Post by thread:[PIC]: Driving high voltage LED from a PIC

Hello, I noticed the following circuit in a friend design of a very cost sensitive product and am curious what the people think about it. The PIC (e.g. 16F73) is to drive an high voltage LED such as digikey 441-1009-ND (http://rocky.digikey.com/WebLib/Sunbrite/Web%20Data/SSP-01TWB9WB12.pdf) . The general idea is to drive the LED directly from the PIC, saving an extra high voltage driver (actually, there are few LED in this circuit, each is driven seperatly). The circuit is as follows: [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. Does this make sense ? How about if the PIC is operating on 3V only ? Thanks, Tal -- http://www.piclist.com hint: PICList Posts must start with ONE topic: [PIC]:,[SX]:,[AVR]: ->uP ONLY! [EE]:,[OT]: ->Other [BUY]:,[AD]: ->Ads

That seems viable; I'm not sure I see the reasoning for the resistor tying the pic's pin to ground. It seems a bit extraneous; after all, you'll be sinking the lion's share of the current through the PIC regardless of how you do it. Mike Hord {Quote hidden}>From: Tal <spam_OUTtalTakeThisOuTZAPTA.COM> >Reply-To: pic microcontroller discussion list <.....PICLISTKILLspam@spam@MITVMA.MIT.EDU> >To: PICLISTKILLspamMITVMA.MIT.EDU >Subject: [PIC]: Driving high voltage LED from a PIC >Date: Tue, 22 Apr 2003 11:35:37 -0700 > >Hello, > >I noticed the following circuit in a friend design of a very cost >sensitive product and am curious what the people think about it. > >The PIC (e.g. 16F73) is to drive an high voltage LED such as digikey >441-1009-ND >(http://rocky.digikey.com/WebLib/Sunbrite/Web%20Data/SSP-01TWB9WB12.pdf) >. The general idea is to drive the LED directly from the PIC, saving an >extra high voltage driver (actually, there are few LED in this circuit, >each is driven seperatly). > >The circuit is as follows: > > > [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. > >Does this make sense ? How about if the PIC is operating on 3V only ? > >Thanks, > >Tal > >-- >http://www.piclist.com hint: PICList Posts must start with ONE topic: >[PIC]:,[SX]:,[AVR]: ->uP ONLY! [EE]:,[OT]: ->Other [BUY]:,[AD]: ->Ads _________________________________________________________________ The new MSN 8: advanced junk mail protection and 2 months FREE* http://join.msn.com/?page=features/junkmail -- http://www.piclist.com hint: PICList Posts must start with ONE topic: [PIC]:,[SX]:,[AVR]: ->uP ONLY! [EE]:,[OT]: ->Other [BUY]:,[AD]: ->Ads

