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'[EE] LED with built in resistor'
2009\07\04@222502 by solarwind

picon face
The most annoying thing about placing LEDs on your breadboard for pin
status indication is that you need a resistor to go along with it. Why
don't they just build the resistor right into the LED so it makes
everyone's life easier? Do such products exist?

-- [ solarwind ] -- http://solar-blogg.blogspot.com/

2009\07\05@011315 by John Coppens

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face
On Sat, 4 Jul 2009 22:24:41 -0400
solarwind <spam_OUTx.solarwind.xTakeThisOuTspamgmail.com> wrote:

> Why
> don't they just build the resistor right into the LED so it makes
> everyone's life easier? Do such products exist?

Yes they do exist.

John

2009\07\05@013655 by Bob Blick

face
flavicon
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solarwind wrote:
> The most annoying thing about placing LEDs on your breadboard for pin
> status indication is that you need a resistor to go along with it. Why
> don't they just build the resistor right into the LED so it makes
> everyone's life easier? Do such products exist?

http://search.digikey.com/scripts/DkSearch/dksus.dll?Detail&name=516-1306-ND

Basically, search for something like "5V LED" and you'll find them.

Be prepared to pay extra and have fewer parts to choose from.

And since the resistor heat is produced within the package, lower
performance.

-Bob

2009\07\05@014718 by Mark Rages

face picon face
On Sun, Jul 5, 2009 at 12:37 AM, Bob Blick<.....bobblickKILLspamspam@spam@ftml.net> wrote:
> solarwind wrote:
>> The most annoying thing about placing LEDs on your breadboard for pin
>> status indication is that you need a resistor to go along with it. Why
>> don't they just build the resistor right into the LED so it makes
>> everyone's life easier? Do such products exist?
>
> http://search.digikey.com/scripts/DkSearch/dksus.dll?Detail&name=516-1306-ND
>
> Basically, search for something like "5V LED" and you'll find them.
>
> Be prepared to pay extra and have fewer parts to choose from.
>
> And since the resistor heat is produced within the package, lower
> performance.
>
> -Bob

Are there any with a built-in current source?

Regards,
Mark
markrages@gmail
--
Mark Rages, Engineer
Midwest Telecine LLC
markragesspamKILLspammidwesttelecine.com

2009\07\05@015218 by Brian B. Riley

picon face

Here is the Mouser Number for the Green LEDS

SSL-LX5093GD-12V

and for the Red 5 VDC;

SSL-LX5093ID-5V


the green LEDs for 12V work fine in 5 volt logic

On Jul 4, 2009, at 10:24 PM, solarwind wrote:

> The most annoying thing about placing LEDs on your breadboard for pin
> status indication is that you need a resistor to go along with it. Why
> don't they just build the resistor right into the LED so it makes
> everyone's life easier? Do such products exist?
>
> -- [ solarwind ] -- http://solar-blogg.blogspot.com/
> --

2009\07\05@021151 by Bob Blick

face
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Mark Rages wrote:
> On Sun, Jul 5, 2009 at 12:37 AM, Bob Blick<.....bobblickKILLspamspam.....ftml.net> wrote:
>> solarwind wrote:
>>> The most annoying thing about placing LEDs on your breadboard for pin
>>> status indication is that you need a resistor to go along with it. Why
>>> don't they just build the resistor right into the LED so it makes
>>> everyone's life easier? Do such products exist?
>> search.digikey.com/scripts/DkSearch/dksus.dll?Detail&name=516-1306-ND
>>
>> Basically, search for something like "5V LED" and you'll find them.
>>
>> Be prepared to pay extra and have fewer parts to choose from.
>>
>> And since the resistor heat is produced within the package, lower
>> performance.
>>
>> -Bob
>
> Are there any with a built-in current source?

I've never seen one. I have used a JFET to do a quick constant current,
works OK but the current is determined by the JFET and not always what
you want, and works best over 5 volts(including LED).

