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  #11  
Old June 27th 19, 12:13 AM posted to alt.comp.hardware.pc-homebuilt
Paul[_28_]
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T. Ment wrote:
On Wed, 26 Jun 2019 17:44:20 -0400, Paul wrote:

Obiwan, use the maths.

Plug in the 2.2V instead of the 1.6V in my example.
The resistor needs to be series-connected to limit
the current flow and reduce LED intensity.


I get the math. But the terminology confuses me. Like forward voltage
drop. What about maximum input voltage. There must be a limit before you
burn it up.


Not with the limiting resistor.

Reverse PIV is another matter (can go avalanche at
high enough potential). But I already mentioned these LEDs
are "perfect" hobby LEDs, because the PIV set by the
designer is... 5V. That means if the LED is reversed,
the reverse potential is not a challenge to the junction.

In the forward direction, it just conducts and tries
to draw as much current as it is allowed. Since we
did our maths, it's "on a leash now" and has to obey us.

Even if I used 100V instead of 5V, and as long as I
don't reverse it by accident, it's all good. At 100V,
I do the maths, and the resistor is probably 20x higher
resistance.

Since the motherboard uses 5V (clever choice), it
doesn't matter whether it's forward or reverse. If
it doesn't light, reverse the leads.

However, if you powered the front panel with 12V,
then there *could* be a reverse PIV issue and
damage to the LED. If you work at voltages above
5V with dime store LEDs like those, take some
care. Take even more care with the powerful LEDs
that cost $20 a pop, as some of those have no
reverse protection. A few have taken to placing
an "antistatic" device inside the LED, and I
suspect that covers the reverse PIV issue.

But never assume anything without a spec sheet,
which has notes about "issues" with usage. For
example, some powerful LEDs have limited
soldering temperatures, and the requirements
are tough enough to meet, you want the LEDs
"mounted" for you on a separate substrate, and
then you solder to the substrate and not the LED.
This is why some of the SMT LEDs are such a pain
in the ass. They're not meant for humans to work with.
Only machines like them. The ones with the little legs
you bought, are perfect for hobbyists.

Paul
  #12  
Old June 27th 19, 12:39 AM posted to alt.comp.hardware.pc-homebuilt
Paul[_28_]
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T. Ment wrote:
On Wed, 26 Jun 2019 21:58:03 +0000, T. Ment wrote:

I get the math. But the terminology confuses me.


No doubt the circuit is simple to you. But my electronics knowledge is
very sparse.

I know a resistor "resists" but what does it reduce? Current, voltage,
or both?

Where is the voltage dropping? At the resistor? At the LED? Or both?


We use Ohms law for resistors and Kirchhoffs circuit laws (KCL, KVL)
to write equations and solve for unknowns.

https://isaacphysics.org/concepts/cp_kirchhoffs_laws

Yes, a resistor resists.

V
--- = I
R

The larger R is, the smaller the current flow would be.
This even works to our advantage with the LED, because
the LED is "very greedy". The diode equation shows "how greedy".

The diode equation is half way down this page. I didn't
show you this, on purpose, because it has nothing to
do with getting our computer LED to work. The LED
diode behavior is invariant enough, we don't have to
solve anything related to this. We just select an
"operating point", like 2.2V at 20mA.

https://www.allaboutcircuits.com/tex...nd-rectifiers/

Knowing that equation is important if the temperature of
the circuit changes. The LED output varies a bit
with temperature. The resistor we're using, is not
a "perfect" implementation of a current source. It's
only an approximation to a current source. The behavior doesn't
vary enough, for us to care. That's why, to a first order
approximation, I wrote the equation I did.

If we spent the evening pondering the actual physics
at the atomic level, we'd probably go crazy :-)

If you want exact solutions for the circuit behavior,
you could try some sort of Newton-Raphson iterative
method. Assuming you can make it converge. The LED is a
non-linear circuit. And strange things can happen
working with those.

