[M]: Re: "Hot unplugging"
| Subject: | [M]: Re: "Hot unplugging" |
| From: | Ray Mote |
| Date: | Mon Apr 13 00:58:41 1998 |
Ive paid minimal attention to this thread, but just enough to know
that there is apparently a capacitor on the supply side of the switch
that may(?) be desirable to discharge when powering down. The comments
about voltage existing at one pin with no ground on the other made no
sense to me -- if you open one wire in a two-wire circuit, there
is *no* power applied, period. (Unless, that is, you have *other*
cables still plugged in to other devices, such that a "ground" still
exists via a rather circuitous route.) But in a simple 2-wire circuit,
disconnecting one wire completely removes the power; it really doesnt
matter which of the two wires you open.
Ive been a bit surprised that no one seems to have mentioned the
primary bugaboo of solid-state circuits: contact "bounce". When
unplugging a cable, you generally dont get a nice clean break in
power. If you doubt that, just take a snapshot of the dc power line
using a digital storage scope. Manually unplugging a connector would
be probably the most likely method of generating multiple spikes on the
power line, but you do find the occasional switch or relay that will
give you severe problems of that type. Same goes for plugging in a
connector. (*Every* switch or electromechanical relay has *some*
contact bounce; its just that some are better designed than others,
and cause less problems.) In view of the above, I think Ill stick
with the switch Meade provided. If I opt to disable the power with
that switch still on, it will be by means of a switch in the AC line
going to the 18 volt power supply. :-)
If Ive missed some information in the thread, it should be obvious to
you by now. In that case, I have great faith that you will ignore my
rantings. :-)
Ray Mote, K5FKT Oxnard, CA
that there is apparently a capacitor on the supply side of the switch
that may(?) be desirable to discharge when powering down. The comments
about voltage existing at one pin with no ground on the other made no
sense to me -- if you open one wire in a two-wire circuit, there
is *no* power applied, period. (Unless, that is, you have *other*
cables still plugged in to other devices, such that a "ground" still
exists via a rather circuitous route.) But in a simple 2-wire circuit,
disconnecting one wire completely removes the power; it really doesnt
matter which of the two wires you open.
Ive been a bit surprised that no one seems to have mentioned the
primary bugaboo of solid-state circuits: contact "bounce". When
unplugging a cable, you generally dont get a nice clean break in
power. If you doubt that, just take a snapshot of the dc power line
using a digital storage scope. Manually unplugging a connector would
be probably the most likely method of generating multiple spikes on the
power line, but you do find the occasional switch or relay that will
give you severe problems of that type. Same goes for plugging in a
connector. (*Every* switch or electromechanical relay has *some*
contact bounce; its just that some are better designed than others,
and cause less problems.) In view of the above, I think Ill stick
with the switch Meade provided. If I opt to disable the power with
that switch still on, it will be by means of a switch in the AC line
going to the 18 volt power supply. :-)
If Ive missed some information in the thread, it should be obvious to
you by now. In that case, I have great faith that you will ignore my
rantings. :-)
Ray Mote, K5FKT