-Bob

2009\07\05@035109 by Vitaliy

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solarwind wrote:
> The most annoying thing about placing LEDs on your breadboard for pin
> status indication is that you need a resistor to go along with it. Why
> don't they just build the resistor right into the LED so it makes
> everyone's life easier?

Resistor value would have to be different depending on the voltage (higher
for 12V compared to 5V). Or you may decide to string a bunch of LEDs
together in series, in which case you may not need the resistor at all.

Different levels of brightness is another consideration.

Etc.

Vitaliy

2009\07\05@044803 by Clint Sharp

picon face
In message <EraseMEa94764e0907041924l2abd8a4aoeccb68560df74646spam_OUTspamTakeThisOuTmail.gmail.com>,
solarwind <x.solarwind.xspamspam_OUTgmail.com> writes
>The most annoying thing about placing LEDs on your breadboard for pin
>status indication is that you need a resistor to go along with it. Why
>don't they just build the resistor right into the LED so it makes
>everyone's life easier? Do such products exist?
>
>-- [ solarwind ] -- http://solar-blogg.blogspot.com/
Yes they do but why don't you just snip one LED leg short and solder the
resistor to the stub to replace the leg?

I have done this for years for my breadboard kit and it works fine, plus
you get to choose your own current, LED colour and style to suit your
purpose.
--
Clint Sharp

2009\07\05@091208 by olin piclist

face picon face
solarwind wrote:
> The most annoying thing about placing LEDs on your breadboard for pin
> status indication is that you need a resistor to go along with it.

Oh gimme a break.  Adding a single resistor to a board is both trivially
easy and cheap.

> Why don't they just build the resistor right into the LED
> so it makes everyone's life easier?

Everyone's?  Are you sure?  Did you ask everyone?  I don't remember you
asking me.  For someone that is early in learning electronics it is a good
idea to start out assuming things are as they are for a reason, and if it
doesn't appear to make sense at first glance that this is probably because
you don't understand the reason rather than everyone else missed something.

LEDs are made with different processes than resistors.  Integrating a
resistor in a LED probably costs more than for you to put a separate
resistor on your board.

Then the market for such a LED would be more limited.  Now you have to drive
it with a specific voltage to get full brightness.  Without a resistor you
can chose the external resistance so that the same LED can be used in 3.3V
circuits, 5V circuits, or others.

You don't always want to drive a LED at its full rated current.  Most
ordinary LEDs are rated for 20mA continuous current to get full brightness.
Today's LEDs are bright enough that I often run them at 10mA or even 5mA
depending on other particulars of the product.  To cater to this and the
variations in drive voltage, a manufacturer would have to have variants of
the same LED with different resistors.  This would drop the volume and
therefore raise the price even more per unique part.

Sometimes LEDs aren't driven with series resistors at all.  The driving
circuit can already be a current source and you don't want a resistor in
series.

> Do such products exist?

Actually they do because in some niches they can make sense.  Go look around
and you will find the selection far more limited than for bare LEDs, and the
price much higher.


********************************************************************
Embed Inc, Littleton Massachusetts, http://www.embedinc.com/products
(978) 742-9014.  Gold level PIC consultants since 2000.

2009\07\05@104204 by Russell McMahon

face picon face
> The most annoying thing about placing LEDs on your breadboard for pin
> status indication is that you need a resistor to go along with it.

Oh gimme a break.  Adding a single resistor to a board is both trivially
easy and cheap.

> Why don't they just build the resistor right into the LED > so it makes
everyone's life easier?

Everyone's?  Are you sure?  Did you ask everyone?  I don't remember you
asking me.  For someone that is early in learning electronics it is a good
idea to start out assuming things are as they are for a reason, and if it
doesn't appear to make sense at first glance that this is probably because
you don't understand the reason rather than everyone else missed something.


