VU Meter - LM3915 IC to control high power LEDs

Discussion in 'The Projects Forum' started by dmckie250, Apr 11, 2011.

  1. dmckie250

    Thread Starter New Member

    Apr 11, 2011
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    Hi,

    I am just starting a project to build a giant 5-band EQ - the first stage of which will be constructing 5 seperate VU meters using LM3915 IC chips.

    I am intending to use the below circuit but with high power 1W LEDS - they are all around 2V and about 500mA of current. I have 12V/1200mA DC power adapter lying around that I was planning on using.

    Can you advise if this will work? It's been about 10 years since I did any electronics in school! I am planning to use the LM3915 in bar mode, so there will be up to 10 1W LEDs operating at the same time.

    Will I have enough power, or should I use a transistor connected to each output of the LM3915 to control a higher powered circuit?

    Thanks,

    Duncan
     
  2. SgtWookie

    Expert

    Jul 17, 2007
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    The LM3915 can only sink about 13mA current per output (yes, it says 1mA-30mA but not when you're in bar mode running 10 LEDs). You would need something like a PNP Darlington transistor (one per each output) to amplify that current enough to drive your LEDs.

    If you are planning on operating 10 of those 1W 500mA LEDs per LM3915, you will need 500mA * 10 = 5A current when all of the LEDs are on for one circuit, just to operate the LEDs. The LM3915 will also require a little bit of current, but trivial compared to the 5A. If you are building five of these for the 5 separate bands, you will need up to 25A to operate all of the LEDs. You should use a lower voltage supply, as otherwise you will have to drop most of the voltage across power resistors.

    Why don't you first try building the circuits using low-power LEDs?
     
    Last edited: Apr 11, 2011
  3. dmckie250

    Thread Starter New Member

    Apr 11, 2011
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    I'm going to build one test circuit according the above schematic using lower powered LEDs, I was just trying to think ahead to the next step before I bought all of the components.

    How would I use a darlington switch? do i need to buy two separate transistors or can i buy an integrated one? I don't suppose you'd have an example that might fit the job? Would I connect each of these transistors to the power supply in parallel or in series?

    For the 5-band EQ I am planning to use a chip such as the BA3812L - will it be ok to connect the 5 outputs from this chip to 5 separate LM3915s? I'm guessing if I'm using these higher currents in the VU meter circuits then I will need a diode or something to stop the current from flowing back into the EQ chip?

    Thanks for your help, this is the first time I've attempted an electronics project and its all quite new to me!
     
  4. dmckie250

    Thread Starter New Member

    Apr 11, 2011
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  5. SgtWookie

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    Jul 17, 2007
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    I see.

    Paddy's using a TIP126; since he's in a land where I don't know what he can obtain, he was kind of on his own. The TIP126 was quite overkill for his needs, but you're driving higher power LEDs.

    I can't tell you that right now; no time to look at the datasheet. However, if the LM3915's can share grounds with the BA3812L, I don't see a problem with it offhand.

    No, the outputs won't affect the inputs.
     
  6. SgtWookie

    Expert

    Jul 17, 2007
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    Yes you are, and yes it should work.

    No time to suggest a schematic at the moment; I'm on my way out the door.
     
  7. dmckie250

    Thread Starter New Member

    Apr 11, 2011
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    Thanks again for your help!

    25amps....that sounds like a lot! Do you think I might be able to salvage a PSU from an old computer? I've been looking at a few online and they seem to be able to put out about 25 Amps at 5V.

    I've put together this schematic - the other thread says something about current limiting resistors for the LEDs and the LM3915, would i need to include one for each (i.e. one between PSU and LED, one between TIP126 and LM3915, and one between PSU an V+ on the IC) or just one for both (i.e. one between PSU and LED)?
     
  8. SgtWookie

    Expert

    Jul 17, 2007
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    Well, try it like this:

    [​IMG]

    I'm only showing one output. Every other output will need to look like that.
    R1 limits the current through the LED; at the present time the value is not calculated. Each LED will need it's own current limiting resistor.

    R2 keeps the TIP126 base pulled high when the LM3915 output is not sinking current, keeping the TIP126 turned off; otherwise the LED may glow with no input.

    The 5v supply can be used for both the LEDs and the LM3915.

    You could use a salvaged (or even a new) ATX form factor computer supply.
    Here's a suitable 400W supply from one of my favorite vendors:
    http://www.mpja.com/prodinfo.asp?number=17981+PS
    $23+shipping.
     
    dmckie250 likes this.
  9. dmckie250

    Thread Starter New Member

    Apr 11, 2011
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    I'm guessing that your R1 resistor doesn't carry any current so it doesn't matter what wattage it is? I've decided to use 0.5W ones as they're pretty cheap. Looks like i had the transistor set up the wrong way too so i've updated my schematics to match your one.

