Measuring 330V dc with MCU

Discussion in 'Embedded Systems and Microcontrollers' started by embedded.world, Apr 10, 2014.

  1. embedded.world

    Thread Starter Member

    Feb 27, 2014
    36
    0
    I need to mesure 400V dc from MCU with reference voltage of 5V.

    So only way I see is to use voltage divider & use factor of 80.

    that is 790K & 10K resistor with it.
    This will make 5V when 400V is applied.

    1. Is my calculations are correct.
    2. Is there any method to do with it.
    3. One problem I see is when voltage is 400V, current through resistors will be .5mA.

    So wattage across 790K resistor = 0.1975W.
    This is near to spec of resistor i.e 0.25W
     
  2. ericgibbs

    AAC Fanatic!

    Jan 29, 2010
    2,503
    380
    You could and should make up the 790K resistor by using 2 or 3 other resistors in series,to make a total of 790K.
    E
     
  3. ErnieM

    AAC Fanatic!

    Apr 24, 2011
    7,394
    1,606
    I did not check your calculations though your assumptions need amends:

    If your A2D tops out at 400V you cannot tell the difference between 400V or 500V... even 400.5 is invisible to your processor.

    My rule of thumb here is to leave 10-20% of the range for over values, or just let my max value have a count near 1000 (in a 10 bit converter).

    Since you don't need to match any exact divider ratio (such as 80) you have more resistor pairs to choose from to get something workable.

    The divider needs calibration, but so does any divider.


    Additionally, when you chop up 400V into 1024 pieces (again assuming 10 bit converter) your resolution is 400 / 1024 = .39 V. You should check if this is acceptable for your secret project.
     
  4. THE_RB

    AAC Fanatic!

    Feb 11, 2008
    5,435
    1,305
    And keep in mind that resistors have a voltage rating!

    Small resistors (like 1/4W size etc) often have a voltage rating of 200v. So you need at least a couple in series.

    AND be aware that even a fraction of a mA at 330v will be a significant amount of power the resistors need to dissipate, so calculate power; P=V*I
     
  5. ErnieM

    AAC Fanatic!

    Apr 24, 2011
    7,394
    1,606
    One way to save power is to add a series switch with the voltage divider: you only close the connection to the high voltage during a measurement. If your measurements are short and only done periodically then you can save up to 15% off on your car insurance and a significant amount of power.

    You still need the voltage rating as RB correctly points out, but you can use a much lower power rating.
     
  6. embedded.world

    Thread Starter Member

    Feb 27, 2014
    36
    0
    Do multi meters also measure such high voltage by resistor divider like 1000V dc?
     
  7. THE_RB

    AAC Fanatic!

    Feb 11, 2008
    5,435
    1,305
    Yes. They do.
     
Loading...