# Opto-coupled EIA-422 receiver (understanding suggested circuits)

I am looking at a standard for communication of marine electronics (NMEA). The standard is electrically essentially EIA-422 but it prescribes opto-coupled differential receiver pins without connection to ground.

• tolerate 15V differential voltage including reverse polarity
• not draw more than 2mA at 2V differential voltage
• not draw more than 3.25mA at 10V differential voltage
• work at 38 400 baud

The standard suggests two circuits that "offer overvoltage, reverse voltage and power dissipation protection for the optoisolator and serve to limit current drawn from the line":

simulate this circuit – Schematic created using CircuitLab

I understand how circuit 2 is supposed to work:

• R3 is limits the current
• D3 offers reverse polarity protection for the optocoupler and Q4
• R4 is a current sense resistor
• which allows Q4 to act current limiter for the optocoupler by providing a bypass when the current through D4 exceeds a threshold

Questions:

1. Why would a separate current limiting for D4 be desirable? Since the total current drawn from the line must not exceed 3.24mA this seems superfluous. The optocouplers I am familiar with usually operate at 20mA.
2. How is circuit 1 supposed to work?

I cannot make sense of it:

• D1 provides reverse polarity protection
• J1 seems to limit current through the optocoupler for "correct polarity"
• in reverse polarity J1 presents no resistance however. Thus R1 is required to limit current in reverse polarity
• but since R1 must be fairly large for this, the current through the optocoupler for "correct polarity" becomes incredibly small. To my understanding too small to be useful: In the 10s to 100s of µA.

I am clearly missing something essential here.

• In party line communication, the specs are designed to limit current over a wide range of voltage yet have sufficient current to drive an opto. So the transfer function has a current controlled input impedance. – Sunnyskyguy EE75 Apr 13 '18 at 15:33