On Tuesday 22 April 2003 11:55 am, Mike Hord wrote: > That seems viable; I'm not sure I see the reasoning for the > resistor tying the pic's pin to ground. It seems a bit extraneous; > after all, you'll be sinking the lion's share of the current > through the PIC regardless of how you do it. I think the point is so you don't draw current through the PIC protection diodes when the LED's OFF. But isn't the leakage current temperature sensitive? I don't know whether this will make a difference or not... {Quote hidden}> From: Tal <.....talKILLspam.....ZAPTA.COM> > > >Reply-To: pic microcontroller discussion list > > <EraseMEPICLISTspam_OUTTakeThisOuTMITVMA.MIT.EDU> To: PICLISTspam_OUTMITVMA.MIT.EDU > >Subject: [PIC]: Driving high voltage LED from a PIC > >Date: Tue, 22 Apr 2003 11:35:37 -0700 > > > >Hello, > > > >I noticed the following circuit in a friend design of a very cost > >sensitive product and am curious what the people think about it. > > > >The PIC (e.g. 16F73) is to drive an high voltage LED such as > > digikey 441-1009-ND > >(rocky.digikey.com/WebLib/Sunbrite/Web%20Data/SSP-01TWB9WB1 > >2.pdf) . The general idea is to drive the LED directly from the > > PIC, saving an extra high voltage driver (actually, there are few > > LED in this circuit, each is driven seperatly). > > > >The circuit is as follows: > > > > > > [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. > > > >Does this make sense ? How about if the PIC is operating on 3V > > only ? > > > >Thanks, > > > >Tal > > > >-- > >http://www.piclist.com hint: PICList Posts must start with ONE > > topic: [PIC]:,[SX]:,[AVR]: ->uP ONLY! [EE]:,[OT]: ->Other > > [BUY]:,[AD]: ->Ads > > _________________________________________________________________ > The new MSN 8: advanced junk mail protection and 2 months FREE* > http://join.msn.com/?page=features/junkmail > > -- > http://www.piclist.com hint: PICList Posts must start with ONE > topic: [PIC]:,[SX]:,[AVR]: ->uP ONLY! [EE]:,[OT]: ->Other > [BUY]:,[AD]: ->Ads -- Ned Konz http://bike-nomad.com GPG key ID: BEEA7EFE -- http://www.piclist.com hint: PICList Posts must start with ONE topic: [PIC]:,[SX]:,[AVR]: ->uP ONLY! [EE]:,[OT]: ->Other [BUY]:,[AD]: ->Ads

part 1 2250 bytes content-type:text/plain; (decoded 7bit) > 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. -- http://www.piclist.com hint: The list server can filter out subtopics (like ads or off topics) for you. See http://www.piclist.com/#topics part 2 1906 bytes content-type:image/gif; (decode) part 3 2 bytes -

> I have used this approach for years. I have never become aware of a > failure. It is a very reliable circuit, providing the parameters are > properly determined. I assume you mean his two resistor circuit. But, unless I am missing some fundamental principle completely, it doesn't work in the example he gave! At 14 volts, either - - The LED gets minimal current and doesn't light in a worthwhile manner or - The LED always receives enough current to turn on enough to glimmer or worse or - The PIC pin sees more than Vdd when turned off. I would genuinely welcome an explanation and/or analysis that shows how such a circuit can sesnibly achieve the objective. Russell McMahon > > > 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. > > -- http://www.piclist.com hint: The list server can filter out subtopics (like ads or off topics) for you. See http://www.piclist.com/#topics

pic microcontroller discussion list <> wrote on Wednesday, April 23, 2003 12:09 PM: >> I have used this approach for years. I have never become aware of a >> failure. It is a very reliable circuit, providing the parameters are >> properly determined. > > I assume you mean his two resistor circuit. > But, unless I am missing some fundamental principle completely, it > doesn't work in the example he gave! > At 14 volts, either - > > - The LED gets minimal current and doesn't light in a worthwhile > manner or > - The LED always receives enough current to turn on enough to glimmer > or worse or > - The PIC pin sees more than Vdd when turned off. And can I add: - if the +14V comes up before the PIC power supply, bad things might happen (latchup etc). That is a problem if the PIC supply is derived from the +14 (is that an automotive +14?), and in lots of other circumstances too. Like Russell, I'm puzzling as to any advantages of this circuit, even if it can be tweaked to work for a particular LED. Nigel -- Nigel Orr, Design Engineer @spam@nigelKILLspamaxoninstruments.co.uk Axon Instruments Ltd., Wardes Road,Inverurie,Aberdeenshire,UK,AB51 3TT Tel:+44 1467 622332 Fax:+44 1467 625235 http://www.axoninstruments.co.uk -- http://www.piclist.com hint: The list server can filter out subtopics (like ads or off topics) for you. See http://www.piclist.com/#topics

