Post by thread:[EE]: Drive bicolor LED w/ one output pin

Hi All, My basic EE skills are lacking so I was hoping someone might suggest a solution to this... I have a bicolor LED - one of the 2-lead types where the green anode is the red cathode and vice-versa. You reverse polarity to change colors. What I need to do is drive this with one output pin on my PIC. I swear I'd do it with two if I had them, but I don't. The green can be on all the time except when the red comes on. So - steady green and then red when my output goes high (or low). Any thoughts? Thanks! -- http://www.piclist.com hint: PICList Posts must start with ONE topic: [PIC]:,[SX]:,[AVR]: ->uP ONLY! [EE]:,[OT]: ->Other [BUY]:,[AD]: ->Ads

If one pin of your LED is driven from a 2.5 volt source, then the other pin can be high, low, or open, giving you red, green, or off. A cheap 2.5 volt source is two 500 ohm resistors in series from ground to 5V. I have used this to drive red-green LEDs from one pin in the past. You don't need a series resistor to the LED with this arrangement, as the impedance of the the resistor pair is already enough. -- Lawrence Lile Senior Project Engineer Toastmaster, Inc. Division of Salton, Inc. 573-446-5661 voice 573-446-5676 fax Matt Redmond <spam_OUTmdredmondTakeThisOuTCHARTER.NET> Sent by: pic microcontroller discussion list <.....PICLISTKILLspam@spam@MITVMA.MIT.EDU> 04/21/2004 05:04 PM Please respond to pic microcontroller discussion list To: PICLISTKILLspamMITVMA.MIT.EDU cc: Subject: [EE]: Drive bicolor LED w/ one output pin Hi All, My basic EE skills are lacking so I was hoping someone might suggest a solution to this... I have a bicolor LED - one of the 2-lead types where the green anode is the red cathode and vice-versa. You reverse polarity to change colors. What I need to do is drive this with one output pin on my PIC. I swear I'd do it with two if I had them, but I don't. The green can be on all the time except when the red comes on. So - steady green and then red when my output goes high (or low). Any thoughts? Thanks! -- 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: PICList Posts must start with ONE topic: [PIC]:,[SX]:,[AVR]: ->uP ONLY! [EE]:,[OT]: ->Other [BUY]:,[AD]: ->Ads

Matt. The simplest scheme is to have a voltage divider between the +5v & 0 lines and connect the led between the centre (or slightly off-centre) connection and the PIC pin. The pin can then be high for one colour, low for the other & defined as an input for both off. Current will be limited by the resistors in the divider. The penalty is that even with both off there will be reasonable current flowing through the divider. This may not be a problem, particularly if you never have the "both off" condition. The tap may be slightly off centre to adjust the relative brightness. A circuit could be derived that reduced the standing current to only switch in when required - but it would add some complexity. It may not be worth it unless current consumption is a critical issue. Richard P Hi All, My basic EE skills are lacking so I was hoping someone might suggest a solution to this... I have a bicolor LED - one of the 2-lead types where the green anode is the red cathode and vice-versa. You reverse polarity to change colors. What I need to do is drive this with one output pin on my PIC. I swear I'd do it with two if I had them, but I don't. The green can be on all the time except when the red comes on. So - steady green and then red when my output goes high (or low). Any thoughts? Thanks! -- 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: PICList Posts must start with ONE topic: [PIC]:,[SX]:,[AVR]: ->uP ONLY! [EE]:,[OT]: ->Other [BUY]:,[AD]: ->Ads

At 10:04 PM 4/21/2004 +0000, you wrote: >Hi All, > >My basic EE skills are lacking so I was hoping someone might suggest a >solution to this... > >I have a bicolor LED - one of the 2-lead types where the green anode is >the red cathode and vice-versa. You reverse polarity to change colors. > >What I need to do is drive this with one output pin on my PIC. I swear >I'd do it with two if I had them, but I don't. > >The green can be on all the time except when the red comes on. So - >steady green and then red when my output goes high (or low). > >Any thoughts? > >Thanks! This one has more output drive and does not load the PIC You get full supply voltage. With 5V you can get 30mA+ through the LEDs. +---|<---+-------+ | | | +--->|---+ .-. | | | | | | |\ | |\ '-' -| >O----+--| >O----------+ |/ |/ SN74LVC2G14 This one only uses half the SN74LVC2G14 and you can turn both LEDs off by tristating the port pin. You get full supply voltage. You must make sure not to load the PIC pin too much so that it can't meet input specs for the SN74LVC2G14. +---|<--+-------+ | | | +--->|--+ .-. | | | | | | | '-' | |\ | ---------+---| >O--------+ |/ 1/2 SN74LVC2G14 The SN74LVC2G14 dual ST inverter is about 0.48 one-off, 0.27 in 100's, and is housed in a compact SOT-23-6, so it won't break the bank. Best regards, Spehro Pefhany --"it's the network..." "The Journey is the reward" .....speffKILLspam.....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: PICList Posts must start with ONE topic: [PIC]:,[SX]:,[AVR]: ->uP ONLY! [EE]:,[OT]: ->Other [BUY]:,[AD]: ->Ads

