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Here is the schematic that I'm working with: http://www.circuitsonline.net/schakelingen/59/audio/led-vu-meter.html

The LED attached to pin 10 is always on. Is there anyway I could get this so that all 10 work as part of the VU-meter?

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2 Answers 2

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That's odd. Are you sure it's the LED on pin 10, and not the one on the other end, on pin 1?

Take a look at page 7 of the datasheet. There's a resistor ladder network creating a series of voltages to compare to, starting with the reference voltage for the LED on pin 10. If the input voltage is higher than this \$V_{REF}\$ the output of the comparator will go low, and the LED will be on. For the other levels the comparator uses a lower voltage each time. So if only the LED on pin 10 is on it would mean that the input voltage is higher than the reference, but NOT higher than the other comparison voltages, which are lower. This just doesn't make sense.
It would make sense if it were the LED on pin 1, then it could be that you have a DC offset in your signal, which is higher than the first threshold.

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I did notice that you use the LM3915 in dot mode (which I commented as odd for a VU-meter; you would rather use bar graph mode there), but I didn't connect this to my previous observation. Of course you'll see only one LED in dot mode! If the LED on pin 10 is the only LED you can get lit your input signal may be too high. Try to short the input to ground, and, alternatively, connect pin 9 to pin 3 and see what happens. If my hunch about a too high level is correct you should see all LED on.

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  • \$\begingroup\$ Neither pin 1 or pin 10 make total sense. Indication is that something additional is happening and that a careful work through of datasheet is needed. If in dot mode then pin 1 should be off when anything higher is on. If it is always on and others are on as well then it's not in dot mode so he's not leaving pin 9 floating so his given circuit isn't what he's shown - hence my request for actual circuit and voltages. // Also - IC could be faulty. \$\endgroup\$
    – Russell McMahon ♦
    Commented Aug 7, 2011 at 7:05
  • \$\begingroup\$ @Russell - one thing with the schematic on page 7 is that it's simplified, it only shows bar graph mode. The parts for dot mode are left out. And the IC could be faulty, but I have to say from personal experience that it's often the user (me!) who made a mistake or wrongly interpreted information. Supposing faulty part is easy if you can't think of anything else, but you order a new part, and it behaves exactly the same way! :-( \$\endgroup\$
    – stevenvh
    Commented Aug 7, 2011 at 7:49
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Summary:

  • The data sheet contains extra data which should be relevant.

  • Knowing the exact circuit you are using will assist. How is pin9 connected? What supply voltage are you using? (Both possibly relevant to the result).

Control of the state of pin 10 / LED10 is complicated and you need to read the LM3915 data sheet carefully to see how it is controlled. There are various parts of the datasheet that may be relevant. There are a number of things that may be affecting what is happening. You need to read the relevant parts of the data sheet and see how what you are doing compares to what it says.

Here are some references to parts of the datasheet that may be relevant. You need to work through these and see which if any is.

  • Note that LED10 is used as part of chaining and you may be seeing permanent device overload. This seems unlikely but may be a factor.

  • The level on pin 9 (which is open circuit in your diagram) is relevant. This pin may do different things depending on how it is connected. See note 5 on page 4.

    • Note 5: Bar mode results when pin 9 is within 20 mV of V+. Dot mode results when pin 9 is pulled at least 200 mV below V+. LED #10 (pin 10 output current) is disabled if pin 9 is pulled 0.9V or more below VLED.

  • See block diagram page 7 and note pins 6 & 7 relationship.

  • See mode pin description pages 8 & 9.

  • Also see bottom of page 9 re possible oscillation.

Understand the above. Try appropriate changes. Report back.

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    \$\begingroup\$ Answers should be able to be self contained for the most part. By reading this answer I have no idea what his answer is. \$\endgroup\$
    – Kellenjb
    Commented Aug 7, 2011 at 0:00
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    \$\begingroup\$ @Kellnjb: The real simple self contained answer is "Read the data sheet and connect the IC correctly based on what it says". As that sort of answer is sometimes less than fully helpful I took the extra time and effort to add some possible leads. I think that was better than the "self contained" answer. No? I've slightly improved the layout BUT life is not about spoon feeding. The LM3915 datasheet was not referenced by the questioner so I don't know if they read it. I think the answer was good enough as was, mayhaps with a little formating, as added. \$\endgroup\$
    – Russell McMahon ♦
    Commented Aug 7, 2011 at 0:43
  • \$\begingroup\$ The LM3915 was not mentioned in the question - only in the circuity referenced, so it is probably but not certainly what they are using. \$\endgroup\$
    – Russell McMahon ♦
    Commented Aug 7, 2011 at 0:45
  • \$\begingroup\$ @Russell McMahon: sorry, but this just sounds silly (pardon my french): the LM3915 is the only major component in the schematic, and you're questioning if he would use it? Besides, I agree with Kellenjb that this is a very vague answer basically just saying "read the datasheet". \$\endgroup\$
    – Federico Russo
    Commented Aug 7, 2011 at 6:51
  • \$\begingroup\$ He nowhere mentions an LM3915. The cct referred to contains one. There are several devices in the family. Lonnnng experience [tm] indicates that anything implied but not stated in concrete terms should be treated with care. I deal with what I think they are using - but am aware that they may not be. The IC is behaving in an unusual manner and is far more complex than it's most simple use may suggest. Pin 9 handling and pin 10 are both areas where special magic applies. As he did not reference the data sheet he may not have read it or be aware of how complex it is. \$\endgroup\$
    – Russell McMahon ♦
    Commented Aug 7, 2011 at 7:13

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