LED Volt. indicator

Discussion in 'Homework Help' started by shseo0315, Apr 19, 2013.

  1. shseo0315

    Thread Starter New Member

    Apr 19, 2013
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    I'm trying to build a simple circuit to indicate the status of my 3.7v lithium ion battery.

    When the battry goes down to 3.0v, the LED should be turned on. It should be turned off when it is above 3.0v.

    Or When the battery goes down to 3.0v, the LED should be turned off. It should be turned on when it is above 3.0v

    So 3.0v is what makes LED different from the original state.

    I've been trying to build this circuit using BJT but it does not seem to be working.

    I'm a bit short of time. Please help. Any tips, links, help will be appreciated.
     
  2. Wendy

    Moderator

    Mar 24, 2008
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    Think LM393 or LM339. When I get home I'll post a circuit.
     
  3. shseo0315

    Thread Starter New Member

    Apr 19, 2013
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    I appreciate it. A bit of explanation too as well? Thanks a lot.
     
  4. WBahn

    Moderator

    Mar 31, 2012
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    How short on time are you? When is the assignment due?
     
  5. shseo0315

    Thread Starter New Member

    Apr 19, 2013
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    By tomorrow.

    It is a part of my project.

    I'm just done with everything else.

    This part just would not work.

    Please help. I have LM393 339.

    Thanks a lot.
     
  6. WBahn

    Moderator

    Mar 31, 2012
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    If we just give you a circuit then that will defeat the purpose of the project, which is for YOU to solve a problem and design a circuit. We are more than will to help point out problems and offer hints, but the onus is on you to do the bulk of the work. So show us your best attempt to date and we will comment on what is causing problems and directions you might look to overcome them.
     
  7. shseo0315

    Thread Starter New Member

    Apr 19, 2013
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    Description of the Circuit:
    D1 is the LED to light on when VCC is below the given limit of 3V.
    When VCC falls below the given limit, Q2 turns off and the LED
    draws current directly through R4. Since the typical voltage drop
    across LED is around 1.4V, there would be 1.6mA available to the
    LED, which is sufficient to light it up. This light‐on current can be
    increased by decreasing R4.

    When VCC is sufficiently high, Q2 turns on and goes into
    saturation, drawing around 3mA from VCC through R4. Under this
    circumstance, there is insufficient bias for D1 to light‐up.

    Q1 together with R2 and R3 provides a Zener‐diode behavior.
    By adjusting R2, a voltage drop of around 2‐2.2V may be obtained
    across the collector and emitter terminals of Q1.
    Another 0.6V drops across the base‐emitter terminals of Q2, and hence the remaining drops across R1.
    By adjusting R1, the minimum base current to obtain saturation in Q2 is
    adjusted, so that the LED does not turn on.

    Questions.
    1. would the circuit behave correctly?
    2. Just to make sure wiring variable resistors for example R2.
    Since there are three terminals for variable resistors like trimpot,
    one should be wired to base of Q1, one to emitter of Q1 and the other to gnd.
    Is that correct?

    Our school offers
    4T Turnpots 1K ohms
    10T Turnpots 1K ohms
    15T Trimpots 1K ohms
    I'm thinking of using one of them.
    Thanks.
     
  8. #12

    Expert

    Nov 30, 2010
    16,252
    6,750
    That might work.
    It's a bit temperature dependent and you might need to put a leakage resistor to the left of R1 (in case Q1 doesn't turn off sharply enough).
     
  9. Wendy

    Moderator

    Mar 24, 2008
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    I will give you a hint, I had not realized this was homework, so I will post the primary schematic day after tomorrow.

    I realized you might have trouble lighting a LED at 2.5V, so I came up with a work around that will work at 2V or less, if you turn this in you will deserve what ever grade your teacher gives you. The back half is a voltage/current convertor.

    [​IMG]

    The thing is, it is a simplified window comparator. With a minor redesign you could modify the front end to do what you want it do, but I strongly suggest you study the data sheet of a LM393, as they have a open collector output. In other words, the output does not actually swing in voltage, it just goes to ground or is open.

    I have similar type schematics that might give you a clue in my blog, and specifically my article.

    Bill's Blog

    LEDs, 555s, Flashers, and Light Chasers
     
  10. shseo0315

    Thread Starter New Member

    Apr 19, 2013
    6
    0
  11. shseo0315

    Thread Starter New Member

    Apr 19, 2013
    6
    0
    I really appreciate your help.
    However, this is not just a homework. This is to show the status of my battery and is also a part of my whole project(semester-long). My device is ready but I just wanted to add this circuit.
    I'm pretty much settled on my schematic above. If this does not work, I'm just going to give up this part. I just can't start on using a different approach. I don't have time plus I'm under a bit of pressure. But thanks for the help. Ill take a look at it.

    I just don't have enough circuit experiences. All I need is just a bit of guidance on making this work or testing procedures on the breadboard based on my schematic above starting with Vcc=3.7 to 3.0v.

    Thanks for your interest. It really helps.
     
  12. WBahn

    Moderator

    Mar 31, 2012
    17,715
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    The term "homework" is generic. It applies to anything that you are doing that results in you receiving a grade or credit for any type of course, certificate, degree, etc.

    I would have come at it differently by creating a diff-amp comparator and using a zener or diode stack for a reference (a real zener, but your effective zener should work, too).

    I would probably also have the LED on when it is above the threshold voltage because I would not want it to be on and further draining the battery if you are already below the threshold voltage. You could use a low current (~2mA LED) and then flash it to further reduce power consumption.
     
  13. toffee_pie

    Active Member

    Oct 31, 2009
    162
    7
    in a nutshell the best way to do this is what is being said

    have a comparator op amp and a reference voltage.

    use the reference voltage on each (-) pin of the op amp and compare it to the incoming voltage source on the (+) pin.

    if the latter is greater than the reference the output will be high and it can drive a led or whatever.


    if its lower there is no output.

    you can set different trigger points by using a voltage divider to adjust voltage points accordingly.
     
  14. Wendy

    Moderator

    Mar 24, 2008
    20,764
    2,535
    Since the deadline has passed here is the way I would do it. It could fail under 3.0 V, as you get closer and closer to the Vf of the LED. This is why I came up with the first overcomplicated monstrosity.

    [​IMG]

    Note that R2 must be test selected to set the voltage divider to the voltage you want to the LED to turn on at. You could replace R2 and R3 with a pot. CR1 is not critical, it is a simple voltage reference.

    .
     
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