Nigel brings up a very good point about the power supply rise time- it's something we forget about often because on the scale WE see it on, it's instantaneous, but to the PIC it usually isn't even linear, much less instantaneous. I lost a number of PICs during one project to an "unexpected" 24-volt input. A solution (although I'm not certain how good it is; feedback DEFINITELY appreciated before I try it myself) would be to stick a FET in there to switch the current through the LED. You have the advantage of minimizing the current flow into the PIC, which also gives you a nice spot of isolation from that 14-V source. With the current design, if that LED goes and the voltage suddenly drops, you'll see a significant voltage/current spike, even with the 50k resistor in place. With the FET, it'll smoke, but the PIC will (probably) be protected. Comments? {Quote hidden}>From: Nigel Orr <KILLspamnigelKILLspamAXONINSTRUMENTS.CO.UK> >Reply-To: pic microcontroller discussion list <RemoveMEPICLISTTakeThisOuTMITVMA.MIT.EDU> >To: spamBeGonePICLISTspamBeGoneMITVMA.MIT.EDU >Subject: Re: [PIC]: Driving high voltage LED from a PIC >Date: Wed, 23 Apr 2003 13:11:39 +0100 > >pic microcontroller discussion list <> wrote on Wednesday, April 23, 2003 >12:09 PM: > > >> I have used this approach for years. I have never become aware of a > >> failure. It is a very reliable circuit, providing the parameters are > >> properly determined. > > > > I assume you mean his two resistor circuit. > > But, unless I am missing some fundamental principle completely, it > > doesn't work in the example he gave! > > At 14 volts, either - > > > > - The LED gets minimal current and doesn't light in a worthwhile > > manner or > > - The LED always receives enough current to turn on enough to glimmer > > or worse or > > - The PIC pin sees more than Vdd when turned off. > >And can I add: > - if the +14V comes up before the PIC power supply, bad things might >happen (latchup etc). > >That is a problem if the PIC supply is derived from the +14 (is that an >automotive +14?), and in lots of other circumstances too. Like Russell, >I'm puzzling as to any advantages of this circuit, even if it can be >tweaked to work for a particular LED. > >Nigel >-- >Nigel Orr, Design Engineer TakeThisOuTnigelEraseMEspam_OUTaxoninstruments.co.uk >Axon Instruments Ltd., Wardes Road,Inverurie,Aberdeenshire,UK,AB51 3TT > Tel:+44 1467 622332 Fax:+44 1467 625235 > http://www.axoninstruments.co.uk > >-- >http://www.piclist.com hint: The list server can filter out subtopics >(like ads or off topics) for you. See http://www.piclist.com/#topics _________________________________________________________________ MSN 8 with e-mail virus protection service: 2 months FREE* http://join.msn.com/?page=features/virus -- http://www.piclist.com hint: The list server can filter out subtopics (like ads or off topics) for you. See http://www.piclist.com/#topics

>A solution (although I'm not certain how good it is; feedback DEFINITELY >appreciated before I try it myself) would be to stick a FET in there to >switch the current through the LED. You have the advantage of minimizing >the current flow into the PIC, which also gives you a nice spot of isolation >from that 14-V source. With the current design, if that LED goes and the >voltage suddenly drops, you'll see a significant voltage/current spike, even >with the 50k resistor in place. With the FET, it'll smoke, but the PIC will >(probably) be protected. If going this way, just use a logic level FET to drive the LED. My current favourite is the Philips PHN210 - 30V, 3.5A logic level input dual FET in SO-8 package 0.66p from RS components in the UK RS stock #198-4195 for those interested. -- http://www.piclist.com hint: The list server can filter out subtopics (like ads or off topics) for you. See http://www.piclist.com/#topics

