R
Rich Grise
Guest
On Tue, 31 May 2005 12:51:54 -0700, robbbbbby wrote:
Have Fun!
Rich
http://www.google.com/search?q=bar+graph+led+array
Have Fun!
Rich
J
Jonathan Westhues
Guest
"Rich Grise" <richgrise@example.net> wrote in message
news
an.2005.06.03.16.42.19.55037@example.net...
The protocol is bit-banged on the parallel port pins, so the interface
circuitry is hardly anything and it fits inside the db25 shell. See for
example
http://www.olimex.com/dev/avr-pg2b.html
at 10 USD each, and I know that one works.
Jonathan
news
an.2005.06.03.16.42.19.55037@example.net...Are you sure? I've seen lots of AVR programmers that look just like that.
The protocol is bit-banged on the parallel port pins, so the interface
circuitry is hardly anything and it fits inside the db25 shell. See for
example
http://www.olimex.com/dev/avr-pg2b.html
at 10 USD each, and I know that one works.
Jonathan
C
cpemma
Guest
robbbbbby wrote:
your own bar in discrete leds of many shapes, sizes & colours).
Some 2-20 led bar modules
http://www.rapidelectronics.co.uk/rkmain.asp?PAGEID=80008 and they'll all
stack for longer bars.
Note there are other bargraph ICs as shown on the page, a lot cheaper than a
3914/5/6 if you only want a 5-high..
For hobbyist use the choice may be a bit more limited (though you can build
your own bar in discrete leds of many shapes, sizes & colours).
Some 2-20 led bar modules
http://www.rapidelectronics.co.uk/rkmain.asp?PAGEID=80008 and they'll all
stack for longer bars.
Note there are other bargraph ICs as shown on the page, a lot cheaper than a
3914/5/6 if you only want a 5-high..
N
neil
Guest
"petrus bitbyter" <pieterkraltlaatditweg@enditookhccnet.nl> wrote in message
news:429ce532$0$154$3a628fcd@reader2.nntp.hccnet.nl...
level ?
- use a second ac signal as a reference, using it to control a two-way
analog mux, using that to route either the true or inverse (via inverting
opamp) ac signal.
No doubt there are suitable circuits available on web.
hth
Neil
news:429ce532$0$154$3a628fcd@reader2.nntp.hccnet.nl...
how about demodulating the signal, and smoothing/processing the resultant dc
level ?
- use a second ac signal as a reference, using it to control a two-way
analog mux, using that to route either the true or inverse (via inverting
opamp) ac signal.
No doubt there are suitable circuits available on web.
hth
Neil
B
Bob Monsen
Guest
Patrick wrote:
AD636. It is a device which will give you a DC output equal to the RMS
value of a waveform. Thus, pass the 2.5V through a capacitor, then run
it into this thingy. You'll need to calibrate it, but it should give you
a good view of the current when coupled with your current sensor.
---
Regards,
Bob Monsen
I'm not sure I understand your problem. However, take a look at the
AD636. It is a device which will give you a DC output equal to the RMS
value of a waveform. Thus, pass the 2.5V through a capacitor, then run
it into this thingy. You'll need to calibrate it, but it should give you
a good view of the current when coupled with your current sensor.
---
Regards,
Bob Monsen
M
Mark VB
Guest
nik-andrejchikov@yandex.ru wrote:
[ÎÉŢĹÇĎ Ď ÜĚĹËÔŇĎÎÉËĹ]
ń ÄŐÍÁŔ ×Ů ×Ů×ĹŰÉ×ÁĹÔ Ë ÎĹĐŇÁ×ÉĚŘÎĎĘ ÇŇŐĐĐĹ. üÔÁ ÇŇŐĐĐÁ Ď ÜĚĹËÔŇĎÎÉËĹ É
ŢĚĹÎŮ ÔĎĚŘËĎ ÇĎ×ĎŇŃÔ ÁÎÇĚÉĘÓËŐŔ ŃÚŮË.
