Exact match. Not showing close matches.
PICList
Thread
'[EE] trouble while driving power leds'
2006\02\24@073558
by
Russell McMahon

>I have confirmed that the correct Vf bin code ranges in 3.51 
>3.75[V] for
> my white Luxeon III Emitter power led. I have to confess that my
> mind a
> little bit confused at this stage. After taking account the proper
> Vf value
> i decided to use max Vf value in my calculations
This should be [EE]. I'll post this to [OT] and [EE] but suggest any
further responses be [EE].
A major problem is that you are "sailing too close to the wind" for
the simple circuit that you are trying to use. You are trying to
approximate a constant current source by using a constant voltage
source and a series resistor. This is a common method and works well
for most applications as long as the percentage variation in voltage
drop across the resistor is small for all possible load combinations.
But when the resistor voltage varies appreciably for various possible
load conditions you will get corresponding substantial changes in
current. This is what happens in your case for all possible
combinations of LED forward voltage.
eg in your case, worst case the LED voltage may be as low as 6 x 3.51
= 21.06 =~ 21V or as high as 6 x 3.75v = 22.5V. With a 24V supply this
would result in a resistor voltage drop of from 3V to 1.5V or a factor
of 2:1. The current would also vary by a factor of 2:1. This would
only hold true if the LED voltage remained constant for widely varying
current, but as you are on steeply increasing part of the LEDs V/I
curve then the LED voltage will change with current. The end result is
that for groups of different LEDs from the same bin you MAY get widely
differing currents.
The two "easy" solutions to your problem are
1. Higher supply voltage  say 30V or more
or
2. Less LEDs in the string (say 5) thus allowing more headroom voltage
across the resistor.
Slightly more complex solutions that can produce an excellent result
are:
3. Active current source. This is the most elegant solution but, as
you only have as little as (2422.5) = 1.5V headroom, the circuit will
need a little care. Given the cost of the 6 x Luxeon LEDs the cost of
the current source is minimal. A common method is to use an LM317 as
a current source but here the available headroom voltage is so
marginal as to be unworkable, Increasing the supply even slightly or
using LEDS from a lower voltage bin would improve the prospects of
using an LM317.
A current source can easily be built with an opamp and a transistor or
two and will happily work with this little headroom.
4. Voltage booster: If you are committed to a 24V supply you could
use a boost converter of some sort to provide either a constant
current LED feed (it then produces just enough voltage to achieve
this) or a constant higher voltage with larger series resistor. Of
these the constant current boost converter is the most elegant.
5. Voltage buck converter with 2 parallel strings of 3 LEDS. A
constant current or constant voltage (see 4.) buck converter can
reduce the 24v to around 13 or 14V to drive 3 LEDs in series and you
then run 2 parallel strings of these. Best is a constant current buck
converter for each 'string' of 3 LEDs but a single converter with
"spreading resistors": in each string will work OK
Of all these solutions I would see the higher supply with larger
series resistor as the easiest if supply voltage can be increased, and
the constant current boost converter as the best if the supply voltage
is fixed at 24V.
RM
{Original Message removed}
2006\02\24@094229
by
Gökhan SEVER

