Chip with simple program for Toy

[Byron A Jeff]

In article <8028c236.0407310039.45737ad0@posting.google.com>,
wylbur37 <wylbur37nospam@yahoo.com> wrote:

When using an LED from a power source whose voltage is higher than the
rating of the LED, a resistor is typically used.

Actually you still need the resistor even when the voltage is at the
rating of the LED albeit a small one. It serves the purpose of current
limitation in addition to dropping voltage.

If more than one LED is to be used (let's say three),
each one would have its own respective resistor,
and the schematic would look something like this ...


+----------+-----------+-----------+--------------
| | | |
| | | |
| LED-1 LED-2 LED-3
power | | |
source | | |
| resistor-1 resistor-2 resistor-3
| | | |
| | | |
+----------+-----------+-----------+--------------

Right. It's the correct way to get uniform brightness among the set.

But if all the LEDs were of the same voltage and amperage
(e.g., if they were all of the typical 3.6V 20mA),
isn't there a simpler way to do this by using only ONE resistor?
In other words, can't it be done like the following ?


+--------------+-----------+-----------+---
| | | |
| | | |
power LED-1 LED-2 LED-3
source | | |
| | | |
| | | |
+---resistor---+-----------+-----------+---

It serves the safety purpose, but then you get variability on the brightness
because each LED will not draw exactly the same current. So some will be
dimmer than others.

The second thing is that the total current through the resistor will be a
factor of the current through each LED. So for your example the resistor
above would need to allow 60ma of current to flow through to feed the LEDs
downline. The problem is that if one LED blows out then that current will have
to be pushed through the other working LEDs, which can cause the others to
fail also.

And what would be the proper value of the resistor?

Single resistor value divided by the number of LEDs across it.

(I'm guessing it would be one third of what it was in the first diagram
because the required voltage reduction would be the same but the
effective current draw of the 3-LED assembly would be three times what it
was before.
So if the power source were 6V and the LEDs were 3.6V and 20mA each,
then the required resistance would be 120 ohms in the first case
and 40 ohms in the second case. Correct?)

Right. But see the caveat above.

A better way to do it is to bump the voltage up and put the LEDs in parallel
like this:

+V -> resistor -> LED1 -> LED2 -> LED3 -> GND

You treat the equation the same as a single LED except that you add the Vf
voltages of the LEDs. So for your example the total voltage of the LEDs
would be 10.8V. So the resistor you'd use with a 12V +V would be:

(12V - 10.8V) / .02A = 60 ohms.

But there are significant differences:

1) Each LED will draw the same current, so you'll get uniform brightness.
2) If a single LED (or the resistor) fails, then the whole string goes out
with only the single failed component, instead of going into cascade failure
due to increasing overload.

When doing a single resistor, that's the better way to handle it.

BTW, thanks for crossposting this message. There are many who don't understand
the concept. But I think I'll limit my replies to basics and misc, which are
appropriate.

BAJ

 

[Watson A.Name - \"Watt Su]

"Byron A Jeff" <byron@cc.gatech.edu> wrote in message
news:ceg5i9$q93@cleon.cc.gatech.edu...
[snip]

A better way to do it is to bump the voltage up and put the LEDs in
parallel
like this:

+V -> resistor -> LED1 -> LED2 -> LED3 -> GND

That is NOT parallel, that is SERIES.

[snip]

BAJ

 

[CFoley1064]

Subject: Convert 110v to 240v
From: "Neil Marsh" neilmarsh@NOJUNKqasolutions.fsbusiness.co.uk
Date: 7/31/2004 9:03 AM Central Daylight Time
Message-id: <ceg9t5$ntf$1@news8.svr.pol.co.uk

Advise sought....

In the early 80's I was able to convert Apple computers (110v US imports) to
run on UK mains by swapping the transformer connections.

Can this procedure be used to convert a 110v cement mixer formerly used on a
building site ? - hence 110v to comply with regulations.

