Chip with simple program for Toy

An easy way, use an in-ear earphone, or other earphone. They've got a
coil of wire inside to pick up EMI, and a jack plug on the end. Tho
they're comparitively lo-impedance but, it's an experiment innit?

Like others have said, put a bit of plasticine over the sound holes.

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

if love is a drug, then, ideally, it's a healing, healthful drug... it's
kind of like prozac is supposed to work (without the sexual side
effects and long-term damage to the brain and psyche)

 

[jalbers@bsu.edu]

I just want to make sure that I am interpreting what you are saying
correctly. The total work that the battery is doing to fully charge
the capacitor with a series resistor is going to be CV^2 not 1/2 CV^2
where V is the voltage across the capacitor or in this case will be the
same as the battery.

The total work that the battery is doing to charge the capacitor to
86.5% of the battery voltage (2 RC time constants) with a series
resistor is going to be C(V*.865)^2 not 1/2 C(V*.865)^2 where V is the
voltage across the capacitor or in this case will be 86.5% of the
battery voltage.

I think I get a slightly different result when calculating the energy
dissipated by the resistor between t=0 and t=2*r*c

Integral(0, 2RC, Vbattery^2/R exp(-2t/RC) dt = (Vbattery^2 C)/2
(1-exp(-4)) which works out to be about .4908421806 Vbattery^2 C

So according to this the total work done by the battery after 2 RC
constants would be .4908 Vbattery^2 C + .5 (Vbattery*.865)^2 C

Is this correct?

 

[Joerg]

Hello Jack,

I had contemplated citing various regulatory requirements but I
thought the OP was after practical reasons.

Yep, you are very true...here. I did wanted to know the practical
reasons.

In other countries you have regulations similar to what the FCC does in
the US. They are just called differently, for example in Germany the
agency's name would be RegTP. EMI limits aren't very different. That's
the practical reason.

As for your question why not LF: LF circuits usually do not radiate
higher frequency harmonics. The ones over 30MHz are the main concern
here. Even a regular digital circuit can generate enough trouble up
there to require at least some shielding.

Regards, Joerg

http://www.analogconsultants.com

 

[Bob Monsen]

On Tue, 08 Nov 2005 07:02:47 -0800, jalbers@bsu.edu wrote:

I think I get a slightly different result when calculating the energy
dissipated by the resistor between t=0 and t=2*r*c

Integral(0, 2RC, Vbattery^2/R exp(-2t/RC) dt = (Vbattery^2 C)/2
(1-exp(-4)) which works out to be about .4908421806 Vbattery^2 C

So according to this the total work done by the battery after 2 RC
constants would be .4908 Vbattery^2 C + .5 (Vbattery*.865)^2 C

Is this correct?

Actually, I got the integration wrong; I was using the voltage across the
cap, rather than the voltage across the resistor.

I believe you are correct about the energy at t=2rc.

---
Regards,
Bob Monsen

Beyond the natural numbers, addition, multiplication, and mathematical
induction are intuitively clear.
- Luitzen Brouwer (1881-1966) (intuitionist)

 

[Ben Miller]

"JosephKK" <joseph2k@lanset.com> wrote in message
news:OFXcf.25905$6e1.22829@newssvr14.news.prodigy.com...

For each 180 degrees of bend you must place a pull box, NEC requirement.
--
JosephKK

Can you cite a reference to the Code that says 180? It is 360 degrees
according to 342.24, 344.26, and others.

Ben Miller
--
Benjamin D. Miller, PE
B. MILLER ENGINEERING
www.bmillerengineering.com

 

[The Real Tom]

On Fri, 11 Nov 2005 07:23:26 GMT, JosephKK <joseph2k@lanset.com>
wrote:

DaveC wrote:

Four AWG #1 conductors in 2-1/2 inch EMT. 50 ft of conduit, four 90 deg.
elbows.

Is this combination reasonable to pull with some lubricant? Or should I
plan some "pull boxes" in this run?

Thanks,
For each 180 degrees of bend you must place a pull box, NEC requirement.

I think most people do that, but the code might say something about
360.

 

[Bud--]

JosephKK wrote:

DaveC wrote:


Four AWG #1 conductors in 2-1/2 inch EMT. 50 ft of conduit, four 90 deg.
elbows.

Is this combination reasonable to pull with some lubricant? Or should I
plan some "pull boxes" in this run?

