How can the LM3914 range 0 - 20 vdc

Discussion in 'The Projects Forum' started by laurich, Jan 22, 2011.

  1. laurich

    Thread Starter New Member

    Jan 22, 2011
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    I have been trying to use the LM3914 to display output up to 20 or 25 volts (a readout for a generator my daughter is building). It seem simple initially, follow the formulas in the data sheet. Not!!! I have a couple of questions.
    1. Does the supply voltage for the IC need to be 1.5vdc greater than my 20 or 25 volts?
    2. After using the formula to determine LED current 12.5/R1, it seems I should use 20/R where R= the sum of R1 & R2 in parallel with the 10K internal divider. Is this correct?
    I had planned to use R1 = 680, R2 = 10.2K, which should give a 0 - 20 volt range, but now I'm not so sure.
    Any suggestions would be greatly appreciated.
    Thanks,
     
  2. SgtWookie

    Expert

    Jul 17, 2007
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    No, it does not. You can scale down the voltage you wish to measure by using a resistive divider. For example, you could use two 10k resistors in series, one end connected to the source you wish to measure, and the other to ground. The junction of the two 10k resistors will be 1/2 of the source voltage.

    Although the LM3914's maximum voltage rating is 25v, you'd be better off to keep it <= 20v.

    Do you have a schematic of your circuit thus far? That will give a starting place of where you are now, and then we can help you figure out where you need to go.

    Posting a schematic of where you are now is generally a requirement - they eliminate ambiguities. You're talking R1, R2, etc - but without a schematic to refer to, the reference designators are meaningless.

    .png format images are preferred, as they are not "lossy" like .jpg images, are compact, and web browsers can view them without additional software.

    Use the "Go Advanced" and "Manage Attachments" buttons to upload the image(s) from your computer.
     
  3. laurich

    Thread Starter New Member

    Jan 22, 2011
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    Thanks for you quick reply!! I am using a diagram from the National spec sheet.
    Do I understand correctly that if my input signal is through a voltage divider and with 20 volts applied and the input on pin 5 of the LM3914 is the middle of the divider so 10 volts are applied to the input, I need the IC supply voltage to be at least 11.5 volts?
     
  4. Audioguru

    New Member

    Dec 20, 2007
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    If the LM3914 is powered from a supply as high as 11.5V then it might get too hot driving 10 LEDs. Divide the input to a lower max voltage and use a lower supply that is 1.5V higher than the max input.
    R1 always has 1.25V across it and its current is also in R2 so you can calculate the reference voltage at pin 7. The LED current is 12.5 times the current in R1 or R2.
     
  5. laurich

    Thread Starter New Member

    Jan 22, 2011
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    You have been a great help. Thank you very much.
     
  6. SgtWookie

    Expert

    Jul 17, 2007
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    Dot mode is preferred for lower power dissipation in the LM3914 and simplicity of construction, as you can use a single resistor to limit maximum current, taking much of the load off the LM3914.

    If you are planning on using bar mode, then things get a bit more complicated.

    Is the LM3914 circuit going to be powered by the voltage that it's monitoring, or will it have a separate supply?

    If a separate supply, what are you planning on using? Or had you considered that aspect?
     
  7. Audioguru

    New Member

    Dec 20, 2007
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    I have an LM3915 VU meter that has been running continuously for 5 years in the bar mode with its LEDs very bright at 26mA each. I used a 10V supply (a 9V wall-wart) because the project is also powered from a 9V (actually 8.4V) rechargeable battery sometimes.

    The 20 red LEDs (2 in series per output) would cause the LM3915 to dissipate 1.5W which is too hot so I use a 10 ohm/1W resistor to share 0.676W and the LM3915 dissipates the remaining 0.824W and all is fine.
     
  8. laurich

    Thread Starter New Member

    Jan 22, 2011
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    I would like to use the circuit in bar mode but dot mode will do if necessary. The power circuit for the leds will be simply a 1 watt 9 vac transformer through a bridge rectifier. I was hoping to get by with a 9 volt battery as the supply for the IC. Assuming 18 volts input across a divider composed of a 20k in series with 2 - 20ks in parallel should give me 6 volts at pin 5. I have a 680 resistor at R1 and a 5k trimmer pot as R2. The circuit is for a science project for my daughter. It will display the voltages of three different generator designs and since the leds would be a little more impressive we chose this over standard panel meters. I really do appreciate all of your help.
     
  9. Audioguru

    New Member

    Dec 20, 2007
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    The LEDs need a voltage of only from 1.8V for red ones to 3.5V for blue ones. The outputs of the LM3914 need an extra 0.8V. The 9VAC might actually be 13VAC then the rectified and filtered LED supply will be very high at 16VDC causing the LM3914 to get much too hot if the LEDs are bright (high current) and in the BAR mode.
    Why not use a lower supply voltage then there will not be so much heat?

    The 680 ohm resistor sets the LED currents at about 18mA. The pot adjusts the reference voltage from 1.25V to about 10.2V.

    In the bar mode and all 10 LEDs are lighted, a 9V alkaline battery will last about 10 minutes when its voltage will be too low at 7.5V.
     
  10. laurich

    Thread Starter New Member

    Jan 22, 2011
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    The transformer is rated at 9VAC 1 Amp with a center tap. Do you think I could create 2 power sources from the center tap, filter and regulate both halves to give me enough vdc to drive the IC from 1 half and the LEDs from the other. I realize I would need to use a voltage divider or something to reduce the voltage ( about 2vdc?) on the LED half.
     
  11. Audioguru

    New Member

    Dec 20, 2007
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    A center-tapped 9VAC/1A transformer would give you + 5.4VDC and -5.4VDC at 0.35A each. Their 0V are connected together.
    If you try to make two positive supplies then each one is only half-wave. Make a full-wave +5.4V supply with 2 diodes and one filter capacitor instead. It doesn't need a voltage regulator.

    The supply voltage for the LEDs must be at least 1.5V higher than the LED voltage for the IC to regulate the LED current.

    The current used by the IC is only 3mA to 8mA.
     
  12. laurich

    Thread Starter New Member

    Jan 22, 2011
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    So I am going to power 3 of these circuits each with 10 LEDs. Keeping my LED voltage at a minimum of about 4 volts (to reduce heat produced by the IC) and total current at 18.3mA per LED for a total of about 550mA (assuming all 30 LEDs lit which should never happen) and since my max input to the IC will be 6 volts, I will need 8 or 9 volt supply at 24mA to for the three LM3914s. Since total current is less than 1A, could I get by with two full wave rectifiers attached to the single transformer and adjust each to the needed supply voltage? One of about 9vdc and one about 4vdc.
     
  13. laurich

    Thread Starter New Member

    Jan 22, 2011
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    'bout ready to test this wammer-jammer, I really do appreciate everyones help, Thank You.
     
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