pic microcontroller discussion list <> wrote on Wednesday, April 23, 2003 2:16 PM: > A solution (although I'm not certain how good it is; feedback > DEFINITELY appreciated before I try it myself) would be to stick a > FET in there to switch the current through the LED. You have the > advantage of minimizing the current flow into the PIC, which also > gives you a nice spot of isolation from that 14-V source. With the > current design, if that LED goes and the voltage suddenly drops, > you'll see a significant voltage/current spike, even > with the 50k resistor in place. With the FET, it'll smoke, but the > PIC will (probably) be protected. > Personally, my generic LED driving circuit when there is a larger supply available (eg 9V, +12V), is usually a transistor+resistor current source. For circuits where there is only +5V, I usually use a FET, Gate to PIC, Source to GND, LED and series resistor from supply to Drain. The current source has PIC output to transistor base, transistor emitter to ground via suitable resistor (see below), + supply to LED anode, LED cathode to transistor collector. It only works if the PIC supply voltage + the voltage required by the load is less than the power supply, so you would need a >10V supply for a 5V PIC driving a 5V LED. Transistor base voltage is 5V, transistor emitter voltage approx 4.3V, current is 4.3/R, where R is the emitter-ground resistor. Have a look at current sources in Horowitz & Hill for more details (and better sources!) Advantages are that it's fairly rugged, if you short out the LED, no more current flows, you can add additional LEDs in series or change to an LED with different Vf and they all get the same current, if the supply changes from 9V to 24V, current remains the same, etc etc. Nigel -- Nigel Orr, Design Engineer RemoveMEnigelTakeThisOuTaxoninstruments.co.uk Axon Instruments Ltd., Wardes Road,Inverurie,Aberdeenshire,UK,AB51 3TT Tel:+44 1467 622332 Fax:+44 1467 625235 http://www.axoninstruments.co.uk -- http://www.piclist.com hint: The list server can filter out subtopics (like ads or off topics) for you. See http://www.piclist.com/#topics

At 02:44 PM 4/23/2003 +0100, you wrote: >The current source has PIC output to transistor base, transistor emitter to >ground via suitable resistor (see below), + supply to LED anode, LED >cathode to transistor collector. It only works if the PIC supply voltage + >the voltage required by the load is less than the power supply, so you >would need a >10V supply for a 5V PIC driving a 5V LED. An additional series base resistor of perhaps 0.3~1K is not a bad idea. Best regards, Spehro Pefhany --"it's the network..." "The Journey is the reward" speffEraseME.....interlog.com Info for manufacturers: http://www.trexon.com Embedded software/hardware/analog Info for designers: http://www.speff.com -- http://www.piclist.com hint: The list server can filter out subtopics (like ads or off topics) for you. See http://www.piclist.com/#topics

> Nigel brings up a very good point about the power supply rise time- it's > something we forget about often because on the scale WE see it on, it's > instantaneous, but to the PIC it usually isn't even linear, much less > instantaneous. I lost a number of PICs during one project to an > "unexpected" 24-volt input. > > A solution (although I'm not certain how good it is; feedback DEFINITELY > appreciated before I try it myself) would be to stick a FET in there to > switch the current through the LED. You have the advantage of minimizing > the current flow into the PIC, which also gives you a nice spot of isolation > from that 14-V source. With the current design, if that LED goes and the > voltage suddenly drops, you'll see a significant voltage/current spike, even > with the 50k resistor in place. With the FET, it'll smoke, but the PIC will > (probably) be protected. I don't know why you guys are making a simple thing so difficult. The easy answer takes two parts and works very well. PIC output goes to base of NPN, emitter to resistor, other end of resistor to ground, collector to LED cathode, LED anode to +V supply. When the PIC output goes to 5V, the transistor becomes a constant current sink at about 4.3V / R. As long as +V - 5V is enough to turn on the LED, this simple little circuit even regulates the current thru the LED. The +V supply can ripple a lot, and the LED brightness will be quite constant. I often use this scheme if the PIC 5V supply is regulated from an unregulated supply of at least 7.5V or so. The current thru the LED is nicely regulated, but it comes from the unregulated supply. Since a few LEDs can easily draw more current than the PIC and associated 5V circuitry, this keeps the regulated current draw down and can allow use of a low power linear regulator. ***************************************************************** Embed Inc, embedded system specialists in Littleton Massachusetts (978) 742-9014, http://www.embedinc.com -- http://www.piclist.com hint: The list server can filter out subtopics (like ads or off topics) for you. See http://www.piclist.com/#topics

