Help with LM3914 Bar Display With Alarm Flasher

Discussion in 'The Projects Forum' started by redblaque, Feb 16, 2009.

  1. redblaque

    Thread Starter New Member

    Jan 30, 2009
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    0
    Hello,

    I've taken a look at the LM3914 datasheet and I want to build one of the applications which is bar display with alarm flasher (attached schematic). I need help with the calculation for the input signal's range of voltage.

    As I've understood in this thread, the R2 & R1 voltage divider determines the max signal input. However, in the attached schematic, is the input signal range 0-4.44 V? With the R1 and C1 junction, why is there a 100 ohms resistor at LED no. 10?

    Which part should I modify to make the input signal range from 0-5V or any higher range than that? What is the suitable voltage supply to power the IC and LED that I would need if so?

    Your help in the circuit calculation is very much appreciated. :)
     
  2. redblaque

    Thread Starter New Member

    Jan 30, 2009
    4
    0
    Anybody? Help?
     
  3. Audioguru

    New Member

    Dec 20, 2007
    9,411
    896
    R1 has 1.25V across it. Its current is in R2. Then their total voltage is 5.2V when the display shows all LEDs turned on.

    An LED has a fixed voltage drop. Then when the 10th LED turns on the voltage at its cathode drops only 1.8V for a red LED. The 100 ohm resistor allows the voltage at the cathode to drop much more when the 10th LED turns on.

    The ratio of R1 and R2 determines the input voltage range.
    The input could have a voltage divider.

    If the power supply voltage is high, the LED currents are high and all lEDs are lighted then the LM3914 IC will get too hot and might melt.

    I have a Sound Level Indicator project that operates with a fairly high supply voltage (9V), very high LED currents (30mA) and I reduce the heating with a series 10 ohms/1W resistor that feeds the LEDs plus a bypass capacitor for the LEDs. The resistor shares the heat.
     
  4. redblaque

    Thread Starter New Member

    Jan 30, 2009
    4
    0
    Thank you for your kind response, AudioGuru.

    Excuse my poor electronics, but how is it that the 100 ohms resistor will cause a greater drop in the LED cathode? Is it that the 100 ohms will consume a bit of the 5V voltage and make the anode lead smaller in volts? Can you picture the math for me?

    Another question, what is the purpose of the 1k resistor in parallel with the diode+100 ohms??
     
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