Chip with simple program for Toy

[Anthony Fremont]

"John Bokma" wrote:

<snipped exactly what I was going to post>

Note that I did this calculation, later checked it with a voltage
meter,
and yet already 2 exotic LEDs died on me. I think it's a bad badge.
(Unless someone can point out my errors).

I came up with the same calculation as you. If LEDs are dieing, then
20mA may be a bit too much current. Since there are two LEDs in series,
one may hog more current than the other resulting in its demise. You
might consider not driving them so hard. There is probably a relatively
insignificant brightness difference between 10mA and 20mA anyway.

 

[Larry Brasfield]

"John Fields" <jfields@austininstruments.com> wrote in message
news:0chi619a9bmorr8pt1pt5qu5ok498dhnv3@4ax.com...

On Fri, 22 Apr 2005 09:42:24 -0700, "Larry Brasfield"
donotspam_larry_brasfield@hotmail.com> wrote:

"John Fields" <jfields@austininstruments.com> wrote in message
news:560i61tfmof8mhjis826fj5q3dcghb53of@4ax.com...
On 22 Apr 2005 05:57:14 -0700, "mjohnson" <crvmp3@hotmail.com> wrote:
....
Before anything can happen we need to find out definitively what's
happening at the buzzer when the alarm goes off, or we need to find a
signal somewhere in the clock which changes state when the alarm goes
off.

The power needed to drive a buzzer will be many
times larger than what needs to be picked off to
activate another circuit, (many mW versus uW).
---
Depends. The OP's advocating using the signal driving the buzzer to
also drive the LED in an opto, which will be milliwatts VS milliwatts.

I was not addressing use of the optoisolator in that
position. However, if the current taken through the
opto LED is limited to a few 100 uA, such usage
would still be a small fraction of the buzzer power.
As you point out, CTR would be reduced, but no
more than a few uA of output would be needed.

All the alarms with buzzers I have heard are very
loud (and annoying). It is hard to imagine getting
that without using many 10s of mW.

I wouldn't use an opto because of the current required for its LED and
the uncertainty of being able to use its transistor output to trigger
the remote.

Some such uncertainty is warranted, but I suggest that
there is reason to believe an opto-isolator will be fine.
The cheap (typically membrane) switches used in many
remotes are not asked to carry much current and, to
conserve battery power, large value pull-{up,down}
resistors are used. If a replacement for the contanct
had to carry more than 100 uA, I would be surprised.
The CTR (current transfer ratio) for opto-isolaters is
often guaranteed to be 100% or better, so a similar
current is all that the LED would need.

---
Not necessarily, CTR falls off quickly as LED forward current
diminishes and there are temperature effects which need to be taken
into consideration which can largely be ignored with a comparator-reed
switch solution. Also, with the reed switch solution there is no
saturation voltage VS LED If problem since it's either on or off.

I agree that the reed relay is simpler to apply. For that
reason alone, it may well be most suitable for the OP's
project. My suggestion about an optoisolator in its
place is more like a feasable alternative than any kind
of compelling improvement. Reed relays are fragile
and, if their leads are not carefully heat-sunk during
soldering, they can fail quickly or slowly as a result.
That, together with a dislike of moving parts, made
me think it might be an attractive alternative.

Finally, the signal sent thru the opto-isolator can be time limited
to just over what is needed for the remote in order to
conserve the battery.
---
Either solution will require the generation of a timed pulse to the
remote, so that's probably a wash.

Yes.

Here's what I see as a much simpler solution, with only the clock
output needing to be defined in order to make it work:

INTERFACE
BATTERY
CLOCK 4.5V REMOTE
BATTERY | BATTERY
3V +---------+---------+ 3V
| | | | |
+--+--+ +--+--+ +--+--+ +--+--+ +--+---+
|CLOCK|---|ALARM|---|1SEC |---|REED |---|REMOTE|
+-----+ | DET | |MONO | |RELAY|---|SWITCH|
+-----+ +-----+ +-----+ +------+

The opto-isolator would plug into that with little
change except reduction of the 4.5V battery drain
(unless my power surmises are completely wrong).

---
Could be. I'll defer judgement and wait until the OP comes back with
something definitive on the buzzer signal to post my design. If he
doesn't, I can always fall back on the acoustic thing I've already
posted.