Thanks,

For each 180 degrees of bend you must place a pull box, NEC requirement.

The NEC allows 360 degrees of bends

bud--

 

Noozer wrote:

These should be fairly simple circuits, but for some reason I can't get my
head around them. Probably do to not knowing what parts out there are
available and what they're called.

First circuit is simple...

My PC has a built in speaker for the soundcard. If I plug
speakers/headphones/etc into the line out port on the PC, the internal
speaker is turned off. Is it possible to keep the internal speaker working
when something is plugged into the line out port?

solder wires to the socket connections on the board, and add a socket
on end of wires. Then the jack switch isnt activated. Support the
mechanics of this properly, or one tug may rip your PCB traces off.

I want to plug in an
extension cable so I can plug headphones into the PC without having to climb
behind my desk. It would be great if the internal speaker were disabled when
I have headphones plugged into the extension cable.

hmm, can be done but you'll need to extend the switch pieces on the
card's socket and use an exntension socket with similar switching.
Might not be the easiest thing to find.

If it matters, this is an integrated SoundMax soundcard on a Dell PC.

Second circuit is a bit more complex...

I have two PC's and one set of quad speakers. One PC has quad output and the
other is just stereo. I plan on building a switchbox that sits on my desk to
control what audio goes where. What I'd like to be able to do is:
- switch the second pair of speakers between the secondary output on PC1
and the stereo output on PC2.

easy nuff, using a 4 pole 2 way switch.

- Plug headphones into a jack on the switchbox and have it disable the
speakers associated with it. There would be a toggle switch to choose
between the front and read of the quad speakers.

suggest using a separate switch to disable speakers. Parts are much
easier to get, and it gives you more control/options.

Any help out there???

xposted to seb


NT

 

[w_tom]

Another way of accomplishing same is to phase lock the
oscillator to AC mains 60 Hz; create the 32 Khz signal from
that phase lock loop. There are other problems that must be
address with this solution including some technical
knowledge. But it does solve the accuracy problem.

Chris Jones wrote:

Others have already supplied you with plenty of possible reasons why the
existing crystal oscillator might have poor stability.

If you have some time to work on this, you could improve the PC's clock by
overdriving the internal crystal oscillator circuit with your own stable
oscillator. Most PCs would use a 32768Hz crystal. The oscillator circuit
typically used to be made from a CMOS unbuffered logic inverter chip which
you should be able to find the datasheet for, using google. Some PCs
probably have the oscillator built into the chipset instead, but the
circuit is probably the same so you should still be able to overdrive it.

I would expect that one pin of the crystal will be connected to the output
of a logic gate and the other pin of the crystal will be connected to the
input of a logic gate. There may be various resistors or capacitors in
series too. If you can create a stable 32768Hz square wave and feed it to
the input of the logic gate which is connected to one pin of the crystal
then this should override the oscillator with your externally supplied
signal. You will have to keep the signal running even when the computer is
off or otherwise the clock will stop.

You might be able to build yourself a stable crystal oscillator, or you
might prefer to buy one.

A temperature compensated crystal oscillator or an ovenised oscillator
should be accurate enough. There are many commercial suppliers of these.

Chris

 

Unconnected inputs are much more likley to read high than low; in some
logic families it's almost reliable.

Some of the parallel port signals are inverted between their electrical
levels on the bits accessable to software.

 

[Peter Andersen]

<cs_posting@hotmail.com> wrote in message
news:1131985487.532763.4780@g49g2000cwa.googlegroups.com...

Unconnected inputs are much more likley to read high than low; in some
logic families it's almost reliable.

I suspect it has something to do with a pull-up resistor inside the port.
Is it possible to force the unconnected input-pin to logic 0 and logic 1
when a voltage source is connected (in this case +5V)?

 

[Roger Hamlett]

"Peter Andersen" <peter@invalid.invalid> wrote in message
news:dla8rp$qin$1@news.net.uni-c.dk...

Hi,

I am trying to capture an input with a status pin on the standard IBM
parallel port.

I looks like all the status pins are set high default when there's no
external voltage source connected.
If I measure the voltage drop across pin 13 (status 4) and pin 25
(ground) I get 5,1 V. Shouldn't it be zero?

I am using FreeBSD and if I read the value of the status pins with
nothing connected I get
Status: 0x7f
Nothing happens if I connect a 5 V voltage source to pin 13 and pin 25.

