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:

GPS400 splitter schematic

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?
  • 3
    \$\begingroup\$ The \$200~\Omega\$ resistor to ground is likely fed from a high impedance transmission line, so it looks like a reasonably big inductance (i.e. RF choke). \$\endgroup\$ Commented Jul 21, 2016 at 21:39
  • \$\begingroup\$ @Captainj2001 Ahhh.. that makes perfect sense. Why don't you make this comment an answer? \$\endgroup\$ Commented Jul 25, 2016 at 21:22
  • \$\begingroup\$ Well, this is only one hypothesis, there are a number of other techniques that could potentially be employed, the one I mentioned previously is just the most likely. There are better biasing implementations using both high impedance lines and radially tapered open stubs on the bias line. \$\endgroup\$ Commented Jul 25, 2016 at 21:34
  • \$\begingroup\$ I will add a complete answer when I can though! \$\endgroup\$ Commented Jul 25, 2016 at 21:40
  • \$\begingroup\$ Alright, that settles it. I need to learn a lot more about transmission lines and RF in general. Thanks for the help. At least now I understand what I should be looking for. \$\endgroup\$ Commented Jul 25, 2016 at 21:43

2 Answers 2


Look up what a bias tee is. It's likely that the 200 ohm load at DC is connected via bias tees. A bias tee is basically a diplexer where one input is at DC and the other is at RF. They can be as simple as a capacitor for the RF path and an inductor for the DC path. So that would basically be what's shown in your schematic, except with inductors (or something equivalent) in series with the resistors.


The DC and AC loads are not in parallel -- they are in series. And the DC load is bypassed with a capacitor, so it is effectively 0 ohms at the RF frequencies of interest. The AC load is the primary of a transformer, so its DC resistance is effectively 0 ohms, as well. Thus, both AC and DC on the feed line see their intended loads in series with 0 ohms.

  • \$\begingroup\$ You didn't address how the 200 \$\Omega\$ and 50 \$\Omega\$ interact \$\endgroup\$ Commented Nov 1, 2016 at 12:03

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