Using the 3914 with High voltages.

Discussion in 'The Projects Forum' started by iONic, Sep 5, 2008.

  1. iONic

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    Nov 16, 2007
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    Can someone verify that I can use an LM3914 with an input signal voltage of 23 - 34V. The datasheet seems to suggest that it can: InputSignalOvervoltage 35V.

    I hate to burn one as they are not easy to find, nor do they cost just a few cents.
     
    Last edited: Sep 6, 2008
  2. SgtWookie

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    Jul 17, 2007
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    I would limit it to under 16v.

    You can always use a voltage divider.

    They are readily available from a number of suppliers for under $3/ea. Try a search on Octopart.com.
     
  3. iONic

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    Really.... I suppose it would be safer. The datasheet shows how to deal with Higher voltage sources (see attachment) but I wasn't sure how to calculate the
    resistance for up to 34V. They use a couple of 3.9K 1W resistors.
     
  4. Audioguru

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    Dec 20, 2007
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    The maximum allowed input signal voltage is plus or minus 35V. The datasheet tells you how to increase it with one resistor.

    The maximum allowed supply voltage is 25V. The supply voltage to the LEDs might need to be reduced to avoid overheating the IC.
     
  5. iONic

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    OK. Simply put.
    The schematic below is how National Semiconductor deals with the 3914 and voltages higher than is recommended for the chip. I do not quite understand what they are doing with it and need to understand this aspect so that I can be sure that when I give the chip my conditions, 23 to 34V, all will be well. In other words, do I need to change the value of the 2 3.9K 1W resistors? I am also not sure how the "3.4V*" comes into play.

    [​IMG]
     
    Last edited: Sep 6, 2008
  6. Audioguru

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    With pin 8 at 0V, then pin 7 is +1.25V. The base-emitter of the transistor is about 0.6 so its emitter is +1.85V. The LED at its emitter is about 1.8V if it is red so the regulated voltage is 1.85V + 1.8V= +3.65V.
     
  7. iONic

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    I need also to understand if I need to change the values of the 3.9K, 1W resistors given my input voltage.

    LM3914: (see my previous post for full image)

    [​IMG]

    Thanks
     
  8. SgtWookie

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    If you decrease both of the 3.9k resistors to 2.7k, you will get about 11mA current through the LEDs when your supply voltage is 34v, and about 7.4mA when your supply current is 24v.
     
  9. iONic

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    Does the two 39K resistors and the one diode evenly divide the voltage?
     
  10. SgtWookie

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    They're not 39K, they are 3.9k.
    No, they don't evenly divide the voltage. You cannot use bar mode with this scheme; there will not be enough current supplied to power more than one LED at a time.

    If you want to be able to use bar mode, then I suggest using a regulator such as a 7810, 7815 or LM317.

    Your signal in (SIG, pin 5) will also need a voltage divider if you're planning on it going near the V+ of the IC. As things are at the moment, SIG would only be able to measure from 0v up to about 1.75v.
     
  11. iONic

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    Sorry about the decimal point. Dot mode is all I was after any ways.
     
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