Chip with simple program for Toy

[ronwer]

Then you want to find a copy of "The Invention That Changed the World" by
Robert Buderi), which is about radar development during WWII, but also
about how that need for radar pushed electronics further (and how
that push helped electronics to move along after the war).

Thanks, I will try Amazon!

You have it wrong, silicon diodes weren't used in radar, radar caused
development of semiconductor diodes so they could be used in radar.

They had to use increasingly higher frequencies to get the fine
detail they needed/wanted, and that stretched the limit of vacuum
tubes at the time. Hence they flashed back to the early days of
radio, looked at "cat's whisker" detectors, and compacted that
into more reliable and small devices. Gone were the finicky adjustment
for the hot spot, they got it right once and then sealed it all up.

But, those early diodes were point contact like the cat's whisker
detectors, and as far as I know, they were germanium.

I read the book "The entangled history of silicon" where they mention the
use of silison in diodes for use in radar. The silicon was imported from the
USA (was it from Bell Laboratories? Don't remember right now and I don't
have the book here).

Indeed, those were point contact diodes.

But they were seeing use not as detectors like in "crystal radios"
but as mixers to get the microwave frequencies down to where tubes
could amplify the signals.

It sure seems like you are looking in the wrong direction, expecting
semiconductor diodes to exist before the war, when really the war
caused them to be developed.

No, they existed before the war too, but it was the demand for better radar
(i.e. higher frequencies) that speeded up the development.

See for instance: http://www.avtechpulse.com/faq.html/IX/

Scratch any piece of equipment that did use semiconductor diodes
in WWII, and you are most likely to find that the development of
the equipment involved development of semiconductor diodes.

After the war you see trends moving away from that scenario, where
general diodes were developed independent of specific use, which in
turn caused electronics to move forward which also in turn caused
the need for semiconductors to develop.

The case has been made that the development of the transistor was
based on that work on semiconductor diodes during the war.

What I would be interested in is as follows:

-type numbers of the diodes
-name/type number of radar/communication equipment
-technical infor on those systems
-info on producers
-pictures of actual diodes, also "in" the circuits
-anecdotal stories about the actual use
-anything else!

The information will be used for an on-going study project related to
practical application of minerals (i.e. quartz) in industry and
technology.

So, since this is an aspect of a broader study, other quartz-related info
would
be most appreciated, especially about early use of piezoelectric
quartz crystals in electronic equipment.

Then you've completely missed the one thing that people will think of
when they think "quartz".

Quartz is used in crystals, ie frequency determining elements. Going
back to at least the 1920s. Blanks of quartz sliced thin and then
ground to resonate at a certain frequency. INitially not that much
more than a novelty, then it saw a lot of use, and it continues today,
even though nowadays it's levelled off as frequency synthesis allows
a single crystal to generate multiple frequencies, unlike WWII or even
into the early seventies where you needed a single crystal for every
frequency you wanted to generate (some of that WWII equipment was
loaded with crystals).

Indeed, you don't hear much about quartz used to make semiconductor
devices, so I'm not even sure if your premise on that account is
correct.

Quartz is extensively used in semiconductor devices as isolator between the
individual components.

But I am basically interested in anything related to the use of quartz in
technology, and therefor also in silicon. They belong together.

Thanks for your reply!

Best regards,

Ronald
Norway

 

[Phil Allison]

"Bob Masta"
"Phil Allison"

The characteristic rise in published THD curves at power levels below
about
1 watt is almost * entirely * due to supply frequency hum and wide band
random noise.

NOT any mysterious x-over region effects !!!!!!!!!!


(I'm not sure why you are calling crossover distortion "mysterious"...
I suspect it is the main source of the residual distortion at the
bottom of the valley around 1 watt or so.)

** The crossover region in most SS amps is around the 10 - 20 mW power
level - a bias setting of 25mA allow class A operation up to 50 mA peak
at 8ohms = 10 mW .

