Hi, I have been using the following circuit with good result: It uses two I/O pins (one bidirectional and one pure output) instead of o= ne. However, the text in the article is wrong. It says that the input signal = to Q3=20 has to be set low in order to activate the pull-up circuit. This is wrong= .=20 Setting the base of Q3 low only turns of the pull-down. In order to activ= ate=20 the pull-up, the input line also has to be switched to an output and set = low.=20 The pull-up will be activated when the clamping resistor has pulled up th= e=20 line higher than 0.6V. So it only is activated if there isn't anything el= se on=20 the line that pulls it low, preventing a short. I also use a lot lower pull up resistor than 4.7k ohms. I think it is dow= n to=20 510 ohms.=20 The output is connected to two 3.3 ohm resistors and inbetween them there= are=20 two shottky diodes (BAT54S), one connected to ground and another connecte= d to a=20 6.8V (breakdown, Vrm=3D5.8V) TVS diode. The TVS is not directly connected= to the=20 1-wire line because of its relatively high internal capacitance. Regards / Ruben > I see the application note on programming the pic for 1-wire. >=20 > But what do I need in the way of line drivers between the pic and the > connector to the outside world? >=20 > Drew > --=20 > http://www.piclist.com PIC/SX FAQ & list archive > View/change your membership options at > http://mailman.mit.edu/mailman/listinfo/piclist >=20 =3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D= =3D=3D=3D=3D=3D Ruben J=F6nsson AB Liros Electronic Box 9124, 200 39 Malm=F6, Sweden TEL INT +46 40142078 FAX INT +46 40947388 ruben@pp.sbbs.se =3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D= =3D=3D=3D=3D=3D --=20 http://www.piclist.com PIC/SX FAQ & list archive View/change your membership options at http://mailman.mit.edu/mailman/listinfo/piclist