If you want a component that will blow your mind, that
would be the tunnel diode. Too bad it isn't all that
useful for hobbyists (I don't think I have one of
those in my parts tray, but we did one or two labs
with them in school). I seem to remember they're used
in Time Domain Reflectometry (TDR), because
of the near vertical edges they can produce.
In the second article, they make a radio transmitter
with nothing but the tunnel diode to make it work.

https://en.wikipedia.org/wiki/Tunnel_diode

http://www.rfcafe.com/references/pop...lectronics.htm

I would need an entire electronics faculty staff
to explain it all, and I'm not that guy. There
are people in sci.electronics who have written
books and/or are professors, if you need that
kind of explaining.

I just know enough to write a LED equation for you.

Paul
  #13  
Old June 27th 19, 01:07 AM posted to alt.comp.hardware.pc-homebuilt
T. Ment
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Posts: 87
Default green led

On Wed, 26 Jun 2019 19:39:53 -0400, Paul wrote:

If you want exact solutions for the circuit behavior,
you could try some sort of Newton-Raphson iterative
method.


I just want to learn enough so I don't burn the house down. I googled
and found questions similar to mine:

https://electronics.stackexchange.co...-the-same-time


  #14  
Old June 27th 19, 03:01 AM posted to alt.comp.hardware.pc-homebuilt
T. Ment
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Posts: 87
Default green led

On Wed, 26 Jun 2019 19:13:51 -0400, Paul wrote:

The ones with the little legs you bought, are perfect for hobbyists.


After some google and Youtube, I think I got it now.

The LED is like a diode, current only flows in one direction, from the
anode to the cathode.

The LED requires a minimum voltage difference between anode and cathode
before it lights up. Mine is rated 2.2v. As Paul said, the input voltage
can be higher. Voltage won't kill it. Current will.

The LED acts like a short circuit once it lights up. It does not limit
the current flowing thru it, so you must add a resistor in series to do
that. Otherwise, it burns up.

The current rating for mine is 20mA, which gives maximum brightness.
Lower current should reduce the brightness. One web site suggested as
little as 1mA.

Higher resistance, lower current. Just beware of the resistor wattage
rating, so you don't burn that up either.


  #15  
Old June 27th 19, 05:16 AM posted to alt.comp.hardware.pc-homebuilt
VanguardLH[_2_]
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Default green led

T. Ment wrote:

My computer case green led failed. I bought some on Ebay, but they are
so bright they light up the room and obscure the other color leds.

Any suggestions for a cheap source of low intensity leds in a 3mm size?
Maybe Mouser, but then $1 worth of parts costs $10 to ship. If I have to
spend that much to fix it, I'll just live without a power led.

Can you add a resistor or something to change the brightness?


You could go through what Paul mentions. Just remember to get some
heatshrink tubing to slide onto the wire, solder the resistor to the
cable wire, slide up the tubing, and heat to coat the bare wires.

Another possibly is scotch tape which is a bit frosted. I don't know
the LED you have fit into the holder in the case. You might be able to
layer a dozen pieces of scotch tape, trim it off using an exacto blade
to match the diameter of the LED's face, and check it still snaps into
the case holder. Else, you could buy some self-stick rubber feet that
are translucent and stick them over the LED opening in the case. The
idea is to attenuate the output via diffusion rather than via voltage
change.
  #16  
Old June 27th 19, 05:41 AM posted to alt.comp.hardware.pc-homebuilt
Paul[_28_]
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Posts: 1,467
Default green led

T. Ment wrote:
On Wed, 26 Jun 2019 19:39:53 -0400, Paul wrote:

If you want exact solutions for the circuit behavior,
you could try some sort of Newton-Raphson iterative
method.


I just want to learn enough so I don't burn the house down. I googled
and found questions similar to mine:

https://electronics.stackexchange.co...-the-same-time



First, let me assure you, that if where those two LED
legs plug in, if you short those two together, the
180 ohm resistor limits the current. By shorting
the legs, the 180 ohm resistor sees a power of
V^2/R or 5*5/180 = 0.138W or less than a quarter
watt (0.250W). Usually, an engineer selects a resistor
suited to being "shorted out by a customer".

There are a few things on a computer which are
actually dangerous. The SPKR 1x4, where that plugs
in, one of the wires is tied to +5V, and if you
short the red wire from SPKR to chassis, the
wire smokes. This usually happens when someone
closes the side on the computer, the red or orange
"SPKR" wire happens to get pinched, and the sharp
metal casing slices the insulation so that the wire
touches ground.

