LM3914 0-5v 20 led gauge Help Please

Discussion in 'The Projects Forum' started by Madmen, Nov 22, 2008.

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  1. Madmen

    Thread Starter Member

    Nov 22, 2008
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    I'm trying to build a gauge using the NS LM3914, I need 20leds , 0.5-4.5v signal input range, 20ma led current and with 14.8v supply. I have read the NS paper on the LM3914, I built the circuit on page 2 and 15 they both worked but I'm not sure how to change the input range from 0-2.4v to 0.5-4.5v signal input. I want to use this gauge to quickly check any 0.5-4.5v sensor in a car with the main supply coming from the battery. If I need to clarify anything please let me know.

    Thank you in advance,
    Bob
     
  2. mik3

    Senior Member

    Feb 4, 2008
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    If you set the Ref out V=4.5V, then you will be able to have an input signal from 0-4.5V. Is that ok?

    To set Ref out V to 4.5V set R2=2.6K and R1=1K.

    With these resistor values the current though each led will be 12.5mA. If you want the current through each led to be 20mA and the Ref out V still be 4.5V set R2=1.6K and R1=625R.
     
  3. Madmen

    Thread Starter Member

    Nov 22, 2008
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    When I had 10leds those where the resistors values I was using. When I tried to make it 20leds with two LM3914 for better resolution, everything went pear shaped. I can't figure out how to adjust the two LM3914 s together for the range I want.

    Thank you for responding, at least I was correct on the first circuit i built.

    Bob
     
  4. mik3

    Senior Member

    Feb 4, 2008
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    Adjust Ref out V=4.5V for the second LM3914 and instead of connecting RLo directly to ground put a 10K resistor on RLo and then to ground.
     
  5. Audioguru

    New Member

    Dec 20, 2007
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    If all ten outputs have 12.5mA and are driving 1.8V red LEDs from a 14.8V supply, then the poor LM3914 might melt because it will be dissipating 1.6W. Add a current-limiting resistor and bypass capacitor to feed the LEDs. 68 ohms/2W.
     
  6. Madmen

    Thread Starter Member

    Nov 22, 2008
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    Can I get a diagram, because now my circuit is non-linear so I must be doing something wrong, though the range is about right, thank you. I really appreciate all the help. :)


    Thank you again,
    Bob
     
  7. mik3

    Senior Member

    Feb 4, 2008
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    How does it behave?

    If you set Ref out V=2.5V for the first IC and Ref out V=5V for the second IC it should work. Check it again and come back.

    The Rlo pin of the second IC has to be connected to ground via a 10K resistor.
     
  8. Madmen

    Thread Starter Member

    Nov 22, 2008
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    I want to run the LM3914 in dot mode not bar so I think I should be ok with current load, correct?

    Bob
     
  9. mik3

    Senior Member

    Feb 4, 2008
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    Yes, you will reduce the power consumption.
    Did it work?
     
  10. mik3

    Senior Member

    Feb 4, 2008
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    A better solution, is to make the RefoutV=5V for the second IC, then disconnect the Rhi of the first IC from RefoutV, then disconnect Rlo of the second IC from ground and connect it to Rhi of the first IC.
    Use the RefoutV of the first IC just to determine the current through the leds.
     
  11. Madmen

    Thread Starter Member

    Nov 22, 2008
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    I think I got it, but I only have a 800 ohm resistor on the Rlo of chip one. I will work up a schematic and try to post it.

    Bob
     
  12. mik3

    Senior Member

    Feb 4, 2008
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    You dont need a resistor on Rlo of IC one.
     
  13. Madmen

    Thread Starter Member

    Nov 22, 2008
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    Here is the schematic please let me know what you think.

    Bob
     
  14. mik3

    Senior Member

    Feb 4, 2008
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    Remove the wire going from pin 9 to pin 11 of IC No2.

    Make R2 of IC No2 equal to 186 ohm.

    Disconnect the wire going from Rlo (No2) to Rhi (No1) from pin 7 of IC No1.

    Remove the 800 ohm resistor from Rlo of IC No1 and connect it directly to ground.
     
  15. Madmen

    Thread Starter Member

    Nov 22, 2008
    20
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    I understand everything you want me to change but which resistor on chip 2.
    So here is a new pdf with the resistors labeled (sorry I over looked it the first time) I was excited to be making headway. I tried removing the 800 ohm resistor and now the first led turns on at 0.21 volts.

    Bob
     
  16. mik3

    Senior Member

    Feb 4, 2008
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    Remove the wire going from pin 9 to pin 11 of IC No2.

    Make R6 of IC No2 equal to 186 ohm.

    Disconnect the wire going from Rlo (No2) to Rhi (No1) from pin 7 of IC No1.

    Remove the 800 ohm resistor (Ρ2) from Rlo of IC No1 and connect it directly to ground.
     
  17. Madmen

    Thread Starter Member

    Nov 22, 2008
    20
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    I made the changes, but now it doesn't work correctly. Led 2 is turned on at 0.17v (the lowest my power supply will go) and led 20 turns on at 1.75v. I have attached a schematic with the changes. I have the schematic in excel format if want me to send it to you.

    Bob
     
  18. mik3

    Senior Member

    Feb 4, 2008
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    I missed something in the datasheet but I think I got it now.
    Disconnect Rhi of IC No2 from Ref out of IC No2 and connect it to the node between R5 and R6.
     
  19. Madmen

    Thread Starter Member

    Nov 22, 2008
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    This is the current state of the circuit, I ran it all day today with the only issue begin a little flicker between voltages but I don't think that will be a issue.

    Bob
     
    Last edited: Nov 27, 2008
  20. Ron H

    AAC Fanatic!

    Apr 14, 2005
    7,050
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    You need to set up your voltages like this. If you need help with resistor values, post here.
     
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