1. Obfuscation & social comment:             If you, in your turn, assumed
such language was a combination of product of youth, northness of border,
turn of phrase, slight difference of culture, lack of US_American as a
second language, general within-norm conversational largess, elegant
variation (a la Fowler) and acceptable linguistic inexactitude to make a
point, then  your following generally very good points would probably have
gone down a treat with said (or unsaid) requester and you two would probably
be one small step for two men further along the path towards being the sort
of friends that you may yet both turn out to become, unlikely as that may so
far seem. As it is, the way the answer is put is liable to achieve the
desired result in terms of recipient reaction but to render the fine
technical content close to worthless. In such, reasonably likely case, you
are persoannly wasting your own time by having your pearls fall trampled &
unnoted.

2.  Technical:              Olin makes, of course, a number of very good
technical points, which are all worth noting. If the question is translated
(as it reasonably may be) as "can you buy LEDs with inbuilt series
resistors?", then the answer is, as he notes, "Yes, they are available, but
the cost is relatively high for what you get, and they are far far less
flexible than a bare LED, and in most cases the inconvenience of having to
add a series resistor is more than offset by the conveniences of being able
to add one that best suits your given application or using an alternative
means of driving, when desired.

3.  Alternative not covered by Olin:    There would indeed by advantages to
having a two-wire leaded optical indicating device (such as a "ballasted
LED" that you describe) when prototyping and thereby not taking up extra
termination points or having to run a nest of wires away from a port. This
can be achieved easily enough by taking a leaded LED and leaded resistor
(value of your choice) and soldering it to the LED in such a way that you
get a 2 lead device which serves the purpose and which can be plugged into a
"breadboard" as required. If you find these useful enough you could make up
a number and keep them for test and development purposes.

I on occasion make just such a device when testing - often on a case by case
basis. The alternative is to place a bank of LEDs and resistors at a
convenient point on a breadboard and then to run plug-in** flying leads as
required. You can buy (here and probably there) thin flexible color coded
flying leads with  solider plug in ends and of differing length, and these
are of immense value* when breadboarding. The colour coding allows you to eg
assign a colour to indicator driving leads, so that you can both trace and
visually note such leads with ease - and remove them easily if their loading
becomes excessive

* & **: Noting that eg VM and others opine that plug-in breadboards cause
far more trouble than they are worth and should never be used, such devices
could be solder-in if desired. I personally find plug-in breadboards of
immense value, although due care does indeed have to be taken to avoid or
allow for the problems they can cause.



       Russell

2009\07\05@115547 by Brian B. Riley

picon face

Here is the Mouser Number for the Green LEDS

SSL-LX5093GD-12V

and for the Red 5 VDC;

SSL-LX5093ID-5V


the green LEDs for 12V work fine in 5 volt logic

On Jul 4, 2009, at 10:24 PM, solarwind wrote:

> The most annoying thing about placing LEDs on your breadboard for pin
> status indication is that you need a resistor to go along with it. Why
> don't they just build the resistor right into the LED so it makes
> everyone's life easier? Do such products exist?
>
> -- [ solarwind ] -- http://solar-blogg.blogspot.com/
> --

2009\07\05@122036 by peter green

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> "can you buy LEDs with inbuilt series
> resistors?", then the answer is, as he notes, "Yes, they are available, but
> the cost is relatively high for what you get
mmm lets see
5V 3mm LED: 20p (ex vat, in one off quantity)
standard 3mm LED: 4.5p (ex vat, in one off quantity)
resistor 0.6p

so the integrated unit costs about 15p more.  but the added conviniance
when breadboarding or bodge modding is more than worth it imo.

The biggest problem I find with such devices is that there is typically
no marking on them, so unless you keep very close track of them it's
kinda tricky to tell them apart from regular LEDs
> 3.  Alternative not covered by Olin:    There would indeed by advantages to
> having a two-wire leaded optical indicating device (such as a "ballasted
> LED" that you describe) when prototyping and thereby not taking up extra
> termination points or having to run a nest of wires away from a port. This
> can be achieved easily enough by taking a leaded LED and leaded resistor
> (value of your choice) and soldering it to the LED in such a way that you
> get a 2 lead device which serves the purpose and which can be plugged into a
> "breadboard" as required. If you find these useful enough you could make up
> a number and keep them for test and development purposes.
>  
might be an idea, you would have to construct it very carefully and
probablly use heatshrink to get a neat result. the last thing I want is
something that creates even more sprawl of stuff over the breadboard
making it even harder to get wires in an out of the required connections.

