LEDs Series-Parallel OK?

[Watson A.Name - Watt Sun, Dark Remover]

Connecting LEDs in series-parallel like the ASCII schem below is okay?
I don't see any reason why it shouldn't work. If a lot of LEDs, like
the 7 below, are paralleled, the total current for each parallel
string should be an average, i.e. some LEDs will draw more, some less,
so they should average out. So the diff between the two strings
should be minimal. But if one parallel string draws a few more
milliamps, the V drop will be less for this string, and more for the
other string. But this diff should also be minimal.

The reason for this configuration is that the PC board is laid out for
ICs, and this connection pattern is more convenient, with less trace
cutting and jumpering.

Typical resistors are 51 ohms, 3.5V across each white LED, about 9.6V
for LED current of 30 mA each.

View with Courier Font

+V
|
+----+----+---------+--- - -
| | |
\ \ \
/ / / Resistors
\ \ \
/ / /
| | | 7 total
_|_ _|_ _|_
\ /-> \ /-> \ /->
--- --- --- LEDs
| | |
| | |
+----+----+---------+--- - -
|
|
+----+----+---------+--- - -
| | |
\ \ \
/ / / Resistors
\ \ \
/ / /
| | | 7 total
_|_ _|_ _|_
\ /-> \ /-> \ /->
--- --- --- LEDs
| | |
| | |
+----+----+---------+--- - -
|
-V





--
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@@t@h@e@@a@f@f@l@u@e@n@t@@m@e@e@t@@t@h@e@@E@f@f@l@u@e@n@t@@

 

[John Stewart]

Connecting LEDs in series-parallel like the ASCII schem below is okay?
I don't see any reason why it shouldn't work. If a lot of LEDs, like
the 7 below, are paralleled, the total current for each parallel
string should be an average, i.e. some LEDs will draw more, some less,
so they should average out. So the diff between the two strings
should be minimal. But if one parallel string draws a few more
milliamps, the V drop will be less for this string, and more for the
other string. But this diff should also be minimal.

The reason for this configuration is that the PC board is laid out for
ICs, and this connection pattern is more convenient, with less trace
cutting and jumpering.

Typical resistors are 51 ohms, 3.5V across each white LED, about 9.6V
for LED current of 30 mA each.

View with Courier Font

+V
|
+----+----+---------+--- - -
| | |
\ \ \
/ / / Resistors
\ \ \
/ / /
| | | 7 total
_|_ _|_ _|_
\ /-> \ /-> \ /->
--- --- --- LEDs
| | |
| | |
+----+----+---------+--- - -
|
|
+----+----+---------+--- - -
| | |
\ \ \
/ / / Resistors
\ \ \
/ / /
| | | 7 total
_|_ _|_ _|_
\ /-> \ /-> \ /->
--- --- --- LEDs
| | |
| | |
+----+----+---------+--- - -
|
-V

--
@@F@r@o@m@@O@r@a@n@g@e@@C@o@u@n@t@y@,@@C@a@l@,@@w@h@e@r@e@@
###Got a Question about ELECTRONICS? Check HERE First:###
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My email address is whitelisted. *All* email sent to it
goes directly to the trash unless you add NOSPAM in the
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Don't be ripped off by the big book dealers. Go to the URL
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http://www.everybookstore.com You'll be glad you did!
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@@t@h@e@@a@f@f@l@u@e@n@t@@m@e@e@t@@t@h@e@@E@f@f@l@u@e@n@t@@

Hi Watson- Not sure if I missed somewhere in your posts but
did you ever run any of your LED projects in the pulsed mode
similar to strobbing as used in 7-segment displays as used in
calculators? Efficiency of LED's goes up significantly in the
pulsed mode. Early HP calculators such as the HP35 Series
went another step farther to improve battery life by using
inductors to discharge into the LED's & avoid losses in resistors.
Still have my HP67 running. Got it in 1975 while at HP.
Part of my time there selling the Opto line.

Cheers, John Stewart

 

[John Fields]

Connecting LEDs in series-parallel like the ASCII schem below is okay?
I don't see any reason why it shouldn't work. If a lot of LEDs, like
the 7 below, are paralleled, the total current for each parallel
string should be an average, i.e. some LEDs will draw more, some less,
so they should average out. So the diff between the two strings
should be minimal. But if one parallel string draws a few more
milliamps, the V drop will be less for this string, and more for the
other string. But this diff should also be minimal.