If I short circuit pin 13 and pin 25 I get
Status: 0x6f

Isn't it supposed to be the other way around or am I doing something
wrong?

Best regards.
Undriven inputs, will 'float' to levels, depending on the logic family

involved. Historically, the TTL inputs used on printer ports, had a small
internal current, that drove them high. If you look at the equivalent
circuit for the original inputs, they have a resistance of about 4KR,
between the input, and the 5v supply rail. Latter chips using CMOS logic,
still emulate the original inputs. Note that the connection to pin 11, is
inverted, while the other default status inputs are non-inverted, which
allows a 'disconnected' status to be detected (this line is called
'Busy/Off Line' for this reason).

Best Wishes

 

[John - kd5yi]

"Peter Andersen" <peter@invalid.invalid> wrote in message
news:dla8rp$qin$1@news.net.uni-c.dk...

Hi,

I am trying to capture an input with a status pin on the standard IBM
parallel port.

I looks like all the status pins are set high default when there's no
external voltage source connected.
If I measure the voltage drop across pin 13 (status 4) and pin 25
(ground) I get 5,1 V. Shouldn't it be zero?

I am using FreeBSD and if I read the value of the status pins with
nothing connected I get
Status: 0x7f
Nothing happens if I connect a 5 V voltage source to pin 13 and pin 25.

If I short circuit pin 13 and pin 25 I get
Status: 0x6f

Isn't it supposed to be the other way around or am I doing something
wrong?

Best regards.

http://www.lvr.com/files/ibmlpt.txt
http://www.lvr.com/jansfaq.htm

 

[Meindert Sprang]

"Peter Andersen" <peter@invalid.invalid> wrote in message
news:dla8rp$qin$1@news.net.uni-c.dk...

Hi,

I am trying to capture an input with a status pin on the standard IBM
parallel port.

I looks like all the status pins are set high default when there's no
external voltage source connected.
If I measure the voltage drop across pin 13 (status 4) and pin 25 (ground)
I
get 5,1 V. Shouldn't it be zero?

I am using FreeBSD and if I read the value of the status pins with nothing
connected I get
Status: 0x7f
Nothing happens if I connect a 5 V voltage source to pin 13 and pin 25.

If I short circuit pin 13 and pin 25 I get
Status: 0x6f

Isn't it supposed to be the other way around or am I doing something
wrong?

The printer port has rather strong pull-up resistors to VCC, so yes, it is
correct that you measure 5 V on the inputs and read all ones when left open.

Meindert

 

[Sjouke Burry]

Peter Andersen wrote:

Hi,

I am trying to capture an input with a status pin on the standard IBM
parallel port.

I looks like all the status pins are set high default when there's no
external voltage source connected.
If I measure the voltage drop across pin 13 (status 4) and pin 25 (ground) I
get 5,1 V. Shouldn't it be zero?

I am using FreeBSD and if I read the value of the status pins with nothing
connected I get
Status: 0x7f
Nothing happens if I connect a 5 V voltage source to pin 13 and pin 25.

If I short circuit pin 13 and pin 25 I get
Status: 0x6f

Isn't it supposed to be the other way around or am I doing something wrong?

Best regards.


Printer port input and outputs act like ttl , and the status pins

are normally high,because older printers could just use a switch
to pull the line to ground(no supply/electronics needed).

 

[Bob Monsen]

On Mon, 14 Nov 2005 17:27:51 +0100, Peter Andersen wrote:

cs_posting@hotmail.com> wrote in message
news:1131985487.532763.4780@g49g2000cwa.googlegroups.com...
Unconnected inputs are much more likley to read high than low; in some
logic families it's almost reliable.

I suspect it has something to do with a pull-up resistor inside the port.
Is it possible to force the unconnected input-pin to logic 0 and logic 1
when a voltage source is connected (in this case +5V)?

Yes. You use a 'pull-down' resistor to gnd, and then a voltage source to
overcome that.

Pullups on PC parallel ports can be as small as 1k, so I'd go for a 220
ohm to ground. However, you can measure it easily by attaching a known
value resistor r to ground, and then Rpullup = r*(5/vmeasured - 1)

Then, choose a pull-down resistor that pulls it down enough to be seen as
a 0 by the port input driver.

---
Regards,
Bob Monsen

Even the greatest of creations start from small seeds.
- Unknown

 

[Chris Jones]

Peter Andersen wrote:

Hi,

I am trying to capture an input with a status pin on the standard IBM
parallel port.