[ÎÉŢĹÇĎ Ď ÜĚĹËÔŇĎÎÉËĹ]
ń ÄŐÍÁŔ ×Ů ×Ů×ĹŰÉ×ÁĹÔ Ë ÎĹĐŇÁ×ÉĚŘÎĎĘ ÇŇŐĐĐĹ. üÔÁ ÇŇŐĐĐÁ Ď ÜĚĹËÔŇĎÎÉËĹ É
ŢĚĹÎŮ ÔĎĚŘËĎ ÇĎ×ĎŇŃÔ ÁÎÇĚÉĘÓËŐŔ ŃÚŮË.
R
Roger Johansson
Guest
Rich Grise wrote:
d-sub-connector, and a connector for the processor board.
It is enough hardware to program an AVR processor.
It is so easy to build that there is no reason to buy one.
The schematic is widespread and well known by AVR users. For example if
you download the BASIC programming program and look in its help file,
there are schematics for this "cable".
The connector used at the processor end is, of course, up to yourself
to choose. The most lightweight solution is no connector at all. Just
solder the wires to the processor circuit during the programming. This
solution takes a minimum of board space on the processor board you are
building/programming.
It is correct to call this "cable" a programmer, as it is the only
hardware necessary.
The programmer software is free to get from the manufacturer and users
of AVR processors.
--
Roger J.
It is not just a cable, It is a cable, a few resistors, a
d-sub-connector, and a connector for the processor board.
It is enough hardware to program an AVR processor.
It is so easy to build that there is no reason to buy one.
The schematic is widespread and well known by AVR users. For example if
you download the BASIC programming program and look in its help file,
there are schematics for this "cable".
The connector used at the processor end is, of course, up to yourself
to choose. The most lightweight solution is no connector at all. Just
solder the wires to the processor circuit during the programming. This
solution takes a minimum of board space on the processor board you are
building/programming.
It is correct to call this "cable" a programmer, as it is the only
hardware necessary.
The programmer software is free to get from the manufacturer and users
of AVR processors.
--
Roger J.
J
Jonathan Westhues
Guest
"Chris" <cfoley1064@yahoo.com> wrote in message
news:1117843865.974772.181170@g47g2000cwa.googlegroups.com...
current through the diode is going to be very small, microamps. The
datasheet for the 1N5817 says that the reverse leakage current is around 100
uA at 4 V though. My simulation doesn't rectify; the diode drop is
negligible, in both directions... (but of course it works fine with a 1N4148
or something, or I guess you could make C bigger and R smaller).
I think R*C should be bigger anyways, because with a not-leaky diode there's
still about 0.3 V of ripple on the output for a 1 V amplitude input. I think
the ripple is roughly given by
dV/dT*T = (I/C)/f
= ((Vout/R)/C)/f
= Vout/(R*C*f)
= 3/(1meg*0.1u*60) = 0.5 V which is kind of close
(assuming that the current through R is constant, and assuming that the
capacitor spends almost all the cycle discharging and hardly any time
charging)
Jonathan
news:1117843865.974772.181170@g47g2000cwa.googlegroups.com...
Will the leakage through the big power Schottky be a problem here? The
current through the diode is going to be very small, microamps. The
datasheet for the 1N5817 says that the reverse leakage current is around 100
uA at 4 V though. My simulation doesn't rectify; the diode drop is
negligible, in both directions... (but of course it works fine with a 1N4148
or something, or I guess you could make C bigger and R smaller).
I think R*C should be bigger anyways, because with a not-leaky diode there's
still about 0.3 V of ripple on the output for a 1 V amplitude input. I think
the ripple is roughly given by
dV/dT*T = (I/C)/f
= ((Vout/R)/C)/f
= Vout/(R*C*f)
= 3/(1meg*0.1u*60) = 0.5 V which is kind of close
(assuming that the current through R is constant, and assuming that the
capacitor spends almost all the cycle discharging and hardly any time
charging)
Jonathan
M
mike
Guest
Patrick wrote:
accomplish?"
If you have a purely resistive load and a pure voltage sinewave, you can
measure about any parameter of the current waveform and know all the rest.
Problem is that most applications are not even close to that ideal.