---
One string _can't_ draw more current than the other, since they're in
series. What will happen is that the resistance of the two will be
different, and the one with the lower resistance will drop less voltage.
---

The reason for this configuration is that the PC board is laid out for
ICs, and this connection pattern is more convenient, with less trace
cutting and jumpering.

Typical resistors are 51 ohms, 3.5V across each white LED, about 9.6V
for LED current of 30 mA each.

Since your supply is 2*((51ohms*30mA)+3.5V) ~ 10V, you will be wasting

1.5V*30mA = 45mW in each resistor.

If you opted for two LED's in series with a jumper replacing one of the
resistors in order to keep the same physical configuration, you'd wind
up with R = E/I = 3V/30mA = 100R, and the power it would dissipate would
be 3V*30mA = 90mW, so the total dissipation would be the same in either
case and since 1/4 watters would be OK in either case, all you'd save
would be the difference in cost between a resistor and a jumper. With
your way, though, if one LED failed, open or shorted, it would share the
extra current/lack of current with the opposing string and all the other
LEDs would be lit. The other way would extinguish two LEDs if one
failed open, and force twice the current through the other one (for a
while) until it also failed, so I'd say your way was better.
---

 

[Watson A.Name - Watt Sun, Dark Remover]

Hi Watson- Not sure if I missed somewhere in your posts but
did you ever run any of your LED projects in the pulsed mode
similar to strobbing as used in 7-segment displays as used in
calculators? Efficiency of LED's goes up significantly in the
pulsed mode. Early HP calculators such as the HP35 Series

Before you make that claim, I suggest you read the truth about pulsing
LEDs on Don K's LED pages. http://members.misty.com/don/ledx.html

"Furthermore, the usual ultrabright white LEDs, as well as gallium
nitride or indium gallium nitride greens, blue-greens and blue LEDs of
wavelength 460-475 nM have maximum efficiency at lower currents. It
works against you to pulse these. Same for "low current red"/GaP
red/"697 nM" red LEDs."

went another step farther to improve battery life by using
inductors to discharge into the LED's & avoid losses in resistors.
Still have my HP67 running. Got it in 1975 while at HP.
Part of my time there selling the Opto line.

Cheers, John Stewart

BTW, my question still remains unanswered.


--
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###Got a Question about ELECTRONICS? Check HERE First:###
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Don't be ripped off by the big book dealers. Go to the URL
that will give you a choice and save you money(up to half).
http://www.everybookstore.com You'll be glad you did!
Just when you thought you had all this figured out, the gov't
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@@t@h@e@@a@f@f@l@u@e@n@t@@m@e@e@t@@t@h@e@@E@f@f@l@u@e@n@t@@

 

[John Fields]

BTW, my question still remains unanswered.

 

[Spehro Pefhany]

Connecting LEDs in series-parallel like the ASCII schem below is okay?
I don't see any reason why it shouldn't work. If a lot of LEDs, like
the 7 below, are paralleled, the total current for each parallel
string should be an average, i.e. some LEDs will draw more, some less,
so they should average out. So the diff between the two strings
should be minimal. But if one parallel string draws a few more
milliamps, the V drop will be less for this string, and more for the
other string. But this diff should also be minimal.

The reason for this configuration is that the PC board is laid out for
ICs, and this connection pattern is more convenient, with less trace
cutting and jumpering.

Typical resistors are 51 ohms, 3.5V across each white LED, about 9.6V
for LED current of 30 mA each.

Yes, it's fine. If you could lay them out without the middle bus it
would be a bit better:

View with Courier Font

+V
|
+----+----+---------+--- - -
| | |
\ \ \
/ / / Resistors
\ \ \
/ / /
| | | 7 total
_|_ _|_ _|_
\ /-> \ /-> \ /->
--- --- --- LEDs
| | |
| | |
| | |
\ \ \
/ / / Resistors
\ \ \
/ / /
| | | 7 total
_|_ _|_ _|_
\ /-> \ /-> \ /->
--- --- --- LEDs
| | |
| | |
+----+----+---------+--- - -
|
-V

The reason being that there's more voltage across the resistor(s) and
therefore the current will tend to be more even, even at the end of
battery life. If you want to see how much margin you have, just put it
on your bench supply and run the voltage down to 7V or 8V. As well as
the lamps getting dimmer, you'll probably see some differences in
brighness between the LEDs.

