It is important to note that the inductor selected must also take power consumption into consideration. The inductor should be sized to handle the maximum anticipated current in addition to the inductance value.

The parallel aggregate impedance should be selected so that the total equivalent impedance at the carrier frequency is Z ≥ 375 Ω. This assumes RS-485 loading with 60 Ω termination. If no termination is used in the application, then the total equivalent impedance at the carrier frequency could be reduced to Z ≥ 60 Ω. These examples assume that termination is used. Equation 1 shows the parallel aggregate impedance equation for inductors L₁ to L_n. Since the inductance value for each node should be the same, it's simple to determine that each node's impedance should be n times the total equivalent impedance. For example, if there are 4 nodes connected to the bus and the equivalent impedance is 375 Ω, then each node impedance should be 1,500 Ω.

Equation 1.

To determine the suggested inductance value, Equation 2 can be rearranged to determine L_n, as shown in Equation 3.

Equation 2.

Equation 3.

ƒ₀ is the carrier frequency (OOK frequency) used. If the previous 1.5 kΩ impedance per node is assumed with a carrier frequency of 1 MHz, the resulting inductance limit is ~240 µH per node. Be aware that this is the minimum suggested value per node. Refer to Figure 9-2 as a quick reference on the minimum inductance value to achieve 375 Ω of total aggregate impedance. This value can be multiplied by the number of nodes on the bus to get the minimum inductance per node. Referring to the previous example, if there are 4 nodes and a carrier frequency of 1 MHz, then the minimum aggregate inductance is about 60 µH, which is 240 µH when multiplied by 4.