    For the different LEDs resistors i've used this: http://led.linear1.org/1led.wiz - its a very hand calculator for us noobs! If other people want to use more than 1 LED then they can use this calculator: http://led.linear1.org/led.wiz

    Wow - this "simple project" i had in mind has gotten complicated pretty quickly & I've already spent about £100 on parts. Still, I'll feel very proud of myself when I have it completed (if I ever get the other 4 bands of the EQ finished...)
     
  10. Audioguru

    New Member

    Dec 20, 2007
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    Usually one of the peak detector circuits shown in the datasheet is used on an LM3915 VU meter so that it holds an LED turned on long enough (about 30ms) so that you can see it. Without a peak detector circuit then the upper LEDs will be a dim blur.
     
  11. dmckie250

    Thread Starter New Member

    Apr 11, 2011
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    sounds like an extra load of work - i think i'll build what I have here first and then see if I need it - don't want to put even more on my plate for the time being!
     
  12. SgtWookie

    Expert

    Jul 17, 2007
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    That is not correct. R1 has the full 500mA current of the LED flowing through it.
    You say your LEDs have a typical Vf of 2v with 500mA current flowing through them. If you are not absolutely certain of that, you better double-check.

    The TIP126 with 500mA collector current will drop about 0.7v from the emitter to the collector; the Vce(sat).

    So:
    R1 >= (Vsupply - (Vf_LED + Vce(sat))) / Desired_Current
    R1 >= (5v - (2v+0.7v))/0.5A
    R1 >= (5-2.7)/.5
    R1 >= 2.3/.5
    R1 >= 4.6 Ohms. This is not a standard value of resistance, but 4.7 Ohms is.
    A decade table of standard resistance values is here: http://www.logwell.com/tech/components/resistor_values.html
    Bookmark that page for future reference. Use the E24 (green) columns for most things.
    So, now the power rating of the resistor needs to be calculated.
    We have a 2.3v drop across the resistor, and 500mA current.
    P=EI, so P=2.3*.5 = 1.15 Watts.

    We double that result for reliability, so 2.3 Watts. This is not a standard wattage rating, but 2 Watts is, and it's about 57% higher than the required power rating, so that will work just fine.

    You do have another alternative; using the 3.3v output from the ATX supply. A 400W supply should have ~26A available at 3.3v.

    So, let's re-calculate with the 3.3v supply:
    R1 >= (3.3v - (2v+0.7v))/0.5A
    R1 >= (3.3-2.7)/.5
    R1 >= 0.6/.5
    R1 >= 1.2 Ohms, which is a standard value of resistance.

    P=EI, so 0.6*0.5 = 0.3 Watts, or 300mW * 2 = 0.6 Watts, or 600mW.
    You could actually use 1/2 Watt 1.2 Ohm resistors, and you would dissipate less heat in the current limiting resistors than you would at 5v.

    OK - but your schematic would be more clear if you had the +5v symbol on top rather than running a wire all around the bottom.

    That calculator is OK, but you have to manually add the Vce(sat) of the Darlington to get the right result; 4.7 Ohms, 2 Watts.

    Gee, that seems like a lot of money.
     
  13. dmckie250

    Thread Starter New Member

    Apr 11, 2011
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    sorry - got R1 and R2 mixed up there - I bought 0.5W 1k resistors for R2 as i assumed there wouldnt be a current flowing through it?

    As for R1 - I didn't realise I had to include the Vce(sat) so I've ordered the wrong ones!

    Just double-checked the datasheets and my green LEDs use 3.5V at 350mA and my amber/red ones use 2.4V at 350mA, so with the extra 0.7V I calculate

    Green - 2.7ohms - 0.28W - use 1W to be safe
    Amber/Red - 5.6ohms - 0.66W - Use 2W to be safe

    Also - does the 1k resistor have to be wired up in via the emitter as shown in your schematic or can it just be wired up in parallel anywhere? On my protoboard it looks like it'll be easier to wire it up like the below schematic:
     
    Last edited: Apr 13, 2011
  14. paddyhughes086

    Active Member

    Jan 30, 2011
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    Hi all paddy here mister overkill lol i do try my best.......wait a min what way do i take this comment...

    here a pic of the way i used my pnp transistors
     
  15. SgtWookie

    Expert

    Jul 17, 2007
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    Just as long as one end is electrically connected to the base, and the other end electrically connected to the emitter, it'll be fine.
     
  16. Audioguru

    New Member

    Dec 20, 2007
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    Notice that you show the negative of the power supply at the top but we show it at the bottom.
    Maybe that is why you show the LED connected upside-down. Our LED's cathode connects to ground but your LED's anode wrongly connects to ground (-).
     
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