Best regards,
Spehro Pefhany

 

[John Stewart]

Before you make that claim, I suggest you read the truth about pulsing
LEDs on Don K's LED pages. http://members.misty.com/don/ledx.html

That is a very good link. Guess I been away too long. I left HP & LED's
in 1984 to join the sales group at R&S. Only LED's I saw after that were
on a front panel!!

"Furthermore, the usual ultrabright white LEDs, as well as gallium

nitride or indium gallium nitride greens, blue-greens and blue LEDs of
wavelength 460-475 nM have maximum efficiency at lower currents. It
works against you to pulse these. Same for "low current red"/GaP
red/"697 nM" red LEDs."


BTW, my question still remains unanswered.

The answer is pretty much straight forward. All you need to know is
the Temperature Coefficient of Forward Voltage. If it is negative as in
a bipolar transistor you will have trouble due to current hogging.
If it is positive as in a power FET then you will be OK.

My recollection is that all the LED's I was selling had a -ve coefficient.
Some of the devices you are working with may be different. Does your
supplier publish any kind of data sheet you could refer too?

Cheers, John Stewart

 

[John Fields]

The answer is pretty much straight forward. All you need to know is
the Temperature Coefficient of Forward Voltage. If it is negative as in
a bipolar transistor you will have trouble due to current hogging.

 

[Watson A.Name \Watt Sun - the Dark Remover\]

John Stewart wrote:
[snip]

The answer is pretty much straight forward. All you need to know is
the Temperature Coefficient of Forward Voltage. If it is negative as in
a bipolar transistor you will have trouble due to current hogging.
If it is positive as in a power FET then you will be OK.

My recollection is that all the LED's I was selling had a -ve coefficient.
Some of the devices you are working with may be different. Does your
supplier publish any kind of data sheet you could refer too?

Supplier won't even tell the buyer who the manufacturer is, let alone
the specs. I guess I shouldn't expect too much for $.25 apiece.

Cheers, John Stewart

--

 

[John Stewart]

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!!

Cheers, John Stewart

 

[Watson A.Name - Watt Sun, Dark Remover]

:

Connecting LEDs in series-parallel like the ASCII schem below is okay?
I don't see any reason why it shouldn't work. If a lot of LEDs, like
the 7 below, are paralleled, the total current for each parallel
string should be an average, i.e. some LEDs will draw more, some less,
so they should average out. So the diff between the two strings
should be minimal. But if one parallel string draws a few more
milliamps, the V drop will be less for this string, and more for the
other string. But this diff should also be minimal.

The reason for this configuration is that the PC board is laid out for
ICs, and this connection pattern is more convenient, with less trace
cutting and jumpering.

Typical resistors are 51 ohms, 3.5V across each white LED, about 9.6V
for LED current of 30 mA each.

View with Courier Font

+V
|
+----+----+---------+--- - -
| | |
\ \ \
/ / / Resistors
\ \ \
/ / /
| | | 7 total
_|_ _|_ _|_
\ /-> \ /-> \ /->
--- --- --- LEDs
| | |
| | |
+----+----+---------+--- - -
|
|
+----+----+---------+--- - -
| | |
\ \ \
/ / / Resistors
\ \ \
/ / /
| | | 7 total
_|_ _|_ _|_
\ /-> \ /-> \ /->
--- --- --- LEDs
| | |
| | |
+----+----+---------+--- - -
|
-V
-- [snip]
went another step farther to improve battery life by using
inductors to discharge into the LED's & avoid losses in resistors.
Still have my HP67 running. Got it in 1975 while at HP.
Part of my time there selling the Opto line.

Cheers, John Stewart

BTW, my question still remains unanswered.

The answer is pretty much straight forward. All you need to know is
the Temperature Coefficient of Forward Voltage. If it is negative as in
a bipolar transistor you will have trouble due to current hogging.
If it is positive as in a power FET then you will be OK.

It's negative, and can be seen as the LED warms up and the V drop
drops. But the 51 ohm resistors give a volt drop at 20 mA, and 1.5V
at 30 mA. So those balance the currents in each parallel array.

My recollection is that all the LED's I was selling had a -ve coefficient.
Some of the devices you are working with may be different. Does your
supplier publish any kind of data sheet you could refer too?