10 mW is 40 dB below 100 watts and so 40 dB below the quoted s/n ratio for
such an amp.

Allow the measured s/n ratio to be 90dB unweighted, then the s/n at 10 mW is
only 50 dB and -50 dB equates to 0.3 %.

The distortion residual at 10 mW output ( as seen on a scope after removing
the fundamental) is usually much lower than 0.3 %.

QED.


........ Phil

 

[Phil Allison]

"Michael Black"

If you have the regulator right next to the large filter capacitor
of a power supply, the input capacitor may not be needed, since the filter
capacitor might do the job of keeping the regulator from oscillating. Or
it might not, since the filter capacitor is there to filter the 60Hz, and
is so large in value that the accumulated inductance completely negates
its value at higher frequency where it would be need to stop oscillation.

** Oh NOOOOOOOOO !!!!!

Not that " all electros are inductive "

FUCKING BULLSHIT AGAIN !!

It's TOTAL BOLLOCKS !!!!


All electros are NON INDUCTIVELY wound.

Go look at the description of one - you ass.



....... Phil

 

[Mr.T]

"Don Bowey" <dbowey@comcast.net> wrote in message
news:C42620BC.B505A%dbowey@comcast.net...

On 4/11/08 6:23 PM, in article
70fb42ae-4253-4427-9d78-95180652e8de@y21g2000hsf.googlegroups.com,
"dpierce.cartchunk.org@gmail.com" <dpierce.cartchunk.org@gmail.com> wrote:

On Apr 11, 7:46 pm, Don Bowey <dbo...@comcast.net> wrote:
On 4/11/08 3:15 PM, in article lh45d5-tt5....@radagast.org, "Dave
Platt"

No "sampling" takes place and nothing digital occurred.

Well, it's "sampled" to the extent that the varying magnetic field,
sensed by the pickup head, consists of the sum of the individual
magnetic fields (vector and magnitude) of the large number of
individual magnetic domains within the ferromagnetic recording layer.

Using the same logic, you would argue that a microphone samples the
sound
that reaches it. That's a stretch beyond common usage.

If, by "sampling" you mean "quantized, then, yes indeed,
that's precisely what happens in a microphone, and yes,
it is a stretch beyond common usage because it's
common usage that's wrong.

No, I did not mean quantize, and that is not what happens in a microphone.

Of course it does.

A microphone "converts" sound pressure to a voltage, or to a change in
capacitance, or into a change in inductance, etc.

The pickup head in an audio tape recorder converts a moving magnetic field
to a voltage.


The total force on any surface, inluding the diaphragm of
a microphone and your ear drum is the net result of individual
discrete collisions of air molucules with that surface. Each
collision is most assuredly discrete. The net force looks
continuous only because we whoose to integrate it over
sufficiently long averaging time, but it is discrete whether
your common usage embraces it or not.

The microphone responds to the combined effects of all the input forces.

Of course, but do you believe electricity is linear below the single
electron level, assuming we could even record or measure to that level?
Do you believe thermal and other noise does not exist in a microphone
signal?
Do you not believe there are finite limits to the linearity of any
electrical device?

I suspect your set of lexicons includes a belief there is no such thing as
a
Direct Current Voltage.

Your right, only a voltage at any given instant, which may be fairly
constant for a given period of time.

And even disregarding that level of quantization, the fact
that one might make the statement that a medium is
"continuous," it does not then follow that the medium
has the properties of infinite resolution as if there were
no quantization going on.

My ear may quantize the result, but the microphone doesn't.

It's obvious you have no idea what "quantise" means in a technical sense. In
this universe everything has finite limits. Analog recordings are no
different, nor the signal from a microphone, or even the variations in air
pressure we call sound.

MrT.

 

[Paul E. Schoen]

"Michael Black" <et472@ncf.ca> wrote in message
newsine.LNX.4.64.0804121653550.18651@darkstar.example.org...