Seems reasonable.

You may want to ask him about the current being conducted by
the remote's switch switch to see whether you can use an opto in
there. An easy way to determine the current would be to jump the
switch contacts with a milli/microammeter...

Yes.

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

 

[Tom Biasi]

"Brilla" <Brilla@mts.net> wrote in message
news:f178a263.0504221616.44260cc0@posting.google.com...

Hi, I want to make a simple LED circuit with 12 LEDs running off a
nine volt battery. I've managed to dig up enough information about
most things, so I know I'll have to wire them in parallel. But the
resistance I should be using still confuses me.
Should I have one (or more) resistors at the beginning of the circuit?
Or one before each LED in the circuit?

The LEDs I'm using have a 3.6v voltage drop, and they are supposed to
get 20mA I beleive.
So that's 12 LEDs off a 9v battery.

I have a bunch of 27 Ohm resistors because that's what I (probably
mistakenly) calculated I should have before each LED. But I have no
idea what i'm doing. So if anyone can help me out on what resistors I
should use and where I should put them that would be greatly
appreciated!

Thanks,
Steve

Sounds like you want the leds to be powered individually. You need a 270 ohm
resister to limit the current to 20 mA.
Each led should get its own resister and then wire the led/resister units in
parallel.
1/4 watt resisters will be fine.
When you put leds in series they will pass the same current but one my have
a larger voltage drop, consequently dissipate more power, because of a
variance of their characteristics.
Regards,
Tom

 

[john jardine]

"bg" <byju.ganga@gmail.com> wrote in message
news:1114157934.811393.159290@l41g2000cwc.googlegroups.com...

i have a dayton shaded pole single phase ac induction motor with
1/150HP, 3000 RPM, 230 V, 60 Hz frequency and Full Load Amps 0.24A . I
would like to know whether there is any equation for troque.(connecting
current/voltage with torque)

ThankYou

For shaded pole motors "speed control" by means of ac voltage, is a bit of a

non starter but ... must admit to doing one for a customer last week .
Similar 55Watt shaded pole as yours (European 230Vac though) and attached to
a small ducted fan.
Motor sits there and buzzes at 10% voltage. Two fingers can hold the shaft
up to 50% voltage. Beyond that and upto 100% voltage, a passing resemblance
to speed control.
Would estimate something like a square law response. Eg, halve the voltage
and torque drops four times. (actually slighly better than this, say T
proportional to V^1.6).

Conversation with customer goes sort of ...
Me. Ye canna break the laws of physics.
Them. But the one you did for the big fan works well.
Me. Yes but that was technically crap and I explained this at the time.
Them. Never mind the physics, just give us a quote.

When will customers ever learn that BumbleBees are incapable of flying?.
regards
john

 

[petrus bitbyter]

"Brilla" <Brilla@mts.net> schreef in bericht
news:f178a263.0504221616.44260cc0@posting.google.com...

Hi, I want to make a simple LED circuit with 12 LEDs running off a
nine volt battery. I've managed to dig up enough information about
most things, so I know I'll have to wire them in parallel. But the
resistance I should be using still confuses me.
Should I have one (or more) resistors at the beginning of the circuit?
Or one before each LED in the circuit?

The LEDs I'm using have a 3.6v voltage drop, and they are supposed to
get 20mA I beleive.
So that's 12 LEDs off a 9v battery.

I have a bunch of 27 Ohm resistors because that's what I (probably
mistakenly) calculated I should have before each LED. But I have no
idea what i'm doing. So if anyone can help me out on what resistors I
should use and where I should put them that would be greatly
appreciated!

Thanks,
Steve

Well,

The resistor voltage will be 9-3.6=5.4V. Using Ohms law the resistor should
be R=5.4/20=270 Ohm. So 27 Ohm is way too low. Using the right resistors
will require your battery to provide 240mA. You will need a pretty big
battery to light your LEDs for even some minutes. Maybe to 4,5V batteries in
series will do.

Most of the energy from the battery is dissipated in the resistors. Really a
waste. You can try to use two LEDs in series. The resistor will be reduced
to (9-2*3.6)/20=90 Ohm. Power efficiency is much better this way. However,
practical 9V batteries tend to loose some voltage pretty fast when in use.
So the current through your LEDs will decrease accordingly.