That's about the largest exposure I know of. The
USB and Firewire ports probably use Polyfuses.

Also, the cooling fan +12V bus, is not fuse protected.
If you have an issue with a fan, the copper trace in
the motherboard will burn out (without a fire issue),
and you lose all cooling fans as a result. I suspect
some mental genius does this on purpose, and uses
an undersized copper track as a zero dollar cost
safety precaution.

You can, of course, use wire adapters and run a lead from
a Molex to the fans, and get power for them that way.
It doesn't mean the computer is "finished" - it just
means you will spend hours poring over catalogs looking
for the right wire set(s) to bring the fan power back :-)
This is some kind of punishment, I would guess.

I have a few bags of connectors here, so that if
this happened to me, I have all the parts I need
to fix it :-) That doesn't say I would appreciate
having to do that though. I've never lost a fan track,
and I've taken my share of chances with goofy
wiring jobs too. It's not like I'm a boy scout on
the wiring :-)

*******

So no, for a fan job, the odds are good there isn't
a fire risk. The mobo resistor is on the +5V side,
and ground is ground. By adding your additional
resistor, the LED is going to get dimmer. Even if
you screw up and short the pins on the LED, that
should not harm it.

Paul
  #17  
Old June 27th 19, 05:46 AM posted to alt.comp.hardware.pc-homebuilt
Paul[_28_]
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Posts: 1,467
Default green led

VanguardLH wrote:
T. Ment wrote:

My computer case green led failed. I bought some on Ebay, but they are
so bright they light up the room and obscure the other color leds.

Any suggestions for a cheap source of low intensity leds in a 3mm size?
Maybe Mouser, but then $1 worth of parts costs $10 to ship. If I have to
spend that much to fix it, I'll just live without a power led.

Can you add a resistor or something to change the brightness?


You could go through what Paul mentions. Just remember to get some
heatshrink tubing to slide onto the wire, solder the resistor to the
cable wire, slide up the tubing, and heat to coat the bare wires.

Another possibly is scotch tape which is a bit frosted. I don't know
the LED you have fit into the holder in the case. You might be able to
layer a dozen pieces of scotch tape, trim it off using an exacto blade
to match the diameter of the LED's face, and check it still snaps into
the case holder. Else, you could buy some self-stick rubber feet that
are translucent and stick them over the LED opening in the case. The
idea is to attenuate the output via diffusion rather than via voltage
change.


There might be a grommet or LED holder in the case,
to hold the LED in place. The Polyolefin tubing
is still a good idea, to make a neat job of insulating.
There aren't many tape products I trust for any long
term stability, and I'd much rather use tubing,
and slide it down into place after some soldering
is done.

The tubing diameter should be about 2x the size of the
solder joint (which would be a bit bigger than the
wires being soldered). Then, when you apply a bit
of hot air from the soldering iron tip (the air
stream off the tip), that can be enough to shrink
the tubing.

Paul
  #18  
Old June 27th 19, 03:04 PM posted to alt.comp.hardware.pc-homebuilt
T. Ment
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Posts: 87
Default green led

On Wed, 26 Jun 2019 23:16:24 -0500, VanguardLH wrote:

Another possibly is scotch tape which is a bit frosted. I don't know
the LED you have fit into the holder in the case. You might be able to
layer a dozen pieces of scotch tape, trim it off using an exacto blade
to match the diameter of the LED's face, and check it still snaps into
the case holder. Else, you could buy some self-stick rubber feet that
are translucent and stick them over the LED opening in the case. The
idea is to attenuate the output via diffusion rather than via voltage
change.


Like the black sharpie marker idea, that leaves the LED running at
higher power, which might mean shorter life. Soldering a resistor is
more work, but seems worthwhile to me.


  #19  
Old June 27th 19, 03:40 PM posted to alt.comp.hardware.pc-homebuilt
T. Ment
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Posts: 87
Default green led

On Thu, 27 Jun 2019 00:46:27 -0400, Paul wrote:

The tubing diameter should be about 2x the size of the
solder joint (which would be a bit bigger than the
wires being soldered). Then, when you apply a bit
of hot air from the soldering iron tip (the air
stream off the tip), that can be enough to shrink
the tubing.


I have a low cost soldering station that includes a hot air gun. I'm
equipped for the job.

