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'[EE] Challenge II:: LED driver'
2007\12\03@225753 by Brian Kraut

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The other challenge brings back something I asked on this list a while back.
I did not get any responses I could use that time, but I like the idea of
presenting it as a challenge.  I will even give a $200 reward for an idea I
can use.

I manufacture a fiber optic instrument light for aircraft.  The illuminator
can be seen at http://www.engalt.com/newpage2.htm.  It is actually much
smaller than that now.  The LED is a 100 mA LED.  I also have a 150 mA LED I
can use, but I am currently only driving the one I use at 75 mA due to heat
dissipation (in the dropping resistors, not the LED).  The inside of the box
is about 3/4" cubed.  I actually have less space to work with than that
because the gland that the fibers go into and the connector takes up space
in the enclosure.  The enclosure is potted with potting compound.

Here is my problem.  The LED is about a 2.5V LED.  I have these units set up
for either 12V or 24V input.  Actually, they run about 13.5V or 26V in use.
With the 12V unit I am dropping about 11V at 75mA which is .825 W of heat I
need to get rid of.  I use two dropping resistors now, one on each leg of
the LED and orient them near the top of the potting compound.  That does a
good job and does not get very warm.  24V is a completely different story.
That gives me 1.76 W of heat to get rid of and the unit just overheats.  I
really want to run the unit at the full 150mA brightness, but that gives me
over 3.5W and that is really too much.

Heat sink is not an option.  External big resistor is how I am having to do
the 24V units now and that is a very bad solution.

A switching regulator is the answer, but one catch is that it has to be
dimmable.  That is easy enough to do with off the shelf parts now, but
presents a problem in this unit.  I only have a positive and negative input
and can not add a third lead for dimming.  What I need is something that can
go in the place of a light bulb and work with the existing dimming rheostat
already in the plane.  I want full brightness about 13V and dimmed to about
nothing at around 3V.  All of the off the shelf driver ICs are made to
output a constant current with a varying voltage and have a third input for
dimming.  I need something that has the output current varying linearly with
the input voltage.  It needs to be inductorless, very small and low parts
count, and cost no more than a buck or three.  I can build the boards with
different value parts for the 12 or 24V units.  Ideal would be something on
a half inch square board or less with no more than one low pin count IC and
a few supporting parts.

If you want to send me your ideas direct my email address is
spam_OUTbrianTakeThisOuTspamengalt.com.  If you think that you can offer a complete solution but
don't want to give it away for a few hundred bucks and think no one else
will either send me a proposal and we will talk.

Brian Kraut
Engineering Alternatives, Inc.
http://www.engalt.com


2007\12\03@232707 by David VanHorn

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Dosen't sound all that hard.  A constant current source, with it's
reference taken from Vin through a divider instead of a zener or
band-gap ref.

2007\12\03@235332 by Apptech

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> The other challenge brings back something I asked on this
> list a while back.
..


Why does it need to be inductorless (RFI?).
I'd say that this was a reasonably simple task to do if
inductors were allowed.
Also feasible inductorless BUT with rather more effort.

Outline:

Buck regulator that gives programmed current to LED
regardless of voltage + current control circuit based on
input voltage. As Vin rises the desired current is raised
and the buck current increases.

If Iled is dropped across a resistor and then compared to a
reference, if the reference is derived by dividing the input
voltage you get almost exactly what you want.

I'd design you one of these for the money offered BUT given
the above description, you or someone else could do it
easily enough.

___

Inductorless you can use a capacitive divider that charges
caps in series and discharges them in parallel.
As 12V works OK, even using a single stage 2 cap system
should do. The 'bottom' cap is always grounded. The "top"
cap has its bottom driven by a totem pole betwixt cap1 + and
ground. output to LED cct is provided only when the two caps
are low (2 diodes and 1 FET or bjt).

You could follow this by an identical stage which would give
Vin/4 - a bit out which should be a useful level to run the
LED from.


>  I will even give a $200 reward for an idea I
> can use.

Hmm. Was that for the idea or an actual circuit? :-).
if the above so inspires you that you want to start waving $
around you could donate it to james for PICList support.
If not then, would you like an actual circuit ? :-)

Note that an inductor based cct would be easier.

       Russell






{Original Message removed}

2007\12\04@001449 by Brent Brown

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> A switching regulator is the answer, but one catch is that it has to be
> dimmable.