--
Remove NOJUNK to email

Regards,

Neil

Doubtful but possible. Find the electric motor, and look on the nameplate for
a wiring diagram. Some AC motors can be wired for either 120VAC or 240VAC. If
so, you might want to have an electrician take a look at things before you
perform the modification. If there's anything else electrical on the mixer
besides a motor, you should also check to see if that will be OK with the mods,
too.

Good luck
Chris

 

[Rich Grise]

Byron A Jeff wrote:

wylbur37 <wylbur37nospam@yahoo.com> wrote:
....
+--------------+-----------+-----------+---
| | | |
| | | |
power LED-1 LED-2 LED-3
source | | |
| | | |
| | | |
+---resistor---+-----------+-----------+---
...
And what would be the proper value of the resistor?

Single resistor value divided by the number of LEDs across it.

Wrong answer. There is no "proper" value for this resistor,
because it's just plain wrong. It _can_ be done, if you want
to spend a day or so matching LEDs, but any perturbation in
LED current will gave a positive feedback effect, so the LEDs
_will_ fail - it's only a matter of time.

Put a resistor in series with each, or with a series string.

Figuring out the resistor value is left as an exercise for the
reader.

Cheers!
Rich

 

[CFoley1064]

Subject: Sensing the flow of current
From: Daniel Rudy
i0n1v2a3l4i5d6d7c8r9u0d1y2e3m4a5i6l7@n0o1p2a3c4b5e6l7l8s9p0a1m2.3n4e5t6
Date: 8/1/2004 7:42 AM Central Daylight Time
Message-id: <az5Pc.5293$AY5.240@newssvr21.news.prodigy.com


Hello Everyone!

I've ran into a situation where I need to sense a very large flow of
current (100A) at 24VDC or so. This is for a motor supply feed. The
problem that I'm running into is isolation because there is a
microcontroller that is part of the control circuit that electrically
needs to stay away from this line as its powered from a different
source. So far, there are two ways that I'm aware of to do this:

1. Use a very low value resistor (I would need a few in parallel),
sense and amplify the voltage across that. Then feed that into a
voltage to frequency converter so it can be sent across an opto-isolator
and then on to a timer on the microcontroller.

2. Wind a couple of turns of the main feed wire onto a torridal core
form. The torrid has a gap cut into it. Inside that gap is a hall
effect device that is epoxied into place. This inherently provides the
galvanic isolation that is required. After signal conditioning, the
analog voltage can be sent directly to a ADC input on the controller.

I would prefer to use #2 as that would be easier to implement, and it
seems to have a lower parts count. Accuracy in the area of 1 amp
resolution or better, if possible, would be good. Any advise, links,
etc. is appreciated. Thanks.

--
Daniel Rudy

Good morning, Dan. If you need isolated sensing of AC current, you would use a
current transformer. For DC current, the best way might be to use a hall
effect sensor. Several manufacturers make hall effect sensors built onto a
zero ohm shunt with integrated electronics. The current is then sensed by the
electrically isolated circuit, which gives you an analog output voltage
proportional to the DC current.

Allegro makes integrated current sensors which are good to up to 100 amps,
available off the shelf. One product you might want to look at is the
ACS750SCA-100, which operates on your microcontroller 5V supply. These are
available from stock at newark.com and other sources.

http://www.allegromicro.com/sf/0750/

Good luck
Chris

 

[Mantra]

"Jacky Luk" <jl@knight.com> wrote in message news:<cedje1$9se3@imsp212.netvigator.com>...

Hi, I'm just a little bit confused how the grounding systems work. First,
when you have a closed circuit, a ground is attached at bottom part of the
circuit (Black, inactive, 0V Potential are given the names. I know that it
is called a reference point sometimes and used to prevent short circuit from
endangering human lives, but how does it work? I ask this question because I
find that the ground symbol that is connected to the common is excessive.
Why won't the current flow to the 0V of the emf while not flowing into the
ground. Also, when an 1000Ohms human touches the active/Red/Fire line, how
does ground systems help to prevent damage to human lives (with fuse)?
Thanks
Jack

http://www.codecheck.com/grounding.htm

 

[J Jensen]

hhc314@yahoo.com (Harry Conover) wrote in message news:<7ce4e226.0408050909.360a3c@posting.google.com>...

jjensen14@hotmail.com (J Jensen) wrote in message news:<c3f2060d.0407230934.7ba8b8e@posting.google.com>...
"~KJPRO~" <KJPRO @ STARBAND.NET> wrote in message news:<%j1Mc.741$VQ1.264@fe25.usenetserver.com>...
"J Jensen" <jjensen14@hotmail.com> wrote in message
news:c3f2060d.0407221116.4fd3587f@posting.google.com...

