I have used this approach for years.  I have never become aware of a
failure.  It is a very rliable circuit, providing the parameters are
properly determined.
----- Original Message -----
From: "Russell McMahon" <apptech@PARADISE.NET.NZ>
To: <PICLIST@MITVMA.MIT.EDU>
Sent: Wednesday, April 23, 2003 3:11 AM
Subject: Re: [PIC]: Driving high voltage LED from a PIC


> > I noticed the following circuit in a friend design of a very cost
> > sensitive product and am curious what the people think about it.
>
> >   [14V]----[A LED C]----[R1]---(A)----[R2]----[GND]
>
> > Where R1 is the current limiting resistor for the LED, point A is
> > connected to a PIC digital output and R2 is a relativly large resistor
> > (about 50K or so). The idea is that to turn the LED off, the PIC output
> > goes to HIGH or TRI_STATE and because of the 'knee' of the LED curve and
> > the resistor R2 that draws a minimal forward current, the voltage at the
> > PIC will not exceed its VDD.
>
> I'd have to say that this doesn't make sense.
> When it's current flow is low the LED will drop LESS voltage than when
> operating.
> Even a white LED is liable to drop say 2.5v at little light output.
> The voltage at the PIC pin is
>
> Vpic =     (Vsupply - Vled) x R2/(R1 + R2)
>
> or  R1 = ((Vs-Vl)/Vp -1) x R2
>
> If Vdd PIC = 5v
> VLED off = 2.5v
> R2 = 50k
>
>     R1 = ((14-2.5)/5 - 1) *50k = 65k
>
> At turn on Iled = (14-3)/65k =~ 0.17 mA
> Not at all bright !
> At turn off current through the LED will still be more than half the
> above!!!
>
> ie it doesn't work (and can't)(according to me :-) ).
>
> Really this is obvious by inspection as lowering and raising the bottom of
a
> resistor + LED chain by 5v when there is 14v at the top end is obviously
> going to have a relatively small effect. If the lower resistor is sized to
> limit swing to 0-5v range then the LED will see a less than 2:1 current
> swing.
> ___________
>
> A better method (that actually works :-) )
>
> A circuit is attached with the same number of components (replace an R
with
> a transistor) produces a near perfect result at very little extra cost.
LED
> current is defined almost solely by (Vgo-0.6)/R1. Vin can be any sensible
> voltage, limited only by transistor breakdown voltage when off and
> transistor dissipation when on. Cost is measured in cents (if using such
> currency :-) ). For a truly cost sensitive design this is probably surface
> mount so the transistor mounting cost is about the same as for a resistor.
>
>
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>
>

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