Cheers, John Stewart

--
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###Got a Question about ELECTRONICS? Check HERE First:###
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Don't be ripped off by the big book dealers. Go to the URL
that will give you a choice and save you money(up to half).
http://www.everybookstore.com You'll be glad you did!
Just when you thought you had all this figured out, the gov't
changed it: http://physics.nist.gov/cuu/Units/binary.html
@@t@h@e@@a@f@f@l@u@e@n@t@@m@e@e@t@@t@h@e@@E@f@f@l@u@e@n@t@@

 

[John Stewart]

:

Connecting LEDs in series-parallel like the ASCII schem below is okay?
I don't see any reason why it shouldn't work. If a lot of LEDs, like
the 7 below, are paralleled, the total current for each parallel
string should be an average, i.e. some LEDs will draw more, some less,
so they should average out. So the diff between the two strings
should be minimal. But if one parallel string draws a few more
milliamps, the V drop will be less for this string, and more for the
other string. But this diff should also be minimal.

The reason for this configuration is that the PC board is laid out for
ICs, and this connection pattern is more convenient, with less trace
cutting and jumpering.

Typical resistors are 51 ohms, 3.5V across each white LED, about 9.6V
for LED current of 30 mA each.

View with Courier Font

+V
|
+----+----+---------+--- - -
| | |
\ \ \
/ / / Resistors
\ \ \
/ / /
| | | 7 total
_|_ _|_ _|_
\ /-> \ /-> \ /->
--- --- --- LEDs
| | |
| | |
+----+----+---------+--- - -
|
|
+----+----+---------+--- - -
| | |
\ \ \
/ / / Resistors
\ \ \
/ / /
| | | 7 total
_|_ _|_ _|_
\ /-> \ /-> \ /->
--- --- --- LEDs
| | |
| | |
+----+----+---------+--- - -
|
-V
-- [snip]
went another step farther to improve battery life by using
inductors to discharge into the LED's & avoid losses in resistors.
Still have my HP67 running. Got it in 1975 while at HP.
Part of my time there selling the Opto line.

Cheers, John Stewart

BTW, my question still remains unanswered.

The answer is pretty much straight forward. All you need to know is
the Temperature Coefficient of Forward Voltage. If it is negative as in
a bipolar transistor you will have trouble due to current hogging.
If it is positive as in a power FET then you will be OK.

It's negative, and can be seen as the LED warms up and the V drop
drops. But the 51 ohm resistors give a volt drop at 20 mA, and 1.5V
at 30 mA. So those balance the currents in each parallel array.

Looks good Watson. In my mind I thought you were going to parallel
LED to LED without benefit of limiting R's. As you are doing should
& does work AOK. Cheers, John

 

[Watson A.Name - Watt Sun, Dark Remover]

If one of the 7 LEDs is substantially lower Vdrop, and the other 6 are
substantially higher, then the lower one could have substantially more
current, and of course that causes escessive heat, and makes the LED's
voltage even lower.. But the 51 ohm resistors help balance out major
differences in each parallel string. Another factor is that the array
is coupled thermally, so LEDs surrounding the hot LED would get warmer
and equalize the current somewhat. But the array really needs even
closer thermal coupling, and they call it a heatsink.



--
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###Got a Question about ELECTRONICS? Check HERE First:###
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Don't be ripped off by the big book dealers. Go to the URL
that will give you a choice and save you money(up to half).
http://www.everybookstore.com You'll be glad you did!
Just when you thought you had all this figured out, the gov't
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@@t@h@e@@a@f@f@l@u@e@n@t@@m@e@e@t@@t@h@e@@E@f@f@l@u@e@n@t@@

 

[Watson A.Name - Watt Sun, Dark Remover]

Yes, it's fine. If you could lay them out without the middle bus it
would be a bit better:



The reason being that there's more voltage across the resistor(s) and
therefore the current will tend to be more even, even at the end of
battery life. If you want to see how much margin you have, just put it
on your bench supply and run the voltage down to 7V or 8V. As well as
the lamps getting dimmer, you'll probably see some differences in
brighness between the LEDs.

I brought it down to under 3V per parallel string, total current was
14 mA, and I couldn't see any diff in brightness between the LEDs. So
the resistors apparently do a good job of evening out the current.

I haven't butchered up the traces yet, so all fourteen are still in
parallel. Thanks.

