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Voltage measurement tip for 12-bit A/D systems (e.g. DL205)

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  • Voltage measurement tip for 12-bit A/D systems (e.g. DL205)

    This only applies to PLC systems where an analog input has a resistance of 250 ohms and converts a 4-20 mA signal into a 12-bit number from 0 to 4095.

    To nicely measure voltages in a 12 - 24 VDC system, you:

    1) Put a 2,310 Ohm resistor between the 12 - 48 VDC supply voltage and an analog input (e.g. one of the inputs of an F2-08AD-1);
    2) Add 1,024 to the 12-bit number from the A/D converter, and
    3) There is your system voltage, with a precision of 0.01 V. All it needs is two decimal places; the HMI will do those.

    For every 0.01 V difference in the voltage, the 2,310 Ohm resistor will provide an exact 1 unit difference in the 12-bit number.
    Effective range: 10.24 - 51.19 VDC.

    Need a 1W 2,310 Ohm resistor above say 32V, otherwise 0.5W minimum.

    Chances are you'll get a resistor that is slightly off. My setup also needed a 1M parallel trim potentiometer for precise adjustment of the reading.
    Last edited by Herman; 06-22-2019, 05:39 PM. Reason: Typo

  • #2
    The 250 ohm input impedance allows measuring 2-20mA from a source of 1-5 V.
    in order to measure a voltage using an analogue current input, first you need to make sure the max voltage will drop to 5v at port connection. The series resistance needs dimensioned such as 20mA x series resistance + 5V= max Voltage.
    however, the minimum voltage measurable will be 4mA x series resistance + 1V = min voltage
    point being that you canít actually measure the entire voltage range.
    for example, to measure 24V you need 5V at port input, meaning you loose 19V over a series resistance of 19V/(20mA)=950ohm. However, the minimum voltage measurable will be 1V+ 4mA x 950 ohm = 4.8V
    in the same logic, the minimum measurable voltage for a 48v source is 9V.