resolving LM3914

Discussion in 'The Projects Forum' started by ramcaress, Mar 3, 2010.

  1. ramcaress

    Thread Starter New Member

    Mar 1, 2010
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    just want to know how to create a circuit for LM3914 within a range of 4.7V to 5.9V, the 10 LEDs should illuminate when the input voltage is at 5.9V and no LEDs will illuminate at 4.7V input, i had already read the data sheet of LM3914 from national instrument but can't have any idea how to construct, thanks,:)
     
  2. R!f@@

    AAC Fanatic!

    Apr 2, 2009
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  3. SgtWookie

    Expert

    Jul 17, 2007
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    On the link Rifaa posted, look at the Typical Application circuit on page 2.

    For the resistor values,
    R1: 1,270 Ohms
    R2: 4,370 Ohms
    You will also need another resistor between pin 4 and ground; a value of 39,167 Ohms.

    These values are approximate, and based on the difference between pins 7 and 8 being 1.25v, and the internal resistive divider measuring exactly 10k Ohms.

    [eta]
    You can't buy resistors with those exact values, but you can use resistors in series and/or parallel to get very close.
    Here is a handy online calculator for series/parallel resistance:
    http://www.qsl.net/in3otd/parallr.html
    Bookmark that page.
    For example, if you select E24 values (commonly available) and ask for 1270 Ohms, you'll get output like:
    910 + 360 = 1270 (0 %) (This means a 910 Ohm and a 360 Ohm resistor in series to get 1270 Ohms, 0% error)
    1600 || 6200 = 1271.795 (0.141 %) (This means a 1.6k and 6.2k resistors in parallel for 1271.795 Ohms; 0.141% error)

    You will need to use a supply of at least 8v.
     
    Last edited: Mar 3, 2010
  4. R!f@@

    AAC Fanatic!

    Apr 2, 2009
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    I still wonder why odd values are used in data sheets
     
  5. ramcaress

    Thread Starter New Member

    Mar 1, 2010
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    0
    got it, its the external resistor combination must be set for the Ref low (4) and Ref high, im goin to reconstruct my old circuit, thanks a lot,:)
     
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