2009\07\05@165901 by Marechiare

picon face
>> Why don't they just build the resistor right into the LED
>> so it makes everyone's life easier?
>
> Everyone's?  Are you sure?  Did you ask everyone?
> I don't remember you asking me.  For someone that is
> early in learning electronics...

It looks like he did not start learning electronics yet. You seem to
have missed his [OT] post:


On Sun, Jul 5, 2009 at 8:49 AM, solarwind wrote:
> People say current kills. So what happens if you push 1 mA through an
> LED @ 1000 V? Will the LED blow?

2009\07\05@180119 by AGSCalabrese

picon face
No, there are not any LEDs with built-in current sources.  There may  
be some with current limiters.   I googled and did not find any.
Gus


{Quote hidden}

2009\07\05@181045 by olin piclist

face picon face
Marechiare wrote:
> It looks like he did not start learning electronics yet. You seem to
> have missed his [OT] post:
>
> On Sun, Jul 5, 2009 at 8:49 AM, solarwind wrote:
>> People say current kills. So what happens if you push 1 mA through an
>> LED @ 1000 V? Will the LED blow?

I don't read OT.  I am surprised however as I thought he was farther along
than this.

Solarwind, look up something called Ohm's Law.  It defines the relationship
between voltage accross a device, the current thru it, and its resistance.
In this case you don't control the characteristics of the LED, so you only
get to chose one of current thru it or voltage accross it and it will pick
the other.  You can't therefore push 1mA thru a LED (or any other device)
while at the same time applying 1000V accross it (unless that device happens
to be a 1Mohm resistor).

At 1mA thru it forwards, you'll get around 2 to 2.5 volts accross most
normal red or green LEDs, and it will light dimly but visible enough in a
typcial office setting.  At 1mA backwards the LED is being driven in a
unintended way so results will vary more widely.  You probably get 5-20V
before it no longer acts much like a diode and allows 1mA of reverse
current, but I haven't tried that so my guess could be off.

Conversely when you apply 1000V in either direction, the LED will draw way
more than 1mA for a very short time before it makes a popping noise and
emits a small mushroom cloud.  What happens after that, including arcing and
possibly catching fire is too dependent on the particulars of the LED,
humidity, and random chance to guess.  Let's just say it would be
"spectacular".  Once.


********************************************************************
Embed Inc, Littleton Massachusetts, http://www.embedinc.com/products
(978) 742-9014.  Gold level PIC consultants since 2000.

2009\07\06@011617 by solarwind

picon face
On Sun, Jul 5, 2009 at 6:10 PM, Olin Lathrop<KILLspamolin_piclistKILLspamspamembedinc.com> wrote:
{Quote hidden}

Thanks for the reply.

Please note: if this is a map -

.  .  . ..           . ..      .
.      . . . . .           ..
.  . .       .. .   .. .  ..
. . ..     ..     ..   .. . .

How do you define "further"?

A                        B

which one's "further"?


The point is - there is no "further" - you just learn different things
at different points. Something may not be "further" than something
else.

2009\07\06@031043 by Vitaliy

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Russell McMahon wrote:
> * & **: Noting that eg VM and others opine that plug-in breadboards cause
> far more trouble than they are worth and should never be used,

Who is this mysterious VM that you speak of? Those are my initials, but I
sure have never said anything like the above.

Vitaliy


2009\07\06@080149 by Marechiare

picon face
solarwind wrote:
> Olin Lathrop wrote:
>> I don't read OT.  I am surprised however as I thought he
>> was farther along than this.
...
> How do you define "further"?

He said "farther", not "further".


> The point is - there is no "further" - you just learn different
> things at different points. Something may not be "further"
> than something else.

You can't start learning to EE without learning Ohm's low first. The
discussion is senseless.

Why didn't you learn Electricity at school? You seem to be more than
capable to get it. Language problem at that age for the immigration?