7. It isn't worthwhile to check on the amount of Freon (or whatever) that
is
in the system -- all that matters is measuring the temperature of the
cold air coming out (say 62 F) and the outside temperature or maybe the
attic temperature.
--Jeff

This has to be some of the purist BS I have read lately.

~kjpro~

This was what an a/c technician told me when I called him out to the
house
with the specific request to check if any Freon had leaked out of the
system over the years, or if it was still operating at 100%. He would
not do it,
although he did check some other things. On the bright side, he did
encourage
me to sign up for yearly maintenance by his company...
--Jeff

Your technician was correct. He would determine if it was operating at
100% though use his gauge set for pressure measurements, and the
temperatures measured on the evaporator and condensor coil surfaces to
approximate this determination. If these are within the prescribed
ranges, there likely is little need to top off the refrigerant charge
in the system.

Yes, after looking at the papers stored with the unit, I see that the
instructions are to measure the outdoor temp, the indoor wet bulb temp,
the suction pressure and the liquid pressure and from this calculate
some quantity called "superheat" and check that against a chart they
give.

The question that most interests me still is whether the unit can possibly
use less current when running at night, or when losing heat to the outside
very efficiently like with water running over the coils. There seems to
be no way to measure this without splicing an high capacity ammeter into
the circut (and I am unwilling to do that since I have a healthy fear
of electricity). I want to say that the resistance of the motor is
constant and can't depend on the difficulty of the motor's task, the power
company is supplying a constant 220 volts at 60 cycles/sec, and that is
all there is to it....?

By the way, does anyone know what are the words that are abbreviated here:
"Min. Brch. Cir. Ampacity" 30 [minimum branch circut amps?]
Br. Cir. -- Max. (Amps) 50
Prot. Rtg -- Recmd. (amps) 50

R.L. Amps -- L.R. Amps 22.0 -- 95

thanks,
Jeff

 

[Keith]

[...]

The question that most interests me still is whether the unit can possibly
use less current when running at night, or when losing heat to the outside
very efficiently like with water running over the coils. There seems to
be no way to measure this without splicing an high capacity ammeter into
the circut (and I am unwilling to do that since I have a healthy fear
of electricity). I want to say that the resistance of the motor is
constant and can't depend on the difficulty of the motor's task, the power
company is supplying a constant 220 volts at 60 cycles/sec, and that is
all there is to it....?

When the condenser temerature is colder, the head pressure the
compressor sees is less, and it doesn't have to pump as hard to condense
the vapor.
To measure the amperage draw, a simple "amp clamp" meter will tell
you what the compressor is drawing, without "splicing in" a meter.

By the way, does anyone know what are the words that are abbreviated here:
"Min. Brch. Cir. Ampacity" 30 [minimum branch circut amps?]

Minimum branch circuit amperage capacity - the size of wire and
overload device for the circuit.

Br. Cir. -- Max. (Amps) 50

The maximum continous load amperage one might expect.

Prot. Rtg -- Recmd. (amps) 50

Protection rating, recommended for the circuit.

R.L. Amps -- L.R. Amps 22.0 -- 95

Runnig load amps - normal full load continous amperage draw.

Locked rotor amperage - if the compressor fails or motor bearings
sieze, or trying to start "dead head", the simple amperage draw of the
windings as a resistive heater.

Keith

thanks,
Jeff

 

J Jensen <jjensen14@hotmail.com> wrote:

The question that most interests me still is whether the unit can possibly
use less current when running at night, or when losing heat to the outside
very efficiently like with water running over the coils. There seems to
be no way to measure this without splicing an high capacity ammeter into
the circut...