As for the resistor equation, one thing I'm still puzzled about.

https://electronics.stackexchange.co...-the-same-time

Says:

So you subtract the LED's forward voltage from the supply voltage,
since those are both fixed voltages, and the result will be the amount
of voltage that must be dropped across the resistor for the whole to
total 9V. So 9V - 2.2V is 6.8V. That is a fixed voltage. The current
you want is fixed too - you have decided on 30mA.


Why does he want to drop all 6.8v? Why not leave a little extra voltage
in the circuit, to be sure enough remains to light the LED?

The concept makes more sense to me in terms of limiting current. In my
case, say I want only 2 - 4 mA thru the LED. Based on your explanations,
I understand how to calculate the resistance for that.

But tell me if I'm understanding this:

The cathode of the LED is connected to ground, so I will see 0 volts
there. The LED needs 2.2v to light up, so the anode will measure 2.2v,
which is the sum of LED forward voltage drop and 0v ground

That's also the output side of the resistor. The supply side is 5v. In
my circuit. those two will never change, no matter what resistor value I
choose. The only thing that can change is the current flowing through
the LED.

So I don't understand why they are talking about resistor voltage drop
above. That seems irrelevant. Or maybe I still don't get it.


  #20  
Old June 27th 19, 07:01 PM posted to alt.comp.hardware.pc-homebuilt
Paul[_28_]
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Posts: 1,467
Default green led

T. Ment wrote:
On Thu, 27 Jun 2019 00:46:27 -0400, Paul wrote:

The tubing diameter should be about 2x the size of the
solder joint (which would be a bit bigger than the
wires being soldered). Then, when you apply a bit
of hot air from the soldering iron tip (the air
stream off the tip), that can be enough to shrink
the tubing.


I have a low cost soldering station that includes a hot air gun. I'm
equipped for the job.

As for the resistor equation, one thing I'm still puzzled about.

https://electronics.stackexchange.co...-the-same-time

Says:

So you subtract the LED's forward voltage from the supply voltage,
since those are both fixed voltages, and the result will be the amount
of voltage that must be dropped across the resistor for the whole to
total 9V. So 9V - 2.2V is 6.8V. That is a fixed voltage. The current
you want is fixed too - you have decided on 30mA.


Why does he want to drop all 6.8v? Why not leave a little extra voltage
in the circuit, to be sure enough remains to light the LED?

The concept makes more sense to me in terms of limiting current. In my
case, say I want only 2 - 4 mA thru the LED. Based on your explanations,
I understand how to calculate the resistance for that.

But tell me if I'm understanding this:

The cathode of the LED is connected to ground, so I will see 0 volts
there. The LED needs 2.2v to light up, so the anode will measure 2.2v,
which is the sum of LED forward voltage drop and 0v ground

That's also the output side of the resistor. The supply side is 5v. In
my circuit. those two will never change, no matter what resistor value I
choose. The only thing that can change is the current flowing through
the LED.

So I don't understand why they are talking about resistor voltage drop
above. That seems irrelevant. Or maybe I still don't get it.


I wasn't joking about the Newton Raphson.

The two components, the LED is "non-linear".

The LED "does not follow Ohms law".

Attempts to use some sort of static voltage divider
behavior, will only lead to hair loss.

The best way to think of it, is there is a "tug of war",
between the diode and the resistor. The resistor has
the upper hand, because you cannot have more than
5V / Rseries amperes of current flowing. The resistor
ensures that no matter what voltage appears across the
LED, not more than that number of amperes (or milliamperes)
can flow.

Now, what can the diode do ?

This is a strawman diode curve I'll just make up.
If we used a "stiff" voltage source, the diode burns
brightly with tiny increases in voltage. The slope
of this curve is the "dynamic impedance".

2.0V 1mA
2.1V 10mA
2.2V 100mA

See how non-linear that is ?

Say that if I asked the resistor for the entire 5V/180ohms = 28mA
by shorting the end of the resistor, what would the LED
do with that much current ? It would hop up to 2.13 volts.

OK, now that this has happened, the ends of the resistor
have (5 - 2.13) volts across it. The current drops to
2.87V / 180ohms = 16mA.