2009\07\06@084138 by olin piclist

face picon face
Vitaliy wrote:
> Russell McMahon wrote:
>> * & **: Noting that eg VM and others opine that plug-in breadboards
>> cause far more trouble than they are worth and should never be used,
>
> Who is this mysterious VM that you speak of? Those are my initials, but
> I sure have never said anything like the above.

Yeah, I've heard people whine about plugin breadboards a lot here too, but I
use them routinely with no problem, including 20MHz PIC crystals.  You have
to know what you're doing and understand the limitations of course, but
that's no different than with any other method.

My ReadyBoard series (http://www.embedinc.com/products) includes a
breadboard area deliberately like the plugin breadboards because I think its
a useful layout for circuits unknown at the time the board is made.  The
ReadyBoards do have a ground plane and the power lines are decoupled at both
ends so you can get away with just a little more sloppiness, but the advice
of making sure you know what you're doing is still sound.


********************************************************************
Embed Inc, Littleton Massachusetts, http://www.embedinc.com/products
(978) 742-9014.  Gold level PIC consultants since 2000.

2009\07\06@085740 by olin piclist

face picon face
solarwind wrote:
>> I don't read OT. I am surprised however as I thought he was farther
>> along than this.
>
> The point is - there is no "further" - you just learn different things
> at different points. Something may not be "further" than something
> else.

I'm assuming this is in reference to the part of my statement I quoted
above.  Some facts about a subject are more fundamental than others in that
new facts can't be understood without first understanding the fundamental
ones.  Explainations of new facts will often be based on these fundamentals
and therefore can't be understood without them.

Ohm's law is one of these fundamentals in electronics.  Think of it as the
F=mA of electronics.  So is Kirkchoff's (sp?) current law, although that one
is more of a formal codification of something that is intuitive for most
people anyway.

The point is not to just memorize them, but to really understand what they
mean.  Fortunately these basic rules of electronics and a few others, like
the voltage change on a capacitor as a function of current, capacitance and
time, or the current change thru a inductor as a function of voltage,
inductance, and time, are very simple to express and grasp the math.  Truly
understanding them (feel the force, Luke) may take a little more, but isn't
hard either.

Just like you can't get far in physics without F=mA, you can't get much into
electronics without these fundamental rules.  If you want to continue with
electronics, you really need to sit down and understand them.


********************************************************************
Embed Inc, Littleton Massachusetts, http://www.embedinc.com/products
(978) 742-9014.  Gold level PIC consultants since 2000.

2009\07\06@092150 by SME

picon face
> You can't start learning to EE without learning Ohm's low first. The
> discussion is senseless.

> Why didn't you learn Electricity at school? You seem to be more than
> capable to get it. Language problem at that age for the immigration?

His major educational field is in the biological sciences. Electronics
is a hobby unrelated to his school curriculum.

That said, Ohm's law is part of the "MRS GREN" of the electrical
world. It is so fundamental to everything that you do that it MUST be
learned to the stage where you are comfortable with the implications
in real world situations without having to go and look things up.

The original question would have benefited from a simple Ohms law
analysis - most of the very useful answers that people have given are
essentially conveying information that would be intuitively obvious
given a basic understanding of the applications of Ohms law in the
real world, and even an hour or less playing with a few cheap meters
and some LEDs and resistors and a power supply. Such an investment in
'basic research' will not only help you to ask better questions and
not attract such adverse comment, but will also - probably more
importantly - provide you with some Eureka moments that can't be
easily replaced by having people tell you stuff. The non linear V/I
relationship of any LED, the reason that LEDs are best current-driven,
the point at which they emit coloured smoke, or stop emitting light,
the interesting results of reverse bias at about zero current as
voltage is increased, and much more, will embed itself usefully in
your brain far more easily, and with a far greater sense of
satisfaction, than it will by getting answers to very basic questions
from other people.

Try it, you'll like it.

