I was hooking up some 10/100BaseT Ethernet cables at a customer site and I noticed that the Voice Over IP phone on a desk had an Ethernet input and an additional RJ45 that serves as a pass-through connection. This was handy when there was just 1 CAT5 cable coming into the cubicle. The incoming cable goes to the phone, and then the pass through jack was connected to their PC via another CAT5 cable.
Do you know what circuit is behind these 2 RJ45 jacks?
I would love to have taken a screwdriver to the phone, but that was beyond the scope of my work plan :)
Do you think it is a transformer? Just a couple RJ45s in parallel (yuk!)? Certainly there are not 2 MAC/PHYS, that seems cost prohibitive for a cheap VoIP phone.
How would one design such a pass though to allow several devices to share a single incoming CAT5 cable?
I've seen this question: Ethernet Without a Switch?
that has a link to this circuit: http://www.zen22142.zen.co.uk/Circuits/Interface/pethhub.htm
Perhaps this is what is inside the VoIP phone?