A clamp-on meter would be a start. I measured a drop from 1050 to 950 W
(including the $10 10 watt Harbor Freight fountain pump) when trickling
rainwater over a window AC, using a $200 Brand Electronics wattmeter.

TURTLE implied that an AC rated at 36K Btu/h at 95 F might make 33K or
31K at 105 F. I'd like to see a manufacturer's COP vs temp diff curve.

Nick

 

[J Jensen]

Keith <keithrol@ixpres.com> wrote in message news:<4114acf6_1@news.vic.com>...

The question that most interests me still is whether the unit can possibly
use less current when running at night, or when losing heat to the outside
very efficiently like with water running over the coils. There seems to
be no way to measure this without splicing an high capacity ammeter into
the circut (and I am unwilling to do that since I have a healthy fear
of electricity). I want to say that the resistance of the motor is
constant and can't depend on the difficulty of the motor's task, the power
company is supplying a constant 220 volts at 60 cycles/sec, and that is
all there is to it....?

When the condenser temerature is colder, the head pressure the
compressor sees is less, and it doesn't have to pump as hard to condense
the vapor.
To measure the amperage draw, a simple "amp clamp" meter will tell
you what the compressor is drawing, without "splicing in" a meter.

Thanks for the information, Keith. I guess I am going to have to do some
more work understanding electric motors. I have a $20 clamp on multimeter
that I got from Radio Shack, but evidently it must not be too good since
all the other ones I've seen cost around $180.

--Jeff

 

[J Jensen]

nicksanspam@ece.villanova.edu wrote in message news:<cf2es9$3r4@acadia.ece.villanova.edu>...

J Jensen <jjensen14@hotmail.com> wrote:

The question that most interests me still is whether the unit can possibly
use less current when running at night, or when losing heat to the outside
very efficiently like with water running over the coils. There seems to
be no way to measure this without splicing an high capacity ammeter into
the circut...

A clamp-on meter would be a start. I measured a drop from 1050 to 950 W
(including the $10 10 watt Harbor Freight fountain pump) when trickling
rainwater over a window AC, using a $200 Brand Electronics wattmeter.

TURTLE implied that an AC rated at 36K Btu/h at 95 F might make 33K or
31K at 105 F. I'd like to see a manufacturer's COP vs temp diff curve.

Nick

Thanks for the information, Nick. I'm just going to have to do the experiment
and see for myself... I notice you keep using the abbreviation COP --
what does that mean?

--Jeff

 

J Jensen <jjensen14@hotmail.com> wrote:

TURTLE implied that an AC rated at 36K Btu/h at 95 F might make 33K or
31K at 105 F. I'd like to see a manufacturer's COP vs temp diff curve.

...and see for myself... I notice you keep using the abbreviation COP --
what does that mean?

The Coefficient Of Performance is the ratio of heat energy moved to
electrical energy used, using the same energy units, eg watts. It's
about 3 for a typical AC, vs 10,000 for a good solar heating system

Nick

 

[bill]

In article <c3f2060d.0408081249.1098817b@posting.google.com>,
jjensen14@hotmail.com (J Jensen) wrote:

Thanks for the information, Keith. I guess I am going to have to do some
more work understanding electric motors. I have a $20 clamp on multimeter
that I got from Radio Shack, but evidently it must not be too good since
all the other ones I've seen cost around $180.

--Jeff

Darwin nominee?

 

[Harry Conover]

nicksanspam@ece.villanova.edu wrote in message news:<cf64nt$4hf@acadia.ece.villanova.edu>...

J Jensen <jjensen14@hotmail.com> wrote:

TURTLE implied that an AC rated at 36K Btu/h at 95 F might make 33K or
31K at 105 F. I'd like to see a manufacturer's COP vs temp diff curve.

...and see for myself... I notice you keep using the abbreviation COP --
what does that mean?

