understanding the use of LM3914

Discussion in 'The Projects Forum' started by david1234, Apr 27, 2014.

  1. david1234

    Thread Starter Member

    Nov 27, 2013
    81
    1
    Hello
    just want to see I understand ;
    I want to monitor a battery of 12V
    I have check with power supply and this is the result I get
    Vin<9.9v- No led is on
    9.9v<Vin<10.6v - the red led is on (LED1,2,3)
    10.6v<Vin<11.2v yellow led is on (LED4,5)
    Vin>11.2 Green LED is on (LED 6,7,8,9,10)

    what do I need to change in the circuit in order to lower Voltage that will light the first LED so in 9.5V the red will go on also?
     
  2. bertus

    Administrator

    Apr 5, 2008
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    Hello,

    You can change the lower limit by changing the resistor on pin 4 in your schematic.
    A lowering the value will lower the voltage of the lower led.

    Have a look at the internal schematic how the comparators are hooked up:

    [​IMG]

    Bertus
     
  3. david1234

    Thread Starter Member

    Nov 27, 2013
    81
    1
    O.k.
    can you show me the formula to calculate the right value?
    I mean - I understand that now I have ~50K between pin 6 and GND (10 inside of the LM3914 and 40 with my resistor)
    but how do I calculate?

    Thanks ,
     
  4. bertus

    Administrator

    Apr 5, 2008
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    Hello,

    You have a voltage divider of 11:1 on the input.
    The reference voltage is 1.25 Volt, the upper led will be 11 X 1.25 = 13.75 Volt.
    You want the lower led to be lit at 9.5 volts (this is 0.86 volts at the input).
    So the current through the voltage divider, between the upper and lower led, is (1.25 - 0.86) / 9 = 0.043 A.
    If you want 0.86 Volt on the lower comparator a resistor of 0.86 / 0.043 = 20 K is needed.
    Intern there is a 1 K, so I would expect a value of 19 K extern on pin 4.

    Bertus
     
  5. david1234

    Thread Starter Member

    Nov 27, 2013
    81
    1
    sorry - but I still don't understand
    so I need to ask again :
    I understand that 0.86 is 9.5/11 ,but why you calculate the current like this - (1.25-0.86)/9 ?
    what does the 9 stand for? doesn't need to be 10 ? the sum of the resistors ?
     
  6. ericgibbs

    AAC Fanatic!

    Jan 29, 2010
    2,503
    380
    hi,
    For 10 LED's there are only 9 intervals/gaps/voltage thresholds between them.

    EDIT:

    There are tolerances in the LM3914 IC, I would advise where possible you make up resistors, using a fixed value and a variable resistor, so that you can trim the voltages for the Rhi and Rlo.
     
  7. bertus

    Administrator

    Apr 5, 2008
    15,647
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    Hello,

    There are 9 resistors between the highest and lowest comparator inputs.
    Each resistor is 1 K, so I use 9 K.
    As said, use a fixed resistor in series with a potentiometer to set the lower limit to your needs.
    I think you can use a 15 K resistor and a 10 K pot for this.

    Bertus
     
  8. david1234

    Thread Starter Member

    Nov 27, 2013
    81
    1
    O.K. - I will try it .

    now another question (maybe I don't understand the circuit right):
    what is the role of the 5.1K resistor? in the R_High (PIN6) ?
    how does it control the high voltage that will light the LED?

    Thanks ,
     
  9. bertus

    Administrator

    Apr 5, 2008
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    Hello,

    The resistor on pin 6 / 7 determines the led current.
    Take a look at the datasheet to determine the value.

    Bertus
     
  10. david1234

    Thread Starter Member

    Nov 27, 2013
    81
    1
    O.k.
    so something doesn't add up in my head
    (I'm applying what you said on the autograph.....)

    Let say that my signal is changing between 9-12V but after the voltage divider I only get
    Vin_max= (12*10)/(100+10) = 1.09V
    Vin_min=(9*10)/(100+10)=0.81V

    how does it manage with the other voltage you explain to me ?

    Thanks ,
     
  11. bertus

    Administrator

    Apr 5, 2008
    15,647
    2,346
    Hello,

    When you want the range from 0.81 to 1.09 Volts, you will have a voltage divider current of:
    (1.09 - 0.81) Volts / 9 KOhms = 0.03111 mA.
    The resistor between the pins 6 (Rhi) and 7 (Vref) will be:
    (1.25 - 1.09) Volts / 0.03111 mA = 5.14 KOhm.
    The resistor on pin 4 to ground will be:
    0.81 Volts / 0.03111 mA = 26.04 KOhm.
    As there is already a resistor of 1 KOhm inside the LM3914, you will need a resistor of 26.04 - 1 = 25.04 KOhm.

    I would take a 3.9 KOhm with a 2.2 KOhm potmeter on the top and
    a 22 KOhm with a 10 KOhm potmeter on the bottom.

    That way you can compensate for component tolerances.

    Bertus
     
  12. david1234

    Thread Starter Member

    Nov 27, 2013
    81
    1
    O.K
    I think I got it -
    I will build and "play" with this and if I have more question I will ask


    Thanks you for the help and the explaining
     
  13. david1234

    Thread Starter Member

    Nov 27, 2013
    81
    1
    well It's working
    I have change it to 20K
    now I want to know is it possible to change the steps of voltage?
    now every 0.4V there is a change
    can I make it 0.5? or even 0.7V?

    Thanks,
     
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