> A circuit could be derived that reduced the standing current to > only switch in when required - but it would add some complexity. > It may not be worth it unless current consumption is a critical issue > > Richard P Which is something I should have mentioned in my reply. Late last year I needed to drive 3 bi-colour LED staus indicators using the fewest number of pins, which turned out to be 1. The LEDs were connected between adjacent o/p pairs of a CMOS shift register, which was driven with a data pattern to make the combinations of red/green on the LEDs. The PIC had to survive on batteries and was mostly powered down, woken fully by a microswitch on the case door PS H-bridge suggestion was a little OTT -- http://www.piclist.com hint: PICList Posts must start with ONE topic: [PIC]:,[SX]:,[AVR]: ->uP ONLY! [EE]:,[OT]: ->Other [BUY]:,[AD]: ->Ads

On Wed, Apr 21, 2004 at 10:04:31PM +0000, Matt Redmond wrote: > Hi All, > > My basic EE skills are lacking You may want to bone up. I've spent quite a bit of time reading the copy of the Art of Electronics I requested from our school library. > so I was hoping someone might suggest a solution to this... Fire away. > I have a bicolor LED - one of the 2-lead types where the > green anode is the red cathode and vice-versa. You reverse polarity to > change colors. Check. > What I need to do is drive this with one output pin on my > PIC. I swear I'd do it with two if I had them, but I don't. Understood. > The green can > be on all the time except when the red comes on. So - steady green and then > red when my output goes high (or low). Good. So it doesn't have to be off. That makes it a lot tougher. > > > Any thoughts? Several. The first is that if you can afford to burn the current you can simply tie the outside leg to a midpoint resistor voltage divider that locks the outside leg at 2.5V (presuming a 5V circuit). Then the PIC output can swing from 0 to 5V, going from one color to the other as it swings. Neither LED in the bipolar should require more than 2V forward voltage. A more active approach could be to inject a opamp voltage follower between the divider and the LED. So you get: PIC pin -> resistor -> LED -> opamp follower -> voltage divider The nice thing about this approach is that you can build the voltage divider with 47K or 100K resistors limiting the amount of current they consume because they no longer have to feed power to the LED. BAJ -- 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

{Quote hidden}>-----Original Message----- >From: Byron A Jeff [EraseMEbyronspam_OUTTakeThisOuTCC.GATECH.EDU] >Sent: 22 April 2004 16:38 >To: PICLISTspam_OUTMITVMA.MIT.EDU >Subject: Re: [EE]: Drive bicolor LED w/ one output pin > > >On Wed, Apr 21, 2004 at 10:04:31PM +0000, Matt Redmond wrote: >> Hi All, >> > >> My basic EE skills are lacking > >You may want to bone up. I've spent quite a bit of time >reading the copy of the Art of Electronics I requested from >our school library. > >> so I was hoping someone might suggest a solution to this... > >Fire away. > >> I have a bicolor LED - one of the 2-lead types where the >green anode >> is the red cathode and vice-versa. You reverse polarity to change >> colors. > >Check. > >> What I need to do is drive this with one output pin on my PIC. I >> swear I'd do it with two if I had them, but I don't. > >Understood. > >> The green can >> be on all the time except when the red comes on. So - steady green >> and then red when my output goes high (or low). > >Good. So it doesn't have to be off. That makes it a lot tougher. > >> >> >> Any thoughts? > >Several. The first is that if you can afford to burn the >current you can simply tie the outside leg to a midpoint >resistor voltage divider that locks the outside leg at 2.5V >(presuming a 5V circuit). Then the PIC output can swing from 0 >to 5V, going from one color to the other as it swings. Neither >LED in the bipolar should require more than 2V forward voltage. > >A more active approach could be to inject a opamp voltage >follower between the divider and the LED. So you get: > >PIC pin -> resistor -> LED -> opamp follower -> voltage divider > >The nice thing about this approach is that you can build the >voltage divider with 47K or 100K resistors limiting the amount >of current they consume because they no longer have to feed >power to the LED. > Even easier is to drive one side of the LED and a CMOS inverter input with the PIC pin, and the other side of the LED with the inverter output. The above solution may be ok, but you are limiting your forward voltage to 2.5volts for a 5v supply rail which may be marginal for some LED's |\ ---+ +-----+ |0-------------+ | | |/ | | | | |----+----+-|>|-+---/\/\/---+ | | | | +-|<|-+ PIC| LED | ---+ Regards Mike ======================================================================= This e-mail is intended for the person it is addressed to only. The information contained in it may be confidential and/or protected by law. If you are not the intended recipient of this message, you must not make any use of this information, or copy or show it to any person. Please contact us immediately to tell us that you have received this e-mail, and return the original to us. Any use, forwarding, printing or copying of this message is strictly prohibited. No part of this message can be considered a request for goods or services. ======================================================================= Any questions about Bookham's E-Mail service should be directed to @spam@postmasterKILLspambookham.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