On Sat, 12 Apr 2008, CS wrote:

Evening,

I am looking at STI Microelectronics L78Sxx 2A positive voltage
regulars to reduce 24 - 31v DC from a boat battery to varying DC
voltages.
It appears that these can be used with no external components, but a
diagram shows an output capacitor of 0.1uF to improve transient
response and an input capacitor of 0.33uF if the regulator is an
appreciable distance from poer supply filter.

Are these really needed and what type of cap?? - I assume a small
ceramic would do.

Yes, they are really needed.

When three terminal regulators were knew, it was easy to dismiss
the capacitors, "why do we need them, it's just a voltage regulator?".
But then they were really great oscillators without the proper bypassing.
Everyone soon learned the necessity of those capacitors.

If you have the regulator right next to the large filter capacitor
of a power supply, the input capacitor may not be needed, since the
filter
capacitor might do the job of keeping the regulator from oscillating. Or
it might not, since the filter capacitor is there to filter the 60Hz, and
is so large in value that the accumulated inductance completely negates
its value at higher frequency where it would be need to stop oscillation.

The old National databook would always list three values for that
input capacitor, depending on the type of capacitor used. The actual
value was less important than the capacitor's effectiveness at bypassing
the higher frequencies.

The 0.1uF on the output would be a ceramic capacitor.

I don't seem to recall having problems with a .1uF ceramic capacitor
on the input of a 3 terminal regulator, but I'm too lazy to dig out
the National databook to find out exactly what they said.

Michael

It's certainly easy and cheap enough to just put a small bypass cap of 0.1
to 1 uF at the input close to the pins. It's also a good idea to have some
sort of output capacitor to handle brief current surges of the load. But if
the output capacitor is much larger than the input capacitor, and
especially if you disconnect the regulator from the main filter capacitor,
it's a good idea to put a diode from output to input to avoid back biasing
the regulator, where the output voltage is higher than the input (by more
than a diode drop).

Paul

 

[Phil Allison]

"BobG"

Yo Phil.... theoretical question.... I have two hi quality sine
generators and a THD meter. If I set generator 1 at 1V 1KHz and
generator 2 at .03V 3khz, thats the only 'distortion component', 3rd
harmonic thats 3% of the fundamental. I guess I could square it and
sqrt it, but since thats the only harmonic, would it show 3% THD on
the meter? Thanks.

** You are no longer talking about waveform distortion - but a spurious
signal.

Once the 1kHz wave is removed, the 3 kHz is left alone.

So it will give a 3% reading, like you say.

But so would some 60 Hz hum or a supersonic signal.



....... Phil

 

[isw]

In article <C42620BC.B505A%dbowey@comcast.net>,
Don Bowey <dbowey@comcast.net> wrote:

On 4/11/08 6:23 PM, in article
70fb42ae-4253-4427-9d78-95180652e8de@y21g2000hsf.googlegroups.com,
"dpierce.cartchunk.org@gmail.com" <dpierce.cartchunk.org@gmail.com> wrote:

On Apr 11, 7:46 pm, Don Bowey <dbo...@comcast.net> wrote:
On 4/11/08 3:15 PM, in article lh45d5-tt5....@radagast.org, "Dave Platt"

No "sampling" takes place and nothing digital occurred.

Well, it's "sampled" to the extent that the varying magnetic field,
sensed by the pickup head, consists of the sum of the individual
magnetic fields (vector and magnitude) of the large number of
individual magnetic domains within the ferromagnetic recording layer.

Using the same logic, you would argue that a microphone samples the sound
that reaches it. That's a stretch beyond common usage.

If, by "sampling" you mean "quantized, then, yes indeed,
that's precisely what happens in a microphone, and yes,
it is a stretch beyond common usage because it's
common usage that's wrong.

No, I did not mean quantize, and that is not what happens in a microphone.

