On Sun, Jun 29, 2003 at 10:29:35PM -0400, Jai Dhar wrote: > This method is something that I actually just thought off after I fired off my > last email!!! Since my 7-seg's are common cathod, I would need some sort of > transistor between the common-cathod pin and ground, correct? What came to mind > was a n-channel MOSFET (low side switch). This seem like a reasonable approach? > BJT's might be more attractive since they are in smaller in through-hole form, > but are there any big differences in what I use?? I'm confused. The last time I checked this thread you said that you wanted to minimize the number of external components. With the solution below you'll need the decoder and the transistors, whereas with the original solution you posted all you'll need is the external resistors with the PIC driving everything. I'm just trying to figure out what's more important, saving I/O, saving board real estate, programming simplicity, or what. Please enlighten me. BAJ > > Thanks, > > Jai > > Quoting Ishaan Dalal : > > > Something that has worked for me in the past is to use one driver (74LS47/48 > > or 4511/4513/4543) and connect its outputs (A...G) to all 4 segments in > > common. That takes up 4 pins of your port. Then you use the other 4 pins to > > drive transistors that source OR sink power from each of the 4 7-segs. > > That's another 4 pins. One port does it! > > > > > ---------------------------------------- > This mail sent through www.mywaterloo.ca > > -- > http://www.piclist.com#nomail Going offline? Don't AutoReply us! > email listserv@mitvma.mit.edu with SET PICList DIGEST in the body -- http://www.piclist.com hint: To leave the PICList mailto:piclist-unsubscribe-request@mitvma.mit.edu