On Thu, Apr 22, 2004 at 05:14:50PM +0100, Michael Rigby-Jones wrote: {Quote hidden}> Even easier is to drive one side of the LED and a CMOS inverter input with > the PIC pin, and the other side of the LED with the inverter output. The > above solution may be ok, but you are limiting your forward voltage to > 2.5volts for a 5v supply rail which may be marginal for some LED's > > |\ > ---+ +-----+ |0-------------+ > | | |/ | > | | | > |----+----+-|>|-+---/\/\/---+ > | | | > | +-|<|-+ > PIC| LED > | > ---+ > > > Regards > > Mike The only problem I see is packaging. You can get opamps two in a 8 pin package no problem. Not sure about inverters. BAJ -- http://www.piclist.com hint: The PICList is archived three different ways. See http://www.piclist.com/#archives for details.

Here you go: http://www-s.ti.com/cgi-bin/sc/family3.cgi?family=INVERTERS Byron A Jeff wrote: {Quote hidden}> On Thu, Apr 22, 2004 at 05:14:50PM +0100, Michael Rigby-Jones wrote: > >>Even easier is to drive one side of the LED and a CMOS inverter input with >>the PIC pin, and the other side of the LED with the inverter output. The >>above solution may be ok, but you are limiting your forward voltage to >>2.5volts for a 5v supply rail which may be marginal for some LED's >> >> |\ >>---+ +-----+ |0-------------+ >> | | |/ | >> | | | >> |----+----+-|>|-+---/\/\/---+ >> | | | >> | +-|<|-+ >>PIC| LED >> | >>---+ >> >> >>Regards >> >>Mike > > > The only problem I see is packaging. You can get opamps two in a 8 pin package > no problem. Not sure about inverters. > > BAJ > > -- > http://www.piclist.com hint: The PICList is archived three different > ways. See http://www.piclist.com/#archives for details. -- http://www.piclist.com hint: The PICList is archived three different ways. See http://www.piclist.com/#archives for details.

Toshiba and Fairchild also make "microgate" parts as well. I used my first Toshiba parts about 5 years ago and have since started using the TI and Fairchild parts also. They're extremely handy in tight SMD layouts where you only need a gate or two. One thing to watch out for is many of the devices only have about half the output drive current capability as their HC or AHC counterparts -- probably die size or package power handling issues, I'm guessing. BTW, I recently had to add a bi-color SMD LED to a circuit that previously was a single LED and had very little room for any external components. I thought about it and scribbled several circuit ideas for a few days and came up with the *exact* same circuit shown below. I used a SO14 74HC14 though, I had six circuits to upgrade and that worked out to the least space required. Matt Pobursky Maximum Performance Systems On Fri, 23 Apr 2004 16:29:23 -0700, Shawn Wilton wrote: {Quote hidden}> Here you go: http://www-s.ti.com/cgi-bin/sc/family3.cgi?family=INVERTERS > > Byron A Jeff wrote: > > > On Thu, Apr 22, 2004 at 05:14:50PM +0100, Michael Rigby-Jones wrote: > > > >>Even easier is to drive one side of the LED and a CMOS inverter input with > >>the PIC pin, and the other side of the LED with the inverter output. The > >>above solution may be ok, but you are limiting your forward voltage to > >>2.5volts for a 5v supply rail which may be marginal for some LED's > >> > >> |\ > >>---+ +-----+ |0-------------+ > >> | | |/ | > >> | | | > >> |----+----+-|>|-+---/\/\/---+ > >> | | | > >> | +-|<|-+ > >>PIC| LED > >> | > >>---+ > >> > >> > >>Regards > >> > >>Mike > > > > > > The only problem I see is packaging. You can get opamps two in a 8 pin package > > no problem. Not sure about inverters. -- http://www.piclist.com hint: The PICList is archived three different ways. See http://www.piclist.com/#archives for details.