PS: Ohm's law always works, by definition. ie I = V/R. R = V/I. V = IR
ie it's essentially a tautology.
BUT as noted - for some substances the ratio of V/I (=R) changes with
V and or I. R can also changes with temperature, pressure, exposure to
EM field etc. For most purposes I = V/R is valid enough to be useful.

For semiconductor devices the ratio of V/I can change inmmensely with
V. Which is exactly what happens in the LED case being discussed on
this thread. Aty say 20 mA an LED may have a Vf of 3.0V, so R = V/I =
150R. However, at 30 mA Vf may be 3.1V. And at 100 mA Vf may be 3.3V.
Giving R's of 103 R and 33 R. Clearly R is VERY dependant on V. We
don't usually think about this example in these erms, but doing so
demonstrates that Ohms law, as it is usually understood, can be easily
'broken' in everyday life.




          Russell.

2009\07\06@163609 by cdb

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:: Think of it as the F=mA of electronics

Oh dear, I'm really going senile, I read this first as Force =
milliamps - and what is worse, I then thought ' I've never thought of
it this way before '.

Sanity has prevailed.

Colin
--
cdb, RemoveMEcolinTakeThisOuTspambtech-online.co.uk on 7/07/2009

Web presence: http://www.btech-online.co.uk  

Hosted by:  http://www.1and1.co.uk/?k_id=7988359







2009\07\06@174910 by Sean Breheny

face picon face
Russell,

I still think that it is misleading to say "Ohm's law always works" It
isn't an arbitrary definition. Current and voltage were each defined
and discovered independent of the concept of resistance. It was then
discovered that, for most materials, the ratio between voltage and
current was a constant for that material (at a particular temperature,
with a particular geometry, etc.) It wouldn't have to be this way - it
could have been that most materials had a highly nonlinear V-I
relationship, in which case, simply defining V=IR would not have made
much sense.

This is more than just semantics or pure theory - it has very
practical implications. If one thinks of V=IR as a fundamental
definition of resistance, then he is likely to try to compute the
"resistance" of a battery, for example. It has a certain V. It has a
certain I, so it should have an R=V/I. But of course, in this case, it
makes much more sense to talk about R=dV/dI, which is often called the
internal resistance of the battery, which is a much more useful
concept for a battery than the simple ratio of V/I.

Sean


On Mon, Jul 6, 2009 at 9:21 AM, SME<spamBeGoneshoemakerselfspamBeGonespamgmail.com> wrote:
{Quote hidden}

>

2009\07\07@015745 by Ruben Jönsson

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face
The electrons in a conductor don't care about that. At any given time, they see
a voltage and a resistance and move accordingly.

/Ruben


{Quote hidden}

> > --

2009\07\07@022005 by John La Rooy

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face

2009/7/7 Ruben Jönsson <RemoveMErubenspamTakeThisOuTpp.sbbs.se>:
> The electrons in a conductor don't care about that. At any given time, they see
> a voltage and a resistance and move accordingly.
>
> /Ruben
>

Ohm's law is a useful for modeling and predicting bulk electron
behaviour, but not so good for very low currents where the discrete
nature of the electrons becomes more apparent.
At one picoamp, you are seeing only 6 or 7 electrons per microsecond
for example.

John La Rooy

2009\07\07@064637 by Ruben Jönsson

flavicon
face
> 2009/7/7 Ruben Jönsson <rubenEraseMEspam.....pp.sbbs.se>:
> > The electrons in a conductor don't care about that. At any given time, they
> > see a voltage and a resistance and move accordingly.
> >
> > /Ruben
> >
>
> Ohm's law is a useful for modeling and predicting bulk electron
> behaviour, but not so good for very low currents where the discrete
> nature of the electrons becomes more apparent.
> At one picoamp, you are seeing only 6 or 7 electrons per microsecond
> for example.
>
> John La Rooy
>

Yes, but in the end it all boils down to fundamental physical nature where the
electrons always act according to ohms law (in one way or another). By that I
mean that ohms law always applies even though the actual resistance may be
nonlinear with respect to voltage, current, pressure, time, temperature or
whatever. A nonlinear resistance may not make it practical to actually use ohms
law to model a behaviour of a circuit but it still applies, since at any given
time a circuit will have a defined potential difference (voltage), current and
resistance. Or...?

