Chip with simple program for Toy

[The Flavored Coffee Guy]

Kim,

Don't you think that it is odd that I first mentioned that I didn't
have the money? So, God, does have a sense of humor. There's Dr.
Frank Norton Stein, giving out perscriptions for pain and pulling teeth
for a living, and you asleep at the keyboard.

The Flavored Coffee Guy.



Kim Sleep wrote:
Ok, lotsa typing and cool graphics....so go ahead and build it.
--
Kim..."A Man Of True Frankenstinean Proportions"

 

[Jonathan Kirwan]

On Sun, 06 Feb 2005 04:57:47 GMT, Steven O. <null@null.com> wrote:
If I follow the intent, I get:

dR/R = a dT
ln R + C1 = a*T + C2
ln R = a*T + C2 - C1
ln R = a*T + C

solve C for initial conditions, which is Rref and Tref

C = ln Rref - a*Tref

Plugging back in, we get:

ln R = a*T + ln Rref - a*Tref
ln R = a*T - a*Tref + ln Rref
ln R = a*(T - Tref) + ln Rref
ln R - ln Rref = a*(T-Tref)
ln (R/Rref) = a*(T-Tref)
R/Rref = e^(a*(T-Tref))
R/Rref = e^(a*T)/e^(a*Tref)

or,

R/Rref = A*e^(a*T), where A = e^(-a*Tref)

I think that puts me in rough agreement with you, to this point. Substituting
1+x for e^x:

R/Rref = (1+a*(T-Tref))
R = RRef * (1+a*(T-Tref))

I suppose if there were two different R's, say R1 and R2, then their ratio would
be:

R1/R2 = [RRef * (1+a*(T1-Tref))] / [RRef * (1+a*(T2-Tref))]
R1/R2 = [(1+a*(T1-Tref))] / [(1+a*(T2-Tref))]

That would look about like your teacher's ratio, wouldn't it?

Jon

 

[cor]

Try RadioShack. Look for a female headphone plug.

Chris Lewis wrote:

I want to build a simple cable that will mute the speakers on my computer
when i plug in headphones. I guess i can do it with a 3.5mm plug and a
coule of 3.5mm sockets.

I need some way of connecting them so that one output loses its connection
when the other is connected.

I tried google but didnt find anything usefull. If anyone could point me in
the direction of a circuit diagram and also somewhere good to buy the parts
in the UK it would be really appreciated.

Thanks

Chris

 

[John Fields]

On Wed, 09 Feb 2005 18:54:39 GMT, "Chris Lewis"
<christopher.lewis@tiscali.co.uk> wrote:

I want to build a simple cable that will mute the speakers on my computer
when i plug in headphones. I guess i can do it with a 3.5mm plug and a
coule of 3.5mm sockets.

I need some way of connecting them so that one output loses its connection
when the other is connected.

I tried google but didnt find anything usefull. If anyone could point me in
the direction of a circuit diagram and also somewhere good to buy the parts
in the UK it would be really appreciated.

---
Click on this:

8ouk01tb0gibdff7guts1nokm4hocr5che@4ax.com


--
John Fields

 

[dB]

"Chris Lewis" <christopher.lewis@tiscali.co.uk> wrote in message news:<P%sOd.217$Ha7.54@newsfe2-gui.ntli.net>...

I want to build a simple cable that will mute the speakers on my computer
when i plug in headphones. I guess i can do it with a 3.5mm plug and a
coule of 3.5mm sockets.

I need some way of connecting them so that one output loses its connection
when the other is connected.

I tried google but didnt find anything usefull. If anyone could point me in
the direction of a circuit diagram and also somewhere good to buy the parts
in the UK it would be really appreciated.

Thanks

Chris

CPC carry two 3.5mm switched jack sockets.

Order code CN00651 panel mount
Order code CN00652 p.c.b mount

Use one of those wired so that the signals a.e connected to the
'speakers when no plug is inserted.

Inserting the 'phones plug will disconnect the 'speakers.

 

[Lord Garth]

Spammers go to hell....

 

Would you have any pictures? I built one about 35 years ago, when I
was studying engineering.

Jerry G.
======

 

[John Fields]

On Thu, 10 Feb 2005 06:46:48 GMT, "Lord Garth" <LGarth@Tantalus.net>
wrote:

It looks like a screw up from J.F.'s post. It should be no big deal for the
OP to go
to ABSE and find it, if it is...I'm hiding now!

---
Yup!

I'm using Agent and what I did was cut and paste the message ID from
the header, after I posted to and read it back from abse, into the
post on seb.

Clicking on the raw message ID in Agent opens a dialogue box where one
of the choices is something like "Message ID?", and clicking on that
gets the message it points to. I guess Outlook Express or ??? handles
it differently.

This seems to work without going through the dialogue box:

news://8ouk01tb0gibdff7guts1nokm4hocr5che@4ax.com


--
John Fields

 

[ILYA]

Be careful, some frequency is dangerous for humans...

