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Best way to get 2.5 volts from somewhere? (Vcc = 5 volts)

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?

Reply to
Tomás Ó hÉilidhe
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You can use another uC to PWM switch a voltage source to an op-amp integrator, since cost/components are not an issue for you.

Reply to
linnix

You don't need a microcontroller pin to generate 2.5V. You just need a constant voltage source. You can use a simple voltage divider off 5V which uses two equal valued resistors that serve as a voltage divider AND current limiters.

It's good if it meets your requirements. Your stated requirement is

2.5V from 5V. An implied requirement is it works. If you try it and it works, it's a good idea. You're the judge of "good".

There is no way. Your datasheet will tell you the pin will either be

0V or 5V if it's an output. It will be at high impedance if it's a an input and your external circuit will determine what voltage it will be.

JJS

Reply to
johnspeth

Predict the current through each LED when it is on, over unit-to-unit variations and temperature. Then see if you think it's a good idea.

Best regards, Spehro Pefhany

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Reply to
Spehro Pefhany

You can get Source/Sink regulators, intended for DDR memory terminators, they are one option. Or an opamp, if the total node power is OK. You might want slightly different from 2.5V, as the LED Vfs are not the same, nor are the uC Pin drivers, and you might want to adjust the 5V power a little.

-jg

Reply to
Jim Granville

Or an RS-485 driver, my first choice. Unfortunately, he wants to use no more than one I/O pin.

Reply to
Jim Stewart

Maybe I'm just dump, but:

Why don't you simply use a 4066 bilateral-switch, a current-limiting resistor plus some kind of inverter (can be as simple as a NPN-tranny in your case)?

It's cheap and just works...

Nils

Reply to
Nils

I think I've found what I want:

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Something along those lines anyway.

Reply to
Tomás Ó hÉilidhe

I am too busy at the moment. Anybody wants to show him the LM317 datasheet?

Reply to
linnix

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?

Reply to
Tomás Ó hÉilidhe

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.

Reply to
linnix

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 to
Jim Granville

LEDs are not compliant with C99.

Vladimir Vassilevsky DSP and Mixed Signal Design Consultant

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Reply to
Vladimir Vassilevsky

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

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Reply to
MK

... snip ...

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.

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Reply to
CBFalconer

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 to
linnix

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 to
Andrew Smallshaw

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

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-jg

Reply to
Jim Granville

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

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Reply to
MK

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

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Reply to
MK

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