On Tue, 8 Jul 2008 21:26:19 -0700 (PDT), rickman <g...@gmail.com>
wrote:

>In your discussion of RS-422 as a current loop without a ground, you
>don't mention what voltage is present on the conductors relative to
>the associated ground, only the current through them. If you ignore
>this voltage, you run the risk of blowing out the chips or at least
>making them work improperly.  A common ground must be established
>between an RS-422 driver and receiver.  How to best do this depends on
>the equipment and the grounding method used.  But to say the common
>ground conductor is not required is a significant
>oversimplification.

The signal ground is required when the system is voltage driven
without any termination resistors. The signal ground will carry the
return current for input transistor bias currents.

Looking at the fail safe termination case, there is a voltage divider
consisting of three resistors at the receiver input, a large
(kilo-ohms) from Vcc to the other input, then there is  the small
(100-120 ohm) termination resistor between the inputs and a large
(kilo-ohms) resistor from the other input to the local power supply
Gnd. Without a connected cable, the input pin voltages are nearly
halfway between Vcc and Gnd. The input transistors are happy, since
they get their bias current through the voltage divider.

Assuming the receiver power supply is floating and a cable from a
transmitter is connected to the receiver inputs, the receiver input
common mode voltage is defined by the Tx common mode voltage.  At the
receiver input, the voltage divider will force the Rx local power
supply Vcc slightly above the Tx common mode voltage and the Rx Gnd
slightly below the Tx common mode voltage. Looking at the situation
from the perspective of the actual Rx chip, the input common mode
voltage range is between the local Vcc and Gnd. However, there may be
problems, if a significant (> 1 mA) leakage current flows from the
floating Rx power supply to the outside world.

Paul