I looks like all the status pins are set high default when there's no
external voltage source connected.
If I measure the voltage drop across pin 13 (status 4) and pin 25 (ground)
I get 5,1 V. Shouldn't it be zero?

I am using FreeBSD and if I read the value of the status pins with nothing
connected I get
Status: 0x7f
Nothing happens if I connect a 5 V voltage source to pin 13 and pin 25.

If I short circuit pin 13 and pin 25 I get
Status: 0x6f

Isn't it supposed to be the other way around or am I doing something
wrong?

Best regards.

TTL inputs tend to float high when they are not connected. The newer chips
would be made to act like the old TTL chips too. This might explain what
you see.

 

[scada]

"Peter Andersen" <peter@invalid.invalid> wrote in message
news:dla8rp$qin$1@news.net.uni-c.dk...

Hi,

I am trying to capture an input with a status pin on the standard IBM
parallel port.

I looks like all the status pins are set high default when there's no
external voltage source connected.
If I measure the voltage drop across pin 13 (status 4) and pin 25 (ground)
I
get 5,1 V. Shouldn't it be zero?

I am using FreeBSD and if I read the value of the status pins with nothing
connected I get
Status: 0x7f
Nothing happens if I connect a 5 V voltage source to pin 13 and pin 25.

If I short circuit pin 13 and pin 25 I get
Status: 0x6f

Isn't it supposed to be the other way around or am I doing something
wrong?

Best regards.

No, it's working as expected. The status port is 5 bits, S7 is normally low
(inverted), S6, S5, S4, & S3 are all normally high. The open circuited byte
in binary would look like 01111111 , or 0x7F. The 3 least significant bits
(LSB S0-S2) are not there, they will allways be high. When you grounded Pin
13 (S4) you got 01101111 or, 0x6F. To change bit S7 you need to pull it high
to +5V (through a resistor - 10K should do it). To change bits S6-S3 you
pull them low (to ground, negative return). Good luck.

 

[Andrew Holme]

Peter Andersen wrote:

Hi,

I am trying to capture an input with a status pin on the standard IBM
parallel port.

I looks like all the status pins are set high default when there's no
external voltage source connected.
If I measure the voltage drop across pin 13 (status 4) and pin 25
(ground) I get 5,1 V. Shouldn't it be zero?

I am using FreeBSD and if I read the value of the status pins with
nothing connected I get
Status: 0x7f
Nothing happens if I connect a 5 V voltage source to pin 13 and pin
25.

If I short circuit pin 13 and pin 25 I get
Status: 0x6f

Isn't it supposed to be the other way around or am I doing something
wrong?

Best regards.

It is working correctly. The inputs have pull-up resistors to 5V. They are
designed to be pulled down by open-collector or open-drain outputs e.g.


VCC
+
|
|
.-.
| | Pull-up
| | Resistor
'-'
|
Pin |
13 | |\
Apply .-----> >------o------| >O-
your | |/
5V |
here ___ |/

----|___|----| NPN
|

10K |
|
===
GND

View in fixed-pitch font.
created by Andy´s ASCII-Circuit v1.24.140803 Beta www.tech-chat.de

 

[DaveM]

"Peter Andersen" <peter@invalid.invalid> wrote in message
news:dla8rp$qin$1@news.net.uni-c.dk...

Hi,

I am trying to capture an input with a status pin on the standard IBM
parallel port.

I looks like all the status pins are set high default when there's no
external voltage source connected.
If I measure the voltage drop across pin 13 (status 4) and pin 25 (ground)
I get 5,1 V. Shouldn't it be zero?

I am using FreeBSD and if I read the value of the status pins with nothing
connected I get
Status: 0x7f
Nothing happens if I connect a 5 V voltage source to pin 13 and pin 25.

If I short circuit pin 13 and pin 25 I get
Status: 0x6f

Isn't it supposed to be the other way around or am I doing something
wrong?

Best regards.

The status inputs on most IBM parallel ports are TTL (or derivatives
thereof) and are active low. This means that an unterminated device input
will float to a logical "1" unless specifically brought to a low state by
external circuitry. A TTL "1" is a voltage of 2.4V or greater (up to a max
of 5V).

Your port and your software are working properly.

--
Dave M
MasonDG44 at comcast dot net (Just substitute the appropriate characters in
the address)

Never take a laxative and a sleeping pill at the same time!!