To make a real good battery powered LED-light you need to go electronic. But
that's a different story.

petrus bitbyter

 

[John Fields]

On Sat, 23 Apr 2005 05:13:57 GMT, "Anthony Fremont"
<spam@anywhere.com> wrote:

"John Bokma" wrote:

snipped exactly what I was going to post

Note that I did this calculation, later checked it with a voltage
meter,
and yet already 2 exotic LEDs died on me. I think it's a bad badge.
(Unless someone can point out my errors).

I came up with the same calculation as you. If LEDs are dieing, then
20mA may be a bit too much current. Since there are two LEDs in series,
one may hog more current than the other resulting in its demise.

---
So, you're saying that because the LEDs may not be identical one may
be drawing more current than the other?
---

You
might consider not driving them so hard. There is probably a relatively
insignificant brightness difference between 10mA and 20mA anyway.

--
John Fields
Professional Circuit Designer

 

[Larry Brasfield]

"aman" <aman.bindra@gmail.com> wrote in message
news:1114272913.501494.238310@l41g2000cwc.googlegroups.com...

When I think of capacitor I can see there are two conducting parallel
plates seperated by an insulator dielectric. So as there are opposite
charges on the inside of the plates they attract each other thus
containing the charge.

Two plates is the minimum, often exceeded.

Here is my question. If there is a perfect insulator used as
dielectric, does it mean that the charge is fully contained and if the
dielectric material gets a little conducting the charge on the plates
gets reduced(cannot be retained fully) and some charge flows in the
dielectric. Am I correct to some extent ?

Yes, although what would be called the charge is fully
contained whether the dielectric conducts or not.

I am asking this because I am constructing a kind of capacitor detector
which is seperated by dielectric which is water(different kind of
samples with different conductivity and am trying to nuetralise the
harmful free ions in water).

If you know your chemistry, you know that water
becomes something like an insulator when you
get the free ion concentration low enough. Are
you using leakage to guage that concentration?
The capacitance will stay just about the same
unless you have very high impurity levels.

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

 

[John Popelish]

aman wrote:

I have seen in some analog circuits that if we need a 2.5 V supply
there is way to do it that generate a 5V square pulse with duty cycle
1/2 and pass it through a low pass filter to get a DC 2.5V.

This method has problems with additional voltage drop through the low
pass filter when the load on the 2.5 volt output uses current.

I was thinking of another way of generating 2.5V. Use a voltage divider
and then use an opamp buffer.
As long as the load doesn't upset the opamp (less current than its

output can deliver without current limiting, and capacitance to ground
that does not make the buffer unstable) this can work fine. You can
also add a capacitor from the divider output to ground to clean up any
high frequency noise that was present on the 5 volt supply. The 2.5
volt output will be less accurate than the 5 volt supply (must include
divider tolerance and drift and opamp offset voltage).

Which of the above 2 ways is a recommended way of generating 2.5V DC?

I prefer the second for most low current needs. There are also fixed
and adjustable regulators that are made for this sort of thing.

For example:
http://www.national.com/ds/LM/LM117.pdf
http://www.national.com/ds.cgi/LM/LM4140.pdf
http://www.national.com/pf/LM/LM4120.html#Datasheet

These may improve on the voltage accuracy of the 5 volt supply.

If the load current is more than a few tens of milliamps, you may want
to look into a switching buck regulator, because of its higher efficiency.

 

[Larry Brasfield]

"aman" <aman.bindra@gmail.com> wrote in message
news:1114275196.548532.65400@l41g2000cwc.googlegroups.com...

So doesnt capacitance depend on conductivity of the dielectric ?

No. The effects act in parallel. When you apply an
electric field to a slightly conductive dielectric, some
bound charges move a limited distance then stop,
and some unbound charges drift continously. We
call the first displacement (or capacitive) current.
We call the second leakage. The movements of
those different charges are largely independent.

I mean
wont a more conductive sample of water give a different capacitance
than less conductive water ?

I don't think you will be able to measure a capacitance
change as you alter the ion concentration until you have
such a large concentration that the dielectric property
of the solute itself is a factor. By that time, measuring
the capacitance would be quite a trick.