Yes and yes, switching will get rid of your dropping resistors and bring your
disspation down (but do remember your LED's still disspate heat), and then
dimmable will be the hard part. The assumption is that the existing dimmer circuit is
a series potentiometer that suits a known load (x number of lamps), so it won't
reduce the voltage the same if you have LED's instead of incandescent lamps due
to lower current plus the switching will really mess things up.

An intelligent (micro based?) switching design design might seek to measure the
source impedance (dimmer potentiometer setting) during switching cycles and then
control the brightness ot the LED's accordingly. A bit vaugue but I think there's a
$200 idea in there somewhere.

--
Brent Brown, Electronic Design Solutions
16 English Street, St Andrews,
Hamilton 3200, New Zealand
Ph: +64 7 849 0069
Fax: +64 7 849 0071
Cell: +64 27 433 4069
eMail:  .....brent.brownKILLspamspam@spam@clear.net.nz


2007\12\04@012108 by William \Chops\ Westfield

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On Dec 3, 2007, at 7:57 PM, Brian Kraut wrote:

> A switching regulator is the answer, but one catch is that it has  
> to be
> dimmable [depending on input voltage only]

Since your load is constant, can you just get the feedback from  
somewhere
on the input side of a standard buck converter, instead of from the  
output?
Essentially, you'd be creating a switching regulator whose output  
voltage
was in a fixed ratio to the input voltage.  You wouldn't get constant  
current
mode (or maybe you could, since the current ratio would be usable as  
well),
but it ought to do a lot better than a linear regulator/resistor.

BillW


2007\12\04@045906 by Alan B. Pearce

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>All of the off the shelf driver ICs are made to output a constant
>current with a varying voltage and have a third input for dimming.

I would have thought that a simple circuit could be designed that allowed
the variation of the input voltage to be applied to the dimming input to
control the brightness.

You do not state what IC has an input like this, but I'll assume that at 0
volts on the dimming input, the LED is out, and at some voltage the LED is
full brightness. Then a constant voltage drop device from the power input,
consisting of an TL431 and a pair of resistors to set the voltage, along
with a suitable load resistor for the dimming control range, would give what
you need.

2007\12\04@091358 by Dave Tweed

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Brian Kraut <brian.krautspamKILLspamengalt.com> wrote:
> I need something that has the output current varying linearly with
> the input voltage. It needs to be inductorless, very small and low
> parts count, and cost no more than a buck or three.

As others have pointed out, it's very straightforward to build a buck
regulator that controls its output current based on its input voltage,
except for the requirement that it be "inductorless".

Where does this requirement come from?

-- Dave Tweed

2007\12\04@094055 by David VanHorn

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> As others have pointed out, it's very straightforward to build a buck
> regulator that controls its output current based on its input voltage,
> except for the requirement that it be "inductorless".
>
> Where does this requirement come from?

Sometimes people avoid inductors out of fear.

2007\12\04@104149 by Rolf

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Why not solve this problem in two parts.

Part 1, use something like a 555 to create PWM in proportion to the
input voltage....
Part 2, create a constant current regulator that outputs 150mA in a
switching/efficient way.

Combine them by using the output of the 555 to power the regulator....
PWM at a few KHz wold be more than enough...

Rolf



Brian Kraut wrote:
{Quote hidden}

2007\12\04@122953 by Brian Kraut

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I don't think that I have the room for an inductor.  It also does not need
to be an efficient buck converter, PWM is fine.  I am also trying to
minimize the parts count.  This is a low cost product that is sold in small
quantities and the boards will be made by hand.  Four or five parts is all I
will have room for.  Efficiency is not really an issue.  Just needs to be
efficient enought to not overheat while potted.

I have seen several ICs in small sizes like 5 pin SOTs that can drive LEDs
with no external parts at all.  Something like that with a two resistor
voltage divider to a dimming input would work.

Something that just was a 1/4 duty cycle PWM would work.  Would not even
need any kind of dimming input, it would always just be putting out 1/4 the
input voltage and that would do the job.  That could easily be done with a
few transistors, caps, and resistors, but again too many parts.

Brian Kraut
Engineering Alternatives, Inc.
http://www.engalt.com

{Original Message removed}

2007\12\04@151144 by Dave Tweed

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Brian Kraut <EraseMEbrian.krautspam_OUTspamTakeThisOuTengalt.com> wrote:
> I wrote:
> > Brian Kraut <brian.krautspamspam_OUTengalt.com> wrote:
> > > I need something that has the output current varying linearly with
> > > the input voltage. It needs to be inductorless, very small and low
> > > parts count, and cost no more than a buck or three.
> >
> > As others have pointed out, it's very straightforward to build a buck
> > regulator that controls its output current based on its input voltage,
> > except for the requirement that it be "inductorless".
> >
> > Where does this requirement come from?
>
> I don't think that I have the room for an inductor.