> >The current source has PIC output to transistor base, transistor emitter to > >ground via suitable resistor (see below), + supply to LED anode, LED > >cathode to transistor collector. It only works if the PIC supply voltage + > >the voltage required by the load is less than the power supply, so you > >would need a >10V supply for a 5V PIC driving a 5V LED. > > An additional series base resistor of perhaps 0.3~1K is not a bad idea. Actually, it is. The emitter resistor will limit the current drawn from the PIC nicely. Additionaly base resistance only decreases the apparent impedence of the current sink driving the LED. The impedence of an ideal current sink is infinite. ***************************************************************** Embed Inc, embedded system specialists in Littleton Massachusetts (978) 742-9014, http://www.embedinc.com -- http://www.piclist.com hint: The list server can filter out subtopics (like ads or off topics) for you. See http://www.piclist.com/#topics

At 10:47 AM 4/23/2003 -0400, you wrote: > > >The current source has PIC output to transistor base, transistor >emitter to > > >ground via suitable resistor (see below), + supply to LED anode, LED > > >cathode to transistor collector. It only works if the PIC supply >voltage + > > >the voltage required by the load is less than the power supply, so you > > >would need a >10V supply for a 5V PIC driving a 5V LED. > > > > An additional series base resistor of perhaps 0.3~1K is not a bad idea. > >Actually, it is. The emitter resistor will limit the current drawn from >the PIC nicely. Additionaly base resistance only decreases the apparent >impedence of the current sink driving the LED. The impedence of an ideal >current sink is infinite. It prevents unwanted oscillation too. Best regards, Spehro Pefhany --"it's the network..." "The Journey is the reward" EraseMEspeffinterlog.com Info for manufacturers: http://www.trexon.com Embedded software/hardware/analog Info for designers: http://www.speff.com -- http://www.piclist.com hint: The list server can filter out subtopics (like ads or off topics) for you. See http://www.piclist.com/#topics

Get rid of R2, and use pin RA4. According to the data sheet, RA4 is an open-collector driver, which means that it will only sink current- never sourcing it. Furthermore, (according to the 18f242 data sheet) the only protection diode on RA4 is to Vss(Gnd). You can put a reletively high voltage on RA4 without worrying about a diode shunting current to Vdd (V+). I believe all the other PICs to have the same essential design. Just like the other ports you need to maintain a current of under 25mA. Enjoy! -Adam Tal wrote: {Quote hidden}>Hello, > >I noticed the following circuit in a friend design of a very cost >sensitive product and am curious what the people think about it. > >The PIC (e.g. 16F73) is to drive an high voltage LED such as digikey >441-1009-ND >(http://rocky.digikey.com/WebLib/Sunbrite/Web%20Data/SSP-01TWB9WB12.pdf) >. The general idea is to drive the LED directly from the PIC, saving an >extra high voltage driver (actually, there are few LED in this circuit, >each is driven seperatly). > >The circuit is as follows: > > > [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. > >Does this make sense ? How about if the PIC is operating on 3V only ? > >Thanks, > >Tal > >-- >http://www.piclist.com hint: PICList Posts must start with ONE topic: >[PIC]:,[SX]:,[AVR]: ->uP ONLY! [EE]:,[OT]: ->Other [BUY]:,[AD]: ->Ads > > > > > -- http://www.piclist.com hint: The PICList is archived three different ways. See http://www.piclist.com/#archives for details.

Can't you emulate an open collector output by changing the port direction to input ? Tal {Quote hidden}> -----Original Message----- > From: pic microcontroller discussion list > [RemoveMEPICLISTspam_OUTKILLspamMITVMA.MIT.EDU] On Behalf Of Alan B. Pearce > Sent: Friday, April 25, 2003 12:38 AM > To: RemoveMEPICLISTTakeThisOuTspamMITVMA.MIT.EDU > Subject: Re: [PIC]: Driving high voltage LED from a PIC > > > >I believe all the other PICs to > >have the same essential design. > > The 16F630/676 do not seem to have an O/C output on any > output. Do not know about any others, but perhaps it is being > dropped from some of the newer designs? > > -- > http://www.piclist.com#nomail Going offline? Don't AutoReply > us! email EraseMElistservspamspamBeGonemitvma.mit.edu with SET PICList DIGEST in the body > -- http://www.piclist.com#nomail Going offline? Don't AutoReply us! email RemoveMElistservKILLspammitvma.mit.edu with SET PICList DIGEST in the body

