Help diagnose LM3914 circuit

Discussion in 'The Projects Forum' started by campeck, Jul 28, 2010.

  1. campeck

    Thread Starter Active Member

    Sep 5, 2009
    194
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    Part two of the project.

    http://www.youtube.com/watch?feature=youtube_gdata&v=GczqAMz1qyI

    As you can see in the video the LED's don't fully turn off. It also seems very sensitive to touch on the wires. This project calls for long lengths of wire unfortunately. The input wires to the chip might be 10 meters long. Anything I can do to reduce noise? Or eliminate wires? Turn the LEDs all the way off? I'm drawing a schematic now.

    Thanks
     
    Last edited: Jul 28, 2010
  2. windoze killa

    AAC Fanatic!

    Feb 23, 2006
    605
    24
    Try putting a 2K resistor from the base to +5V. This will switch the transistor hard off when there is no output from the chip.

    Try using twisted pair for your wiring and to make it even better shielded twisted pair. CAT5E ethernet cable is a nice cheap way of running multiples. It has 4 pair in it and each pair has a slightly different twist so they "shouldn't" interfere with each other.
     
  3. campeck

    Thread Starter Active Member

    Sep 5, 2009
    194
    3
    Great suggestions! I will do that.
    Can cat5e handle 5 volts and 600 ma?

    Thanks you!
     
  4. Audioguru

    New Member

    Dec 20, 2007
    9,411
    896
    Use shielded audio cable for the input.
    Use a 47uf supply bypass capacitor at the LM3914 IC.

    The input pin 5 needs at most a 1M resistor to ground or it floats to a positive DC voltage which turns on all the LEDs.

    Why do you need the transistor to destroy the LEDs with too much current? The LED current is about 45mA but ordinary LEDs have a max current of 30mA.

    The output currents of the LM3914 are regulated by the current from pin 7 to ground. But your pot changes the voltage of pin 7 which changes the output current of each output. Maybe the pot should be at the input to the LM3914 so that the output currents do not change.
    Each output can easily supply up to 30mA to an LED.
     
  5. campeck

    Thread Starter Active Member

    Sep 5, 2009
    194
    3
    The input is just shown floating. It goes over to a 100k pot with a 100k pot above and below it for lower and upper level setting for a total of 300k.

    The LED's I am using have a Vf of 3v and use 20ma.
    so 5 - .7 = 4.3 - 3.0 = 1.3v
    1.3/.02 = 65 ohm

    I am also using the transistors because in about 2 weeks I will be switching to 300mA 1W LED's.

    What are you saying I should do to pin 7? Oh..you think thats my control pot. No that pot is just there because I couldn't work out the values of the two resistors that the data sheet mentions. So I made it adjustable.

    And do I even need the base resistors on the transistors?

    Also the two 10k pots in the below schematic, would there be a benefit in making them 1k? or doing away with them since the transistor is controlling current to the LEDs not the chip?

    Thanks

    Modified:
     
    Last edited: Jul 29, 2010
  6. Audioguru

    New Member

    Dec 20, 2007
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    Pin 7 of the LM3914 is the output of its voltage reference. It is set with two resistors exactly like the voltage of an LM317 voltage regulator is set.

    The voltage from pin 7 to pin 8 (the ADJ pin) is 1.25V so if pin 8 is grounded then pin 7 is 1.25V.
    If pin 7 has two 470 ohm resistors in series and the lower one connected to ground and pin 8 connects to their junction then pin 7 is 2.5V.
    If pin 7 connects to a 220 ohm resistor that connects to a 680 ohm resistor to ground and pin 8 connects to their junction then pin 7 will be 5.1V if the supply is 1.5V higher or more.

    If you replace your pot with two resistors and set a suitable output current that is determined by the resistance from pin 7 to ground then your transistors do not need series base resistors.

    Your transistors do not "control" the LED current. The transistors are simply on-off switches. The 68 ohm resistors control the LED currents.
     
  7. campeck

    Thread Starter Active Member

    Sep 5, 2009
    194
    3

    Ok. Pin 7 has me confused. I read the datasheet and it didn't help.
    So pin 7 is 1.25v over pin 8. How the heck does adding resistors to ground bring pin 7 up to 5v? And why do they show current coming out of pin 8? I thought it would go from pin 7 to 8...

    I saw where they ground pin 8 and then the current out of pin 7 defines the output currents. that makes sense.

    page 6 of the datasheet just isn't helping me understand. Is there another way to put it Audioguru?

    also what should I do with unused output pins?
     
    Last edited: Jul 29, 2010
  8. Audioguru

    New Member

    Dec 20, 2007
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    I don't know which manufactuer's datasheet you are looking at. The calculation for the two resistors that determine the reference voltage is extremely simple. Natrional Semi invented the LM3914 so I look at their datasheet.

    Unused outputs are the collectors of transistors that do nothing so leave the outputs not connected to anything.
     
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