---
do not forget Tesla "Philadelphia project"

 

[Larry Brasfield]

"R.Spinks" <rspinks1@wowway.com> wrote in message
news:18KdnQUSkNmul43fRVn-qQ@wideopenwest.com...

"John Fields" <jfields@austininstruments.com> wrote in message
news:qqov01lvrh95ek6kddf6q0c17d8fb33l61@4ax.com...
On Sun, 13 Feb 2005 18:21:24 -0500, "R.Spinks" <rspinks1@wowway.com
wrote:

I would like to see the math involved in calculating the current in the
4k
resistor and the current in the 12k resistor. I have tried to do a node
current at the top -right of 1k resistor and top of all parallel items)
which I have called 'node A' (Va) but I keep getting the wrong answer.
There
is 8mA in the 4K and 2.66mA in the 12k (this means the voltage at 'node
A'
must be 32V. (I hand drew the 50mA and 90mA current sources so that's
what
they are supposed to be -- if there was a symbol I missed it). I took the
current at node A (top) to be:

((Va-8)/1k)-50mA - (Va / 12k) - (Va/6k) + 90mA - (Va/4k) = 0 but that
does
not yield 32v. Can someone point an error I've made (as I assume I've set
something up wrong) or correctly indicate the math for me? Thanks.




1K
___
|-|___|-----------------------------------|--------
| | | | | |
| | | | | |
| | .-. .-. |90ma .-.
/+\ | | | | | | | |
8v ( ) 50ma (\|/) 12K | | | | 6K (/|\) | | 4K
\-/ | '-' '-' | '-'
| | | | | |
---------------------------------------------------
(created by AACircuit v1.28.4 beta 13/12/04 www.tech-chat.de)

---
Since the 12K, the 6K, and the 4K resistors are all in parallel, their
total resistance becomes:


1
Rt = -------------------------------- = 2000 ohms
1 1 1
------- + ------- + ------
12000R 6000R 4000R


Now, if we redraw your schematic with that in mind, we'll have:


+8V>--[1K]--+---E2
|
[2K]
|
0V>---------+


and we can say:

8V * 2K
E2 = --------- = 5.333V
1K + 2K




If we now look at the original schematic:


+8V>--[1K]---+------+------+----5.333V
| | |
[12K] [6K] [4K]
| | |
0V>----------+------+------+

we can see that 5.333V will be across all of the parallel resistors,
so the current in any one of them will be:

E
I = ---
R

Thus, for the 12000 ohm case,:

5.333V
I = -------- ~ 0.000444A ~ 444ľA
12000R

for the 6000 ohm case,:

5.333V
I = -------- ~= 0.000888A ~ 888ľA
6000R


and for the 4000 ohm case:

5.333V
I = -------- ~ 0.00133A ~ 1.33mA
4000R


--
John Fields

Thanks for your help, John. I am actually not in school, but this problem is
from a book. The current in the 4k resistor is 8mA not 1.33mA (the solution
is given). Also, the current in the 12k is 2.66mA (not given but power is
given at 85.3mW -- works backwards to 2.66mA at 32v which also establishes
the 8mA in the 4k). In both cases, node A , where you have derived 5.33V is
actually 32V to establish those voltages. Any ideas?

Not counting the 1K resistor, the other resistors are equivalent
to a single 2K resistor. The nodal impedance at Va is therefor
1K || 2K == (2/3)K. The voltage divider from the 8V source
to Va is 2K/(1K+2K) == 2/3. Applying superposition from
the 3 sources to get 3 terms:
Va == 8V * (2/3) + 90mA * (2/3)K - 50mA * (2/3)K
Va == 16/3 V + 80/3 V == 96/3 V == 32 V.

You can get any of the currents in the parallel collection
by applying Ohm's law using 32V and the resistance.

If you are trying to learn this stuff on your own, I have
to applaud the effort. My father did that and, after a
long haul and much difficulty, became a professional
engineer. We still argue about which way current is
best considered to flow. He keeps talking about
electrons.

--
--Larry Brasfield
email: donotspam_larry_brasfield@hotmail.com
Above views may belong only to me.

 

[Howard Long]

"Thomas A. Horsley" wrote in message

what kind of experience they had with
limited run board services like ExpressPCB?

I have used them a number of times here in the UK. Their turnaround times
are about the same as for European manufacturers. The quality of the boards
is excellent, especially the production service.

The only problems are (a) it's not cheap and (b) as far as I know you're
limited to using ExpressPCB for your designs - there is no way to export to,
say, Gerber files, or generate your own transparencies. This second part is
quite a smart business model - few will want to redesign their entire PCB
all over again in a new package.

Regards, Howard

 

[JeffM]

the ascii program types used here to text draw the schematic?
R.Spinks

Andy's ASCII Circuit by Andreas Weber
http://www.tech-chat.de/download.html is a nice tool.

 

[Terry Pinnell]

"Terry Vacha" <photosail@adelphia.net> wrote:

I'm a little confused with the results I'm getting when I do a search on
Google for this. Perhaps someone here can help.

