How can the GPS400 signal splitter provide a \$200~\Omega\$ DC load and \$50~\Omega\$ RF load to GPS receivers on all four ports?
Most GPS receiver modules require an active antenna. As a result, most receiver modules provide a DC bias on the RF connector to power said active antenna. These modules typically implement an "antenna fault" detector circuit that checks for a DC load. The assumption is that if there is a load, the the active antenna is present.
When working with a simple:
GPS module <-> cable <-> Active antenna
everything works as expected.
The problems come up when there are multiple GPS receivers that all need to use a single active GPS antenna. In my case this would the antenna mounted outside the building with a single cable running in to the lab.
I was able to find the GPS400 splitter to do just what I need. The GPS module that will power the antenna is connected on port 1. Ports 2,3,4 are then presented with a dummy 200ohm load to prevent an "antenna fault" detection on the remaining modules.
What I understand:
- It uses a Wilkinson Splitter to split the RF signal while preserving the 50ohm input impedance.
- The use of DC blocks on ports 2,3,4 to prevent the DC bias from passing through to the input port.
- The use of a \$200~\Omega\$ resistor to ground which presents a DC load for ports 2,3,4.
The closest to a schematic from the datasheet is this:
What I do not understand:
- Why does this configuration not affect the output impedance on ports 2,3,4?
- If there is a \$200~\Omega\$ resistor to ground in parallel to the \$50~\Omega\$ antenna, shouldn't the output impedance be \$40~\Omega\$ instead?