Discovery Checksum Problem Analysis

[sjackman]

A fundamental assumption made above is that the transmission medium acts as a wire-and, such that the receiver sees a one if both transmitters transmit a one, and sees a zero otherwise. DMX uses a differential transmission medium, and so a one is transmitted as

D+ = V+
D- = V-

and a zero is transmitted as

D+ = V-
D- = V+

If we assume that the more negative voltage is seen by the receiver -- I'm not entirely positive this assumption is valid, but it seems reasonable -- and transmitter A is transmitting a one, and transmitter B is transmitting a zero, the receiver will see

D+ = V- of B
D- = V- of A

The datasheet for the SN75176 differential transceiver says that if the absolute differential voltage is less than 0.2 V

|D+ - D-| < 0.2 V

the output is indeterminate. In practice, if V- of B is more positive than V- of A, the receiever will see a one, otherwise the receiver will see a zero. The result being that in the case of a collision, the bit the receiver sees depends more on the characteristics of the transmitting transceivers than the bits being transmitted.

Cheers,
Shaun Jackman
Pathway Connectivity