I had already assumed that was the case. Had I read the entire thread more carefully, I would have seen you did exactly that. I'll know more tonight when I get it apart. Does anybody have a resistor pack unsoldered that can measure the resistance? Better yet, what's the spec on the pack?

IIRC, 561 = 560ohms.

I believe it will require 1 (one) 140 ohm resistor. Brian... what is your opinion as to whether this single resistor would work instead of the multiple resistor pack?

I looked around last night and saw where a SMT manufacturer/supply company was in Austin. It looks easy enough to spec out what anything you need but probably available only in quantities. Of course, if a standard resistor with leads could be fitted, those would be available anywhere.

Technically you could do that and the single resistor would be the same value as one of the resistors in the original design. If you did put in all four, which would be pointless in that situation, you would need each resistor to be 4*R.

Now for the reason you don't want to do that. Each LED will not drop the exact voltage, leading in one pulling more current than the others. As time goes on that may effect that diode increasing the current draw until premature failure. So in an engineering stand point you don't want to do that. My two cents.

So the LEDs will stop working in 100 years instead of 900 years?

You want at least 560 ohms, and to be safest use 1000 ohms so you don't blow the micro port.

Allrighty then. I believe you are saying that if I were to put in a single resistor it should be at least 560 ohms and preferably 1000 ohms. Another question (purely curiosity), is the voltage to ground on the trace 12V or 5V? I sent you an order but in the meantime, I was going to experiment around with some LEDs and see how it worked.

Hmm... I don't understand why the LED's would draw (slightly) different voltages... I am assuming you're wiring the LED's in parallel. In series, yes, I understand how that could be a problem. I just don't know. I'm going to have to break out the old breadboard and scope on that one.

Then again, it's been years since I've actually had to do the math.

Mason

No, the voltage across all LEDs would be the same as they're in parallell, but since LEDs are highly non-linear devices, how much current each draws might vary considerably due to quite small differences in forward voltage.

Say that the LEDs total current consumption and the resistor causes the voltage across the LEDs to be 2.0V. We aimed for a target current through the LEDs of 10mA each, but one LED might only draw 5mA due to it having a slightly higher forward voltage, while another might draw 20mA due to it having a slightly lower forward voltage...

/Michael

How is that solved by individual resistors?

That's exatly what I don't understand. It SEEMS like exactly the same thing to me. Keep in mind I was a pre-med/MIS major/minor so I don't pretend to know what I'm talking about.

With one resistor, if one of the LEDs fails, then the wrong voltage will be going through all of the remaining LEDs, causing them to fail prematurely. Before the failure, it was (voltage-resistance)/4. After the failure, it becomes (voltage-resistance)/3. Too much voltage through the three remaining LEDs.

With multiple resistors, the failure of one LED does not change the amount of voltage going to the other LEDs.

The two cases are not the same, as in the case with a single resistor the voltage across all LEDs will be forced to be identical, as the anode (+ end) are all tied together at the resistor and the cathode (- end) are tied together at ground) but with the individual resistors it can differ as the anodes are separated from each other.
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Say we have a 5V supply voltage, nominally 2V LEDs @ 10mA current through each LED.

Case 1, One single resistor

Total current through resistor is 40mA -> R = (5-2)/0.04 = 75 Ohm
All LEDs will have the same voltage over them 5-(75*I_total) (75*I_total is the voltage dropped over the resistor)
Unfortunately 2 of the LEDs aren't 2V @ 10mA, instead one is 1.95V @ 10mA which might translate to 2V @ 18mA and the other is 2.05V @ 10mA which might translate to 2V @ 2mA (note: using canned numbers to simply total math, and due to being to lazy to dig up a real LED UI graph).

Now, the LED with 18mA going through it will be noticeably brighter than the one with 2mA going through it.

Note: In reality differences might be too small to notice.


Case 2, individual resistors

Through each resistor we have a current of 10mA and we need to drop 3 (5-2) volts over it -> R = (5-2)/0.01 = 300 Ohms (which is 4x the value of case 1, surprise surprise...)

