> My overall power supply would be 9 V coming from a square battery, but > I'll be putting it thru a voltage regulator to give me out 2 V.Please ignore my last post, since you are changing spec on me. You said you were getting 2.5V out of 5V with the LM317.
Best way to get 2.5 volts from somewhere? (Vcc = 5 volts)
Started by ●May 13, 2008
Reply by ●May 13, 20082008-05-13
Reply by ●May 13, 20082008-05-13
Tom�s � h�ilidhe wrote:> I have a question... > > If I have an LED that has about 2 volts across it, then is it OK to > put a 2 volt power supply across it without a current limiting > resistor? > > My overall power supply would be 9 V coming from a square battery, but > I'll be putting it thru a voltage regulator to give me out 2 V. > > I'll then be putting the 2 V across the LED. > > Can I leave out the LED's current-limiting resistor, or is there still > a chance of there being too much current that would fry components?Go to the data sheet. Find the V/I curves, and include the temperature coefficent. Now find the MIN and MAX specs, and draw those load-lines for 2V drive. The real world is a little more forgiving than the corner cases, and you will find LEDs within a batch match better than random scattering MIN-MAX. [SMD leds on a tape, are actually very well Vf matched] Your practical problems will be brightness matching (well before your current variations hit damage levels), and thermal tracking. -jg
Reply by ●May 14, 20082008-05-14
Tom=E1s =D3 h=C9ilidhe wrote:> I have a question... >=20 > If I have an LED that has about 2 volts across it, then is it OK to > put a 2 volt power supply across it without a current limiting > resistor?LEDs are not compliant with C99. Vladimir Vassilevsky DSP and Mixed Signal Design Consultant http://www.abvolt.com
Reply by ●May 14, 20082008-05-14
"Tom�s � h�ilidhe" <toe@lavabit.com> wrote in message news:205cd1d5-1f20-4828-88d7-855cb4efe7a0@d1g2000hsg.googlegroups.com...> > I've got a bi-colour LED. It has two pins. Internally it consist of > two LED's in parallel except they face in different directions. > > I'm looking into ways of using one micrcontroller pin to control the > LED as follows: > Pin High = Light up Red > Pin Low = Light up Green > Pin as Input = Nothing lights up > > A friend of mine suggested to me today to connect one of the LED pins > to the microcontroller, and the other to 2.5 V. That way, if the uC > pin is high, it will source current from 5 volts to 2.5 volts. If it's > low, it will source current from 0 volts to 2.5 volts. (Of course I'd > have a resistor somewhere). > > So the only question is how I'd put one of the pins at a constant 2.5 > volts. My first thought was to use a zener diode, i.e. take a pin from > the LED, put into one side of the zener, and tie the other side of the > zener to ground. I'm not entirely sure if this will work though. > Another complication would be that I'd need two zeners in parallel > facing the opposite direction in order to let current flow in both > directions. > > Do you think the whole 2.5 volts idea is good? What's the best way of > getting one of the LED pins to sit at 2.5 volts?Tomas - This has been said before - please get a processor with more pins - you are draining the worlds supply of engineering resource by asking us to think up single pin solutions to two pin problems - this may in fact be the cause of global warming !! However - to address your problem - as always soemone got there before you: en.wikipedia.org/wiki/Th%C3%A9venin%27s_theorem Connect 2 resistors of equal value (R) in series from the 5V supply to 0V and the mid point is the 2.5V you need. Connect the LED from the mid point to your processor pin and drive high/low/open. The current limiting you *NEED* for leds comes free because the source resistance of the 2.5V is R/2. I'm going to let you do the sums to work out the ideal resistor value for the LED that you have. Michael Kellett www.mkesc.co.uk
Reply by ●May 14, 20082008-05-14
MK wrote:>... snip ...> > Connect 2 resistors of equal value (R) in series from the 5V > supply to 0V and the mid point is the 2.5V you need. Connect the > LED from the mid point to your processor pin and drive > high/low/open. The current limiting you *NEED* for leds comes > free because the source resistance of the 2.5V is R/2. > > I'm going to let you do the sums to work out the ideal resistor > value for the LED that you have.Of course the on voltage of the LED, and the variance of that, will not affect the values in the least. Nor will the variance in current drive needed for the two LED colors. -- [mail]: Chuck F (cbfalconer at maineline dot net) [page]: <http://cbfalconer.home.att.net> Try the download section. ** Posted from http://www.teranews.com **
Reply by ●May 14, 20082008-05-14
On May 14, 6:27 am, CBFalconer <cbfalco...@yahoo.com> wrote:> MK wrote: > > ... snip ... > > > Connect 2 resistors of equal value (R) in series from the 5V > > supply to 0V and the mid point is the 2.5V you need. Connect the > > LED from the mid point to your processor pin and drive > > high/low/open. The current limiting you *NEED* for leds comes > > free because the source resistance of the 2.5V is R/2. > > > I'm going to let you do the sums to work out the ideal resistor > > value for the LED that you have. > > Of course the on voltage of the LED, and the variance of that, will > not affect the values in the least. Nor will the variance in > current drive needed for the two LED colors. >So, we need 2 resistors at the uC end, and 2 resistors at the voltage regulator end. Oh, great, we eliminated 2 LED resistors with a regulator and 4 other resistors.