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

 

[John Fields]

On Sat, 23 Apr 2005 00:31:26 -0700, "Larry Brasfield"
<donotspam_larry_brasfield@hotmail.com> wrote:

"John Fields" <jfields@austininstruments.com> wrote in message
news:0chi619a9bmorr8pt1pt5qu5ok498dhnv3@4ax.com...
On Fri, 22 Apr 2005 09:42:24 -0700, "Larry Brasfield"
donotspam_larry_brasfield@hotmail.com> wrote:

"John Fields" <jfields@austininstruments.com> wrote in message
news:560i61tfmof8mhjis826fj5q3dcghb53of@4ax.com...
On 22 Apr 2005 05:57:14 -0700, "mjohnson" <crvmp3@hotmail.com> wrote:
...
Before anything can happen we need to find out definitively what's
happening at the buzzer when the alarm goes off, or we need to find a
signal somewhere in the clock which changes state when the alarm goes
off.

The power needed to drive a buzzer will be many
times larger than what needs to be picked off to
activate another circuit, (many mW versus uW).
---
Depends. The OP's advocating using the signal driving the buzzer to
also drive the LED in an opto, which will be milliwatts VS milliwatts.

I was not addressing use of the optoisolator in that
position. However, if the current taken through the
opto LED is limited to a few 100 uA, such usage
would still be a small fraction of the buzzer power.
As you point out, CTR would be reduced, but no
more than a few uA of output would be needed.

---
That's not the point. As you've already stated, large value pullups
or pull-down resistors may be used in the remote in order to conserve
battery power during switching, and it's precisely that which makes
using an opto in other than a saturated mode problematical. Consider:

+V
|
[100k]
|
+----->Eout
|
O |
|<-
O |
|
GND

In this case, if the load on Eout is insignificant, (CMOS, say) Eout
will be either +V or, assumong GND is at 0V, 0V.

However, in this case:

+V
|
[100k]
|
+----->Eout
|
C A
B <--[LED]
E K
|
GND

Eout will depend on the collector-to-emitter resistance of the
transistor, and if it isn't driven low enough (if the current through
the LED isn't high enough) Eout may not cross the switching threshold
of the driven device.

The problem can also be exacerbated by the use of cheap conductive
rubber switches which require a substantially lower value pullup.

Additionally, if the remote used pull-down resistors, the opto's
transistor would be operating as a follower which, with only a 3V
supply available in the remote, would complicate matters even more.

All of these problems go away with the mechanical contacts of a reed
relay.
---

All the alarms with buzzers I have heard are very
loud (and annoying). It is hard to imagine getting
that without using many 10s of mW.

I wouldn't use an opto because of the current required for its LED and
the uncertainty of being able to use its transistor output to trigger
the remote.

Some such uncertainty is warranted, but I suggest that
there is reason to believe an opto-isolator will be fine.
The cheap (typically membrane) switches used in many
remotes are not asked to carry much current and, to
conserve battery power, large value pull-{up,down}
resistors are used. If a replacement for the contanct
had to carry more than 100 uA, I would be surprised.
The CTR (current transfer ratio) for opto-isolaters is
often guaranteed to be 100% or better, so a similar
current is all that the LED would need.

---
Not necessarily, CTR falls off quickly as LED forward current
diminishes and there are temperature effects which need to be taken
into consideration which can largely be ignored with a comparator-reed
switch solution. Also, with the reed switch solution there is no
saturation voltage VS LED If problem since it's either on or off.

I agree that the reed relay is simpler to apply. For that
reason alone, it may well be most suitable for the OP's
project. My suggestion about an optoisolator in its
place is more like a feasable alternative than any kind
of compelling improvement. Reed relays are fragile
and, if their leads are not carefully heat-sunk during
soldering, they can fail quickly or slowly as a result.

---
Yeah, right! I can just see millions of through-hole reed relays
going through wave-solder machines with little heat sinks attached to
their leads, LOL. Worse yet, millions of surface-mount units going
through soldering ovens with no heat sinks attached...
---

That, together with a dislike of moving parts, made
me think it might be an attractive alternative.

Finally, the signal sent thru the opto-isolator can be time limited
to just over what is needed for the remote in order to
conserve the battery.
---
Either solution will require the generation of a timed pulse to the
remote, so that's probably a wash.