In most cases, you have to actually measure the parameter of interest.
Some typical parameters might be peak, average, RMS values. Phase
may or may not be important.
If it's a current sine wave and you can determine the zero crossing
with a comparator and you know the frequency, you know exactly were
the peaks are and you can sample there.
If it's not a current sine wave, you can peak detect and subtract off
the 2.5V either with an op-amp or software.
What do you want to do about noise?
How fast do you need your loop response?
All comes back to, "what do you really want to know about this current
waveform?"
mike
--
Return address is VALID but some sites block emails
with links. Delete this sig when replying.
..
Wanted, PCMCIA SCSI Card for HP m820 CDRW.
FS 500MHz Tek DSOscilloscope TDS540 Make Offer
Wanted 12" LCD for Compaq Armada 7770MT.
Bunch of stuff For Sale and Wanted at the link below.
MAKE THE OBVIOUS CHANGES TO THE LINK
ht<removethis>tp://www.geocities.com/SiliconValley/Monitor/4710/
I know people are tired of hearing me ask, but "What are you trying to
accomplish?"
If you have a purely resistive load and a pure voltage sinewave, you can
measure about any parameter of the current waveform and know all the rest.
Problem is that most applications are not even close to that ideal.
In most cases, you have to actually measure the parameter of interest.
Some typical parameters might be peak, average, RMS values. Phase
may or may not be important.
If it's a current sine wave and you can determine the zero crossing
with a comparator and you know the frequency, you know exactly were
the peaks are and you can sample there.
If it's not a current sine wave, you can peak detect and subtract off
the 2.5V either with an op-amp or software.
What do you want to do about noise?
How fast do you need your loop response?
All comes back to, "what do you really want to know about this current
waveform?"
mike
--
Return address is VALID but some sites block emails
with links. Delete this sig when replying.
..
Wanted, PCMCIA SCSI Card for HP m820 CDRW.
FS 500MHz Tek DSOscilloscope TDS540 Make Offer
Wanted 12" LCD for Compaq Armada 7770MT.
Bunch of stuff For Sale and Wanted at the link below.
MAKE THE OBVIOUS CHANGES TO THE LINK
ht<removethis>tp://www.geocities.com/SiliconValley/Monitor/4710/
B
Bob Monsen
Guest
Chris wrote:
I agree that a simple scheme may work properly. One thing to factor into
the discussion is that a microprocessor pin is a good way to zero a peak
detector. He can use a shottky diode and cap, like you suggest, but skip
the drain resistor. Before each sample, just use another pin to drain
the cap. Then, make the pin an input, wait a few cycles, and take the
reading.
Another thing to consider is that some micro ADCs want fairly low input
impedance; the PIC 12 and 16 series, for example, want 10k.
Yet another issue is that this won't get him the right value if that
2.5V center voltage drifts around. He can fix that by using a big cap to
make the 5VAC signal be centered at zero. Then, his answer will be
between 0 and 2.5V, minus the drop of the diode.
If he wants to add a couple of cheap opamps, he can eliminate that diode
drop as well by building a simple peak detector. However, that will
involve more external hardware.
None of this really works, however, if the current isn't a sine wave.
---
Regards,
Bob Monsen
<snip suggestion>
I agree that a simple scheme may work properly. One thing to factor into
the discussion is that a microprocessor pin is a good way to zero a peak
detector. He can use a shottky diode and cap, like you suggest, but skip
the drain resistor. Before each sample, just use another pin to drain
the cap. Then, make the pin an input, wait a few cycles, and take the
reading.
Another thing to consider is that some micro ADCs want fairly low input
impedance; the PIC 12 and 16 series, for example, want 10k.
Yet another issue is that this won't get him the right value if that
2.5V center voltage drifts around. He can fix that by using a big cap to
make the 5VAC signal be centered at zero. Then, his answer will be
between 0 and 2.5V, minus the drop of the diode.
If he wants to add a couple of cheap opamps, he can eliminate that diode
drop as well by building a simple peak detector. However, that will
involve more external hardware.