Thanks for clarifying the subject. The terminology is getting mysterious for
me. I think, i have to dig in the power electronics tomes. Due to the fact
i dont have much background information related to smps nor any practical
design experience. Especially, i will concentrete on "constant current boost
converters". The powerled design issues are completely on my shoulder in
our company. Although i'm a new electronics engineer, i have to master on
power, optical, control and pcb design areas except the mechanical parts :)
(But we are planning to integrate our pcb tool with cad/cam tools for
manufacturing better products.) In short i have to become proficient in
every detail of led powered products' from beginning parts to the last.
2006/2/24, Russell McMahon <spam_OUTapptechTakeThisOuTparadise.net.nz>:
{Quote hidden}>
> >I have confirmed that the correct Vf bin code ranges in 3.51 
> >3.75[V] for
> > my white Luxeon III Emitter power led. I have to confess that my
> > mind a
> > little bit confused at this stage. After taking account the proper
> > Vf value
> > i decided to use max Vf value in my calculations
>
> This should be [EE]. I'll post this to [OT] and [EE] but suggest any
> further responses be [EE].
>
> A major problem is that you are "sailing too close to the wind" for
> the simple circuit that you are trying to use. You are trying to
> approximate a constant current source by using a constant voltage
> source and a series resistor. This is a common method and works well
> for most applications as long as the percentage variation in voltage
> drop across the resistor is small for all possible load combinations.
> But when the resistor voltage varies appreciably for various possible
> load conditions you will get corresponding substantial changes in
> current. This is what happens in your case for all possible
> combinations of LED forward voltage.
>
> eg in your case, worst case the LED voltage may be as low as 6 x 3.51
> = 21.06 =~ 21V or as high as 6 x 3.75v = 22.5V. With a 24V supply this
> would result in a resistor voltage drop of from 3V to 1.5V or a factor
> of 2:1. The current would also vary by a factor of 2:1. This would
> only hold true if the LED voltage remained constant for widely varying
> current, but as you are on steeply increasing part of the LEDs V/I
> curve then the LED voltage will change with current. The end result is
> that for groups of different LEDs from the same bin you MAY get widely
> differing currents.
>
> The two "easy" solutions to your problem are
>
> 1. Higher supply voltage  say 30V or more
> or
> 2. Less LEDs in the string (say 5) thus allowing more headroom voltage
> across the resistor.
>
> Slightly more complex solutions that can produce an excellent result
> are:
>
> 3. Active current source. This is the most elegant solution but, as
> you only have as little as (2422.5) = 1.5V headroom, the circuit will
> need a little care. Given the cost of the 6 x Luxeon LEDs the cost of
> the current source is minimal. A common method is to use an LM317 as
> a current source but here the available headroom voltage is so
> marginal as to be unworkable, Increasing the supply even slightly or
> using LEDS from a lower voltage bin would improve the prospects of
> using an LM317.
>
> A current source can easily be built with an opamp and a transistor or
> two and will happily work with this little headroom.
>
> 4. Voltage booster: If you are committed to a 24V supply you could
> use a boost converter of some sort to provide either a constant
> current LED feed (it then produces just enough voltage to achieve
> this) or a constant higher voltage with larger series resistor. Of
> these the constant current boost converter is the most elegant.
>
> 5. Voltage buck converter with 2 parallel strings of 3 LEDS. A
> constant current or constant voltage (see 4.) buck converter can
> reduce the 24v to around 13 or 14V to drive 3 LEDs in series and you
> then run 2 parallel strings of these. Best is a constant current buck
> converter for each 'string' of 3 LEDs but a single converter with
> "spreading resistors": in each string will work OK
>
> Of all these solutions I would see the higher supply with larger
> series resistor as the easiest if supply voltage can be increased, and
> the constant current boost converter as the best if the supply voltage
> is fixed at 24V.
>
>
> RM
>
>
> {Original Message removed}
2006\02\24@162836
by
William Chops Westfield
>> A major problem is that you are "sailing too close to the wind" for
>> the simple circuit that you are trying to use. You are trying to
>> approximate a constant current source by using a constant voltage
>> source and a series resistor.
Russell said it better than I did. The use of a series resistor
and constant voltage supply to approximate a constant current
source is only going to work "well", if the voltage drop across
the resistor is "large" compared to the variation in voltage drop
across the other components with changing conditions and parts.
If you were driving three luxeons (at, say, Vf = 3 to 4 V), you'd
have about R=15 ohms and your current would vary from 1A to 0.8A,
or about 20%. Driving 5 series luxeons you'd use a 9 ohm resistor
and your current would vary from 1A to .44A, or close to 60% (and
it'd be worse with 6 luxeons.)
You don't say which parts of your design are fixed, or how cost
sensitive things are. The current you're trying to get (1A) is
close to the limits of "simple" LM317based linear current regulators
unless you go to a fair amount of trouble in heat management.
That would hold true for any linear regulation scheme, of course,
even resistors. (With 3 series LEDs and 15 ohm resistor, you
dissipate 15W and probably want at least a 20W resistor!)
Switching regulator design has always seemed complex to me, but
it's getting to the point where you can buy simpletouse chips
designed for driving power LEDs and use their suggested circuits.
Those tend to be relatively expensive (but on the third hand, you
may be able to replace the 24V regulated supply you're using now
with something cheaper...)
BillW
2006\02\24@192435
by
David VanHorn
A boost reg configured as a constant current source would be pretty easy.
VinVout is almost arbitrarily low, and you could use a 0.1 ohm sense
resistor if you amplify it to compare with your Vref.
2006\02\24@200954
by
Harold Hallikainen
> A boost reg configured as a constant current source would be pretty easy.
> VinVout is almost arbitrarily low, and you could use a 0.1 ohm sense
> resistor if you amplify it to compare with your Vref.
>
I did that in a dental curing light. An LTC boost converter with a Maxim
current sense amp driving the voltage feedback input of the boost
converter. Also added a zener between the output of the current sense amp
and the top of the LED so an open LED would not cause the boot converter
output to go to infinite voltage. Instead, in this failure mode, the
output pulls up the current sense output through the zener, limiting the
output voltage to the reference voltage plus the zener voltage.
Harold

FCC Rules Updated Daily at http://www.hallikainen.com
More... (looser matching)
 Last day of these posts
 In 2006
, 2007 only
 Today
 New search...