Yes.

Here's what I see as a much simpler solution, with only the clock
output needing to be defined in order to make it work:

INTERFACE
BATTERY
CLOCK 4.5V REMOTE
BATTERY | BATTERY
3V +---------+---------+ 3V
| | | | |
+--+--+ +--+--+ +--+--+ +--+--+ +--+---+
|CLOCK|---|ALARM|---|1SEC |---|REED |---|REMOTE|
+-----+ | DET | |MONO | |RELAY|---|SWITCH|
+-----+ +-----+ +-----+ +------+

The opto-isolator would plug into that with little
change except reduction of the 4.5V battery drain
(unless my power surmises are completely wrong).

---
Could be. I'll defer judgement and wait until the OP comes back with
something definitive on the buzzer signal to post my design. If he
doesn't, I can always fall back on the acoustic thing I've already
posted.

Seems reasonable.

You may want to ask him about the current being conducted by
the remote's switch switch to see whether you can use an opto in
there. An easy way to determine the current would be to jump the
switch contacts with a milli/microammeter...

Yes.

--
John Fields
Professional Circuit Designer

 

[Larry Brasfield]

"John Fields" <jfields@austininstruments.com> wrote in message
news:qluk61lf2c2m6u277u034u9654lclp41fo@4ax.com...

On Sat, 23 Apr 2005 00:31:26 -0700, "Larry Brasfield"
donotspam_larry_brasfield@hotmail.com> wrote:

"John Fields" <jfields@austininstruments.com> wrote in message
news:0chi619a9bmorr8pt1pt5qu5ok498dhnv3@4ax.com...
On Fri, 22 Apr 2005 09:42:24 -0700, "Larry Brasfield"
donotspam_larry_brasfield@hotmail.com> wrote:

"John Fields" <jfields@austininstruments.com> wrote in message
news:560i61tfmof8mhjis826fj5q3dcghb53of@4ax.com...
....
The power needed to drive a buzzer will be many
times larger than what needs to be picked off to
activate another circuit, (many mW versus uW).
---
Depends. The OP's advocating using the signal driving the buzzer to
also drive the LED in an opto, which will be milliwatts VS milliwatts.

I was not addressing use of the optoisolator in that
position. However, if the current taken through the
opto LED is limited to a few 100 uA, such usage
would still be a small fraction of the buzzer power.
As you point out, CTR would be reduced, but no
more than a few uA of output would be needed.

---
That's not the point. As you've already stated, large value pullups
or pull-down resistors may be used in the remote in order to conserve
battery power during switching, and it's precisely that which makes
using an opto in other than a saturated mode problematical. Consider:

+V
|
[100k]
|
+----->Eout
|
O |
|<-
O |
|
GND

In this case, if the load on Eout is insignificant, (CMOS, say) Eout
will be either +V or, assumong GND is at 0V, 0V.

However, in this case:

+V
|
[100k]
|
+----->Eout
|
C A
B <--[LED]
E K
|
GND

Eout will depend on the collector-to-emitter resistance of the
transistor, and if it isn't driven low enough (if the current through
the LED isn't high enough) Eout may not cross the switching threshold
of the driven device.

I've seen commercial uses of photo-transistors
used as switches with 1M pullups. The transistors
themselves do fine at such levels, suffering only
slight beta reduction. Saturation resistance is
very nearly inversely proportional to excess
base current (or the equivalent photocurrent
for a phototransistor), so I see no reason to
expect the problem you allude to here.

As for using the phototransistor in a mode
other than saturated (or nearly off), I have
not suggested that. It might be useful, when
the receiver can deal with a non-switching
input, but that is not the case for the position
I suggested the opto for.

The problem can also be exacerbated by the use of cheap conductive
rubber switches which require a substantially lower value pullup.

I thought pullup values would go up in that case,
at least when expressed in Ohms. What makes
you say they would go lower?

Additionally, if the remote used pull-down resistors, the opto's
transistor would be operating as a follower which, with only a 3V
supply available in the remote, would complicate matters even more.

The output transistor can be, and often is, used as a two
terminal switch. Even if a base-emitter resistor is added
to control switching speed or leakage, (not needed here),
it can be used that way. Whether it pulls high or low is
not a complication and no follower need be created.