Reply by ●May 14, 20082008-05-14
On 2008-05-13, Tom�s � h�ilidhe <toe@lavabit.com> wrote:> > A friend of mine suggested to me today to connect one of the LED pins > to the microcontroller, and the other to 2.5 V. That way, if the uC > pin is high, it will source current from 5 volts to 2.5 volts. If it's > low, it will source current from 0 volts to 2.5 volts. (Of course I'd > have a resistor somewhere).Potential problem here. What voltage is present on the MCU pin? 5V strongly suggests TTL compatible inputs/outputs to me which are _not_ 0V and 5V. From memory low is up to 0.8V and high is at least 2.0V. There is a possibility that your pin could be 'high' and delivering 2.0V which is still _less_ than the 2.5V on the other end of the LED. This is of course the worst case scenario, but the LED's barrier voltage is also conspiring against you. The exact value varies depending on the device but typically around 1.7V is needed for the LED to conduct. 2.5V+1.7V=4.2V which is a big ask from a 5V device. -- Andrew Smallshaw andrews@sdf.lonestar.org
Reply by ●May 14, 20082008-05-14
Tom�s � h�ilidhe wrote:> I've got a bi-colour LED. It has two pins. Internally it consist of > two LED's in parallel except they face in different directions. > > I'm looking into ways of using one micrcontroller pin to control the > LED as follows: > Pin High = Light up Red > Pin Low = Light up Green > Pin as Input = Nothing lights up > > A friend of mine suggested to me today to connect one of the LED pins > to the microcontroller, and the other to 2.5 V. That way, if the uC > pin is high, it will source current from 5 volts to 2.5 volts. If it's > low, it will source current from 0 volts to 2.5 volts. (Of course I'd > have a resistor somewhere). > > So the only question is how I'd put one of the pins at a constant 2.5 > volts. My first thought was to use a zener diode, i.e. take a pin from > the LED, put into one side of the zener, and tie the other side of the > zener to ground. I'm not entirely sure if this will work though. > Another complication would be that I'd need two zeners in parallel > facing the opposite direction in order to let current flow in both > directions. > > Do you think the whole 2.5 volts idea is good? What's the best way of > getting one of the LED pins to sit at 2.5 volts?Or, if you really want to have LOTS of LED drive capacity :), and low OFF power, then you CAN drive two power outputs, from a single pin this way : (These device have a special 3 State sense ) I think it could also drive either 2 or 3 terminal LEDs http://www.intersil.com/cda/deviceinfo/0,1477,ISL6615,0.html http://www.intersil.com/cda/deviceinfo/0,0,ISL6609,0.html -jg
Reply by ●May 15, 20082008-05-15
"linnix" <me@linnix.info-for.us> wrote in message news:f95b16db-d9ae-4f86-b5c5-27b959d18012@q24g2000prf.googlegroups.com...> On May 14, 6:27 am, CBFalconer <cbfalco...@yahoo.com> wrote: >> MK wrote: >> >> ... snip ... >> >> > Connect 2 resistors of equal value (R) in series from the 5V >> > supply to 0V and the mid point is the 2.5V you need. Connect the >> > LED from the mid point to your processor pin and drive >> > high/low/open. The current limiting you *NEED* for leds comes >> > free because the source resistance of the 2.5V is R/2. >> >> > I'm going to let you do the sums to work out the ideal resistor >> > value for the LED that you have. >> >> Of course the on voltage of the LED, and the variance of that, will >> not affect the values in the least. Nor will the variance in >> current drive needed for the two LED colors. >> > > So, we need 2 resistors at the uC end, and 2 resistors at the voltage > regulator end. Oh, great, we eliminated 2 LED resistors with a > regulator and 4 other resistors.You've got some extra resistors and a voltage regulator in here somehow - my suggestion to Tomas was that he use 2 resistors and the 5V supply already available for the processor. My reference to Thevenin (which you snipped) was to give him a hint as to how to work out the values. Given the constraints of the original question I am interested in your alternative suggestions. Michael Kellett www.mkesc.co.uk
Reply by ●May 15, 20082008-05-15
"CBFalconer" <cbfalconer@yahoo.com> wrote in message news:482AE8D6.8333F276@yahoo.com...> MK wrote: >> > ... snip ... >> >> Connect 2 resistors of equal value (R) in series from the 5V >> supply to 0V and the mid point is the 2.5V you need. Connect the >> LED from the mid point to your processor pin and drive >> high/low/open. The current limiting you *NEED* for leds comes >> free because the source resistance of the 2.5V is R/2. >> >> I'm going to let you do the sums to work out the ideal resistor >> value for the LED that you have. > > Of course the on voltage of the LED, and the variance of that, will > not affect the values in the least. Nor will the variance in > current drive needed for the two LED colors. > > -- > [mail]: Chuck F (cbfalconer at maineline dot net) > [page]: <http://cbfalconer.home.att.net> > Try the download section. > > > ** Posted from http://www.teranews.com **Hello Chuck, I think you mis-typed. The forward voltage of the LEDs will affect the current. If the two LEDs need different drive currents the resistors can be set to different values. The OP wanted a simple solution and set some constraints in his question. Obviously a simple linear, passive solution has limits. Michael Kellett www.mkesc.co.uk