You can get some pretty substantial shielded power inductors in a 7mm x 7mm
SMT package. Is that too big?

> It also does not need to be an efficient buck converter, PWM is fine.

You're up against some basic physical limitations here. If you don't have
the means to store energy, then you must dissipate it one way or another.
Can your LED really handle the direct application of 24-28V, no matter how
briefly? You'll probably need some sort of current-limiting device even
with PWM, and that can either be a resistor or an inductor. The inductor
is a big win in terms of efficiency, and is probably no larger than the
resistor you'd need anyway.

-- Dave Tweed

2007\12\04@161640 by Brian Kraut

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I am dealing with some pretty tight quarters left in the enclosure.  I can
squeeze in a board about 15mm square and no more than 6mm deep.

I never really thought about LED damage.  I was kind of figuring that 100mA
average current would determine its life and it would not really matter if
it is getting high current short duration pulses.

I guess that an inductor is O.K. if we can still come up with something
small and with a low parts count and not microscopic components that I will
have to place by hand.

The enclosure is plastic and I do need something with very low EMI.  It will
be mounted behind an instrument panel with other avionics.

Brian Kraut
Engineering Alternatives, Inc.
http://www.engalt.com

{Original Message removed}

2007\12\04@180054 by Apptech

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Comment on one aspect:

> It also does not need
> to be an efficient buck converter, PWM is fine.

This alas is a fallacy.
If you use PWM to provide a given output current at a low
voltage from a higher voltage then the output power will
indeed be I x Vlow, as desired, BUT the net power dissipated
will be I x Vhigh and the difference will be dissipated just
as with a linear regulator.
If this sounds unintuitive (and it usually is) work through
the arithmetic while ensuring that ALL real world effects
are accounted for. eg if you apply 20 volt to a 5 volt
capacitor to charge it to 10 volts account for either some
series resistance and finite charging time OR zero
resistance and infinite currents.

Note that infinite current^2 x R tends to spoil your day.

ONLY with an energy transformer will you get a net reduction
in power dissipation.

Unfortunately.


       Russell

2007\12\04@180054 by Apptech

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> I guess that an inductor is O.K. if we can still come up
> with something
> small and with a low parts count and not microscopic
> components that I will
> have to place by hand.

There are modern LED driving ICs that are so small they are
almost an inhalation hazard and work at 1 MHz plus. Using
chip inductors gets the size way down. OR for ultimate
shielding a gratifyingly small toroid has theoretically no
and actually very very very little EMI generated.


       Russell


2007\12\04@213950 by Peter van Hoof

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With a single capacitor this would be true , if you would charge a
1 uf cap from the 20 volt and subsequently charge a  3 uf cap from
this charge you end up with 5ish volts at a fairly high efficiency only
the loss in the two switches and the series resistance of the caps
would matter

Peter van Hoof

{Original Message removed}

2007\12\04@234245 by Apptech

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> With a single capacitor this would be true , if you would
> charge a
> 1 uf cap from the 20 volt and subsequently charge a  3 uf
> cap from
> this charge you end up with 5ish volts at a fairly high
> efficiency only
> the loss in the two switches and the series resistance of
> the caps
> would matter

No, alas.
What you say about which resistances matter is true, but the
consequences are not what you expect.

To amend Watergate's 'Deep Throat's advice - 'follow the
current'.

ie work out what the current MUST do to get to its
destination and what must happen to it along the way.
Realise that a zero resistance switch is an infinite current
switch.

If you posit zero loss connections then when you parallel
two capacitors you get infinite current. The losses tend to
be high :-).

If you posit a finite resistance, even very small,  then you
can do the arithmetic to derive actual currents and losses
and the result is "not good" [tm].

As a demonstration of the outcome.
Charge a capacitor C to Voltage 2V.
Discharge into an equal cap C so combined voltage is now V.
Original energy is 0.5CV^2 = 0.5 x C x (2V)^2 = 2V^2C.
Final energy is 2 x 0.5xCxV^2 = V^2C.
*HALF* your energy has vanished!!!!
Q: Where to?
A: Into the necessary resistance that allowed you to carry
out the energy transfer.

Caps CAN be charged resistively at good efficiency by
limiting the voltage excursions to a small percentage of the
voltages involved. People make very efficient convertyers
and multipliers doing tghis. But, discharge or charge the
cap resistively by a substantial percentage of it's
greatests voltage and efficiency will drop badly.