All of these problems go away with the mechanical contacts of a reed
relay.

Most or all of them go away on their own.

I agree that the reed relay is simpler to apply. For that
reason alone, it may well be most suitable for the OP's
project. My suggestion about an optoisolator in its
place is more like a feasable alternative than any kind
of compelling improvement. Reed relays are fragile
and, if their leads are not carefully heat-sunk during
soldering, they can fail quickly or slowly as a result.
---
Yeah, right! I can just see millions of through-hole reed relays
going through wave-solder machines with little heat sinks attached to
their leads, LOL. Worse yet, millions of surface-mount units going
through soldering ovens with no heat sinks attached...

Ok, that's funny. I was thinking of the bare reed
switches which, due to their glass envelope, are
fragile. Obviously, they are not so fragile when
packaged.

That, together with a dislike of moving parts, made
me think it might be an attractive alternative.

I guess that's not so funny.

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

 

[Lord Garth]

"CNM" <charles.macleod@gmail.com> wrote in message
news:19123424.0504230929.efc95c8@posting.google.com...

Why does the effciency of my low voltage transformer very much
decrease (to eg 10%) when increasing the frequency to around 40000hz.
Why does this increase in frequency affect the efficiency value so
much?

The impedance of the transformer is Xl=2*PI*F*L if F go up to 40KHz, Xl also
goes way up. Your power transformer was designed to work efficiently at
line
frequency, not 40KHz.

 

[Peter Bennett]

On 23 Apr 2005 04:41:48 -0700, Tommi-Vogel@gmx.de (Thomas Vogel)
wrote:

You´re right, but for me it had been a big puzzle work, to know how I
must construct the elements, if I don`t have seen the schematic with
the elements in the right order, here while I have answear you.

What are comparators? I don´t have find it in my dictornary. Can you
explain it?

A comparator is an integrated circuit which has two inputs (called +
and -), and one output. When the + input is more positive than the -
input, the output will be high. If the - input is more positive, the
output will be low.

And what is (COIL) and (1N4148)? Is the last one a diode?

(COIL) is the coil of the relay, and (1N4148) is a diode. The "K"
indicates the cathode end of the diode.

And who are these conected? * *

| | +-|+\ | | | | | O--> |
| | | | >-+------|-|+\ | | |
[BAT] +-[0.1ľF]-+-----|-|-/ | | | >--+-------+ +----------->NO
| | | | | +-|-/


And what did this mean:

--|+\
| >---
--|-/

That is a comparator

You will have to go to the National Semiconductor website (or google
for "LMV393 datasheet") to get the datasheet for the comparator that
John suggested - The LMV393 is a dual comparator - there are two
separate comparators in the package. I would suggest using an LM393
instead, as the LMV393 is in a tiny surface mount package which will
be hard to handle.


--
Peter Bennett, VE7CEI
peterbb4 (at) interchange.ubc.ca
new newsgroup users info : http://vancouver-webpages.com/nnq
GPS and NMEA info: http://vancouver-webpages.com/peter
Vancouver Power Squadron: http://vancouver.powersquadron.ca

 

[John Larkin]

On 23 Apr 2005 10:29:17 -0700, charles.macleod@gmail.com (CNM) wrote:

Why does the effciency of my low voltage transformer very much
decrease (to eg 10%) when increasing the frequency to around 40000hz.
Why does this increase in frequency affect the efficiency value so
much?

Core loss and skin loss. Cheap, thick steel laminations in the core
work fine at 60 Hz but have lots of eddy-current loss at hf. The wire
itself has skin and proximity effects, too. I'm guessing the core loss
is the biggie.

Besides, if it works at 60 Hz, it has far too many turns to be optimum
at 40K. A ferrite torroid or pot core would be much better up there.

John

 

[Larry Brasfield]

"CNM" <charles.macleod@gmail.com> wrote in message
news:19123424.0504230929.efc95c8@posting.google.com...

Why does the effciency of my low voltage transformer very much
decrease (to eg 10%) when increasing the frequency to around 40000hz.
Why does this increase in frequency affect the efficiency value so
much?

The most likely reason is that eddy current losses
in the core go up and magnetizing current goes up.
In effect, the shunt inductance of the transformer
goes down as frequency goes up when eddy
currents become signficant. Those currents are
(nearly) in phase with the applied voltage and
represent a loss.