/Ruben
==============================
Ruben Jönsson
AB Liros Electronic
Box 9124, 200 39 Malmö, Sweden
TEL INT +46 40142078
FAX INT +46 40947388
EraseMErubenspampp.sbbs.se
==============================

2009\07\07@082919 by Marechiare

picon face
> Fortunately these basic rules of electronics and a few others,
> like the voltage change on a capacitor as a function of current,
> capacitance and time, or the current change thru a inductor
> as a function of voltage, inductance, and time...

Yes, that's true, there is no sense to hunt for PIC32 without the
understanding how it will drive the load. The output is current
limited, the load is resistive/inductive/capacitive. To meet the specs
you need to trade the parameters carefully.

2009\07\07@095028 by Sean Breheny

face picon face
What is the "that" which the electrons do not care about?

The electrons only know about the electric field they see, as well as
their semi-random thermal motion, and their collisions with each other
and the material structure. In a typical conductor, this results in a
linear relationship between V and I. However, in a semiconductor, the
differences in electron (and hole) concentration in different areas
means that diffusion and drift (E field force) both play significant
and sometimes counter-acting roles. The result is a very non-linear V
and I relationship.

Similarly, in many situations, there are other charge carriers besides
just electrons (like ions in a solution). This makes the situation
more interesting, too.

To say that "Ohms law still applies" just because one can always
divide V by I is silly. I could make up a "law" called Sean's Law,
where I define W as "wonkyness", and define that W=P/L, where P is air
pressure and L is light level. I can always divide and come up with a
value for W, but that is pretty meaningless unless I can show that,
for a significant set of relevant conditions, there is a relationship
between P and L and it is linear.

Sean


2009/7/7 Ruben Jönsson <RemoveMErubenEraseMEspamEraseMEpp.sbbs.se>:
> The electrons in a conductor don't care about that. At any given time, they see
> a voltage and a resistance and move accordingly.
>
> /Ruben
>

2009\07\07@105724 by Ruben Jönsson

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face
> What is the "that" which the electrons do not care about?

Nonlinear resistance

{Quote hidden}

When is U NOT equal R*I?

/Ruben
==============================
Ruben Jönsson
AB Liros Electronic
Box 9124, 200 39 Malmö, Sweden
TEL INT +46 40142078
FAX INT +46 40947388
RemoveMErubenspam_OUTspamKILLspampp.sbbs.se
==============================

2009\07\07@120417 by Sean Breheny

face picon face
2009/7/7 Ruben Jönsson <RemoveMErubenTakeThisOuTspamspampp.sbbs.se>:
> When is U NOT equal R*I?

If you define U=R*I, then, as you are saying, they are always equal.

The point I was trying to get across is that Ohm's Law is much more
than a definition. It is an empirical observation that V and I are
usually linearly related. This is actually a rather amazing thing
(resistance in conductors is one of the most linear phenomena - it is
rare that you find something which is almost perfectly linear over at
least 12 orders of magnitude).

Using Ohm's Law in circumstances where there is highly nonlinear
conduction is only useful when done in a linearized sense (that is,
for small signal changes about an operating point). Here, the
definition is different (R=dV/dI not R=V/I). It allows you to use
relationships which were derived for linear circuits in some
restricted nonlinear situations. This is doable not because of
tautological definitions, and not because of inherent linearity in
nonlinear electrical circuits, but simply because of the mathematical
property that even rather nonlinear functions look relatively linear
for small deviations.

Sean


{Quote hidden}

>

2009\07\07@122829 by M.L.

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face
On Tue, Jul 7, 2009 at 2:20 AM, John La Rooy<RemoveMEpiclist.jlrKILLspamspamlarooy.com> wrote:
> Ohm's law is a useful for modeling and predicting bulk electron
> behaviour, but not so good for very low currents where the discrete
> nature of the electrons becomes more apparent.
> At one picoamp, you are seeing only 6 or 7 electrons per microsecond
> for example.