I want a program that will allow me to draw solid state or mechanical
relays, diodes, 555 chips, 7404 chips, light bulbs, etc. I'd like the
program to have a library of the common symbols. I'd like the symbols for a
555 or 7404 to look like those on the back of a Radio Shack package. In
other words -- little rectangle showing the pinouts with the schematic
structure showing within the rectangle that is appropriate for that device.
I'd like also simple devices available in the "library" such as mechanical
relays with a simple schematic showing the switch and magnetic coil and
voltage inputs.

I downloaded Eagle Layout Editor and can't seem to make a simple magnetic
relay.

Any suggestions for some programs to draw circuits?

Terry
tvacha*remove*@yahoo.com

See my notes and links to some 60 ECAD programs at
http://www.terrypin.dial.pipex.com/ECADList.html

Terry Pinnell
Hobbyist, West Sussex, UK

 

[dB]

"Terry Vacha" <photosail@adelphia.net> wrote in message news:<qZidncBk79OoB5LfRVn-hw@adelphia.com>...

I'm a little confused with the results I'm getting when I do a search on
Google for this. Perhaps someone here can help.

I want a program that will allow me to draw solid state or mechanical
relays, diodes, 555 chips, 7404 chips, light bulbs, etc.

Any suggestions for some programs to draw circuits?

Terry
tvacha*remove*@yahoo.com

If you only want to draw, and nothing else, use MS Paint. You can make
up a library of shapes then copy and paste them to another instance of
Paint opened at the same time.

 

[John Larkin]

On Mon, 14 Feb 2005 23:51:42 GMT, "Genome" <ilike_spam@yahoo.co.uk>
wrote:

That's quite sweet.

It's Valentine's Day!

John

 

[loedown]

These sort of applications are ideal for PIC chips, simple to use, require
almost no auxilliary circuitry, and are easy to program. Very steep learning
curve, and the programmer will set you back somewhere in the order of $200
AUD, but the PIC 18F672 has provision for RS232 and multiple outputs (22).

Paul

 

[Lord Garth]

"Chris W" <1qazse4@cox.net> wrote in message
news:LxuQd.63925$jn.19302@lakeread06...

Ruediger wrote:

loedown wrote:
snip
Also does someone make some kind of IC chip that takes say 3 bit binary
number for input on 3 pins and then makes one of 8 output pins go high
based on the input binary number?

The 74HC138 is such a chip but the outputs go low based upon the input.
You can do this function with any EPROM if you wish.

 

[Don Kelly]

"R.Spinks" <rspinks1@wowway.com> wrote in message
news:tsCdna2X3v4dQ5LfRVn-sw@wideopenwest.com...

I would like to see the math involved in calculating the current in the 4k
resistor and the current in the 12k resistor. I have tried to do a node
current at the top -right of 1k resistor and top of all parallel items)
which I have called 'node A' (Va) but I keep getting the wrong answer.
There
is 8mA in the 4K and 2.66mA in the 12k (this means the voltage at 'node A'
must be 32V. (I hand drew the 50mA and 90mA current sources so that's what
they are supposed to be -- if there was a symbol I missed it). I took the
current at node A (top) to be:

((Va-8)/1k)-50mA - (Va / 12k) - (Va/6k) + 90mA - (Va/4k) = 0 but that
does
not yield 32v. Can someone point an error I've made (as I assume I've set
something up wrong) or correctly indicate the math for me? Thanks.




1K
___
|-|___|-----------------------------------|--------
| | | | | |
| | | | | |
| | .-. .-. |90ma .-.
/+\ | | | | | | | |
8v ( ) 50ma (\|/) 12K | | | | 6K (/|\) | | 4K
\-/ | '-' '-' | '-'
| | | | | |
---------------------------------------------------
(created by AACircuit v1.28.4 beta 13/12/04 www.tech-chat.de)


Taking + currents into node A you should have (8-Va)/1k) rather than

(Va-8)/1k).
--
Don Kelly
dhky@peeshaw.ca
remove the urine to answer

 

[cpemma]

Glenn Ashmore wrote:

I have a 1500 watt 12VDC motor ... I also want to slow it down
... I figure adding a PWM routine to the PIC program would be the
way to do it.

The questions are, am I on the right track and if so, what do I use
to drive the MOSFETs?

Download the kit instructions
http://www.oatleyelectronics.com/kits/k098.html

There's a 20A 12V circuit that looks upgradeable.

 

[Peter Michelson]

Yes, a flip-flop would do it and someone is sure to suggest a PIC, both
requiring a fair amount of extra components to actually work. Have you
looked for alternate action switches or relays? It would run your
lights and not require a degree to get it to work.
Glenn Gundlach

Is that because flip-flops require capacitors to store state information? I
would be interested to know the minimum complement of components necessary
to support a single flip-flop. Perhaps there is a good link to this kind of
information (?)

I have a similar request as the original poster. I would like to use a
momentary switch to cycle among three LEDs. Is the circuit design for this
using flip-flops complex? Is there an easier way to generate the same
functionality? It seems that there must be some ICs out there that provide
this kind of functionality out of the box. Does anyone know of any?

Thanks a lot!
Peter