You probably have a transformer designed for
line frequency with the lamination thickness set
accordingly. At 40 KHz, that thickness is way
too large.

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

 

[Anthony Fremont]

"John Fields" wrote

On Sat, 23 Apr 2005 05:13:57 GMT, "Anthony Fremont"

I came up with the same calculation as you. If LEDs are dieing, then
20mA may be a bit too much current. Since there are two LEDs in
series,
one may hog more current than the other resulting in its demise.

---
So, you're saying that because the LEDs may not be identical one may
be drawing more current than the other?

Perhaps "dissipate more power" would have been more appropriate than
"hog more current".

 

[John Fields]

On 23 Apr 2005 04:41:48 -0700, Tommi-Vogel@gmx.de (Thomas Vogel)
wrote:

John Fields <jfields@austininstruments.com> wrote in message news:<2bjf61l7clur0educ44sj4q5ngpiqjlubp@4ax.com>...
On 21 Apr 2005 07:06:35 -0700, Tommi-Vogel@gmx.de (Thomas Vogel)
wrote:

John Fields <jfields@austininstruments.com> wrote in message news:<agta611dbh5utnvos47ggtfdve6ufglr48@4ax.com>...
On 19 Apr 2005 11:03:51 -0700, Tommi-Vogel@gmx.de (Thomas Vogel)
wrote:

Hi,

no, it´s not for a bomb! I just need it for my music station to woke up in
the morning. My new music station has no timer and I can´t switch it with a
timer. So I want to construct a circuit, which switch by an alarm clock over
a relay a remote control. Pleas email me the schematic+parts list.

---
Here ya go:


+---+---------------+------+------+---+---+-------+
| | | | | | | |
| [10k] [1M] [100K] [10K] | |K | O------>C
| | | | | |[1N4148] [COIL]- -|
| | +-|+\ | | | | | O--> |
| | | | >-+------|-|+\ | | |
[BAT] +-[0.1ľF]-+-----|-|-/ | | | >--+-------+ +----------->NO
| | | | | +-|-/
| | | | |+ | |
| [MIC] [100k] [10K] [10ľF] [1M] |
| | | | | | |
+---+---------+-----+------+------+---+


The microphone is a Panasonic WM-61A, the comparators are part of an
LMV393, the battery is 3X 1.5VAA, and the relay is a COTO 9007-05-40.

Here's how it works:

After you finish building and testing it, put it in a soundproof box
with your alarm clock and connect the relay contacs (C and NO, above)
to the remote control. Voila!, when the alarm clock goes off you
won't hear it, bit it'll turn on your music station. (Radio???)

Or, you could forget the whole thing and just let the alarm clock wake
you up.




hi,

thank you for this schematic, but please send it at my e-mail adress
in a better format as a picture, that I can better realize it.

No. I prefer to keep discussions like this here, so that everyone can
participate in arriving at a solution for you, if they choose.

Writing schematics in ASCII is a way to use very little bandwidth to
convey a lot of information, and also to make sure that when the
article is archived the information stays with it essentially forever.
If you want to play, you need to learn how to use the toys. If the
schematic is garbled, view it using a non-proportional font like
Courier, and if you don't understand the operation of the circuit
post your questions here and one or some of us will be happy to help
you.


You´re right, but for me it had been a big puzzle work, to know how I
must construct the elements, if I don`t have seen the schematic with
the elements in the right order, here while I have answear you.

What are comparators? I don´t have find it in my dictornary. Can you
explain it?

---
http://home.cogeco.ca/~rpaisley4/Comparators.html
---

And what is (COIL) and (1N4148)? Is the last one a diode?

---
[COIL] is the coil of the reed relay, and [1N4148] is a diode.
---

And who are these conected? * *

| | +-|+\ | | | | | O--> |
| | | | >-+------|-|+\ | | |
[BAT] +-[0.1ľF]-+-----|-|-/ | | | >--+-------+ +----------->NO
| | | | | +-|-/

Means this:

\
--+
/

that two ways are conected to one?

