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'[EE] Constant current schematic any recommendatio'
2005\04\08@015347 by Andre Abelian

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Hi to all engineers,

I am looking for constant current schematic to drive power led
any circuit recommendation? I came up op amp based circuit but I am not
sure if this is the best way to go see attachment.
Any ideas?

Thank you

Andre  





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2005\04\08@023001 by Charles Craft

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Here's a circuit I built in SwitchCAD then breadboarded to verify.

Don't remember the current draw but it was pretty closer to what the simulator came up with.


{Original Message removed}

2005\04\08@042200 by Russell McMahon

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I'll duplicate this at the top as it's the main point

       R2 ~= (Vbase-0.6)/ILED

_______________


> Here's a circuit I built in SwitchCAD then breadboarded to verify.

That's a good starting point.
The original shows 3 current sources.

I've copied a portion of the circuit to show 1 only current source for
explanation so you can adapt it if you wish.

The circuit works by placing a constant voltage across the emitter
resistor. This causes (almost) the same current to flow in the LED as
in the emitter resistor. As originally shown the current is about
1.2/Re = 1.2/50 = 24 mA in this case.

The transistor has ABOUT 0.6v drop between base and emitter = Vbe.
The diodes have about 0.6v forward drop each = Vd.
So Vr2 is (3 x Vd - Vbe) = 3 x 0.6 - 0.6 = 1.2v.

R1 = 10k is not too critical - it simply provides enough current to
forward bias the diodes and also provide base current for the
transistor. Base current has to be more than (LED current) /
(transistor current gain). As current gain is typically 100 or more
for most small signal transistors then the value shown for R1 shold be
fine.

If desired you can replace the 3 diodes with a zener diode or other
constant voltage source. If the supply voltage was constant (eg 5
volts) then you could use 2 resistors to form a voltage divider to
bias the base. eg in this case if Vsupply = 5 volts, then to get 1.8v
at the base you could replace the diodes with a 1.8/(5-1.8) x 10k =
5k6 resistor.

__________

ILED = Ir2
Ir2 = Vr2/R2
Vr2 = Vbase - Vbe
Vbe ~= 0.6
so ILED ~= (Vbase-0.6)/R2

       so             R2 = (Vbase-0.6)/ILED

which is enough to design with.




           RM


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2005\04\08@045219 by Michael Rigby-Jones

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>-----Original Message-----
>From: spam_OUTpiclist-bouncesTakeThisOuTspammit.edu [.....piclist-bouncesKILLspamspam@spam@mit.edu]
>Sent: 08 April 2005 06:53
>To: 'Microcontroller discussion list - Public.'
>Subject: [EE] Constant current schematic any recommendation?
>
>
>Hi to all engineers,
>
>I am looking for constant current schematic to drive power led
>any circuit recommendation? I came up op amp based circuit but
>I am not sure if this is the best way to go see attachment.
>Any ideas?
>
>Thank you
>
>Andre  

What is the intended function of D2?  The circuit would not work well,
if at all with D2 as it stands. The voltage on pin 2 of the opamp will
not be well defined so the diode may be forwards or reverse biased.  If
you added a resistor from pin 2 of the opamp to the +ve rail to ensure
the diode was forwards biased, then it has the function of adding a
~0.6volt error term into the feedback path, reducing the LED current.
However, the diode forward voltage will vary with both temperature and
any supply voltage changes, making this not such a good idea.  If you
need to reduce the current in the LED, then either choose a larger
current sense resistor (R1) or pot down the reference voltage with a
couple of resistors.

Apart from that, it's a classic voltage controlled current source.  With
a good reference you should get a stable current that moves very little
with temperature etc.  You could use a simple transistor based current
source if your accuracy and stability requirements are not great, but
they are quite temperature sensitive.

Regards

Mike

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2005\04\08@074657 by Dominic Stratten

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Personally I use an LM317 with a resistor - easiest way of doing it
(especially for the LEDs I use which require 350ma max regulated).

Not getting into circuit diagrams but its pretty easy to knock up - have a
look at the LM317 datasheet for an example.

Dom

{Original Message removed}

2005\04\08@104331 by Harold Hallikainen

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How much power for the power LED? Here are a few approaches I've taken:

1. I used a Linear Tech boost converter to drive an LED in a dental
application (1.5A, 7V). I used a 50 milliohm current sense resistor
driving a Maxim current sense amplifier. The output of the amplifier drove
the voltage feedback input of the LT converter. In addition, a 10V zener
was between the output of the boost converter and the feedback input so an
open LED did not cause the voltage out of the boost converter to go high
enough to blow out all the other parts.

2. In another application where I'm driving photoflash LEDs (1.25A), I
just used an LM317 to drive the high side of the LED. A resistor between
the output of the LM317 and the high side of the LED developed a voltage
proportional to the current. The adjust input of the LM317 was connected
to the load side of that resistor. The LM317 adjusts itself until there is
1.25V between the output and the adjust pin. In this application, I then
used an FET on the low side of the LED to flash the LED. The FET was
driven (of course) by a PIC.

3. LED driver chips from Supertex. They have a variety of simple switchers
designed to drive high power LEDs.

Harold


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2005\04\08@111044 by rosoftwarecontrol

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too much many component.

one lm317z can give perfect 5-100mA contant current,
by adding a resistance. low cost.






----- Original Message -----
From: "Michael Rigby-Jones" <Michael.Rigby-JonesspamKILLspambookham.com>
To: "Microcontroller discussion list - Public." <.....piclistKILLspamspam.....mit.edu>
Sent: Friday, April 08, 2005 4:51 AM
Subject: RE: [EE] Constant current schematic any recommendation?


>
>
> >{Original Message removed}

2005\04\08@113256 by Harold Hallikainen

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> too much many component.
>
> one lm317z can give perfect 5-100mA contant current,
> by adding a resistance. low cost.
>
>


"The ideal design has zero parts."



--
FCC Rules Updated Daily at http://www.hallikainen.com

2005\04\08@135127 by Andre Abelian

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Mike,

Thank you for your replay you are right d2 makes no sense because
of another resister was missing see attachment2 again.

Andre




{Original Message removed}

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