A microphone "converts" sound pressure to a voltage, or to a change in
capacitance, or into a change in inductance, etc.

The pickup head in an audio tape recorder converts a moving magnetic field
to a voltage.


The total force on any surface, inluding the diaphragm of
a microphone and your ear drum is the net result of individual
discrete collisions of air molucules with that surface. Each
collision is most assuredly discrete. The net force looks
continuous only because we whoose to integrate it over
sufficiently long averaging time, but it is discrete whether
your common usage embraces it or not.

The microphone responds to the combined effects of all the input forces.

I suspect your set of lexicons includes a belief there is no such thing as a
Direct Current Voltage.

There isn't -- unless you measure it for an infinite time.

And a related question: What is the low-frequency response limit of a DC
amplifier?

Isaac

 

[Richard Seriani]

"Bill Bowden" <wrongaddress@att.net> wrote in message
news:fe9e1f0a-0029-4df9-9b29-c9583ca49075@r9g2000prd.googlegroups.com...

I have a stepper motor from Airpax, modle C82710 that has 6 wires
connected to 2 windings with a center tap. It's rated at 12 volts, 36
ohms from center tap to the end of the winding, and 7.5 degrees per
step.

I'm not too familiar with stepper motors, but understand the shaft can
be moved in either direction one step at a time. I read some articles
on Google but couldn't find any that showed the necessary signals and
timing to rotate the motor continously in one direction.

I played around with it, and found I could move the motor shaft one
step at a time by just alternating the connection to either side of
one of the windings with the center tap common. But it only goes so
far and stops, and the second winding was not being used.

So, the question is, what is a proper polarity and timing sequence on
the various connections to continuously move the motor in the same
direction?

-Bill

Bill,

Here are a couple of links that may help. The first will show the drive
sequence required for unipolar and bipolar operation (among other things);
the second will show some methods of doing it.

Lots more on the web. Many manufacturers of driver IC's if you don't want to
use transistors to drive the motors. Also, much good info about controlling
these with microcontrollers or discrete logic.

http://www.shinano.com/xampp/stepper-motors.php
http://www.techlib.com/electronics/stepper.html

Have fun playing with these.

Richard

 

[Calab]

| Astronauts come back a few seconds younger. So, what's 200uS back in
time?


Actually, they don't come back younger. They just come back less older than
everyone else.

 

[Bob Masta]

On Sun, 13 Apr 2008 10:49:15 +1000, "Phil Allison"
<philallison@tpg.com.au> wrote:

"Bob Masta"
"Phil Allison"

The characteristic rise in published THD curves at power levels below
about
1 watt is almost * entirely * due to supply frequency hum and wide band
random noise.

NOT any mysterious x-over region effects !!!!!!!!!!


(I'm not sure why you are calling crossover distortion "mysterious"...
I suspect it is the main source of the residual distortion at the
bottom of the valley around 1 watt or so.)


** The crossover region in most SS amps is around the 10 - 20 mW power
level - a bias setting of 25mA allow class A operation up to 50 mA peak
at 8ohms = 10 mW .

10 mW is 40 dB below 100 watts and so 40 dB below the quoted s/n ratio for
such an amp.

Allow the measured s/n ratio to be 90dB unweighted, then the s/n at 10 mW is
only 50 dB and -50 dB equates to 0.3 %.

The distortion residual at 10 mW output ( as seen on a scope after removing
the fundamental) is usually much lower than 0.3 %.

QED.

Got it! Thanks, Phil. For decades I have apparently been laboring
under a misconception that there was always some residual crossover
distortion, but now that you point it out it seems totally
unjustified... and I'm more than glad to be rid of it!

Best regards,


Bob Masta

DAQARTA v3.50
Data AcQuisition And Real-Time Analysis
www.daqarta.com
Scope, Spectrum, Spectrogram, FREE Signal Generator
Science with your sound card!