---
If you see a circuit that looks like this:

+V
|
[R1]
|
+---->V2
|
[R2]
|
gnd

This is what it means:

+V <--- A positive voltage, a power supply
|
[R1] <--- A resistor
|
A connection --> +---->V2 <--- An output. In this case, a voltage
|
[R2] <--- A resistor
|
GND <--- The power supply return

---

And what did this mean:

--|-|+\
| |
+-|-/

---
This is a symbol for an opamp or a comparator:


-|+\
| >--
-|-/


And this is what it means:


Non-inverting input--> -|+\
| >-- <--- Output
Inverting input -----> -|-/

---

in the middel?

---

I use the brackets to indicate components, with the values contained
within. For example:


--[100k]-- is a 100,000 ohm resistor

--[0.1ľF]-- is a 0.1ľF capacitor
---

And what is with the K over the(1N4148)?

---
That indicates the cathode end of the diode.
---

and to what go the arrows at
the right side?

---
Those are the outputs of the circuit; the Normally Open (NO) and
the Common (C) contacts of the relay.

And what mean there the [ | ] ?

---
???

If this is what you're asking about:

O
|
O--> |

those are the relay contacts.

Also,

--------------------- indicates a horizontal wire,



|
|
|
|
|
|
|
|

indicates a vertical wire,


|
|
|
----+
|
|
|

indicates a connection, and


|
|
|
----------
|
|
|

or

|
|
|
----|-----
|
|
|

indicate wires crossing without being connected.

--
John Fields
Professional Circuit Designer

 

[Larry Brasfield]

"aman" <aman.bindra@gmail.com> wrote in message
news:1114276814.684474.157420@z14g2000cwz.googlegroups.com...

In C= KA/d

Usually, 'K' refers to the relative dielectric "constant" and
would be multiplied by the permittivity of free space in
the above formula.

if K does not depend on conductivity, what makes a
dielectric have different dielectric constant K. What are the
parameters on which K depends on?

It is affected by how easily bound charges can be
displaced and the density of those charges. For
example, in water, the bound charges are at
sort of opposite ends of the H2O molecule and
displacement occurs as they become polarized
in an E field.

Your question covers a lot of territory.

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

 

[John Bokma]

Anthony Fremont wrote:

"John Bokma" wrote:

snipped exactly what I was going to post

Note that I did this calculation, later checked it with a voltage
meter,
and yet already 2 exotic LEDs died on me. I think it's a bad badge.
(Unless someone can point out my errors).

I came up with the same calculation as you. If LEDs are dieing, then
20mA may be a bit too much current. Since there are two LEDs in
series, one may hog more current than the other resulting in its
demise.

And where does it store that extra current?

You
might consider not driving them so hard. There is probably a
relatively insignificant brightness difference between 10mA and 20mA
anyway.

In my case: 30 mA is max, so feeding them 22 mA shouldn't be that bad.
Moreover, the LED died when the circuit at
http://johnbokma.com/pet/scorpion/detection-using-uv-leds.html

was connected to 9V, or maybe even 7V (less then 12V anyway).



--
John MexIT: http://johnbokma.com/mexit/
personal page: http://johnbokma.com/
Experienced programmer available: http://castleamber.com/
Happy Customers: http://castleamber.com/testimonials.html

 

[Larry Brasfield]

"John Fields" <jfields@austininstruments.com> wrote in message
news:qluk61lf2c2m6u277u034u9654lclp41fo@4ax.com...

[Brasfield had suggested an optoisolator in parallel with

a button on a remote.]

That's not the point. As you've already stated, large value pullups
or pull-down resistors may be used in the remote in order to conserve
battery power during switching, and it's precisely that which makes
using an opto in other than a saturated mode problematical. Consider:
[Points made and responded to elsewhere cut.]

Brasfield once wrote:
... I suggest that there is reason to believe an opto-isolator will be fine.
and later wrote:
... I agree that the reed relay is simpler to apply. For that
reason alone, it may well be most suitable for the OP's project.

It occurs to me now that the most problematical attribute
of the optoisolator in that position is its capacitance. If
that had an effect only at button pushing speeds, it would
hardly matter. But it is quite likely that the buttons on the
remote at in a matrix and scanned via pulses applied to
the matrix and responses sampled shortly after changes.
The commonly large capacitance of the phototransistors
used in optoisolators could interfere with such scanning.

The reed relay is more certain to work for that reason.

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