Connect with us

Help needed; analyze omron output plc circuit

Discussion in 'General Electronics Discussion' started by toradatanglagi, Dec 8, 2012.

Scroll to continue with content
  1. toradatanglagi


    Dec 8, 2012
    hello all

    I'm need help analyzing this circuit of an output module from omron.. the output module is of model cj1m-od211 ..

    I'm using 24VDC as input signal to the input module.. so how to explain the circuit operation (current flow, voltage drop etc) from the internal circuit until 24VDC is produced at the output of the module (Jxx_Ch1_Outxx) ?


    thx in advance

    Attached Files:

  2. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

    Nov 28, 2011
    This is a very straightforward digital output circuit.

    Starting from the left of the top diagram, the "internal circuits" represents the inside of the PLC. When the PLC wants to activate an output, it applies a voltage between the middle and bottom wires coming out of that "internal circuits" box. This causes current to flow through the resistor and the LED in the optocoupler - the first dotted box, which contains an LED and a phototransistor.

    When the phototransistor in the optocoupler sees the light from the LED, it conducts, and current flows from its collector to its emitter. Its collector is connected to the externally provided power supply, which you say is +24V. The transistor therefore pulls its emitter voltage upwards towards +24V, which provides positive gate bias on the N-channel MOSFET (the second dotted box) and makes it conduct as well.

    The MOSFET conducts between its drain and source; this completes the output circuit and supplies current to the load, which is connected between the external positive supply rail and the drain (top terminal) of the MOSFET, on the Jxx_CH1_Outnn pin.

    When the PLC wants to turn the output OFF, it removes the voltage that was supplying the optocoupler, the optocoupler LED goes out, the optocoupler transistor stops conducting, and the resistor at the bottom left corner of the MOSFET pulls the MOSFET's gate voltage down to zero (relative to the source, which is the bottom terminal of the MOSFET, which connects to the COM rail).

    The MOSFET turns OFF, and stops passing current in its drain-source path, interrupting current to the load and turning the load off.

    This design is intended to drive non-inductive loads. If you use an inductive load such as a relay coil, you need to connect a reverse-biased diode across the coil. Otherwise, when the MOSFET turns OFF, the back EMF from the inductance of the relay coil will cause the MOSFET's drain voltage to shoot upwards, and this could damage the MOSFET, unless it includes a feature to protect it against this voltage spike. Or the MOSFET may be protected by other components which are present in the module but are not shown in this diagram. This is probably just a representative diagram to show you the general idea of how the output circuit works.

    The MOSFET also includes two back-to-back zener diodes between its gate and its source, shown inside the dashed box, to protect the MOSFET's gate against overvoltage. The protection for the optocoupler's transistor, if there is any, is not shown.

    To use this output, you need to provide a DC supply of 12~24V, with its negative connected to the COM rail and its positive connected to the +V rail. Loads must be connected between the output pin (Jxx_Ch1_Outnn) and the positive rail.
Ask a Question
Want to reply to this thread or ask your own question?
You'll need to choose a username for the site, which only take a couple of moments (here). After that, you can post your question and our members will help you out.
Electronics Point Logo
Continue to site
Quote of the day