How to terminate branches in a transmission line

Question

some research has unfortunately not led to a clear result. Branches in transmission lines are not desirable, but if they are unavoidable the question arises how to properly terminate the individual branches. Assume I have a data signal that I want to transmit to several listeners in a star topology and the impedance of the lines in the star distribution is 50 ohms. Does that mean that each end of the line has to be terminated with 50 ohms?

Answer 1

A distinction is usually made between stubs and branches in transmission lines. A stub is a short section for "tapping" a transmission line and should not have a termination resistor. If a long branch is needed, a line splitter should be used to match the impedances for all three branches (or 4 if there are that many.) Do not simply join the transmission line branches without a line splitter. Each branch should be terminated at its end with an appropriate terminator (usually a resistor matching the characteristic impedance of the transmission line).

In the case you describe, the characteristic impedance is 50 ohms, so all branches should be terminated with 50 ohms, and you need a 50 ohm line splitter.

Edit:

There are a number of different splitter/coupler topologies. Each one has their advantages and disadvantages. Without knowing more about which transmission lines can source signals, which ones need to receive and other details such as these, one cannot tell which topology is "best". However, in @SteveSh answer, he mentions a topology in which different transmission lines are connected via a resistive star. Here is an example.

^{simulate this circuit – Schematic created using CircuitLab}

The terminals are represented by R4, R5, and R6. There could, for example be high impedance transceivers connected in parallel to these resistors.

I claim that the impedances "seen" by R4, R5, and R6 are all 50 \$\Omega\$.

To see this, we just look at "the rest of the circuit" for each of these resistors. Choosing R4, the rest of the circuit consists of

R2 and R5 in series, all in parallel with R3 and R6 in series, all in series with R1. That is,

$$Z(4) = (( R2+R5) || (R3 + R6)) + R1$$

Assuming that all branches are "peers", and none has any special treatment, R1 = R2 = R3,

and, assuming that all branches are terminated with the characteristic impedance of the transmission line, \$Z_0\$

$$Z(4) = ((R1 + Z_0) || (R1 + Z_0)) + R1 = \frac{R1 + Z_0}{2} + R1 = \frac{3 R1 + Z_0}{2}$$

If we want the impedance "seen" by R_4 to also equal Z_0, then we have

$$Z_0 = \frac{3 R1 + Z_0}{2}$$

or $$2Z_0 = 3R1 + Z_0$$

$$Z_0 = 3R1$$

which is true if \$Z_0 = 50 \Omega\$ and \$R_1 = 16.67\Omega\$

By similar reasoning, if there were N transmission lines connected via such a resistive star, we would arrive at the formula

$$\frac{Z_0 + R1}{N-1} + R1 = Z_0$$

$$Z_0 + R1 = (Z_0 - R1)(N-1)$$ $$Z_0 + R1 = NZ_0 - NR1 - Z_0 + R1$$ $$Z_0 = NZ_0 - NR1 - Z_0$$ $$NR1 = NZ_0 - 2Z_0$$ $$R1 = Z_0 - \frac{2}{N}Z_0$$

Answer 2

I have a data signal that I want to transmit to several listeners in a star topology and the impedance of the lines in the star distribution is 50 ohms. Does that mean that each end of the line has to be terminated with 50 ohms?

If you are only sending data to listeners then, if you have a powerful enough driving source with sufficiently low drive impedance, you can treat all the limbs of the star network as separate entities. This means, you can either terminate at each listener with a parallel termination or, series terminate at the send end multiple times.

Answer 3

A lot depends on your topology (how many branches, how long they are relative to the main line) and what your frequency of concern is (edge rate), but in general the answer is NO, you don't want to terminate each branch.

One reason is that if you have N branches, each one terminated with a parallel resistor R, the load seen by the driver chip DC-wise is R/N. This may be too low for the driver to reliably handle.

Another reason is that if the main trace and each branch are designed to be 50 ohms, at the junction of the branches (the node of the star), the 50 ohm main trace sees a trace impedance of 50/N, which in itself is a bad mismatch.

Your best approach to making something like this work is to insert a series resistor (say 33 ohms for starters) in each leg of the star. Then you have to simulate the design, using your favorite SI tool and see if your waveforms are acceptable.

This is a heuristic approach (some would call it a hack).

Another approach is to treat this as a RF problem and use something like Wilkinson splitters (power dividers). But you usually have to limit the bandwidth of you signal for these to work (they are not DC to light devices).

EDIT 1 - Added reference below

Some of these same points are discussed here: Why longer stubs cause more intense ringing?