 

[Calab]

"W. Watson" <wolf_tracks@invalid.com> wrote in message
news:lhuMj.1419$7Z2.1317@newssvr12.news.prodigy.net...

I bought the dvd player in the Subject. It has two sockets called A/V
in/out and coaxial. the take what looks like the same plug. I can take the
RCA 3 plug to a single plug and put it in either socket. If I do this with
my Panasonic dvd player I cannot see any image on the portable's screen. I
do this to checkout the Panasonic, which has TV to connect it to. What's
the difference between the two sockets, and particularly what is coax
really about? A socket that's both in and out?

See specs at
http://reviews.cnet.com/portable-dvd-players/lg-dp781-portable-8/4507-6498_7-32466958.html
and general info at
http://reviews.cnet.com/portable-dvd-players/lg-dp781-portable-8/4505-6498_7-32466958.html>.

Did you read the review that you posted the link to?

Coax is the digital audio output.

The two A/V plugs are most likely composite video out and composite video
in.

 

[geoff]

Michael R. Kesti wrote:

Don Bowey wrote:

It's obvious you are an argumentative idiot who would like to count
angels on the head of a pin.

Regardless of the topic of discussion or the correctness of your
position, Don, you just lost the debate.

Must be very satisfying for the sad fuck who started this thread to see his
profound achievement...

geoff

 

[geoff]

Michael R. Kesti wrote:

Don Bowey wrote:

It's obvious you are an argumentative idiot who would like to count
angels on the head of a pin.

Regardless of the topic of discussion or the correctness of your
position, Don, you just lost the debate.

Must be very satisfying for the sad fuck who started this thread to see his
profound achievement...

geoff

 

[Mr.T]

"Don Bowey" <dbowey@comcast.net> wrote in message
news:C426BB21.B521A%dbowey@comcast.net...

It's obvious you have no idea what "quantise" means in a technical
sense. In
this universe everything has finite limits. Analog recordings are no
different, nor the signal from a microphone, or even the variations in
air
pressure we call sound.


It's obvious you are an argumentative idiot who would like to count angels
on the head of a pin.

Is that your pathetic way of admitting you are wrong then?
(in my opinion there will also be a finite number of angels on the head of a
pin, if any

MrT.

 

[Bob Eld]

"Ante Maljkovic" <antem_makniovo@post.htnet.hr> wrote in message
news:fu0b3h$ba$2@ss408.t-com.hr...

Do you know for replacement transistor that I can use instead of 2N1711
transistor, which I can't find anywhere near me...

Thanks...

Try a 2N2219A.

 

[Bob Eld]

"Ante Maljkovic" <antem_makniovo@post.htnet.hr> wrote in message
news:fu0b3h$ba$2@ss408.t-com.hr...

Do you know for replacement transistor that I can use instead of 2N1711
transistor, which I can't find anywhere near me...

Thanks...

Try a 2N2219A.

 

[Bob Eld]

"Eeyore" <rabbitsfriendsandrelations@hotmail.com> wrote in message
news:4803BD36.98CE285@hotmail.com...

Bob Eld wrote:

"Ante Maljkovic" <antem_makniovo@post.htnet.hr> wrote

Do you know for replacement transistor that I can use instead of
2N1711
transistor, which I can't find anywhere near me...

Try a 2N2219A.

Is that likely to be any more available ?

The voltage rating is somewhat lower too.

Graham

What, You want it availavle too? Boy some people want everything. BTW, a
2N2219A is in the April 2008 Mouser catalog, pg 478.

 

[Phil Allison]

<chesemonkyloma@gmail.com>

My problem is this: You are distributing 100 watts. You start out with
10 volts, 10 amps. You increase the volts by a factor of 2 through a
transformer. Now you must have 20 volts, 5 amps right? Using Ohm's
Law, 10 volts = 10 amps * 1 ohms, in the first case there is 1 ohm.
But, in the next case you have 20 volts = 5 amps * 4 ohms. So why did
the ohms change? When you use a transformer to increase voltage,
doesn't that mean ohms must increase if amps decrease? I am thoroughly
confused. I know I am thinking about it the wrong way, can someone
explain it to me? If you want me to clarify, let me know. Basically my
question is this: How do you create more voltage without creating more
current?