Now, for the 1.95V @ 10mA LED there will be a slighty larger voltage drop across the resistor (worst case: assume all the extra 0.05V ends up at the resistor; in reality it will be divided over LED and resistor)
I = U/R = 3.05 / 300 = 10.16mA

Which is a far slighter difference than we could get in case 1.

Same sort of calculation can be made for the 2.05V LED -> I = 9.83mA

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Another point is if one LED were to fail open - due to a bad solder joint or whatever.

In case 1 we need to drop 3V over a 75 Ohm load. That was accomplished with a 40mA current that now needs to go through 3 LEDs, ie 13.3mA (in reality slightly less, as the voltage over the LEDs would increase slightly) each instead of 10mA. Not that that's likely to hurt them in any practical way, but still.
In case 2 we'd just get a non-lit LED, the others would remain unaffected,

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All in all, it's a better design to use individual resistors, but LEDs of today are probably consistent enough for it not to matter... (famous last words... )

/Michael

That was what I was saying but just realized I wasn't thinking. You want 560/4 to 1000/4, so 250 would be the safest bet.

Ok.... I get it. Thanks.

Thanks.

Can anyone give me a size (physical dimensions) of the SMT resistor pack as shipped by Brian?

I'm no EE, but if I'm understanding this correctly, having all four run off of one resistor will lead to failure do to uneven draw?
Would the damage occur after one fails and the resistor is doling out that extra 1/4 to the 3 remaining lights?
If the damage were to occur after failure, would it be possible to just replace the broken led asap?

Just speculating, I'd -love- to have illuminated buttons, and my experience with electronics is severely limited. I would be willing to spend time practicing for this, as it seems the idea of one resistor would be a lot easier to put together than having to put that resistor pack on the board. *scary*

thanks

Can anyone give me a size (physical dimensions) of the SMT resistor pack as shipped by Brian?

Yes, it's exactly 15 angstroms wide. The distance between each of the contact pads is 3 angstroms. I know this from experience.

What's that in micro furlongs?

Rob

Well, since a furlong seems to be 1/8 mile (http://www.nass.usda.gov/ok/bulletin00/page100.pdf)
which in meters equals 1609/8 = 201.25 m

* 1 Ångström is 0.1 nanometer

* 1 microfurlong is 0.000201125 meters or 201.125 micrometer

* 15 Ångström is 1.5 nanometer

This gives (if I got it right - and that's a pretty big if right now) that 15 Ångström
equals (roughly) 7.458E-6 microfurlong or 7.458 picofurlong

/Michael

Part number EXB-V8V102JV, look at attached pdf.

Where's Doug?

Where's Doug?

Tied and gagged. You should see him struggling against the bonds. He can't stand not being able to do the measurement conversions. I feel sorry for him, but it's for his own good, you know.

Should I get one of those huggy feely thick shirts with extra long sleeves and lots of nice buckles and straps next time I go shopping

/Michael

Hmm, Doug is trying to say something. Can't understand him through the gag, but I think it's "hectares".

Sheesh, then he's not even in the right dimension... (as that's a surface unit, hekto (100) ar = 100 ar = 100 x 100m² = 10 000m² (square meters))

And please don't ask how much that is in square AU's...

/Michael

For the record. Part No. EXB-V8V102JV appears to be 3.2mm x 1.6mm. Center to Center on the pads appears to be 0.8 mm.

Did that say '10 gig SPARE' in your sig?

Greetings!

Yup.

I was one of the original Mark2_ purchasers. I paid $2,000 for a 40GB empeg at the time. When they went on fire sale, I picked up four more. One went to a friend of mine, I upgraded two of the 2a's to 120GB, sold the original Mark2 and have the last 2a boxed as a spare. I am not the most dexterous person in the world, and I have an extra in case one of my boxes suffered from a bout of "deceleration trauma".