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

Discussion in 'Embedded' started by Tomás Ó hÉilidhe, May 13, 2008.

  1. 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?
     
    Tomás Ó hÉilidhe, May 13, 2008
    #1
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  2. Tomás Ó hÉilidhe

    linnix Guest

    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.
     
    linnix, May 13, 2008
    #2
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  3. Tomás Ó hÉilidhe

    johnspeth Guest

    I've got a bi-colour LED. It has two pins. Internally it consist of
    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
     
    johnspeth, May 13, 2008
    #3
  4. 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
     
    Spehro Pefhany, May 13, 2008
    #4
  5. Tomás Ó hÉilidhe

    Nils Guest

    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
     
    Nils, May 13, 2008
    #5
  6. Tomás Ó hÉilidhe, May 13, 2008
    #6
  7. Tomás Ó hÉilidhe

    linnix Guest

    linnix, May 13, 2008
    #7
  8. 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?
     
    Tomás Ó hÉilidhe, May 13, 2008
    #8
  9. Tomás Ó hÉilidhe

    linnix Guest

    My overall power supply would be 9 V coming from a square battery, but
    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.
     
    linnix, May 14, 2008
    #9
  10. LEDs are not compliant with C99.

    Vladimir Vassilevsky
    DSP and Mixed Signal Design Consultant
    http://www.abvolt.com
     
    Vladimir Vassilevsky, May 14, 2008
    #10
  11. Tomás Ó hÉilidhe

    linnix Guest

    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.
     
    linnix, May 14, 2008
    #11
  12. 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, May 14, 2008
    #12
  13. I thought that I replied to this yesterday but I can't see it here
    so I'll post again...

    There's a potential problem here. Consider what voltage you will
    be getting from the MCU pin. It will _not_ be 0V, 5V, or Hi-Z.
    Most 5V devices aim for TTL compatibility. From memory that allows
    a low to be up to 0.8V and a high to be as low as 2.0V. Therefore
    it is possible for your 'high' voltage to be below the 2.5V centre
    voltage.

    This is of course the worst case scenario, but the LED barrier
    voltage is also conspiring against you. This varies from device
    to device but is typically around 1.7V. 2.5 + 1.7 = 4.2V needed
    from the pin in its high state, which is asking a lot from a 5V
    device, particularly when you are drawing current from it.

    Do yourself a favour. Put an H-bridge in there.
     
    Andrew Smallshaw, May 15, 2008
    #13
  14. I thought that I replied to this yesterday but I can't see it here
    so I'll post again. Having news server issues at the moment.

    There's a potential problem here. Consider what voltage you will
    be getting from the MCU pin. It will _not_ be 0V, 5V, or Hi-Z.
    Most 5V devices aim for TTL compatibility. From memory that allows
    a low to be up to 0.8V and a high to be as low as 2.0V. Therefore
    it is possible for your 'high' voltage to be below the 2.5V centre
    voltage.

    This is of course the worst case scenario, but the LED barrier
    voltage is also conspiring against you. This varies from device
    to device but is typically around 1.7V. 2.5 + 1.7 = 4.2V needed
    from the pin in its high state, which is asking a lot from a 5V
    device, particularly when you are drawing current from it.

    Do yourself a favour. Put an H-bridge in there.
     
    Andrew Smallshaw, May 15, 2008
    #14
  15. Tomás Ó hÉilidhe

    linnix Guest

    And for the rest of us dumb old timer (in the OP's mind), we just
    drive LEDs low with separate pins. Sometimes, we need to KISS up.
     
    linnix, May 15, 2008
    #15
  16. Tomás Ó hÉilidhe

    Tom Guest

    From p.135 of the PIC16F684 datasheet:
    Voh min = Vdd - 0.7V at 3.0 mA and Vdd=4.5V
    Vol max = 0.6V at 8.5 mA and Vdd=4.5V
    So no problem getting 4.3V with a 5V supply. Actual values at room temperature
    are even better than this.

    A bigger concern is how he plans to multiplex the LEDs. The maximum current
    into/out of Vdd or Vss is 95mA. So if he's using 25mA per LED and has more
    than three LEDs connected in this manner, all it takes is one small glitch
    somewhere to turn them all on at once and there goes the CPU. This might be an
    acceptable risk for his hobby project but not something I would ever do in a
    commercial product.
     
    Tom, May 17, 2008
    #16
  17. At 3.0mA. I would expect the voltage to drop further as the LED
    draws, say, 25mA. You may end up with LEDS that are lit but unusably
    dim.
     
    Andrew Smallshaw, May 17, 2008
    #17
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