My suggestion re caps which are charged in parallel and
discharged in series used physical space division to perform
energy translation.
ie Charge C+C in series to 2V. Now parallel them so you get
2C at V. An energy analysis will show that for ideal
components no energy is lost when changing from 2V to V.


       Russell












>
> Peter van Hoof
>
> {Original Message removed}

2007\12\05@074345 by Alan B. Pearce

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>The inductor is a big win in terms of efficiency, and is
>probably no larger than the resistor you'd need anyway.

and almost certainly no larger than the resistors the OP said were being
used currently.

2007\12\05@101836 by Brian Kraut

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The resistors now are mounted by their leads between the connector and the
LED and are bent to be just under the level of the potting compound.  Just
complicates things to do inductors off the PCB.  Again, if I could fit the
inductor and everything else on a board 15mm square and 6mm tall it would
fit.

Brian Kraut
Engineering Alternatives, Inc.
http://www.engalt.com

-----Original Message-----
From: @spam@piclist-bouncesKILLspamspammit.edu [KILLspampiclist-bouncesKILLspamspammit.edu]On Behalf
Of Alan B. Pearce
Sent: Wednesday, December 05, 2007 7:43 AM
To: Microcontroller discussion list - Public.
Subject: Re: [EE] Challenge II:: LED driver


>The inductor is a big win in terms of efficiency, and is
>probably no larger than the resistor you'd need anyway.

and almost certainly no larger than the resistors the OP said were being
used currently.

2007\12\05@104444 by David VanHorn

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On Dec 5, 2007 10:13 AM, Brian Kraut <RemoveMEbrian.krautTakeThisOuTspamengalt.com> wrote:
> The resistors now are mounted by their leads between the connector and the
> LED and are bent to be just under the level of the potting compound.  Just
> complicates things to do inductors off the PCB.  Again, if I could fit the
> inductor and everything else on a board 15mm square and 6mm tall it would
> fit.

I would think that would be pretty trivial. At 1 MHz the inductors are
very small, as others have said, if you can fit a 1206 SMD resistor,
then you can fit the inductor.

You asked for advice, you've been given a fair bit.
:)

2007\12\05@163309 by Ruben Jönsson

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Here is a link to a LED driver circuit that just turned up in a newsletter.

<http://www.claremicronix.com/datasheets/MXHV9910%20Data%20Sheet.pdf>

/Ruben
==============================
Ruben Jönsson
AB Liros Electronic
Box 9124, 200 39 Malmö, Sweden
TEL INT +46 40142078
FAX INT +46 40947388
spamBeGonerubenspamBeGonespampp.sbbs.se
==============================

2007\12\05@171018 by David VanHorn

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On Dec 5, 2007 4:32 PM, Ruben Jönsson <TakeThisOuTrubenEraseMEspamspam_OUTpp.sbbs.se> wrote:
> Here is a link to a LED driver circuit that just turned up in a newsletter.
>
> <http://www.claremicronix.com/datasheets/MXHV9910%20Data%20Sheet.pdf>

They win an award for badly written data sheet.
What's the max LED current?

Also, 1mA idle drain at 450V?  Yikes!

2007\12\05@173405 by Spehro Pefhany

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Quoting Ruben Jönsson <RemoveMErubenspamTakeThisOuTpp.sbbs.se>:

> Here is a link to a LED driver circuit that just turned up in a newsletter.
>
> <http://www.claremicronix.com/datasheets/MXHV9910%20Data%20Sheet.pdf>
>

It claims to be "dialectically" isolated.

Would it be isolated from Hegelian dialectics, Marxian dialectics or both?

http://en.wikipedia.org/wiki/Dialectical_materialism

Best regards,
Spehro Pefhany
--
"it's the network..."                          "The Journey is the reward"
s...EraseMEspam.....interlog.com             Info for manufacturers: http://www.trexon.com
Embedded software/hardware/analog  Info for designers:  http://www.speff.com

2007\12\05@173537 by Brian Kraut

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That one requires at least an 8V input.  I need something to work down to at
least 3V.

Brian Kraut
Engineering Alternatives, Inc.
http://www.engalt.com

{Original Message removed}

2007\12\05@180336 by David VanHorn

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>
> It claims to be "dialectically" isolated.

"challenged" rather than isolated, maybe.. :)


'[EE] Challenge II:: LED driver'
2008\03\27@121729 by Apptech
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To:    Brian Kraut

email sent offlist on this topic.
If not received check your spam folder.


       Russell McMahon

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