None of this really works, however, if the current isn't a sine wave.
---
Regards,
Bob Monsen
R
Roger Johansson
Guest
julian814@hotmail.com wrote:
The keyboard gave up a long time ago.
I built a matrix of wires, the horizontal ones lifted up by small
pieces of foam plastic, from a table tennis racket, by the way.
Under it all I placed a picture of the keyboard somehow, so I could
orient myself.
Each contact point between a horisontal and vertical wire is a key on
the keyboard.
When I touch a wire I push it down one millimeter so it comes into
contact with the wire under it.
The rubber pads lift the wire again if I don't push it down too hard,
so it works quite well. Occasionally I need to move a wire when it gets
stuck but it is good enough for playing a little with this old museum
piece.
The first thing you need is a big picture of the keyboard.
I have a vague memory that I cut a picture of the keyboard into
horisontal strips and moved them slightly in relation to each other, so
the pictures of the keys would fit better into a perpendicular matrix.
The key rows on the Spectrum are placed diagonally over each other.
Attach the picture to a base plate, wood or metal, or pcb board
material.
Lay out bare copper wires in a matrix, fix them with glue and prop up
the horisontal wire between each key with some flexible material.
--
Roger J.
I have a ZX Spectrum, bought 1984.
The keyboard gave up a long time ago.
I built a matrix of wires, the horizontal ones lifted up by small
pieces of foam plastic, from a table tennis racket, by the way.
Under it all I placed a picture of the keyboard somehow, so I could
orient myself.
Each contact point between a horisontal and vertical wire is a key on
the keyboard.
When I touch a wire I push it down one millimeter so it comes into
contact with the wire under it.
The rubber pads lift the wire again if I don't push it down too hard,
so it works quite well. Occasionally I need to move a wire when it gets
stuck but it is good enough for playing a little with this old museum
piece.
The first thing you need is a big picture of the keyboard.
I have a vague memory that I cut a picture of the keyboard into
horisontal strips and moved them slightly in relation to each other, so
the pictures of the keys would fit better into a perpendicular matrix.
The key rows on the Spectrum are placed diagonally over each other.
Attach the picture to a base plate, wood or metal, or pcb board
material.
Lay out bare copper wires in a matrix, fix them with glue and prop up
the horisontal wire between each key with some flexible material.
--
Roger J.
M
Mark VB
Guest
quadran@telenet.be wrote:
Serial/Parallel Shift register like the 74164 and a
Parallel/Serial-Shift register like the 74165.
HTH,
Mark Van Borm
As I understand your question, you're not searching for a fifo, but a
Serial/Parallel Shift register like the 74164 and a
Parallel/Serial-Shift register like the 74165.
HTH,
Mark Van Borm
M
Mark VB
Guest
PYCTAM wrote:
battery-contacts. There's also another possibility by means of a
resistor (approx. 1K) between the T-contact of the battery and Ground.
That way the phone "thinks" there is a battery.
HTH,
Mark Van Borm
You can't power it up via the connector, but you can when using the
battery-contacts. There's also another possibility by means of a
resistor (approx. 1K) between the T-contact of the battery and Ground.
That way the phone "thinks" there is a battery.
HTH,
Mark Van Borm
J
John Bokma
Guest
ShamShoon wrote:
+-----+-----+-----+
| | | |
V R R I
| | | |
R | | |
| | | |
+-----+-----+-----+
2
?
Has been ages and ages, but I would say that 1 and 2 are both branches
that don't belong to the left and the right loops.
--
John MexIT: http://johnbokma.com/mexit/
personal page: http://johnbokma.com/
Experienced programmer available: http://castleamber.com/
Happy Customers: http://castleamber.com/testimonials.html
1
+-----+-----+-----+
| | | |
V R R I
| | | |
R | | |
| | | |
+-----+-----+-----+
2
?
Has been ages and ages, but I would say that 1 and 2 are both branches
that don't belong to the left and the right loops.