Which isn't really an issue because electrical engineers seldom deal
with picoamps. Active devices don't do much below the nano-amp level
because of leakage.

And it's certainly possible to measure down to the femto-amp level
without too much trouble if that's what you need to do. But I agree
that quantum physicists don't deal with Ohm's law very often.
--
Martin K.

2009\07\07@132558 by Ruben Jönsson

flavicon
face
> 2009/7/7 Ruben Jönsson <rubenSTOPspamspamspam_OUTpp.sbbs.se>:
> > When is U NOT equal R*I?
>
> If you define U=R*I, then, as you are saying, they are always equal.

That is just my point. Ohms law always apply, as (if I remember correctly now)
was the original question. Although, I completely agree with you that it is not
practical to use it in all circumstances.

{Quote hidden}

/Ruben

==============================
Ruben Jönsson
AB Liros Electronic
Box 9124, 200 39 Malmö, Sweden
TEL INT +46 40142078
FAX INT +46 40947388
spamBeGonerubenSTOPspamspamEraseMEpp.sbbs.se
==============================

2009\07\07@140708 by Sean Breheny

face picon face
Ok, but I am claiming that Ohm's Law is not a definition, but rather
the statement that V=I*R where R is a constant over all V and I. R may
vary with temperature, etc., but not with changes in V or I alone. If
we take this as Ohm's Law, then it applies only to relatively good
conductors (mostly metals).

In this sense, Ohm's Law is neither a definition nor a universal law.
It is an empirical observation which applies in a restricted, albeit
wide, range of circumstances and materials.

Sean


2009/7/7 Ruben Jönsson <KILLspamrubenspamBeGonespampp.sbbs.se>:
>> 2009/7/7 Ruben Jönsson <EraseMErubenspamEraseMEpp.sbbs.se>:
>> > When is U NOT equal R*I?
>>
>> If you define U=R*I, then, as you are saying, they are always equal.
>
> That is just my point. Ohms law always apply, as (if I remember correctly now)
> was the original question. Although, I completely agree with you that it is not
> practical to use it in all circumstances.
>

2009\07\07@165651 by Barry Gershenfeld

picon face
30+ messages, all on account of someone wanting to light up an L.E.D. :)
Ohm's law is useful where it applies, and it's that mysterious attribute
called "experience" that allows you to know when it can be applied.

Every "law" can be picked apart until you make it fail. Just like Newton's
(I'm referring to Sir Issac, rather than James), which falls apart at the
speed of light, so, too, will this.  And it isn't just the "law", but the
conditions.  You neglected the resistance of the wire, because it doesn't
matter.  Except when you're supplying high currents, or doing milliohm
measurements, suddenly it does.

LED's don't normally create the most demanding of calibration problems, and
yet they are nonlinear, so the this problem actually does become
interesting.  It's worth the exercise of looking at running an LED from the
mains.  You'll find several surprises, dealing with estimation techniques,
quasi-constant currents, heat dissipation, etc.

2009\07\07@191217 by solarwind

picon face
On Tue, Jul 7, 2009 at 4:56 PM, Barry Gershenfeld<@spam@gbarry42@spam@spamspam_OUTgmail.com> wrote:
> 30+ messages, all on account of someone wanting to light up an L.E.D. :)

It happens whenever I post a topic.

2009\07\08@010401 by Xiaofan Chen

face picon face
2009/7/8 Ruben Jönsson <spamBeGonerubenspamKILLspampp.sbbs.se>:
>> 2009/7/7 Ruben Jönsson <.....rubenspam_OUTspampp.sbbs.se>:
>> > When is U NOT equal R*I?
>>
>> If you define U=R*I, then, as you are saying, they are always equal.
>
> That is just my point. Ohms law always apply, as (if I remember correctly now)
> was the original question. Although, I completely agree with you that it is not
> practical to use it in all circumstances.
>

This one is not bad.
http://en.wikipedia.org/wiki/Ohm%27s_law



--
Xiaofan http://mcuee.blogspot.com

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