LM339 set-points

Discussion in 'General Electronics Chat' started by Dritech, Dec 1, 2013.

  1. Dritech

    Thread Starter Well-Known Member

    Sep 21, 2011
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    Hi,

    How can I calculate the values for the resistors which will be used in a voltage division configuration to set 6 different voltage set-points?

    So basically I know Vcc, the set-points voltages and need to determine the resistors' values.

    Thanks in advance.
     
  2. kubeek

    AAC Fanatic!

    Sep 20, 2005
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  3. tubeguy

    Well-Known Member

    Nov 3, 2012
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    What voltages do you need ?
    You can calculate a quick divider string by thinking of it as 1000 ohms per volt (for example).
     
    Last edited: Dec 1, 2013
  4. Dritech

    Thread Starter Well-Known Member

    Sep 21, 2011
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    Thanks for the replies.

    But that can only be done when having 1 voltage output right? In this case I need 6 outputs

    The voltages needed are 50mV, 200mV, 250mV, 300mV, 450mV, 500mV.
     
  5. tubeguy

    Well-Known Member

    Nov 3, 2012
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    What is your supply voltage and available current?
    And is the supply voltage well regulated ?
     
    Last edited: Dec 1, 2013
  6. Dritech

    Thread Starter Well-Known Member

    Sep 21, 2011
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    The supply voltage will be 12V which will be regulated to 10V using a zener diode. Is the available current of the supply? If so, I will be using a bench top power supply so the current can go up to 5A.
     
  7. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    You can use this voltage divider.
    And for R7 you can use 180K + 10K or 180K + 51K (precision trim potentiometer). For 3K you can use 3x1K resistor or two 1.5K resistors.
     
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  8. #12

    Expert

    Nov 30, 2010
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    I did it with the 1 ma theory.
     
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  9. Dritech

    Thread Starter Well-Known Member

    Sep 21, 2011
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    Thanks for the reply :) Can you please tell me how the values were determined? Is the value of R1 assumed?
     
  10. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    Yes you can assume R1 value and then use Ohm's law to find the current
    I = 50mV/1K = 50uA And next use Ohms low and find the remaining resistors.
    R2 = (200mV - 50mV)/50uA = 150mV/50uA = 3K and so on.
    Or you can assume current as #12 did. But he made a small error.
    He assume 1mA.
    So from then R1 = 50mV/1mA = 50R;
    R2 = 150; R3 = 50; R4 = 50; R5 = 150; R6 = 50; R7 = (10V - 0.5V)/1mA = 9.5K = 9.1K + potentiometer
     
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  11. tubeguy

    Well-Known Member

    Nov 3, 2012
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    That's where thinking of the divider as 1000 ohms per volt is very useful.
    Ohms law states that 1volt/1000ohms = 1ma. 1ma is a convenient low current which works well in circuits like this.
    The 1ma current flows through the whole resistor string and each resistor has a voltage across it based on it's resistance.
    So, 1000 ohms = 1 volt(1000mv) drop, 50 ohms = 50mv drop etc,,,

    One thing that can easily trip you up is that you need to calculate the drop individually across each resistor in the stack.
     
    Last edited: Dec 1, 2013
  12. Dritech

    Thread Starter Well-Known Member

    Sep 21, 2011
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    Thanks for the replies Jony130 and tubeguy.
     
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