--
John MexIT: http://johnbokma.com/mexit/
personal page: http://johnbokma.com/
Experienced programmer available: http://castleamber.com/
Happy Customers: http://castleamber.com/testimonials.html
L
Larry Brasfield
Guest
"ShamShoon" <mhelshou@yahoo.com> wrote in message news:1117989523.499267.103060@g47g2000cwa.googlegroups.com...
you mention "the two parallel resistors". I get all three
in parallel from the above description. For discussion, I
propose the following diagram. (View with fixed-width font.)
B1 B12 B2
.--------o--------o-------o-------.
| | | | |
/+\ .-. .-. .-. / \
( ) | | | | | | ( | )
Es \-/ R1| | R3| | R2| | Is \^/
| '-' '-' '-' |
| | | | |
'--------o--------o-------o-------'
and pretending R3 does not exist, one set of independed loops is:
(Es B1 R1), (R1 B12 R2), (R2 B2 Is).
Another set would be:
(Es B1 B12 R2), (Is B2 B12 R1), (R1 B12 R2).
Another set would be:
(Es B1 B12 B2 Is), (Es B1 R1), (Is B2 R2).
There are a few more. (factorial(3) altogether)
Let's consider just the first set, which is the one you
appear to find confusing.
The (Es B1 R1) loop is at least one independent loop.
The (R1 B12 R2) loop is independent because it contains at
least one branch (B12 or R2) which is not part of any other
independent loop.
The (R2 B2 Is) loop is independent because it contains at
least one branch (B2 or Is) which is not part of any other
independent loop.
There are no more because all the possible branches
have been included in the loops already listed.
Let's consider the members in the order that leaves
you wondering about the definition.
The (Es B1 R1) loop is at least one independent loop.
The (R2 B2 Is) loop is independent because it contains at
least one branch (Is, B2 or R2) which is not part of any
other independent loop.
The (R1 B12 R2) loop is independent because it contains
at least one branch (B12) which is not part of any other
independent loop.
B12. Branches B1 and B2 are redundant with Es
and Is, so you can eliminate them from consideration.
But branch B12 is not uniquely associated with any
circuit element, so you can eliminate it only if you are
willing to forget what a branch is.
Another clue to your difficulty is this: The order I
first elaborated satisfies the definition even if the B?
branches are left out. Yet the order you mentioned
does not when the B? branches are left out. So it
must be the case that branches and circuit elements
are not the same thing.
--
--Larry Brasfield
email: donotspam_larry_brasfield@hotmail.com
Above views may belong only to me.
That is a commonly accepted definition.
That description mentions three resistances. Yet below
you mention "the two parallel resistors". I get all three
in parallel from the above description. For discussion, I
propose the following diagram. (View with fixed-width font.)
B1 B12 B2
.--------o--------o-------o-------.
| | | | |
/+\ .-. .-. .-. / \
( ) | | | | | | ( | )
Es \-/ R1| | R3| | R2| | Is \^/
| '-' '-' '-' |
| | | | |
'--------o--------o-------o-------'
Using the above diagram so we can refer to resistors easily,
and pretending R3 does not exist, one set of independed loops is:
(Es B1 R1), (R1 B12 R2), (R2 B2 Is).
Another set would be:
(Es B1 B12 R2), (Is B2 B12 R1), (R1 B12 R2).
Another set would be:
(Es B1 B12 B2 Is), (Es B1 R1), (Is B2 R2).
There are a few more. (factorial(3) altogether)
Let's consider just the first set, which is the one you
appear to find confusing.
The (Es B1 R1) loop is at least one independent loop.
The (R1 B12 R2) loop is independent because it contains at
least one branch (B12 or R2) which is not part of any other
independent loop.
The (R2 B2 Is) loop is independent because it contains at
least one branch (B2 or Is) which is not part of any other
independent loop.
There are no more because all the possible branches
have been included in the loops already listed.
Let's consider the members in the order that leaves
you wondering about the definition.
The (Es B1 R1) loop is at least one independent loop.
The (R2 B2 Is) loop is independent because it contains at
least one branch (Is, B2 or R2) which is not part of any
other independent loop.
The (R1 B12 R2) loop is independent because it contains
at least one branch (B12) which is not part of any other
independent loop.