The Coefficient Of Performance is the ratio of heat energy moved to
electrical energy used, using the same energy units, eg watts. It's
about 3 for a typical AC, vs 10,000 for a good solar heating system

What is the value for a good solar A/C system?

Harry C.

 

[Rich Grise]

pil wrote:

Hi, I am still having trouble with my amplified ear. I noticed that my
headphones have a dc resistance of 16ohms per ear piece. The ones
specified in amplified ear has impedances of 32 ohms wired in series to
create a 64ohm load on Q4.

Could this be why my circuit does not work?

It _could_ be, but it could also be any number of other things. You
haven't shown any component values here. Do you even have a meter?
Is the mic known good? Have you double-checked all your connections?

We need much more to go on here.

Good Luck,
Rich

 

[Rileyesi]

These fans will be placed in a way that they
blow air at the sails, thus propelling the boat.

The 'forward' force of the wind upon the sails will be offset by the 'backward'
force generated by the fan and the boat will not move...in theory. In
practice, I would guess that it would actually move in a mostly backward motion
because the force induced on the boat by the fan's wind on the sails would be
less than the force induced on the boat by the fan itself.

 

[~^Johnny^~]

On Mon, 09 Aug 2004 04:00:16 GMT, bill <tsurber@columbus.rr.com> wrote:

In article <c3f2060d.0408081249.1098817b@posting.google.com>,
jjensen14@hotmail.com (J Jensen) wrote:

Thanks for the information, Keith. I guess I am going to have to do some
more work understanding electric motors. I have a $20 clamp on multimeter
that I got from Radio Shack, but evidently it must not be too good since
all the other ones I've seen cost around $180.

--Jeff

Darwin nominee?

Go to hell.




--
-john
wide-open at throttle dot info

 

[John Miller]

Peter Sammon wrote:

I guess from the beginning of time itself humanity has shown its
strengths and weaknesses sometimes with very little subtlety.
441 lines snipped

Whew! So, what was your point, again?

--
John Miller
Email address: domain, n4vu.com; username, jsm

A little experience often upsets a lot of theory.

 

[dan]

What's that Lassie? You say that gothika fell down the old
sci.electronics.basics mine and will die if we don't mount a rescue by
Fri, 20 Aug 2004 19:30:31 -0500:

You could easily convert that emergency light to rechargables.
I've got a couple I did this with.
Just wire in coaxial power jack and use a wall wart to keep the
rechargables in them charged up to peak.
I use the six volt flashlights that normally use the 6v lantern
battery and have subbed various 6v nicad packs(old style 6v batteries
for camcorders mostly.) instead.
This leaves plenty of room for wiring in the coax jacks.

Look for a SLA lantern battery Part # UB5-6S
At: http://www.batterystation.com
--

Dan

 

[Watson A.Name - \"Watt Su]

"Rich Grise" <null@example.net> wrote in message
news:eA8Sc.985$EQ5.215@nwrddc03.gnilink.net...

pil wrote:

Hi, I am still having trouble with my amplified ear. I noticed that
my
headphones have a dc resistance of 16ohms per ear piece. The ones
specified in amplified ear has impedances of 32 ohms wired in series
to
create a 64ohm load on Q4.

Could this be why my circuit does not work?

It _could_ be, but it could also be any number of other things. You
haven't shown any component values here. Do you even have a meter?
Is the mic known good? Have you double-checked all your connections?

We need much more to go on here.

Here is the original circuit.
http://www.redcircuits.com/Page38.htm

I built this and found that the compressor doesn't work very well. Of
course it has only a volt and a half to work with. But it does amplify.

You should check the voltages, and if your earphones have about .2VDC
across them then the output stage is feeding it a current of about 6 mA,
which is most of the current for the whole amplifier.

You might disconnect D1 and the base of Q3 from C3 and P1. This will
shut off the compressor circuit. Then see what happens. BTW, I found
that for R2, 1 megohm was a bit high, I think I used 560k or so. But it
depends on the gain of the preamp transistor Q1.

I built mine into the handset of a telephone. I sawed off the
mouthpiece. The DC resistance of the handset is more like 150 ohms.

Good Luck,
Rich