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LM3914 --help :/

Discussion in 'Electronic Design' started by [email protected], Dec 2, 2005.

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  1. Guest

  2. Most likely your circuit starts to oscillate when that last LED comes on.
    Carefully decouple the power supply (pins 2 and 3.) Let's say 10uF elco //
    100nF ceramic. Make sure the power traces are wide (thick) enough.
    Especially the GND trace should have a very low resistance.

    petrus bitbyter
  3. Guest

    petrus bitbyter napisal(a):

    You did not understand me. The circuit does oscilate as it should, but
    I do not understand the idea of oscilating, I know that whe the last
    LED begins to ligth then something is going on with capacitor and ...?
    why does it blink ?
  4. John B

    John B Guest

    First, you need to look at the block diagram of the chip on page 6 of
    the data sheet. You will see that the comparator chain comes out on the
    'Rhi' pin. In the circuit on page 12 the 'Rhi' pin is connected to one
    end of the capacitor via a 470 ohm resistor and the other end of the
    capacitor to the last LED driver pin. The capacitor acts as feedback
    and as LED 10 lights it discharges the capacitor and causes the
    comparator chain input to be modulated thus turning all the LED's off
    again. Then the cycle repeats.
  5. Guest

    but exactly how does it modulate the chain inputof comarators, to
    disable all leds the + input of comparators should become higher than
    -... am I rigth ?
  6. John B

    John B Guest

    You need to understand how an Eccles-Jordan Multivibrator works, then
    all will become clear.

    Study this page:
  7. I looked at the datasheet and came to the same conclusion. I don't
    understand how it works, either.

  8. But it seems to be going the wrong way! When the final led turns on,
    the voltage drops. This is coupled through the capacitor to Rhi. This
    *reduces* the voltage accross the resistor chain so the indication
    will go overrange. I would expect that to turn off all the lights the
    indication would need to go to zero instead.

    However, looking at the block diagram more closely, it appears that
    the "load" on the reference also determines the LED brightness. So
    presumably injecting a current into it switches the leds off!
  9. Jim Thompson

    Jim Thompson Guest

    The LM3914 schematic is "simplified" ;-)

    ...Jim Thompson
  10. John B

    John B Guest

    On 02/12/2005 the venerable Jim Thompson etched in runes:

    Now, how does he know that? My guess is that he had a hand in the
    design somewhere.
  11. Jim Thompson

    Jim Thompson Guest

    Nope, wasn't me ;-)

    I think it was Leroy Long or <first name lost in my head :> Davis.

    ...Jim Thompson
  12. Guest

    So... which version is the one ? that with Eccles-Jordan Multivibrator
    or with the "load" on the reference which determines the LED brightness

    Best regards
  13. Yes, there's a 10:1 current mirror thing going on with the comparator
    outputs vis-a-vis the reference output current.

    But why the 100 ohm resistor?

    Best regards,
    Spehro Pefhany
  14. I guess it drives the capacitor with the full 5V instead of 3V or so?
  15. Okay, yes- increases the swing. Could be as low as ~1.6V with an old
    GaAs red LED.

    Best regards,
    Spehro Pefhany
  16. Guest

    Spehro Pefhany napisal(a):

    so all in all what does make all the leds blinking while the last one
    is ligthing ?

  17. Bob Monsen

    Bob Monsen Guest

    The output current for the LEDs is a 10x current mirror of the output
    of the REF OUT pin. When the signal input goes above the reference
    voltage of 1.25, The last LED turns on. Because of C1, this causes MORE
    current to be output by the ref out pin, causing more current in the
    output of all the LEDs, and making the voltage across C1 drop even more
    due to the 100 ohm resistor. Over time, the voltage at the REF OUT pin
    climbs back up to 1.25, and the current out of REF OUT diminishes. If you
    recall, this current is amplified 10x by the mirror, so this decrease
    causes the voltage at pin 10 to start to increase, because the voltage
    across the 100 ohm resistor depends on the current. This voltage drop
    decreases the current out of REF OUT even more; thus, this positive
    feedback turns off the output at pin 10, causing the voltage to spike up
    on the REF OUT pin. This turns off the output completely, because now the
    ref high pin is > the signal input. While the voltage at the REF OUT node
    is higher than the reference, the current out of the ref out pin is
    basically 0. This is what shuts off all the LEDs.

    The voltage decays for a while through the 1.2k resistor. When it gets
    down to the signal level, the LED at pin 10 is turned on again, causing
    the CAP to pull the voltage down, restarting the cycle...

    The off time depends on the cap and the resistance to ground it drains
    through when it spikes, I think. The on time depends on how quickly the
    REF OUT pin can charge the 100uF cap.

    Bob Monsen

    The question of the ultimate foundations and the ultimate meaning of
    mathematics remains open; we do not know in what direction it will find its
    final solution or even whether a final objective answer can be expected at
    all. "Mathematizing" may well be a creative activity of man, like language
    or music, of primary originality, whose historical decisions defy complete
    objective rationalization.
    - Hermann Weyl in 1944
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