LED indicator

Discussion in 'The Projects Forum' started by timplett, Aug 2, 2009.

  1. timplett

    Thread Starter Member

    Jan 20, 2009
    19
    0
    I think this circuit is what I need, but I need some help modifying it to suit me. What I am wanting to do is to connect a 40-50 step voltmeter to the subwoofer in my car, so that the bar would go up as the speaker output does. The PDF sheet the circuit was on said it could be expanded to 100 steps but I don't know to do that. The subwoofer is 300w RMS at 2 ohms, therefore 24.5 volts, but peak output is double, so 49 volts peak. Speaker output is AC so it will need to be rectified first. I've got basic electronics skills but not enough to design the circuit myself.

    Thanks in advance,
    Tim
     
  2. beenthere

    Retired Moderator

    Apr 20, 2004
    15,815
    282
    Welcome to AAC.

    I have moved you post to its own thread. The thread you placed your question in is almost a year old - http://forum.allaboutcircuits.com/showthread.php?t=12580

    A voltmeter display may not be satisfactory. Something more like a VU meter should be better. here is one example - http://sound.westhost.com/project60.htm

    By the way, there is no need to monitor the signal driving the speakers. The amp has a fixed gain, so monitoring the signal into the amp's input is much easier.
     
  3. timplett

    Thread Starter Member

    Jan 20, 2009
    19
    0
    Okay so I've done some looking around online and decided to use the schematic included in the LM3916 datasheet (http://www.datasheetcatalog.org/datasheet/nationalsemiconductor/DS007971.PDF). My only question is in regards to limiting the ~14 volts (this is going in a car) down to the proper voltage for the LEDs. I know how to find the proper resistor value for individual LEDs and parallel arrays and such, but I'm not sure how this works seeing as the number of LEDs that is lit varies. These are the LEDs I'm using (http://www.futurlec.com/LED/LED5BULB.shtml) What value resistor will I need for this?

    Thanks,

    Tim
     
  4. timplett

    Thread Starter Member

    Jan 20, 2009
    19
    0
    Okay I thought this out some more, and I think I need to knock it down to the correct voltage for one, with a resistor capable of the milliamp draw of all of them. Using this information I'm getting a 39 ohm 3.5 watt resistor. Is this correct?
     
  5. beenthere

    Retired Moderator

    Apr 20, 2004
    15,815
    282
  6. timplett

    Thread Starter Member

    Jan 20, 2009
    19
    0
    I've got the datasheet already, and building it isn't an issue, I just want to make sure I'm using the correct resistor to knock down the voltage for the LEDs. The datasheet doesn't get specific, and the project shown at that link uses a different power supply so it doesn't help as far as that goes.

    Tim
     
  7. Audioguru

    New Member

    Dec 20, 2007
    9,411
    896
    The LM3914, LM3915 and LM3916 regulate the current to the LEDs by the amount of load on pin 7 of the IC.

    You do not "knock down the voltage" to LEDs, you limit the current and the LEDs knock down the voltage to whatever voltage they need. The IC limits the current by regulating the current.

    Post your schematic for us to make recommendations.
     
  8. timplett

    Thread Starter Member

    Jan 20, 2009
    19
    0
    The circuit I'm using is the basic one from the datasheet. Forgive my confusion about the voltage issue. The schematic does say however that the voltage to the LEDs should 7 volts or less to limit power dissipation from the IC. Do I not need to worry about this?

    Tim
     
  9. Audioguru

    New Member

    Dec 20, 2007
    9,411
    896
    You must be aware that the output transistors in an LM39xx create heat when they have a voltage across them and the LED current through them. The datasheet spec's a max allowed IC dissipation of 1.37W.

    If you have a 12V supply, the LED currents are set to 25mA each and 2V red LEDs are used then each output transistor dissipates 0.25W and the IC will dissipate 2.5W if all 10 LEDs are lighted. The IC will melt.
    A resistor can be added in series with the IC to share the heat as is shown in the datasheet.
     
  10. timplett

    Thread Starter Member

    Jan 20, 2009
    19
    0
    Ok I see what you are saying about the IC setting the LED amperage. However, when I do the math I get 16 mA @ 14v (usual car voltage). 14volts / 8700ohms * 10 = 16mA output. The LEDs I'm using have a voltage drop of 3.5V at 12mA (if I'm reading that correctly). Using 3.5v as the voltage drop at 16mA (to be on the safe side) each output transistor will have 11.5v flowing through it at 16mA. 11.5v * 16mA = 0.184w * 10 = 1.84w, still too much. If the LED input voltage were to be limited to say 5v rather than 14v, the LED would drop 3.5v leaving 2v @ 16mA across the output transistor, equalling a total IC dissipation of 0.32w. Much better.

    Now for a resistor to accomplish this with. I want a voltage drop of 9v @ 16mA. 9v / 16mA = 562.5ohms. 9v * 16mA = 0.144w, but this value is for one LED only, so I multiply by 10 to get a resistor rating of 1.44w.

    So to sum this all up, I need a 560 ohm, 1.5 watt resistor (or a 560ohm, 1/4watt resistor for each LED).

    Did I get it right this time? :confused:

    Tim
     
Loading...