Alright, what does the LED do now ? The voltage
drops to maybe 2.11V. Notice how we're asymptotically
converging to a "final answer".

If we could solve the equations in closed form, I
could give a rather ugly formula that would be
"accurate" but meaningless to a computer builder.

Suffice to say, that the resistor performs a "current limiting"
function. The diode presents a "relatively low impedance".
The impedance is dynamic (changes with current flow),
but is still present, and in my description of the
LED behavior, you can see that "stepping along the
dynamic diode curve causes hardly any voltage change".

Using my Newton-like method, I got 2.13, then 2.11, and
after a few iterations, it'll be somewhere between
say 2.10 and 2.11.

The situation is also relatively insensitive. If the
temperature changes, that affects the diode equation,
but won't measurably affect your perception of what
the LED is doing. The current doesn't change much.
The light doesn't change much. Nothing burns.
The LED lasts for 25000 hours at 16mA or so.

Using a much higher resistance, like putting an 820 ohm
total in the circuit, just pushes us further down the
diode curve. The resistor is the boss (as it's a relatively
high impedance and a "pseudo current source"). The diode
cannot do much about the situation - it's been
whipped into shape by its resistor neighbor.

The diode is very greedy. If you connected it directly
to +5V, it's life would be very short, and the bond
wire near the anvil would burn out. The diode is "stiff".
The resistor is like "floating on air", and the
resistor is the boss in the circuit. The diode
"can hardly move" in this situation. If we were
writing a non-linear solver, it would converge
in a half dozen iterations.

*******

I've seen this "low impedance" behavior of LEDs
in a home project here.

I made my own LED bicycle lights, which I run from a
3 Watt bottle generator (spins against the back wheel
of the bicycle). I can promise you that LEDs "are
very stiff". It's so bad in fact, that a length of
lamp cord that leads from the generator on the back
wheel, to the front light assembly, results in the
front light assembly being noticeably dimmer. That's
16ga wire, and you'd think the resistance is close
to zero. But the dynamic impedance of the multiple
LEDs says that the wire is actually a substantial
source of resistance, and makes my front light weaker.
Both the front and back lamp assemblies use the
same number of LEDs, but as the LEDs are all in
parallel they "fight amongst themselves" to see
who gets the current. This is referred to as
"current hogging". With LEDs, it's a "dog eat dog"
world. Every LED is out for itself. It was only by
carefully matching the LEDs (examining the diode
curve for every one), that I was able to build
a nicely working lamp assembly.

So I've seen just how stiff LEDs can be. I never
even considered the power cable would be a problem.
It even affects how much the lights flicker.
(Not shown in the circuit, is the schottky bridge
rectifier and the filter caps.) By putting two
diodes in series, and matching the Vfb sum of
every diode pair, you can prevent any one "column"
from taking more than its fair share. The LED summations
are matched to within about 20mV or so. It took me
two hours of measuring LEDs in a jig, and assigning
each LED a serial number, to put this idea together.
The LED has to thermally stabilize, to get decent
(20mV level) accuracy. You watch the meter for 30 seconds
to a minute, until it stops changing.

5VDC +---+-- gen ----- 16ga wire------+---+ 4.9 VDC
| | | |
LED LED LED LED (only enough LEDs
| | | | drawn in diagram
LED LED LED LED to show the concept)
| | | |
+---+------------ 16ga wire -----+---+

The circuit has *no* resistors :-) The generator
is known to be a current source, which is why
resistors are not needed in this case. I actually
did a simulation in LTSpice of my bicycle light!
There's a model available for "gen" that someone
cooked up, and it captures the generator behavior nicely.

As the bicycle goes faster (about 30MPH), the
light output actually drops. Going faster doesn't
actually burn out the LEDs. Unlike when I used
to have a sealed beam incandescent, and one night
I went down a steep hill, and that same bottle generator
burned out my GE sealed beam. And two minutes later
a cop pulled me over for "no light on a bicycle" :-/

Some white LEDs are around 3V, while these happen
to be 2.5V each. Two in series takes 5V. Each
column carries the same (tiny) current. The generator
puts out a precise 500mA (at 6V, but the diodes
chop that down to 5V and the generator gets a little
warm). Using 3V LEDs would have been a better match
for the project.

Paul
 




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