** You are confusing yourself by * failing to separate * load ohms with the
ohms in the current carrying cable going to the load.

High voltage transmission is ALL about reducing the percentage of power lost
in the cables that deliver electrical energy across a country. .

http://en.wikipedia.org/wiki/Electric_power_transmission#Losses



....... Phil

 

[Joel Koltner]

<chesemonkyloma@gmail.com> wrote in message
news:e89f08c4-b769-4600-8fb1-b2bc0adf4658@24g2000hsh.googlegroups.com...

How do you create more voltage without creating more
current?

OK, step back a second and I'll attempt to explain this...

Here's the deal: You're building a (really tiny) power grid, and you'd like to
distribute *up to* that 100 watts you specified. Assume that you've already
decided that your "customer" will be getting 10V... this means they're allowed
to pull anywhere from 0 to 10A, or, in resistance, anything from an infinite
load (open circuit) down to a 1 ohm resistor.

Now, let's figure out how to get that 100W to the customer. First try the
"easy" approach -- just ship them 10V directly! OK, but now let's assume
there's a long power line between you and the customer that has a resistance
of 1 ohm. If your only has a 10V/1W load (100 ohm or 100mA) -- then you now
have a series circuit with 1 ohm resistance in the power line and 100 ohms in
the customer's load. That's 101 ohms total, so the load actually draws
10V/101 ohm = 99mA and dissipates 99ma^2*100 = 980mW -- close enough that the
customer won't complain. But what if they want to plug in a 10V/100W load?
That implies the load's resistance is 1 ohm, but... oh oh... we now have a
series circuit with 2 ohms total resistance and, as you can calculate, the
customer only gets 5A going through their load or an open-circuit voltage of
5V. In many cases, 5V will be far too low to correctly operate what was
designed as a 10V/100W load. Additionally, you can calculate that we're
dumping 25W of power into the power line, and *the customer doesn't pay for
that* (the power meter only monitors their load), so you're just losing a lot
of money if you're the power company.

This problem goes away if we add a transformer to boost the power line voltage
to, say, 1kV. Now, to transport 100W over the line, we only need 100mA rather
than 10A. Worst case, in the power line we dissipate 100mA^2*1 ohm (the power
line resistance) = 10mW -- utterly negligible.

Does this make sense?

The basic idea is that customers get to connect a load *up to* a specified
wattage to your power line and that you want to be sure that, at the maximum
specified load, only a tiny percentage of the total power gets dumped into the
power line and the vast bulk of it goes to the customer.

BTW, the losses in the power lines are referred to as "I-squared-R" losses
since that's how you calculated the power lost in them (I^2*R). Since
doubling the voltage halves the current (for the same load power), the losses
are quartered due to the I^2 term. Nice, huh? 10 times the voltage has 100
times less loss, etc...

---Joel

 

[Paul E. Schoen]

"Ante Maljkovic" <antem_makniovo@post.htnet.hr> wrote in message
news:fu0h66$h16$1@ss408.t-com.hr...

Eeyore wrote:

Ante Maljkovic wrote:

Do you know for replacement transistor that I can use instead of 2N1711
transistor, which I can't find anywhere near me...

Does it need to be in TO-39 metal can ? What's the application ? It's a
truly ancient device !

I guess it doesn't need to be in TO-39 metal can.
It's for repairing some old welding device...

Would that be a "Duffers" current meter? We have been replacing them (in
old circuit breaker test equipment), and there might be some old PC boards
around. I seem to remember the boards bristling with old style transistors.
I might even be able to get hold of a complete working unit, but shipping
would probably be costly.

Paul