LEDs Series-Parallel OK?

[Watson A.Name \"Watt Sun]

John Fields wrote:

On Wed, 29 Oct 2003 20:28:21 -0500, John Stewart
jh.stewart@sympatico.ca> wrote:

A common occurence, since a -ve temperature coefficient of voltage will
reduce the Vf as the temp rises. Whichever LED in a paralleled circuit has
the lowest Vf to begin will take the most current & temp will rise as a
result.
If you are running near the limit of If as Watson is, temp will rise at that

junction & more If will flow in the hottest LED. It's a good example
of +ve FB. Destruction follows in a little while.

You might be successful if the LED's Vf's were carefully matched.

-ve temperature coefficient of voltage was a real problem for designers in
the
Ge transistor era. You needed all manner of "save your ass" circuits.
Si is somewhat more forgiving, but lookout!!

---
You must have missed the part where I asked about current hogging in
Watsons _circuit_. You may notice that he has current limiting
resistors in series with each LED, which should prevent current hogging
quite nicely.

Taking a look at

http://www.spoerle.de/binaries/1063272285520-TLCW%205100.pdf

and assuming that it's a typical white LED reveals a Vf tempco of
-4mV/K°, so a change in temp from 25°C to 100°C (Tj max) would result in
a Vf change of -300mV. Subtracting this from the nominal 3.5Vf results
in a new Vf of 3.2V, which leaves 1.8V to be dropped across the 51 ohm
series resistor, so the new current into the LED will be I = E/R =
1.8V/51R = 35mA. Over the limit, but not due to current hogging.

Then there's another snag, the tempco of the series R.

Looking at

http://industrial.panasonic.com/www-data/pdf/AOA0000/AOA0000CE12.pdf

we find a typical tempco of resistance of -250ppm/C° from room temp to
100°C.

Unfortunately, we don't have any thermal resistance figures for the
resistor, but let's assume that with 1.8V across it and 35mA flowing
through it (63mW) we get a temperature rise of 25°C over ambient, to
50°C. That will cause the resistance to change by -6250ppm, or -0.625%
from its nominal 51 ohms to 50.7 ohms. No big deal, considering that
the tolerance of the resistor itself is +/-5%.

But that 35mA _is_ 5mA over the top, so what to do? Change the value of
the resistor to limit the current through the LED considering its
highest permissible junction temp.

We know that with a tempco of -4mV/K° we can expect the drop across the
LED to be 3.2V, leaving 1.8V to be dropped across the R, so for 30mA
through the LED and the R, R = E/I = 1.8V/30mA = 60 ohms. The closest
higher value standard 5% resistor is 62 ohms, which ought to do it.

But, considering the 5% tolerance on the resistor, 62 ohms could wind up
being 58.9 ohms on the low end, which would allow 31mA to flow. Still
over the top, so sticking with standard 5% parts would result in the
next higher value being 68 ohms, which would allow 28mA to flow with a
low end resistance of 64.6 ohms and 25mA to flow with a high end
resistance of 71.4 ohms.

We haven't considered the variation in Vf due to everything _but_ the
LEDs tempco, but perhaps that's best saved for another day, since the
point here is that current hogging is _not_ an issue with Watson's
circuit.

I appreciate the concerns. I think, from reading the other LED
"overclockers" on such blogs as Candlepower Forums, that the 30 mA issue
isn't really a problem. The higher current is probably going to shorten
the life of the LED, but not dramatically. Like they don't seem to be
concerned about running an average of 30 mA, with substantially more
current when the batteries are fresh, maybe 40 or 50 mA. The amount of
time that the LED is at this higher current isn't a lot, it tapers off
quickly as the batteries get 'unfresh'. In the case of the little
keychain squeeze lights, they don't even use a resistor, just the
battery internal resistance.



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[DarkMatter]

On Thu, 30 Oct 2003 04:30:07 GMT, "Jeff" <levy_jeff@hotmail.com> Gave
us:

Looks good Watson. In my mind I thought you were going to parallel
LED to LED without benefit of limiting R's.

This works better then expected, although not great, due to the LED internal
resistance.

As you are doing should
& does work AOK. Cheers, John


One is much better off placing an LED in series with a resistor.

Make several. Place them on a pair of rails that are fed by a
constant voltage source that is regulated.

VIOLA!

 

[John Fields]

On Thu, 30 Oct 2003 20:17:28 -0500, John Stewart
<jh.stewart@sympatico.ca> wrote:

Here is more, altho maybe we have "enough already!"

---
Obviously not <g>
---

This page taken from an HP Opto Electronics Division (OED)
publication by McGraw-Hill in 1977. These guys were real hot
shots & regularly beat everything to death.

It does show some LED die actually connected in parallel
without benefit of current sharing resistors.

---
"Current limiting" resistors don't you mean? I believe the resistors
shown _are_ current sharing resistors.
---

However, I recall
the dice were sorted into many categories, including Vf which
they would need to do in this application.

---
Totally unnecessary in _this_ application, since a separate current
limiting resistor is provided for each LED.
---

Another sort is in respect to color. Turns out the eye can ID
very small changes, which regularly occurred in production
at that time. If my recollection is still on track, seems to me
the biggest color problem we had was with yellow. Apparently
the eye is more sensitive to minute changes of wavelength
in that region.

---
So what's your point?

--
John Fields