Best regards,
Spehro Pefhany

--
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###Got a Question about ELECTRONICS? Check HERE First:###
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My email address is whitelisted. *All* email sent to it
goes directly to the trash unless you add NOSPAM in the
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Don't be ripped off by the big book dealers. Go to the URL
that will give you a choice and save you money(up to half).
http://www.everybookstore.com You'll be glad you did!
Just when you thought you had all this figured out, the gov't
changed it: http://physics.nist.gov/cuu/Units/binary.html
@@t@h@e@@a@f@f@l@u@e@n@t@@m@e@e@t@@t@h@e@@E@f@f@l@u@e@n@t@@

 

[Watson A.Name - Watt Sun, Dark Remover]

I believe it should work fine. I have done similar.

The first white LED illuminator I built had some LED failures
after it ran a while. I operated the LEDs near their max
current spec. I attributed the failures to that, which may have been
wrong - but white LED illuminators I've made since then running
at lower currents have had no problems. Maybe I had a bad
batch of LEDs, or maybe the whites are more sensitive than others.

Yeah, heat's a killer, and that concerns me, since this array gets
hot. But it's inexpensive, so it won't cost but a quarter if I have
to replace an LED.

Steve J. Noll | Ventura California |
| The Used High-Tech Equipment Dealer Directory
| http://www.big-list.com
| The Peltier Device Information Site:
| http://www.peltier-info.com

Thanks.

--
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###Got a Question about ELECTRONICS? Check HERE First:###
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Don't be ripped off by the big book dealers. Go to the URL
that will give you a choice and save you money(up to half).
http://www.everybookstore.com You'll be glad you did!
Just when you thought you had all this figured out, the gov't
changed it: http://physics.nist.gov/cuu/Units/binary.html
@@t@h@e@@a@f@f@l@u@e@n@t@@m@e@e@t@@t@h@e@@E@f@f@l@u@e@n@t@@

 

[Watson A.Name - Watt Sun, Dark Remover]

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.

I don't think destruction is possible with the current balancing
resistors. But yeah, the added heat will shorten the life as compared
to the other LEDs. But then, what's a couple thousand hours when they
usually last for a hundred thou?

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

Well, then I could get rid of the resistors?

-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!!

Not silicon here, gallium Arsenide! Or some other more exotic
combination of indium, phospor, aluminum, etc.

I heard that some blue LEDS (white is a blue LED die) have a substrate
of sapphire or something like that. So maybe they do have some
sillycon!

Cheers, John Stewart

--
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###Got a Question about ELECTRONICS? Check HERE First:###
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goes directly to the trash unless you add NOSPAM in the
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Don't be ripped off by the big book dealers. Go to the URL
that will give you a choice and save you money(up to half).
http://www.everybookstore.com You'll be glad you did!
Just when you thought you had all this figured out, the gov't
changed it: http://physics.nist.gov/cuu/Units/binary.html
@@t@h@e@@a@f@f@l@u@e@n@t@@m@e@e@t@@t@h@e@@E@f@f@l@u@e@n@t@@

 

[Jeff]

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.

 

[Watson A.Name - Watt Sun, Dark Remover]

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

I stuffed aniother PCB while watching Startrek, with 10 LEDs, 5 in
parallel over 5 in parallel. Instead of every other hole, i used
every third hole, which put more space between the LEDs so they are
running a bit cooler. Resistors were again 51 ohms. I had to make
only one trace cut and one jumper. Again, the PC board was part of a
Radio Snack 276-168 predrilled board. The PS is cranked up to 9.46V
at 150 mA, or about 30 mA per LED. This works out nicely witht he
9VDC 200 mA wall wart I have. 7 LEDs in parallel would've been 210
mA, and they would've been closer together and hotter.



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[John Fields]

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 5100.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.

 

[Watson A.Name \Watt Sun - the Dark Remover\]

On Wed, 29 Oct 2003 20:28:21 -0500, John Stewart



---
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

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

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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###Got a Question about ELECTRONICS? Check HERE First:###
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My email address is whitelisted. *All* email sent to it
goes directly to the trash unless you add NOSPAM in the
Subject: line with other stuff. alondra101 <at> hotmail.com
Don't be ripped off by the big book dealers. Go to the URL
that will give you a choice and save you money(up to half).
http://www.everybookstore.com You'll be glad you did!
Just when you thought you had all this figured out, the gov't
changed it: http://physics.nist.gov/cuu/Units/binary.html
I speak only for myself, no one else, & they're my opinions
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