Your point about an open line at high impedence makes sense.

However, as I understand it, current loop originated with teletypes. I was
under the impression that the whole thing was electromechanical and the
circuit was open when the 'switch' was open. Did they have a resistor in
parallel with the switching element, or something like that?

~ Bob Ammerman
RAm Systems

> -----Original Message-----
> From: piclist-bounces@mit.edu [mailto:piclist-bounces@mit.edu] On Behalf
> Of John Ferrell
> Sent: Tuesday, December 02, 2014 9:05 PM
> To: Microcontroller discussion list - Public.
> Subject: Re: [EE] Line drivers
>=20
> We may be thinking about different conditions. In the 20 ma teletype loop
a
> mark was 4 ma, a space was 20 ma. A line that is 20 ma and open has unsaf=
e
> transients due to the unknown inductance in the line. Neither the 20ma
> condition or the 4ma condition is high impedance or saturated logic. The
> implementation is analog, but the data is digital.
> This is likely another time I should have remained silent!
> Your suggestion to implement with optoisolators is a winner and the actua=
l
> current values are not important.
>=20
> On 12/2/2014 7:56 PM, Bob Ammerman wrote:
> > Just a plain (20ma) current loop. The 4-20 thing is for analog signals.
> > Maybe use an optoisolator at the receiving end for each direction.
> >
> > ~ Bob Ammerman
> > RAm Systems
> >
> >
>=20
> --
> John Ferrell W8CCW
>     Julian NC 27283
>   It is better to walk alone,
> than with a crowd going the wrong direction.
>                    --Diane Grant
>=20
>=20
> --
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