# Lesson 2 : Section 6 : S-Parameters and multiple driven ports

### Section 6: (Advanced topic) S-Parameters and multiple driven ports

S-Parameters assume one port is driven and the rest are terminated- if this condition isn't met, then these gain numbers won't directly apply to the observation. However, you can still use S-Parameters to calculate what will happen when multiple ports are driven. Consider the four-way power divider in Figure 2.5.1, where port 1 is the input (the cable TV signal goes into here) and ports 2-5 are the output (these signals are sent to TVs around the house). If port 1 is driven, it makes sense that port 2 will have around a quarter of the power coming out (-6dB less than the input). But if we turn this around, and Ports 2-5 are driven, obviously the output from Port 1 is not going to go down to quarter power (-6dB gain): if we drive multiple ports, we expect more power coming out then we put into any single port- we expect power to be conserved. We expect four times the power of any single port. The problem is we are breaking that S-Parameter condition. Two approaches can be taken.

**Excess Loss**

The quick approach is the concept of excess loss: how much loss do we see beyond the expected power divider loss. For example, the output of a four-way divider should be -6dB lower than the input. In reality, the device will have some additional loss in the metal wires and the dielectrics; the actual divider might measure -7dB. The 1dB difference between the theoretical loss and the measured loss is the excess loss, and it is the power burned up inside the device and/or reflected by the inputs (that S_{11} return loss keeps some power from even entering the divider).

A quick way to approximate the combined power coming out of Port 1 with Ports 2-5 driver is to simply sum the power going into the combiner (e.g., 1W into four ports is 4W, or 6dBW) and subtract out the 1dB of excess loss (so 5dBW is at the combiner port, or 3.2W).

**Real Math**

Perhaps you want a more exact answer, or don't trust "engineering" math. Doing the math for real isn't hard.

Equation 2.6.1

The voltage wave coming out of port 1 (the summed port) is equal to the weighted sum of all of the

incident waves. V_{1}^{+} is equal to zero (this is the output port), and assuming all ports are driven equally then.... that's just the standard power / voltage / resistance relationship.

Equation 2.6.2

To get the power coming out of the system... it's the same power relationship again.

Equation 2.6.3

So, doing it for real really isn't so hard. If your intuition is throwing a red flag, go through the math:

For a four-way combiner, from the port 1 single input to the ports 2-5 outputs, the transmission S-parameters are S

_{21}= S_{31}= S_{41}= S_{51}= 0.5 (S-Parameters are voltage ratios, and the square of the voltage ratio is the power ratio, so the power ratio is 0.25)The sum of incident power is S

_{21}^{2}+ S_{312}+ S_{41}^{2 }+ S_{51}^{2}= 1The incident voltages are 7.07V for 1W incident power and a 50-ohm system.

The output voltage wave (V

_{1}^{-}) is 28.28VThe output power is 4W.

**Main Point**

Even though S-Parameters don't individually apply directly when multiple ports are driven, it's easy to get expected power output from the S-Parameters. Please remember that these voltage waves are complex vectors, keep track of phase as you do the math!

Copyright 2010, Gregory Kiesel