A star topology connects N devices in a completely symmetrical, peer-to-peer fashion. Each device in the star connects to a central node where all the wires directly join together. The star topology looks like a great way to interconnect a collection of remote devices, but a big problem exists.
Every signal arriving at the central hub encounters the other N–1 wires all connected together in parallel. Assuming all wires have the same characteristic impedance, Z0, the apparent load impedance of the hub equals Z0/(N–1), producing a reflection coefficient of (N–2)/N. Large values of N produce large reflections that wreak havoc on your signal quality.
A small star might work satisfactorily if the reflections settle quickly. However, you will never achieve first-incident wave switching with full-amplitude received signals if the connecting wires, and, thus, the signal-propagation times, are long compared with the rise or fall time of your signals .
You can, however, obtain excellent signal quality if you permit a certain degree of signal attenuation. As a general rule, you can always insert damping into any old hairball network to stop the reflections but at the cost of a diminished received amplitude. If you are using LVDS (low-voltage-differential-signaling) logic, then the diminished amplitude may not hurt you. LVDS logic, like many other differential-logic families, is built to handle plenty of signal attenuation.
Figure 1 illustrates a single-ended attenuating network that you can use with a star topology. The figure assumes that all of the inactive endpoints end-terminate with impedance Z0. A differential network uses the same circuit, just replicated for the other element of the differential pair.
For the case N=3, set R1 equal to Z0/3. That solution produces no reflections, making the entire hub electrically transparent except for a fixed gain of ½. In essence, the circuit trades half of your signal amplitude to eliminate a major source of reflections. In general, a star having N devices requires R1=Z0(N–2)/N and produces a gain of 1/(N–1).
For the special case of N=3, you may transform the network of Figure 1 into a delta configuration having a resistor with value Z0 spanning between each pair of signal inputs.
In all cases, the entire network of resistors at the central hub must remain physically small compared with the rise time of your signals, so that the resistor network behaves in an ideal lumped-element fashion.
If you unplug one device from the star, you can no longer depend on that
device's termination to maintain a stable impedance at the hub. For example, in
a hub designed for N=3, disconnecting one device changes the impedance of
the hub by roughly one part in three. The only way to stabilize the impedance of
the hub in this system is to add even more attenuation to your coupling
network. To accomplish this feat, you could design the R1 resistor values in Figure 1 for, say, N=10, but then install only a maximum of three
devices. In this case, you must install a fixed resistor, R2=(R1+Z0)/7, to model the
effect of the seven missing devices. Removing one device now affects the circuit
by only one part in 10.
 Johnson, Howard, "To tee or not to tee?" EDN, Feb 2, 1998, pg 24