> Can't you emulate an open collector output by changing the port > direction to input ? In general, no!, unfortunately. This is why RA4 was mentioned. All other pins (and all pins on some PICs?) have high side protection diodes connected to Vdd. If an input pin rises to Vdd + about 0.6v the diode will clamp to Vdd. RA4 does not have such a diode as it is used as the high voltage 9+12v ish) programming voltage pin. Using the simple emitter follower current source that I posted a few days ago is arguably the simplest and cheapest way of getting a very capable high voltage driver - and it only takes 1 resistor and 1 transistor. RM -- http://www.piclist.com#nomail Going offline? Don't AutoReply us! email listservSTOPspamspam_OUTmitvma.mit.edu with SET PICList DIGEST in the body

On Fri, Apr 25, 2003 at 09:19:30PM +0200, Wouter van Ooijen wrote: > > RA4 does not have such a diode as it is used as the high > > voltage 9+12v ish) programming voltage pin. > > A bit confused? HVP uses /MCLR, not RA4. Correct. Also I've been meaning to get into this thread. Even though RA4 doesn't have a clamp diode, every datasheet that I've seen recently does have a maximum voltage specification for RA4. IIRC I've seen as low as 8V max. I just took a quick peek at the 16F62X datasheet and it specifies 14V max. Doom lies down the path of those who violate datasheet specifications. Is this project so cost sensitive that it cannot afford a nickel transistor and a penny resistor? BAJ -- http://www.piclist.com#nomail Going offline? Don't AutoReply us! email EraseMElistservEraseMEmitvma.mit.edu with SET PICList DIGEST in the body

At 10:47 AM 4/23/03 -0400, Olin Lathrop wrote: > > >The current source has PIC output to transistor base, transistor >emitter to > > >ground via suitable resistor (see below), + supply to LED anode, LED > > >cathode to transistor collector. It only works if the PIC supply >voltage + > > >the voltage required by the load is less than the power supply, so you > > >would need a >10V supply for a 5V PIC driving a 5V LED. > > > > An additional series base resistor of perhaps 0.3~1K is not a bad idea. > >Actually, it is. The emitter resistor will limit the current drawn from >the PIC nicely. Additionaly base resistance only decreases the apparent >impedence of the current sink driving the LED. The impedence of an ideal >current sink is infinite. 2 reasons why the series base resistor is a good idea: 1) eliminate parasitic oscillation 2) keep current sourced from PIC to a reasonable value if LED supply drops low enough that the transistor saturates. Note that the emitter current is going to remain constant whether or not the collector can supply the current - that extra current will come from the base. Like you, I've been using this technique for years. I mostly use tiny linear regulators (LP2950A) or simple zener regulated supplies and simply can't tolerate an unexpected large current just because the current source dropped out of regulation. Adding that one resistor in series with the base fixes the problem. Having that resistor also stop a nasty spike at 100 MHz is an added bonus. dwayne -- Dwayne Reid <spamBeGonedwaynerKILLspamplanet.eon.net> Trinity Electronics Systems Ltd Edmonton, AB, CANADA (780) 489-3199 voice (780) 487-6397 fax Celebrating 19 years of Engineering Innovation (1984 - 2003) .-. .-. .-. .-. .-. .-. .-. .-. .-. .- `-' `-' `-' `-' `-' `-' `-' `-' `-' Do NOT send unsolicited commercial email to this email address. This message neither grants consent to receive unsolicited commercial email nor is intended to solicit commercial email. -- http://www.piclist.com hint: To leave the PICList .....piclist-unsubscribe-requestspam_OUTmitvma.mit.edu>