You found the definition you quoted.
Apparently, you have mssed the presence of branch
B12. Branches B1 and B2 are redundant with Es
and Is, so you can eliminate them from consideration.
But branch B12 is not uniquely associated with any
circuit element, so you can eliminate it only if you are
willing to forget what a branch is.
Another clue to your difficulty is this: The order I
first elaborated satisfies the definition even if the B?
branches are left out. Yet the order you mentioned
does not when the B? branches are left out. So it
must be the case that branches and circuit elements
are not the same thing.
--
--Larry Brasfield
email: donotspam_larry_brasfield@hotmail.com
Above views may belong only to me.
J
John Bokma
Guest
Larry Brasfield wrote:
e.g.:
+-----+-----+-----+
| | | |
V R R I
| | | |
R | | |
| | | |
+-----+-----+-----+
--
John MexIT: http://johnbokma.com/mexit/
personal page: http://johnbokma.com/
Experienced programmer available: http://castleamber.com/
Happy Customers: http://castleamber.com/testimonials.html
Yup, the 1st one is "voltage source and a resistance in series"
e.g.:
+-----+-----+-----+
| | | |
V R R I
| | | |
R | | |
| | | |
+-----+-----+-----+
--
John MexIT: http://johnbokma.com/mexit/
personal page: http://johnbokma.com/
Experienced programmer available: http://castleamber.com/
Happy Customers: http://castleamber.com/testimonials.html
L
Larry Brasfield
Guest
"John Bokma" <john@castleamber.com> wrote in message
news:Xns966C8D28DD283castleamber@130.133.1.4...
--
--Larry Brasfield
email: donotspam_larry_brasfield@hotmail.com
Above views may belong only to me.
news:Xns966C8D28DD283castleamber@130.133.1.4...
That's certainly a more sensible way to interpret those words.
--
--Larry Brasfield
email: donotspam_larry_brasfield@hotmail.com
Above views may belong only to me.
J
John Bokma
Guest
Larry Brasfield wrote:
--
John MexIT: http://johnbokma.com/mexit/
personal page: http://johnbokma.com/
Experienced programmer available: http://castleamber.com/
Happy Customers: http://castleamber.com/testimonials.html
I had to read it a few times, and draw it on paper
--
John MexIT: http://johnbokma.com/mexit/
personal page: http://johnbokma.com/
Experienced programmer available: http://castleamber.com/
Happy Customers: http://castleamber.com/testimonials.html
J
John Fields
Guest
On Thu, 02 Jun 2005 23:02:00 +0100, Terry Pinnell
<terrypinDELETE@THESEdial.pipex.com> wrote:
You're in luck! He did, so I posted the circuit description and a
corrected schematic (sigh...) to abse.
--
John Fields
Professional Circuit Designer
<terrypinDELETE@THESEdial.pipex.com> wrote:
---
You're in luck! He did, so I posted the circuit description and a
corrected schematic (sigh...) to abse.
--
John Fields
Professional Circuit Designer
J
Joerg
Guest
Hello Bob,
deal to design it "pop free". I had an FM radio that never popped. It
looks at the noise content on the audio signal to determine squelch
action. I wonder why the wireless mikes couldn't do that.
When I repaired radios I sometimes looked at the schematics to see why
they popped. Usually the designers didn't fully study the DC levels.
There cannot be any DC shift or cap charge/discharge when the squelch
kicks in. Other times they ignored the charge injection of the FET path
used for muting.
Regards, Joerg
http://www.analogconsultants.com
Tone squelch is a great concept but even without it isn't such a big
deal to design it "pop free". I had an FM radio that never popped. It
looks at the noise content on the audio signal to determine squelch
action. I wonder why the wireless mikes couldn't do that.
When I repaired radios I sometimes looked at the schematics to see why
they popped. Usually the designers didn't fully study the DC levels.
There cannot be any DC shift or cap charge/discharge when the squelch
kicks in. Other times they ignored the charge injection of the FET path
used for muting.
Regards, Joerg
http://www.analogconsultants.com