Don Wilhelm wrote:

close to it when the computer is not powered. If you check the R-232 signal levels you will find that the region around zero volts (+3v to -3v) is "undefined" - the system is designed that way to include noise

Actually, for some modem control lines, it is well defined, and, if you use standard line receivers, it is well defined for all inputs. Zero volts is required to read as OFF. E.g the Fairchild 1489A is guaranteed to be OFF for any input less than +750mV.

immunity - the RS-232 receiver output might be either ON or OFF when its

The plus and minus three volts are driver specifications. Typical receivers have a rather narrow transition region, but with a rather wider hysterisis band, which is still only about 150mV (1489A), or less.

input is at zero volts - it is not predictable and the real state depends on the behavior of the particular LT1039ACN driver/receiver chip

It should not, the positive going threshold should be definitely greater than zero volts.

in the K3. Couple that with the fact that the RS-232 'no signal' or 'Space' state must be between +3 and +15 volts to assure an OFF

RS232 uses positive logic for control signals and negative logic for data, so +3 volts is *ON* and SPACE, when using normal line receivers. This is on the line side. Typical drivers, and receivers, are inverting, so the TTL side has negative logic controls and positive logic data.

condition at the output of the RS-232 receiver - there is just no way to avoid the 'problem'.

As a result, if your K3 is set for PTT-KEY from the RTS or DTR lines, it would be prudent to power the computer on before the K3 so these input signals are at a defined state (hopefully 'space'). Even unplugging the

The zero voltage state is MARK, not space, but I suspect that you may actually be referring to the CW MARK state and actually representing it by an RS232 SPACE state. If you actually have RS232 MARK set to CW mark, you will get problems from unpowered drivers.

cable will not help because that will also result in zero volts at the input.


--
David Woolley
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