2 LED-3
source | | |
| | | |
| | | |
+---resistor---+-----------+-----------+---


And what would be the proper value of the resistor?

(I'm guessing it would be one third of what it was in the first
diagram
because the required voltage reduction would be the same but the
effective current draw of the 3-LED assembly would be three times what
it
was before.
So if the power source were 6V and the LEDs were 3.6V and 20mA each,
then the required resistance would be 120 ohms in the first case
and 40 ohms in the second case. Correct?)

 

[Mark (UK)]

Hi!

Or you could use LEDs with built in resistors?

Yours, Mark.

si wrote:

There's no guarantee the LEDs will share the current equally due to internal
differences (even between LEDs of a similar batch). You will at best get one
or two at different brightness, at worse one or more LEDs will be totally
off. Hence the preference to use the seperate resistors.

Si.


"wylbur37" <wylbur37nospam@yahoo.com> wrote in message
news:8028c236.0407310039.45737ad0@posting.google.com...

When using an LED from a power source whose voltage is higher than the
rating of the LED, a resistor is typically used.
If more than one LED is to be used (let's say three),
each one would have its own respective resistor,
and the schematic would look something like this ...


+----------+-----------+-----------+--------------
| | | |
| | | |
| LED-1 LED-2 LED-3
power | | |
source | | |
| resistor-1 resistor-2 resistor-3
| | | |
| | | |
+----------+-----------+-----------+--------------


But if all the LEDs were of the same voltage and amperage
(e.g., if they were all of the typical 3.6V 20mA),
isn't there a simpler way to do this by using only ONE resistor?
In other words, can't it be done like the following ?


+--------------+-----------+-----------+---
| | | |
| | | |
power LED-1 LED-2 LED-3
source | | |
| | | |
| | | |
+---resistor---+-----------+-----------+---


And what would be the proper value of the resistor?

(I'm guessing it would be one third of what it was in the first diagram
because the required voltage reduction would be the same but the
effective current draw of the 3-LED assembly would be three times what it
was before.
So if the power source were 6V and the LEDs were 3.6V and 20mA each,
then the required resistance would be 120 ohms in the first case
and 40 ohms in the second case. Correct?)

 

[CWatters]

"wylbur37" <wylbur37nospam@yahoo.com> wrote in message
news:8028c236.0407310039.45737ad0@posting.google.com...

When using an LED from a power source whose voltage is higher than the
rating of the LED, a resistor is typically used.

.... and if the voltage is the same as the "rating of the LED" you have a
slight problem (eg you can't connect them without a resistor).

If more than one LED is to be used (let's say three),
each one would have its own respective resistor,
and the schematic would look something like this ...


+----------+-----------+-----------+--------------
| | | |
| | | |
| LED-1 LED-2 LED-3
power | | |
source | | |
| resistor-1 resistor-2 resistor-3
| | | |
| | | |
+----------+-----------+-----------+--------------


But if all the LEDs were of the same voltage and amperage
(e.g., if they were all of the typical 3.6V 20mA),
isn't there a simpler way to do this by using only ONE resistor?

Well if the power supply voltage is high enough you can connect the LED in
series and use one resistor in series with the lot.

In other words, can't it be done like the following ?


+--------------+-----------+-----------+---
| | | |
| | | |
power LED-1 LED-2 LED-3
source | | |
| | | |
| | | |
+---resistor---+-----------+-----------+---

Not a good idea because if the LED are not matched exactly you get a big
difference in brightness. They may also age differently - so it may start
out ok but in 3 years time one might be brighter than another.

Think about what happens if one LED is 3.59 and the other 3.61V

So if the power source were 6V and the LEDs were 3.6V and 20mA each,
then the required resistance would be 120 ohms in the first case
and 40 ohms in the second case. Correct?)

Colin