Applying SWR to Circuit Efficiency - RF Cafe Forums

 RF Cafe Forums closed its virtual doors in 2010 mainly due to other social media platforms dominating public commenting venues. RF Cafe Forums began sometime around August of 2003 and was quite well-attended for many years. By 2010, Facebook and Twitter were overwhelmingly dominating online personal interaction, and RF Cafe Forums activity dropped off precipitously. Regardless, there are still lots of great posts in the archive that ware worth looking at. Below are the old forum threads, including responses to the original posts. NOTICE: The original RF Cafe Forum is available again for reading, and the new RF Cafe Blog is an active board. -- Antennas -- Systems

lucycling

Post subject: Applying SWR to Circuit Efficiency

Unread postPosted: Mon Mar 14, 2005 2:27 pm

Offline

Lieutenant

Joined: Mon Mar 14, 2005 1:00 pm

Posts: 1

Location: Pittsburgh, PA

Understanding and applying SWR measurements.

I am matching two circuits to a 50Ohm load at 915MHz, using a vector network analyzer. My matching network is a shunt capacitor followed by series inductor, a standard L-matching network. Both circuits have the same topology but use a different diode. While matching the circuits, I also record SWR and use this for efficiency calculations.

After the circuits are matched, I connect each to an RF signal generator (50Ohm impedance) which is set to 915MHz, 10dBm and then measure the output voltage of my circuit. Voltage is taken across a load resistor.

One of my goals is to calculate the circuit efficiency. I calculate efficiency as the output power of the circuit divided by input power to the circuit. Input power is set by the signal generator and output power is calculated by V^2 over R (my load). I am also interested in output voltage (and hence output power) alone. So I have two measures for how to rate my circuits, efficiency and output power.

What I have seen is that a circuit with lower output voltage but high SWR seems to be more efficient than a circuit with higher output voltage but low SWR. When looking at efficiency with SWR applied I am confused by the results. I apply SWR to Pin as a reduction of this value. SWR tells me that the input power to the circuit is the incident power (10dBm) minus the reflected power. I am confident that this part of my calculation is correct. Here is some data from my tests:

Circuit 1

SWR, Pin, Pout, Eff w/o SWR, Eff w/ SWR

3.00, 10dBm, 8.09dBm, 64.5%, 86.0%

Circuit 2

SWR, Pin, Pout, Eff w/o SWR, Eff w/ SWR

1.35, 10dBm, 8.78dBm, 75.5%, 77.2%

What does it mean that Circuit 1 can operate more efficiently (when applying the SWR) but have a lower output voltage than the second circuit? Does it make sense that I should be able to increase the efficiency of Circuit 2 while maintaining the low SWR? If I could do this, then the output power/voltage should increase.

The efficiency calculations that do not account for SWR make sense to me but the other calculations do not. Circuit 1 may have a higher efficiency but the power of my source still needs to be 10dBm for this to happen, it is not like I can reduce the power with a better match (lower SWR) and have the same efficiency.

Any thoughts?

Thank you,

Kevin

Top

Profile

j_almira

Post subject:

Unread postPosted: Mon Apr 04, 2005 1:44 am

are you measuring SWR at Pin =10dBm or based on small signal measurement?

Regards,

Jean

Top

Guest

Post subject: Efficiency

Unread postPosted: Fri Apr 08, 2005 12:47 pm

Hello!

What kind of circuit are you working with?

You mention a diode: is this a frequency multiplier, a parametric amplifier, or some other circuit?

Nonlinear circuits do not work the same as linear circuits.

Good Luck!

Posted  11/12/2012